249497952-AircraftPerformance-Keith-Williams.pdf

1000 Questions ANSWERS MID EXPLANATIONS FOR JAR IICPLAND CPL

AIRCRAFT PERFORMAWOE

1000 Questions ANSWERS AND EXPLANATIONS FOR JllR ATPL 11HO CPL

AIRCRAFT PERFORMANCE

Keith Williams

transmitted, in any form or/byany means, electronic, mechani the prior permission of the author. This publication shall not, by out 01otherwise circulated without the p The information contained this publication is for private stud to ensure its accuracy and validity, no responsibility is

, recording or otherwise, without

it

Typeset by

Printed and Bound Gopsons Papers Ltd, ISBN 81-7002-088-

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I A Himalayan Books ~ i r iPublished t in 2005 by

Distributed by The English Book Store -17-L, Connaught Circus, New Delhi 110001 Tel: 2341 7126,2341 5031,2341 '?936 Fax: 091-11-23417731. E-mail: info@englishbookst~re.in

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effort has been made

CONTENTS

Section 1

Key Facts and Equations

Section 2

Questions

Section 3

Summary of Answers

Section 4

Answers and Explanations

INTRODUCTION

The purpose of this book is to assist students preparing to undertake the JAR ATPL and CPL Aircraft performance examinations. The majority of the questions are based upon feedback provided by students who have undertaken the JAR ATPL examination. By drawing feedback from the widest possible range of sources, this book provides a more comprehensive range of questions than can be achieved by any single ground school~consultingonly its own students. Other questions are intended to assist students in developing the required level of understanding of the more important aspects of each part of the syllabus. Because of the very low incidence of examination questions requiring the use of graphs in the CAP 698, this book does not include any such exercises. Examinations frequently include questions relating to key aspects of the shapes of the CAP 698 graphs, however, so questions of this form are included in this book. The explanations provided in Section 3 indicate the degree of understanding required of students undertaking these examinations. The JAR CPL Aircraft Performance syllabus is similar to that for the KrPL, with the exceptions of matters relating to the MRJT 1 aircraft in the CAP 698. Students preparing for the CPL exarr~inationsshould therefore ignore all questions relating to this subject. Students wishing to comment on the content or format of this book may contact the author at [email protected]

SECTION 1 KEY FACTS AND EQUATIONS

Effects of variables on take-off, climb, and landing performance. Definitions and limiting conditions for various speeds. Effects of clirr~bingto crossover altitude at constant IAS and above crossover altitude at constant Mach number. Effects of descending to crossover altitude at constant Mach number and below crossover at constant IAS. Commonly used equations.

KEY FACTS AND EQUATIONS

Effect of variables on take-off, climb and landing performance. The tables below indicates the effect of increasing the value of variables.

Altitude

v~~~

v~

Speed

I

Pressure

v~~~

,

v~

Decrease Decrease Increase Increase Increase Decrease

No effect No effect

v~~~

"R

Increase Decrease

v2

v~

'4

Increase

Increase

Decrease No effect Decrease

Decrease

Decrease

Increase No effect Increase

Decrease

Decrease

Increase

No effect Increase

Density

No effect

Increase

Temperature

No effect

lncrease

lncrease

Mass

Increase

Decrease Decrease lncrease No effect No effect Increase

Increase

Increase

Decrease No effect Decrease Increase Increase Increase

Runway upslope

No effect

No effect

No effect Increase

No effect

No effect

No effect No effect No effect

Headwind

No effect

No effect

No effect Increase

No effect

No effect

Flap angle

Decrease

No effect

No effect Decrease

Decrease

Decrease

Decrease Decrease Decrease

I

No effect 1 No effect No effect

Performance

VMBE

Ground speed

TORR

TODR

LDR

ASD

Climb Field Limited TOM Limited TOM

Altitude

Decrease

Increase

Increase

Increase

Increase

Increase

Decrease

Increase

Decrease

Pressure

Increase

Decrease

Decrease

Decrease

Decrease Decrease Increase

Density

Increase

Decrease

Decrease

Decrease

Decrease Decrease Increase

Increase

Temperature

Decrease

lncrease

lncrease

lncrease

lncrease

lncrease

Decrease

Decrease

Increase

Increase

Not applicable Not applicable

Mass

Increase

Increase

Increase

Runway upslope Increase

No effect

Increase

Increase

Decrease Decrease No effect

lncrease

Decrease

Decrease

Decrease

Decrease Decrease

1 Increase

Decrease

Decrease

Decrease

Decrease Decrease

Headwind F l a anale ~

(

, Decrease

Definitions and Limiting Conditions for Various Speeds Speed VSR

'SRO

and S' O

Limiting conditions The reference stalling speed V ,, replaces V, for class A aircraft. It is not less than the l g stalling speed. May not be less than 2 knots or 2% (whichever is greater) above the speed at which the stick pusher (if fitted) operates. This means that the stick pusher operates at not more than 2% or 2 .,, -. . knots below V The stalling speed or minimum steady flight speed with flaps in the landing setting.,,V, applies to Class A aircraft certificated under JAR 25 and Vsoapplies to Class B aircraft certificated under JAR 23. The minimum CAS at which lift equal to weight can be generated in the clean configuration at less than the stalling angle of attack. ,V ,, applies to Class A aircraft certificated under JAR 25 and V, applies to Class B aircraft certificated under JAR 23.

4


vsw

v

~

c

~

v

~

c

~

'MCL

v~~ v,

, which must be at least: Class A aircraft stall warning systems must activate at a speed,,V by reduced handling qualities (JAR indicated stall is a. 5 knots or 5% above the speed at which 125.201(d)). .,, b. 3 knots or 3% greater than V c. High enough to enable a recovery to be made when initiated not less than 1 second after stall warning onset in slow 1.59 turns decelera..ing a.: 2 knots per second, when trimmed for flight at 1.3 VSR. The minimum CAS at which control can be rnaintair ed, using primary flying controls (not nose wheel steering), on or close to the ground, follywing a critical engine failure. It is usually, but not alwhs, less than.,,V , The minimum CAS at which control can be ma ntaived, in the air in the take-off configuration, It is usually, but not always, folloyving a critical engine failure. Sometimes referred to as V., greater than VMc,. The minimum CAS at which control can be ma ntairsd in the air in the landing configuration, follovding a critical engine failure. The GAS at which it is assumed the critical encine v,rill fail whsn calculating take-off performance. Must always be greater than, ,V , and less tha i V,. Taka-off decision speed. The maximum CAS at whiqh it is popsible to abort a take-off safely fo~~o+ing an engine failure. And ever less than VMcnor greater than Nev$r less than the speed attained 2 seconds fter

I

B.

\,

vGO

vSTOP

v~

1

F

I

i

The lowest decision speed (CAS) at which it is possbble to continue a take-off safely following a critical engine failure. The highest decision speed (CAS) at which it i:, pos::,ible to at ort a take-off safely. The rotation speed, V, is the speed at which a? airqraft must be rotated into the take-off attitude. Must never be less than: a. V, b. 705%.,V, c. That required to attain V, by screen heigh

The fihal take-off speed must be achieved by th tim~? the airc-aft achieves the en-route (flaps and gear up) configuration.

p

Must be not less than: a. b.

'4

11.18.V ,, Tihe speed necessary to achieve the minimdm J44R25.143 manoeuvre capability. CAS @twhich best angle of climb is achieved. It is sli!;]htly less than VIM, for propeller aircraft and is VIM, f4r jets.

v~

CAS $t which best rate of climb is achieved. It i sligt~tlyhigher than VIM, for propeller aircraft and slightlb, higher than VIM, for jets.

v~~~

The dAS at which power required is minimum. \,',is I

I I

always Icwer than V .,,


5

The CAS at which total drag is minimum. It provides the best L:D ratio. ,, for propeller aircraft and 1.6 ,V ,, for jets. Approximately 1.3 ,V The CAS at which the ratio of CAS : drag is maximum. ,, or 1.7 ,V ,, for propeller aircraft and 2.1 ,V , for jets. Approximately 1.32 V Ensures that an adequate climb performance is available in the event of a discontinued approach. climb speed Must be a. 1.08 V ,, for Cengine aircraft in which the power increases give a significant reduction in stall speed. ,, for all other Class A aircraft. b. 1. I 3 V c. Not less than,,V, nor more than V,for Class A aircraft. d. Not exceeding 1.3 V, for all class B aircraft.

v~~~ V ~Max/ ~ Landing

The target threshold speed is the CAS that should be achieved at screen height during a landing. VAT,and VAT,are the threshold target speeds with all engines operating and one engine inoperative.

A 'T

The target threshold speed with all engines operating. Must be not less than 1.23V, steady descent to screen height at a descent gradient of not more than 5%. The target threshold speed with one engine inoperative. Must be not less than .V ,,

VAT,

following a

The reference speed is the CAS to be achieved at screen height when landing in any given configuration. Not less than the greater of VMcLor 1.23 ,V , for class A aircraft. Not less than the greater of VMcLor 1.3V,, for class B aircraft. The maximum brake energy velocity is the maximum CAS at which it is possible to stop an aircraft within the available ASDA, in the event of an aborted take-off, without exceeding the maximum energy absorption capabilities of the wheel brakes. The energy to be absorbed is proportional to ,, Increasing airfield altitude or air mass multiplied by TAS2, so increasing mass decreases.,V temperature both increase the TAS at any given CAS, thereby decreasing VMBE.

R ' EF

"ME,

Effects of Climbing to Crossover Altitude at Constant IAS, then to Higher Altitude at Constant Mach Number. IAS I

TAS

Effect of climbing CONSTANT to crossover altitude This means that %pVremains at constant IAS constant

INCREASING Because TAS:IAS ratio increases with increasing altitude

Effect of climbing at constant mach number above the crossover altitude

DECREASING Because LSS decreases up to 36,000 ft

DECREASING Because LSS decreases and TAS:IAS ratio increases

1

AofA

I

I

1

Pitch Attitude

I

I

Climb Gradient

I

CONSTANT To maintain constant lift at constant %pV

DECREASING To maintain constant A of A as gradient decreases

DECREASING Because thrust available decreases as altitude increases INCREASING INCREASING INCREASING To maintain To inctpase A of A Because both constant lift to maihtain constanl pitch and A of A as %pV lift as %pV are increasing decreases with IAS decreaSes with IAS

Effects of Descending to Crossover Altitude at Constant Mach Number, then td Lower Altitude at Constant IAS 1 IAS

Aof A

TAS

I

\

2

Effect of descending at constant mach number to the crossover altitude. This assumes that idle power is set, so the aircraft is effectively gliding.

INCREASING Because LSS increases as altitude decreases below 36,000 ft. And because IAS:TAS ratio increases as

I

INCREASING DECREASING Increasing IAS Because LSS increases as altitud decreases below 36,000 ft would reduce speed, so A of A must decrease to reduce drag, so thal

~

Pitch Attitude

Descent Gradient

INCRE~SING To decrease A of A the node down pitch attitude( must be gradually increased as altitude decreases

INCREASING As the nose is pushed gradually downwards and AofA decreases, the descent gradient increases

I I

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at constant IAS TAS:IAS ratio

effectively gliding.

I

Commonly Used Equationp

Climb gradient (in still air) = ROC / TAS

GRADIENT WITH CHANGES and it is estimated using the equation: IN MASS ted using the equation:

at any given speed does not vary w equation for profile drag is: D, = C ,,

Fey Facts and Equations

~

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INDUCED DRAG

VARIATION OF STALLING SPEED WITH MASS CHANGES

The coefficient of induced drag is proportional to C,2 SO induced drag at any given air speed I increases with increasing weight. The equation for induced drag is: Dl = C,,%pVS. Dl is proportional to IW, Where V is the EAS. Where V is the EAS. C, is proportional to IN4, C, is also proportional to C,2. C, = kc:/ pA Where k is the induced drag factor (k is 1 for elliptical plan forms) A is the aspect ratio. Stalling speed is proportional to the square of mass so the new stalling speed following a change in mass can be estimated using the following equation: V, (at new weight) = V, (at previous weight) X d(new weight / ol weight) This method also works for other speeds such as V, V ,, V ,,,, .V,, For small changes in weight the percentage increase in stalling peed is approximately half the percentage increase in weight. For example, a 10% increase in weight gives 5% increase in st lling speed and 5% increase in .V ,,

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SECTION 2 QUESTIONS

Atmospheres Airfields Lift and drag Flaps Clirr~bingand descending Power required and power available Curves Speeds Ta ke-off En-route Landing

ATMOSPHERES

ATMOS 1. If pressure altitude is 30000 ft amsl and mach number is 0.84, what is the IS/ TAT?

ATMOS 2.

I

If indicated TAT is-10, pressure altitude is 30000 ft amsl, and mach number is 0. 14, what is the temperature deviation?

ATMOS 3. If pressure altitude is 40000 ft amsl, indicated TAT is -29, what is the mach numbr r assuming ISA conditions.

ATMOS 4. If mach number is 0.88 and TAT is 4,what is the pressure altitude in the iqternati nal standard atmosphere? a. b. c. d.

22000 ft. 24000 ft. 26000 ft. 28000 ft.

ATMOS 5. If pressure altitude is 30000 ft, indicated TAT is -10, mach number is 0.82, wl at i s the density altitude? a. b. c. d.

31000 ft. 30472 ft. 30573 ft. 30674 ft.

12

I

Atmospheres

ATMOS 6.

I

If QNH is 999 hPa, what is the pressure altitude at an elevakion of 2500 3 ft? a. b. c. d.

25100 ft. 25200 ft. 25300 ft. 25400 ft.

ATMOS 7. If pressure altitude is 22800 ft, at an elevation of 22000 ft,

J

hat s QNH'I

ATMOS 8. The wind at an airfield is reported to be 330130. What will be he win:j component along and across runway 04? a. b. c. d.

10 Kts headwind 10 kts tailwind 28 kts headwind 28 Kts tailwind

28 Kts crosswind from 28 Kts crosswind from 10 kts crosswind from 10 Kts crosswind from

ATMOS 9. can take-off is 10 Kts. What

The maximum factorised cross wind that in which a particul is the acceptable wind at 330 in which the aircraft can a. b, c. d.

2 Kts. 32 Kts. 12 Kts. 22 Kts.

ATMOS 10.

F1

The reported wind at an airfield is 270130. What along track nd cross ,:rack wind components must be used in the take-off calculation for an MRJT using runway 2 a. b. c. d.

28 kts tailwind 28 Kts headwind 10.5 Kts headwind 10.5 Kts tailwind

10.5 Kts from the right. 10.5 Kts from the left. 28 kts from the right. 28 Kts from the left.

ATMOS 11. If field elevation is 4000 ft amsl and QNH is 900 mb, what is

3

altitude?

I

Atmospheres

J

ATMOS 12.

If field elevation is 3500 ft amsl and QFE is 1020 mb, what is the pressure altit de?

ATMOS 13. If pressure altitude is 3700 ft amsl and QNH is 1000 mb, what is filed elievatior a. b. c. d.

3310. 3210. 390. 490.

ATMOS 14.

e~ e

Take-off and landing performance must be calculated based on ... and ... resp ctively? a. b. c. d.

Forecast temperatures Forecast temperatures Actual temperatures Actual temperatures

Forecast temperatures. Actual temperatures. Actual temperatures. Forecast temperatures.

ATMOS 15.

1

If the OAT at a pressure altitude of 5000 ft amsl is 1O°C what is the temperatur deviation?

ATMOS 16. Density altitude is? a. b. c. d.

The altitude at which the existing density would occur in the ISA! The density at which the existing temperature would occur in the ISA. The elevation at which the existing density would occur in the ISA. The pressure altitude corrected for density deviation. I

ATMOS 17. The pressure altitude of the field can be found by? a. b. c. d.

Setting QNH on the altimeter subscale. Setting QFE on the altimeter subscale. Setting 1013 mb on the altimeter subscale. From an ADC only.

13

14

Atmospheres

ATMOS 18. If field pressure altitude is 5000 ft amsl and OAT is 25OC, wt a. 5000 + 118(25 - (15 - (5 x b. 5000 - 118(25 - (15 + (5 x C. 5000 + 118(25 + (15 - (5 x d. 5000 - 118(25 + (15 + (5 x

le der ity altitude?

1.98))) = 7348.2 ft. 1.98))) = 4988.2 ft. 1.98))) = 8551.8 ft. 1.98))) = 10888.2 ft.

ATMOS 19. If QFE is 1022 hPa what is the pressure altitude of the field? a. b. c. d.

270 ft amsl. -270 ft amsl. 30660 ft amsl. 500 ft amsl.

ATMOS 20. If QNH is 1000 hPa and field elevation is 4500 ft amsl, what

-. -7

ATMOS 21. Pressure altitude is? a. b. c. d.

The altitude above sea level. The altimeter indication when QFE is set on the subThe altimeter indication when QNH is set on the sub The altimeter indication when 1013.25 hPa is set on

0-scal

ATMOS 22. Which of the following cause air density to decrease? a. b. c. d.

lncreasing humidity, increasing altitude, increasing tt lncreasing humidity, increasing altitude, decreasing t lncreasing humidity, decreasing altitude, increasing 1 Decreasing humidity, increasing altitude, decreasing

ATMOS 23. If QNH changes from 1013 hPa to 1022 hPa will? a. b. c. d.

Increase field elevation. Decrease field elevation. Not affect field elevation. Decrease QFE.

~ture. ature. ature. ratur~

Atmospheres

ATMOS 24. .

If QFE changes from 1013 hPa to 1022 hPa will? a. b. c. d.

lncrease field elevation. Not affect QNH. lncrease QNH. Decrease QNH.

ATMOS 25. EAS is? a. b. c. d.

CAS corrected for adiabatic compression. IAS corrected for adiabatic compression. IAS corrected for instrument errors. IAS corrected for pressure sensing errors.

ATMOS 26. As pressure altitude increases when climbing at constant IAS? a. b. c. d.

1

CAS will increase. CAS will decrease. TAS will increase. TAS will decrease.

ATMOS 27. As pressure altitude increases when climbing at constant mach number? a. b. c. d.

CAS will increase. CAS will decrease then remain constant. TAS will increase. TAS will decrease then remain constant.

ATMOS 28. As pressure altitude increases? a. b. c. d.

Temperature decreases. Temperature increases. Temperature increases then remains constant. Temperature decreases then remains constant.

ATMOS 29. When descending at constant CAS? a. b. c. d.

Dynamic pressure increases. Dynamic pressure decreases. Dynamic pressure remains constant then decreases. Dynamic pressure remains constant.

15

16

Atmospheres

ATMOS 30. At a fixed pressure altitude an increase in temperature will? a. b. c. d.

Decrease density but increase density altitude. Decrease density altitude. Not affect density altitude. Increase density but decrease density altitude.

ATMOS 31. When descending through an inversion at constant TAS? a. b. c. d.

Mach number increases. Mach number decreases. Mach number remains constant. CAS decreases.

ATMOS 32. When climbing through an inversion at constant TAS? a. b. c. d.

Mach number increases. Mach number decreases. Mach number remains constant. CAS increases.

A-TMOS 33. When descending through an inversion at constant CAS? a. b. c. d.

TAS increases. Mach number increases. Mach number remains constant. TAS decreases.

ATMOS 34. When climbing through an inversion at constant CAS? a. b. c. d.

TAS increases. Mach number increases. Mach number remains constant. TAS decreases.

ATMOS 35. When climbing through an inversion at constant mach num a. b. c. d.

CAS increases. TAS decreases. TAS remains constant. TAS increases.

Atmospheres

ATMOS 36. When descending through an inversion at constant mach number? a. b. c. d.

TAS increases. TAS decreases, TAS remains constant. CAS increases.

ATMOS 37. When climbing through an inversion at constant mach number? a. b. c. d.

CAS decreases. LSS decreases. TAS remains constant. TAS increases.

ATMOS 38.

1

When descending through an inversion at constant mach number? a. b. c. d.

CAS increases. LSS increases. LSS remains constant. TAS decreases.

ATMOS 39. When climbing through an inversion at constant CAS? a. b. c. d.

TAS increases. TAS decreases. TAS remains constant. Mach number increases.

ATMOS 40. When descending through an inversion at constant CAS? a. b. c. d.

TAS increases. TAS decreases. TAS remains constant. Mach nurr~berincreases.

ATMOS 41. When descending through an isothermal layer at constant TAS? a. b. c. d.

Mach number increases. Mach number decreases. Mach number remains constant. CAS decreases.

I

17

18

Atmospheres

ATMOS 42. When climbing through an isothermal layer at constant a. b. c. d.

Mach number increases. Mach number decreases. Mach number remains constant. CAS increases.

ATMOS 43.

r

When descending through an isothermal layer at constant C S? a. b. c. d.

Mach number increases. Mach number decreases. Mach number remains constant. TAS increases.

ATMOS 44.

i

When climbing through an isothermal layer at constant CAS. a. b. c. d.

Mach number increases. Mach number decreases. Mach number remains constant. TAS decreases.

I

ATMOS 45.

When climbing through an isothermal layer at constant mach nu ber? a. b. c. d.

TAS increases. TAS decreases. TAS remains constant. CAS increases.

I

ATMOS 46. When descending through an isothermal layer at constant m a. b. c. d.

umber

TAS increases. TAS decreases. TAS remains constant. CAS decreases.

ATMOS 47. When climbing through an isothermal layer at constant mach nur ber? a. b. c. d.

CAS increases. CAS decreases. CAS remains constant. TAS decreases.

Atmospheres

ATMOS 48. When descending through an isothermal layer at constant CAS? a. b. c. d.

LSS increases. LSS decreases. LSS remains constant. TAS increases.

ATMOS 49. When climbing through an isothermal layer at constant CAS? a. b. c. d.

TAS increases. TAS decreases. TAS remains constant. Mach number decreases.

ATMOS 50. When descending through an isothermal layer at constant CAS? a. b. c. d.

TAS increases! TAS decreases. TAS remains constant. Mach number increases.

ATMOS 51. If aerodrome elevation is 4000 ft amsl, and QNH is 1025 hPa, what is a. b. c. d.

3540 ft. 3640 ft. 3740 ft. 3840 ft.

ATMOS 52. If pressure increases whilst temperature increases in the ISA? a. b. c. d.

Density will increase. Density will decrease. Density might increase or decrease, Density will remain constant.

ATMOS 53.

I

If pressure increases whilst temperature increases in a non-standard airnosph re? a. b, c. d.

Density will increase. Density will decrease. . Density might increase or decrease. Density will remain constant.

19

20

Atmospheres

ATMOS 54. If pressure remains constant as temperature increases? a. b. c. d.

Density will increase, causing the CAS : TAS ratio to Density will increase, causing the CAS : TAS ratio to Density will decrease, causing the CAS : TAS ratio tc Density will decrease, causing the CAS : TAS ratio tc

se. ase. ase. kase.

A-TMOS 55. Density altitude is? 1. The elevation at which the prevailing density occurs 2. The pressure altitude at which the prevailing temper; 3. The pressure altitude at which the prevailing density a. b. c. d.

SA. CCUrs ; in th

the ISA. SA.

I

1. 1 and 2. 1 and 3. 2 and 3.

N M O S 56. As altitude increase in the ISA? a. The effects of decreasing pressure outweigh tho performance decreases. b. The effects of decreasing temperature outweigh performance decreases. c. The effects of decreasing density outweigh those c aircraft performance decreases. d. The effects of increasing TAS : CAS ratio outweigh tt performance decreases.

3ecre ;ing temperature, so aircraft of dc reasing pressure so aircraft easin pressure and temperature so

. pres re and temperature so aircraft

ATMOS 57. What is density altitude? a. b. c. d.

Altitude at which prevailing conditions would occur in Altitude at which prevailing density would occur in t h ~ Altitude at which ISA conditions occur in the real atrr Altitude found by dividing ambient density by actual (

A.

re. lapst ate.

ATMOS 58. If the temperature is 12" C at a pressure altitude of 10000 ft, a. b. c. d.

11993ft. 12993 ft. 13993 ft. 14993 ft.

s the

?nsityaltitude?

ATMOS 59. If pressure altitude is 0 ft and ambient temperature is 10° C, what is the

altitude?

a. 593 ft. b. - 593 ft. c. 1593 ft. d. - 1593 ft.

ATMOS 60. If field elevation is 1000 ft amsl, QNH is 1025 mb, and QFE is 9916 7 db, wha is the pressure altitude at the field? a. - 640 ft. b. 1640 ft. c. 640 ft. d. -1640 fi.

AI RFIELL

FlELD 1. Maximum allowable clearway length is limited by? a. b. c. d.

50% of TODA. 50% of TORA. Minimum acceptable load bearing strength. 150% of Stopway.

FlELD 2. Maximum allowable stopway length is limited by? a. b. c. d.

50% of TODA. 50% of TORA. Minimum acceptable load bearing strength. 50% of runway.

FlELD 3. Stopway slope must be? a. b. c. d.

Same as the runway. Not more than 125% of that of runway. Less than that of runway. More than that of runway.

FlELD 4. Stopway semi-width must be? a. b. c. d.

At At At At

least 150% that of the runway. least 95 m. least GO m. least the same as the runway.

FlELD 5. A balanced field is one where? a. b. c. d.

ASDA= EMDA. ASDA= T O W . ASDA = TODA. ASDA = LORR.

Airfields

FlELD 6. The maximum TODA is equal to? a. b. c. d.

150% of T O M . 125% of TORA. From BRP to the first obstacle. EMDA;

FlELD 7. The maximum TODA is equal to? a. 150% of EMDA. I b. 125% of T O M . 1 c. From the first point at which the take-off can commence, to the first ta ke-off. d. ASD.

FlELD 8. Minimum width at the ends of the clearway must be? a. b. c. d.

180m. 150m 75 m 90 m. 76 m 90 m. The same as the runway.

FlELD 9. TODA is? a. b. c. d.

The take-off run plus the clearway. The take-off run minus the clearway. The take-of run plus the stopway. The take-off run minus the stopway.

FlELD 10. A field is unbalanced when? a. b. c. d.

There is no clearway or stopway. The clearway length is equal to that of the stopway. The stopway is longer than the clearway. The clearway is harder than the stopway.


There is no clearway or stopway. The clearway length is equal to that of the stopway. The clearway is longer than the stopway. The clearway is harder than the stopway.

23

---__-____-__I

24

Airfields

ITI I

I


There is no clearway or stopway. The clearway length is equal to that of the stopway. There is a clearway but no stopway. There is a stopway but no clearway.

FlELD 13. A field with a stopway but no clearway? a. b. c. d.

Is not permitted. Is unbalanced. Is more operationally flexible. Is balanced.

FlELD 14. A balanced field? a. b. c. d.

Improves climb gradient. Is more operationally flexible. Makes calculation of take-off performance more simple. Is shorter.

FlELD 15. Runway 27 threshold elevation is 200 ft amsl and that of runwa and the EMDA is 1400 m for both runways, what is the slope a. b. c. d.

1.27% up. 1.27% down. 4.1 % up. 4.1% down.

FlELD 16.

amsl. If the TORA is 1200 m

:r k

Runway 27 threshold elevation is 275 ft amsl and that of runwa 0 is 310 amsl. Runway 27 TORA is 900 m and the EMDA is 1100 m. Runway 09 TORA is 1100 m an th EMD is 1200 m. What is the slope of runway 09? a. b. c. d.

1.2% up. 1.2% down. 2.3% up. 2.3% down.

FlELD 17. Runway 04 threshold is 300 ft amsl and that of runway 22 is 320 msl. For bo. h runways the TORA is 4000 m and the EMDA is 4500 m. what is the slope of runway 22? a. b. c. d.

0.1 5% down. 0.1 5% up. 2.5% down. 2.5% up.

Airfields

25

FlELD 18. A contaminated runway is? a. One on which more than 25% is covered with more than 3 mm 1 to an equivalent depth of more than 3 mm. b. One on which more than 33% is covered with at least 3 mm de equivalent depth of at least 3 mm. c. One on which more than 25% is covered with up to 15 mm of d. One on which more than 20% is covered with more than 2 slush.

pth of

ater, or by slush or snow

of w

?rof slush or snow to an

I

~ter,st ~v or slush. i m of tanding water, snow or

FlELD 19. Water equivalent depth is? a. b. c. d.

Equal to the actual depth multiplied by the specific gravity of tt conta Equal to the actual depth multiplied by the density of the contz inant. Equal to the amount of water in the contaminant. Not relevant to snow ice because these contain no free water.

nant.

FlELD 20. A contaminated field is? a. One on which more than 25% is covered with compacted snob b. One on which more than 33% is covered with at least 3 mm de h o f w ar of slush or snow to an equivalent depth of at least 3 mm. c. One on which more than 25% is covered with up to 15 mm o f ' ~ter,SI w or slush. d. One on which more than 20% is covered with more than 2 nm of tanding water, snow or slush.

FIELD 21. A damp runway is one on which? a. More than 33% is covered with at least 1.5 mm depth of wate~ depth of at least 1.5 mm. b. More than 25% is covered with up to 1.5 mm of water, snow o c. More than 20% is covered with more than 2 mm of standing w d. The surface is not dry but has insufficient moister to make its !

f slusl )r snow to an equivalent lush. 3r, snc or slush. .face i pear shiny. I

FlELD 22. A damp runway? a. b. c. d.

Does not affect take-off performance calculations. Does not affect take-off performance calculations except wher Does not affect take-off performance calculations except wher Always affects take-off performance calculations.

Irass. .ontan ated.

26

Airfields

FlELD 23. A dry runway? a. Includes those with grooves or a porous surface that re! moisture is present. b. Includes those that are not contaminated. c. Includes those with less than 3 mm of water or equivaler d. Never includes grass surfaces.

ffecti

y dry braking action when

h of:

)w or slush.

of col

~minantor with sufficient tc

FlELD 24. A wet runway is? a. One covered with less than 15 mm of water. b. One covered with less than 3 mm of water. c. One covered with less than 3 mm of water or equivalent 1 cause the surface to appear reflective. d. One covered with more than 1.5 mm but less than 3 mm

ter.

FlELD 25. A flooded runway is? a. b. c. d.

One covered with more than 15 mm of water. One covered with less than 3 mm of water. One covered with more than 3 mm over more than 50% One covered with more than 3 mm of water over more tt

;urfac % of

surface.

FlELD 26. Depth of snow and slush are measured? a. b. c. d.

Every 300m between 5 m and 10m from the centreline. Every 500m between 5m and 1Om from the centreline. Along the entire length of the runway, at its centreline. Along the entire length of the runway only if its depth is (

'

thar

mm.

FlELD 27. Maximum allowable runway slope in Europe is?

FlELD 28. Runway alignment reduction is? a. b. c. d.

Errors in compass systems caused by reinforcing strips in Errors made in the alignment of runways due to constructi The distance required by certain large aircraft types to linc The distance required by certain large aircraft types to tax

ete ri CCU T i

r takc e run

vays. 3s. iff. ~yafter landing.

Airfields

FIELD 29. A clearway? a. b. c. d.

Must be capable of supporting the weight of the aircraft using i. Must be clear of obstructions above a slope of 1.25% from the l d of 1 le T O W . Must be free of obstructions above a slope of 1.255 from the e I of th TODA. Must be longer than the stopway.

FIELD 30. Which of the following equations is true? a. b. c. d.

TODA = EMDA = TORA + Clearway. TODA = EDA = TORA + Clearway TORA = EMDA = TODA + Stopway. ASDA = EMDA = TORA + Stopway.

FIELD 31. Which of the following equations is true? a. b. c. d.

TODA = TORA + Clearway. TODA = EDA+ Clearway TORA = TODA + Stopway. ASDA = TORA + Clearway.

FlELD 32. When taking account of runway alignment reduction? a. b. c. d.

The reduction in T O W , TODA and ASDA are the same. The reduction in TORA is greater than the reductions in TODA nd AS The reduction in TODA is greater than the reductions in TORA nd AS The reduction in ASDA is greater than the reductions in TORA nd TC

FIELD 33. When taking account of runway alignment reduction the ASDA is take to cor mence at the? a. b. c. d.

Nose wheel position. Main wheel position. C of G position. Nose position.

FIELD 34. When taking account of runway alignment reduction the TODA is take to cor mence at the? a. b. c. d.


27

28

Airfields

FlELD 35. When taking account of runway alignment reduction the TORA a. b. c. d.

tal



Runway length. Runway plus stopway length. Runway plus stopway plus clearway length. Runway plus clearway length.

FlELD 37. Increasing ambient temperature will? a. b. c. d.

Decrease field length limited TOM. Increase field length limited TOM. Increase climb limited TOM. lncrease.,V ,,

FlELD 38. The ACN of an aircraft landing on a newly constructed concrete a. b. c. d.

J n\

Must never be greater than PCN. Must never be greater than 125% of PCN. Can exceed PCN by up to 10% without prior ATC appro\ I. Can exceed PCN by 10% only with prior ATC approval.

FlELD 39. If the stopway of an airfield is extended to a length equal to its c an a. b. c. d.

Maximum TOM and V, will both increase. Maximurrl TOM will be unchanged but maximum V, will i! :re, Maximum TOM will increase but maximum V, will be unc 3n! Maximum TOM and V, will both decrease.

FlELD 40. The maximum TODA is? a. b. c. d.

Declared runway length plus stopway length. Declared runway length plus clearway length. Declared runway plus stopway plus clearway. Declared runway length minus runway alignment reducti

1.

bmmence at the?

Airfields

29

FlELD 41. The TODA for a particular runway is?

-

a. b. c. d.

The same for all aircraft that use it. The same for all aircraft using it.at the same mass. Sometimes dependent upon aircraft mass. Sometimes dependent upon aircraft type.

FlELD 42. An airfield has a declared runway length of 4000 m and the first obstac runway. What is the maximum TODA for a class A aircraft?

DO m from the end of the

FlELD 43. An airfield has a declared runway length of 4000 m and the first obsta of the take-off run. What is the maximum TODA.

I00 meters from the end

FlELD 44. The length of the clearway is limited by? a. b. c. d.

The first obstacle capable of damaging an aircraft after it has t 150% of the TODA. 150% of the TORA. 50% or the TODA.

FlELD 45. The clearway starts? a. b. c. d.

At the beginning of the TORA. At the end of the TODA. At the beginning of the stopway. At the end of the stopway.

FlELD 46. A field is unbalanced? a. b. c. d.

Only when the clearway is longer than the stopway. When the clearway is equal to the stopway. When there is a stopway but no clearway. When thereis neither a clearway nor a stopway.

rborne.

30

Airfields


Never more than 50% greater than the TORA. Never less than 50% greater than the TORA. Never more than the stopway plus the clearway. Never more than runway length plus stopway length.

I i

Ii

FlELD 48. The purpose of the stopway is to? a. b. c. d.

lncrease the TORA. lncrease the TODA. lncrease the ASDA. Prevent obstacles striking aircraft after take-off. i

FlELD 49.

i

1

What is distance A to B in the diagram? a. b. c. d.

TORA. TODA. ASDA. EMDA.

i A

I1 i

FlELD 50. What is distance A to C in the diagram? a. b. c. d.

TORA. TODA. ASDA. LDA.

FIELD 51.

-

What is distance A to D in the diagram? a. b. c. d.

TORA. TODA. ASDA. EMDA.

FlELD 52. What is distance B to C In the diagram? a. b. c. d.

Stopway. TODA. ASDA. Clearway.

...

i

.:: B

C

.

..: ...

:

D

Airfields

31

FlELD 53. What is distance B to D In the diagram? a. b. c. d.

Stopway. TODA. ASDA. Clearway.

1

I

A

i

I

C

D

FlELD 54. What is the maximum length of B to D in the diagram? a. b. c. d.

50% of AB. 50% of AD. 150% of AB. 100% of AB.

FlELD 55. What is the gradient of runway 27 in the diagram? A = 500 ft Elev. B = 550 ft Elev. C = 600 ft Elev. A to B = 2000 m. A to C = 2500 m. a. b. c. d.

0.76% down. 0.76% up. 2.5% up. 2.5% down.

FlELD 56. BRP is 1500 ft amsl, runways slope is -2% and the TODR for a give elevation of the first obstacle is 2000 ft amsl, what is its height above a. b. c. d.

596.8 ft. 696.8 ft. 500 ft. 168.2 ft.

FlELD 57. BRP is 1500 ft amsl, runways slope is 2% and the TODR for a give elevation of the first obstacle is 2000 ft amsl, what is its height above a. b. c. d.

303.2 ft. 196.8 ft. 696.2 ft. 500 ft.

aircraft is 3000 m. If the

32

Airfields

FlELD 58. If the obstacle in question FlELD 57 is 1000 m from the end of th path that will just touch its top?

: is the gradient of a flight

FlELD 59. If the obstacle in question FlELD 56 is 10000 m from the end of tt path that will just touch its top?

t is the gradient of a flight

'F

FIELD 60. For an unbanked Class A aircraft the screen is assumed to be ... a. b. c. d.

50ft 50ft 35 ft 35ft

At the end of the clearway. At the end of the stopway. At the end of the TODR. At the end of the T O W .

FlELD 61. The minimum acceptable semi-width for a stopway is? a. b. c. d.

The same as that of the runway. 90 m. 70 m. 75 m.

FlELD 62. Which of the following are limited to 50% of the runway length? a. b. c. d.

Stopway. Clearway. TORA. TODA.

FlELD 63. Where does the clearway start? a. b. c. d.

The end of the TODA. The end of the TORA. The end of the runway. The end of the stopway.

ited at the?

Airfields

33

FIELD 64. Which of the following would not limit the length of the stopway? a. b. c. d.

A ditch or depression. A reduction in load bearing strength. 50% of the T O M . An obstacle.

FIELD 65. What percentage of PCN would be considered as an acceptable only? a. b. c. d.

Over 50%. Over 60%. Over 150%. Over 160%.

FIELD 66. What might be the T O M on the diagram below if CD is the stopway?

a. A-B. b. B-C. C. B-D. d. B-E.

FIELD 67. What might be the TODA on the diagram below if BC is the runway?

a. b. C. d.

A-B. B-C. B-D. B-E.

emergency operations

34

Airfields

~

FIELD 68. What might be the ASDA on the diagram below if CE is the clea#ayt

a. b. C. d.

A-B. 5-C. B-D. B-E.

W

FIELD 69.

What might be the EMDA on the diagram below if BC is the run ay?

1

i

.. ...

. ...

A

B

:

J

:

:

C

3

E

a. A-B. b. B-C. C. 5-D. d. B-E.

FIELD 70.

11 i

Over what distance on the diagram below must the surface be ap ble of upporting the weight of an aircraft if B c is the runway?

Airfields

35

FlELD 71. The load bearing strength of a stopway must be at least? a. b. c. d.

That of the associated runway. 75% that of the associated runway. 50% that of the associated runway. 30% that of the associated runway.

FlELD 72. fficient of friction for.. .?

When the mu-meter readings fall below ... pilots must be informeid of I

a. b. c. d.

0.4 each 113 of the runway length. 0.3 each '/z of the runway. 0.4 all of the runway. 0.3 all of the runway.

n

FlELD 73.

What is the minimum depth of dry snow that needs to be taken 'into accou performance? a. b. c. d.

10 mm. 15 mm. 3mm. 25 mm.

FlELD 74.

I

What is the maximum permissible depth of wet snow for take-off and a. b. c. d.

95 mm. 75 mm. 15 mm. 35 mm.

FlELD 75. What is reference zero? a. b. c. d.

An An An An

imaginary horizontal plain commencing at the end of the imaginary vertical plain commencing at the end of the imaginary horizontal plain commencing at the end of imaginary horizontal plain commencing at the end of

FlELD 76. Which of the following describes the length of the TORA? a. b. c. d.

The TODA minus the clearway. The ASDA minus the clearway. The TODA minus the ASDA. The clearway minus the stopway.

I

when calculating take-off

36

Airfields

FlELD 77. Which of the following need not be capable of supporting the wei a. b. c. d.

The TODA minus the clearway. The ASDA minus the clearway. The clearway minus the stopway. The TORA minus the stopway.

FlELD 78. Which of the following is an unbalanced field? a. b. c. d.

TORA 3000 m T O W 3000 m TORA3400 m T O W 3400 m

ASDA 3000 m ASDA 3400 m ASDA 4000 m ASDA 3400 m

FlELD 79. For an aircraft with tyre pressures of 150 psi landing on a wet run a. b. c. d.

94.3 Kts. 92.3 Kts 110.2kts. 101.2 Kts.

FlELD 80. For an aircraft with tyre pressures of 150 psi landing on a runway Vp will be? a. b. c. d.

with wet snow (SG = 0.5)

158.6 Kts. 188.6 Kts. 94.3 Kts. 79.3 Kts.

FlELD 81. For an aircraft with tyre pressures of 150 psi taking-off from a wet a. b. c. d.

94.3 Kts. 92.3 Kts. 110.2 Kts. 120.2 Kts.

FlELD 82. For an aircraft with tyre pressures of 150 psi taking-off from a run 0.5) Vp will be? a. b. c. d.

202.5 Kts. 220.5 Kts. 188.6 Kts. 108.6 Kts.

ated with wet snow (SG =

Airfields

FIELD 83.

I

For any given aircraft mass, tyre pressure and contaminant VA at to1 that in the event of a rejected take-off? a. b. c. d.

i don

will be

37

...Compared with

The same. Higher. Lower. Higher or lower depending on the runway condition.

FIELD 84. What is measured performance? a. The average performance of a type of aircraft when tested. b. The average performance of a type of aircraft, reduced suc that : individual aircraft will be capable of achieving it. c. The performance that any given aircraft can be predicted to a ieve. d. The performance that a type of aircraft can achieve when he\n r in a as-new condition.

FIELD 85. A runway is flooded when

...% of it is covered with more than ... rnm

FIELD 86. The provision of a stopway will a. b. c. d.

lncrease maximum V, at any given mass and increase ASDA Decrease maximum V, at any given mass and increase A$D/ lncrease maximum V, at any given mass and decrease A$DP Decrease maximum V, at any given mass and decrease ~ S D

FIELD 87. The provision of a clearway will? a. b. c. d.

lncrease minimum V, at any given mass and increase T O ~ A . Decrease minimum V, at any given mass and increase TqDA lncrease maximum V, at any given mass and decrease T ~ D L Decrease maximum V, at any given mass and decrease T/3D

wate ?

LIFT AND DW

LD 1. Which of the following statements is most accurate? a. b. c. d.

Dl is proportional to Dl is proportional to Dl is proportional to Dl is proportional to

1 I(EAS)2 1 I(IAS)2 1 I (RAS)2 1 I (TAS)2

LD 2. Which of the following statements is most accurate? a. b. c. d.

Dl is proportional to L. Dl is proportional to LN. Dl is proportional to IIL. Dl is proportional to VIL.

LD 3. If a 50000 Ibf aircraft requires 25000 Ibf of thrust maintain unacl what is its L:D ratio?

;ed str ight and level flight at 250 Kts

LD 4. What effect does increasing load factor have on power requirc a. b. c. d.

ny gii 'n aircraft weight and TAS?

No effect. Increases in direct proportion to load factor. Decreases in direct proportion to load factor. Power required is inversely proportional to load factor.

LD 5. What effect will increasing weight have on Dl, D, and , ,D a. b. c. d.

Increase, Increase, Decrease, Decrease,

No significant effect, Decrease, Increase, No significant effect,

at : Inc Nc Nc In(

len lo;

j

factor and airspeed?

!cant Ffect. 'icant Ffect.

Lift and Drag

39

I

LD 6. What effect will increasing aircraft weight have on minimum drag bpe d (VMD)and speed stability. a. b. c. d.


Increase. Decrease. Increase. Decrease.

LD 7. What effect will lowering the landing gear have on VMDand a. Increase, b. Increase, C. Decrease, d. Decrease,


LD 8. nstant low IAS .

What is the effect on Dl and D, when an aircraft climbs to a. b. c. d.

Increase, Increase, No change, Decrease,

Increase. Decrease. No change. Decrease.

'

LD 9.

1

What would be the effect on Dl, D, and speed stability if the tra~lindedge flaps )fan aircraft were lowered to the 10 degree setting while maintaining constant speed and wipgs l p l ? l

a. Increase, b. Increase, C. Decrease, d. Decrease,

Increase, Increase, Increase, Decrease,


LD 10. What would be the effect on Dl, D, and speed stability if the trailin to the 40 degree setting while maintaining constant speed and a. b. c. d.


Increase, Increase, Increase, Decrease,


LD 11.

)fan aircraft were lowered

~b

P

What will be the effect on drag if humidity is increased if air te pe-ature, ressure and TAS remain constant?

i

a. b. c. d.

Increase. Decrease. Remain constant. Decrease or increase depending on Mach number.

40

Lift and Drag I

LD 12.

1

When flying at best L:D ratio what is the ratio of D,: Dl? a. b. c. d.

2:l. 1:l. 1:2. Depends on aerofoil section.

LD 13. Which of the following occur at ? V ,, a. b. c. d.

Minimum power required and best L:D ratio. Minimum drag and greatest L:D ratio. Minimum angle of attack and best rate of climb. Minimum drag and greatest jet propeller aircraft range.

I

I

LD 14. In what direction does the drag force act? a. b. c. d.

Parallel to relative airflow. Parallel but opposite to the direction of flight. Parallel to lift. Parallel to weight.

LD 15. a wing flying at zero lift angle

Which of the following statements are true of the total reaction of attack. a. b. c. d.

It equals D,. It equals Dl. It equals weight. It is twice D,.

LD 16. In what direction does the total reaction act when a wing is flyi a. b. c. d.

"

Parallel but opposite to the direction of flight. Vertically aft through the C of G. Vertically upward through the aerodynamic centre. Vertically aft such that it produces a nose up pitching m me t.

LD 17. What is the relationship between D, and D, at speeds below V a. Dl is greater than D,. b. D, is less than D,. c. Dl = D.,

0

lift angle of attack?

~ 1 -

~

LD 18.

~

I 1

Lift and Drag

................ increases and to CLWlx

Complete the f~llowingstatement. As airspeed changes From V, .............. decreases. I I l

~

I

LD 19. Which of the following definitions of D, is the most accurate?

l

a. b. c. d.

It is made up of form, friction and induced drag. It is made up of friction, induced and shock drag. It is made up of friction, form and interference drag. It is made up of interference, shock and form drag.

~~

LD 20.

I

I

1

1 ccurate?

Which of the following statements concerning the generation o a, b. c. d.

I

I

~

Lit? is generated by a cambered aerofoil. Lit? is generated by an aerofoil at a positive angle. Lit? is generated by high speed airflow. Lift is generated by the downward acceleration of air.

I I

i

,I

LD 21. Which of the following causes induced drag (Dl)? a. b. c. d.

Shock waves above and below the wing. Friction due to the air passing over the wing. Rotating airflow caused by wing tip vortices. The upwash of air caused by wingtip vortices.

LD 22. Which of the following causes induced drag (Dl)? a. b. c. d.

Shock waves above and below the wing. Friction due to the air passing over the wing. Downwash of aitflow over the trailing edge caused by Upwash of airflow over the trailing edge caused by I

LD 23.

d

Which of the following is responsible for the creation of induce a. b. c. d.

Angle of attack, camber, wing area and airspeed. Pitch angle, camber, wing area and airspeed. Pitch angle, camber, wing area and angle of attack. Airspeed, wing area and pitch angle.

I I

i

41

42

Lift and Drag

LD 24. What proportion of total drag is made up of induced drag wher lyir

at V, ?

LD 25. If TAS is increased from 300 Kts to 400 Kts with no change in ltiti le, co 'iguration or weight, by what percentage will power required change? a. b. c. d.

Decrease by 135% Decrease by 35% lncrease by 135% lncrease by 235%

LD 26. All other factors being equal, minimum drag is..........? a. b. c. d.

Constant. Proportional to weight. A function of density altitude. A function of pressure altitude.

LD 27. If air density is reduced by a factor of 4, by what factor will dra! a. b. c. d.

?

Decrease by a factor of 2. Decrease by a factor of 4. lncrease by a factor of 4. Decrease by a factor of 16.

LD 28. If indicated airspeed is maintained constant while air density dt re; es by alf what will be the effect on total drag? a. b. c. d.

lncrease by a factor of 2. Decrease by a factor of 2. lncrease by a factor of 4. Remain unchanged.

LD 29. If IAS is increased from 100 Kts to 200 Kts, by what factor wou

C be multiplied? D ~nd

Lift and Drag

43

LD 30. In what direction does the weight of an aircraft act? a. b. c. d.

At right angles to the flight path. Opposite lift. Straight down. Depends on rate of turn.

LD 31. In what direction does lift act? a. b. c. d.

Straight up. At right angles to the flight path. At right angles to the relative airflow. At right angles to thrust.

LD 32. The majority of lift is produced by? a. b. c. d.

High pressure below the wing. Low pressure above the wing. lncreased velocity below the wing. lncreased density below the wing. I

LD 33.

How does static pressure below a wing at low positive angles of atjack l static pressure? a. b. c. d.

3 with

the local ambient

Higher. The same. Lower. Higher or lower depending on speed.

LD 34. Which of the following would give minimum glide gradient?

b' C,2 / L' C. C,4 / C,

d'

D '

Max. Min.

Max.

A low speed aircraft climbs from sea level to 40000 feet pressure alti will its profile drag be multiplied if all other factors remain unchange

at con ant TAS. By what factor

44

Lit7 and Drag

LD 36, What will be the effect of raising the undercarriage of an ai lncreased induced drag due to more efficient li!? Decreased induced drag due to less efficient lift lncreased profile drag and greater speed stability. Decreased profile drag and lower speed stability.

a. b. c. d.

LD 37. What causes wing tip vortices? a. b. c. d.

Pressure differences in front of and behind the wing High pressure air leaking from below the wings. Spanwise flow from tip to root under the wings. spanwise flow from root to tip above the wings.

t . 1

LD 38. How does aspect ratio affect wingtip vortex strength? a. b. c. d.

Shorter tip chord length. Longer tip chord length. Shorter root chord length. Higher energy airflow.

LD 39. In what ways are D,and D, proportional to V? a. b. c. d.

Inversely. v3, V2, IN2,

1 ~ 1N 2 . V2.

~

.

I

LD 40. What happens to total drag when increasing speed to V,? a. b. c. d.

Decreases approximately with the square of speed. Increases approximately with the square of speed. Decreases approximately with the inverse of the squ Remains constant.

LD 41. If IAS decreases by a factor of 5 how would drag vary?

Lift and Drag

LD 42. If IAS decreased by a factor of 5 how would drag vary?

LD 43. How does total drag vary with air density? a. b. c. d.

Constant. Directly. Inversely. Conversely.

LD 44. Which of the following are true of ? V ,, a. b. c. d.

Lower than.V ,, Gives best L:D ratio. Gives best endurance in a propeller aircraft. Gives best range in a jet aircraft.

LD 45.

er

How is drag affected if pressure decreases with TAS and temperatur co stant? a. b. c. d.

Increases. Decreases. Constant. lncrease or decrease depending on altitude.

1

LD 46.

How does retraction of flaps affect induced drag if IAS remains constbnt? a. b. c. d.

Increases. Decreases. Constant. lncrease or decrease depending on speed.

LD 47. At low angles of attack the major component of total drag is ............ a. b. c. d.

Induced. Vortex. Shock. Profile.

45

46

Lift and Drag

LD 48. At high angles of attack the major component of total drag is.............drag? a. b. c. d.

Induced. Vortex. Shock. Profile.

LD 49.

.........causes induced drag? a. Angle of attack. b. Aspect ratio. c. Boundary layer separation. d. The generation of lift.

LD 50. Induced drag is directly proportional to ............? a. b. c. d.

Angle of attack. (Lift f ~ r c e ) ~ . (True air~peed)~. Aspect ratio.

LD 51. Induced drag is caused by.. ......? a. b. c. d.

Upwash. Tip tanks. High tailplanes. Wing tip vortices.

LD 52. Lowering the undercarriage will ........? a. b. c. d.

lncrease induced drag and nose down pitching Decrease induced drag and nose down lncrease profile drag and nose down Decrease profile drag and nose

LD 53. Doubling IAS multiplies drag by ....?

Lift and Dmg

LD 54. Tripling IAS multiplies drag by.. ..?

LD 55. C, varies ....?

a. b. c. d.

Directly with C: and inversely with aspect ratio. Directly with speed and aspect ratio. Directly with wing area and C,. Directly with angle of attack and speed.

LD 56. Increasing camber will.. ..? a. b. c. d.

lncrease the gradient of the lift slope. Decrease the gradient of the lift slope. lncrease C, Max. Decrease induced drag.

LD 57. lncreasing camber will.. ..? a. b. c. d.

lncrease the gradient of the lift slope. Decrease the gradient of the lift slope. Decrease C, Max. Reduced stalling speed.

LD 58. lncreasing air temperature will.. ..? a. b. c. d.

lncrease the gradient of the lift slope. Decrease C, max. Decrease lift at any given C, and TAS. lncrease profile drag.

LD 59. lncreasing humidity will.. .....? a. b. c. d.

lncrease C, max. lncrease required C,. Decrease stalling speed. lncrease profile drag.

47

48

Lift and Drag

LD 60. lncreasing humidity will ..... ..? a. b. c. d.

lncrease C, max. Decrease required C,. lncrease stalling speed. lncrease profile drag.

LD 61. To maintain constant angle of climb, when climbing at constant TAS ......? a. b. c. d.

Angle of attack must increase. Angle of attack must decrease. Angle of attack must remain constant. IAS must increase.

LD 62. Which of the following statements are true? 1. 2. 3. 4.

lncreasing aspect ratio reduces induced drag. lncreasing sweep back angle reduces induced drag. lncreasing EAS increases induced drag. lncreasing C, increases induced drag.

a. b. c. d.

1 and 4. 1, 2 and 3. 1, 3 and 4. 2, 3 and 4.

LD 63. Which two of the following statements are most accurate? 1. 2. 3. 4.

C ,, is approximately constant and D, is proportional to V2. C, is proportional to IN4and D, is proportionai to V2. The V in a and b above is TAS. The V in a and b above is EAS.

a. b. c. d.

1 and 4. 1 and 3. 1 and 2. 2 and 3.

LD 64. At 1.3VSC, would be .......% of C,

Max?

Lift and Drag

LD 65. .......... Altitude at constant TAS will ......... D,? a. b. c. d.

Increasing, Increasing, Decreasing, Decreasing,


LD 66. Induced drag is increased by ......... weight, ......... airspeed and ........ aspect ratio? a. b. c. d.

Increasing, Increasing, Increasing, Decreasing,

Increasing, Decreasing, Decreasing, Increasing,

Increasing. Decreasing. Increasing. Decreasing.

LD 67. Increasing load factor will .......... D, a. b. c. d.

Increase, Decrease, Not affect, Not affect,

.............V,, and ...........speed stability?

Increase, Decrease, Increase, Decrease,

Increase. Decrease. Decrease. Increase.

LD 68. At 0.9 V, the C, will be ............ C, Max and C, will be ........... than at 1.I V? , a. b. c. d.

Greater than, Greater than, Less than, Less than,

Greater than. Less than. Less than. Greater than.

LD 69. 7 C, is proportional to ............

49

FLAPS

FLAPS 1. What is the effect of deploying trailing edge flaps? a. b. c. d.

lncreases C, and V, Decreases C, and V, lncreases C, and decreases V, Decreases C, and increases V,

FLAPS 2. What trailing edge flap angle will give the minimum stalling speed? a. b. c. d.

Maximum deflection. Zero degrees. 20 degrees. 30 degrees.

FLAPS 3. What trailing edge flap angle will give best L : D ratio? a. b. c. d.

Zero angle. Maximum angle. 20 degrees. 30 degrees.

FLAPS 4. Deployment of trailing edge flaps in straight and level flight will a. b. c. d.

Not affect. Decrease. Increase. Increase or decrease depending on flap angle selected.

FLAPS 5. Which of the following will reduce L:D ratio most? a. b. c.

6.

..... , ......induced drag?

15" trailing edge flap. 30" trailing edge flap. 45" trailing edge flap. 15"slat.

Flaps

FLAPS 6. Deployment of flaps ....... C,? a. b. c. d.

Increases. Decreases. Increases then decreases. Decreases then increases.

FLAPS 7 . Deployment of trailing edge flaps ........stalling angle and...... ?,,C , a. b. c. d.

Increases, Increases, Decreases, Decreases,

Increase. Decreases. Decreases. Increases.

FLAPS 8. Maximum speed for extending flaps is.. ...? a. b. C. d.

V., V .,, VF. V.,

FLAPS 9. Maximum speed with extended flaps is.....? -.

a. V., b. V .,, c. v,. d. V.,

FLAPS 10. Raising slats too soon after take-off might........? a. b. c. d.

Increase stalling angle. Cause stalling. Prevent flutter. Increase rate of climb.

FLAPS 11. 7 Flap deployment improves C, most ...........

a. b. c. d.

During the first few degrees. During the last few decrees. At high subsonic speeds. During the take-off roll.

51

52

Flaps

FLAPS 12. Failure of trailing edge flaps to deploy on landing will? a. b. c. d.

Decrease landing speed. Decrease landing roll. lncrease nose up attitude. lncrease angle of attack.

FLAPS 13. Retracting trailing edge flaps whilst leaving slats deployed in a climt a. b. c. d.

Reduce drag. Reduce lift, drag and L:D ratio. lncrease !ift and reduce drag. Reduce lift and drag and increase L:D ratio.

FLAPS 14. Retraction of slats prior to flaps might ........ ? a. b. c. d.

lncrease L:D ratio. lncrease stalling angle. Reduce stalling speed. Cause stall.

FLAPS 15. Trailing edge flap deployment .....? a. b. c. d.

lncreases stalling speed and stalling angle of attack. Decreases stalling speed and angle of incidence. Decreases stalling speed and stalling angle of attack. Improves stability.

FLAPS 16. Spoiler deployment? a. b. c. d.

lncreases L:D ratio. Decreases L:D ratio. Decreases separation over flaps. Is not possible with flaps deployed.

FLAPS 17. Flap deployment ...... D, and a. Increases, b. Increases, C. Decreases, d . Decreases,

...... D,?

Increases. Decreases. Decreases. Increases.

Flaps

FLAPS 18. Landing configuration is usually ........7. a. b. c. d.

Slats and flaps fully deployed. Slats only deployed. Flaps only deployed. Slats fully deployed and flaps at minimum deflection angle.

FLAPS 19. Flap deployment ..... the landing run? a. b. c. d.

Increases. Decreases. Does not affect. lncreases or decreases depending on landing speed.

FLAPS 20. Flap deployment ..... the take-off run? a. b. c. d.

Increases. Decreases. Does not affect. lncreases or decreases depending on deployment angle.

FLAPS 21. Trailing edge flap deployment a. b. c. d.

......... power required?

Increases. Decreases. Does not affect. lncreases or decreases depending on C of G position.

FLAPS 22. Use of small angles of flap deflection .... ? a, b. c. d.

lncreases D, more than D., lncreases D, more than Dl. Reduces Dl and increases D,. lncreases Dl and reduces D.,

FLAPS 23. Maximum flap deployment speed is .... ? a. V., b. V., C. v ., d. V.,

53

54

Flaps

FLAPS 24. Full flap deployment in take-off will a. b. c. d.

..... ?

lncrease climb performance. lncrease take-off distance required. lncrease acceleration rate. lncrease climb gradient.

FLAPS 25. Full flap deployment in landing will ..... ? a. b. c. d.

Require a shallower approach. Permit a steeper approach. lncrease landing speed. lncrease landing run.

FLAPS 26. Stalling angle is typically degrees?

........ degrees with

plain flaps deployed vhile that with split flaps is

.........

CLIMBING AND DESCENDING

CLIMB 1. What speed is requiredto achieve maximum endurance in a piston engine powered and jet engine powered aircraft respectively?

CLIMB 2. Select the approp~iatewords to complete the following statement. Fuel flow in a piston engine aircraft is proportional to .......... whilst that in a jet powered aircraft is proportional to ......... Thrust output of a jet engine ..................with increasing airspeed whilst that of a piston engine ............

I. Is approximately constant 2. Thrust 3. Reduces rapidly 4. Power 5. RPM

CLIMB 3. Select the correct words to complete the following statement. To achieve the maximum possible glide range it is necessary to fly at ...........This is achieved by flying the aircraft in a ....................condition and at ......... a. b. c. d.

BEST L:D RATIO BEST L2:D RA1-10 BEST L:D RATIO BEST L2:D RATIO

200FLAPS

VMp

FLAPS UP FLAPS UP FLAPSUP

'MD 'MD -

YAP

56

Climbing and Descending

CLlMB 4. What speed is required to achieve maximum angle of climb in a jet ai raft and a piston aircraft respectively and for what purpose might this be required?

a. V~~ b. "MP

Minimum safe speed Minimum safe speed

c. Minimum safe speed

M 'D

d- V~~

Minimum safe speed

Ob: lcle clearance after take-off. Ob: lcle clearance after take-off. Gai height rapidly. Gai height rapidly.

CLlMB 5. An aircraft of mass 200000 Kg is able to achieve a maximum climb ! adient of 5%. At what mass would it be able to achieve a maximum gradient of 4%? a. b. c. d.

250000 Kg. 150000 Kg. 300000 Kg. 275000 Kg.

CLlMB 6. What happens to the range between minimum and maximum flig : speeds for a subsonic aircraft as altitude increases? a. b. c. d.

It increases. It decreases. It remains constant. It decreases then increases.

CLlMB 7. Select the appropriate words to complete the following statement. Thrust horsepower output of a propeller aircraft ...... with increasin< iirspeed whilst that of a jet ... 1. 2. 3. 4. 5.

Is approximately constant. Is unchanged. lncreases rapidly then decreases rapidly. Reduces slowly. lncreases approximately linearly.

CLlMB 8. When flying at VMDan aircraft has a C, of 0.45 and a CDof 0.0225. 1 ts engines fail when flying at 36000 feet what will be its maximum glide range? a. b. c. d.

100 nm. 200 nm. 120 nm. 175 nm.


57

CLlMB 9. When flying at VMDan aircraft weighing 400000 Ibf has a C, of 0.45 and a CDof 0.0225. If its engines fail when fljling at 36000 feet what will be its maximum glide range if the pilot immediately dumps 100000 Ibf of fuel? What effect will the reduced weight have on VMD? a. 100 nm b. 200 nm. c. 120 nm. d. 120 nm.

Reduce VMDby 25%. Increase VMDby 25%. No change in VMD. Reduce VMD.

CLlMB 10. What is the available rate of climb at service ceiling for a piston and jet aircraft respectively? a. b. c. d.

500 fpm 100 fpm 50 fpm 250 fpm

100 fpm. 500 fpm. 100 fpm. 500 fpm.

CLlMB 11. What speed is required to achieve best rate of climb in a jet and piston aircraft respectively? a. b. C. d.

VMD VMp Between VMpand VMD Greater than VMD

VMp. VMDBetween VMDand V., Less than V .,,

CLlMB 12. How is the absolute ceiling indicated on a power available i power required graph for a piston and jet aircraft respectively? a. The power available curve will be tangential to the power required curve. b. The distance between power available and power required will be just sufficient to give rates of climb of 100 fprn and 500 fprn respectively. c. For a piston aircraft the power available curve will be tangential to the power required curve. A thrust available i thrust required curve must be used for a jet aircraft. d. At VMDand VMprespectively.

CLlMB 13. What will be the effect of a headwind on glide range and glide angle respectively? a. b. c. d.

Decrease No effect Decrease lncrease

Decrease. Increase. Increase. No effect.

58


CLIMB 14. What will be the effect of a tailwind on glide range and rate of des a. b. c. d.

Increase No effect Decrease Increase

nt respectively?

Decrease. Increase. Increase. IVo effect

CLIMB 15. Which of the following best describes the effect of flap deploymer a. lncreased lift at any given speed, reduced L;D ratio, increas 1 take-off speed and landing speeds, increased angle of climb. b. lncreased lift at any given speed, increased L;D ratio, reduc 1 take-off speed and landing speeds, reduced angle of climb. c. lncreased lift at any given speed, reduced L;D ratio, reducl take-off speed and landing speeds, reduced angle of climb. d. lncreased lift at any given speed, reduced L;D ratio, increa5 1 take-off speed and landing speeds, no effect on angle of climb.

CLIMB 16. Best endurance for a piston aircraft is achieved at V ,,, is this so?

whilst that fa

I jet

aircraft is achieved at (V,).

Why

a. Fuel consumption in a piston aircraft is proportional to pow output, whereas that of a jet aircraft is proportional to thrust and hence drag. b. Fuel consumption in a piston aircraft is proportionalto rpm, wt -easthat of a jet aircraft is proportional to power and hence drag. c. Fuel consumption in a piston aircraft is proportional to tl rst, whereas that of a jet aircraft is proportional to power and hence drag. d. Fuel consumption in a piston aircraft is proportional to pow output, whereas that of a jet aircraft is proportional to rpm.

CLIMB 17. Which of the following best describes the effect of increasing altitt a. Both power available and power required increase with in them narrows so rate of climb reduces. b. Both power available and power required decrease with altit so rate of climb reduces. c. Power available increases with altitude whilst power requi and hence rate of climb increases. d. Power available reduces with altitude whilst power require and hence rate of climb decreases.

3

on maximum rate of climb?

2asing altitude but the gap between le but the gap between them narrows

1 reduces so the gap between them increases so the gap between them


59

CLIMB 18. Which of the following best describes the manner in which best climb speed varies with altitude? a. Best climb €AS increases whilst best climb TAS reduces with increasing altitude for both piston and jet aircraft. The effect on TAS is of the same magnitude in both cases. b. Best climb EAS and best climb TAS both increase with increasing altitude for both piston and jet aircraft. The effect on TAS is greater for a piston than for a jet. c. Best climb EAS and TAS both decrease with increasing altitude for both piston and jet aircraft. The effect on TAS is greater for a jet than for a piston. d. Best climb EAS reduces whilst best clirr~bTAS increases with increasing altitude for both piston and jet aircraft. The effect on TAS is greater for a piston than for a jet.

CLIMB 19. How will a headwind and a tailwind respectively affect best range glide speed? a. b. c. d.

Decrease Decrease lncrease lncrease

lncrease Decrease Decrease lncrease

CLIMB 20. If an aircraft enters a banked turn whilst climbing to cruising altitude what effect will this have on rate of climb at constant power setting? a. Climbing and turning occur in different planes and hence are not related, so rate of climb will remain constant regardless of bank angle. b. Banking will increase load factor and hence induced drag. This will reduce the rate of climb. c. Although banking wiil increase load factor and hence induced drag, rate of climb depends only on excess thrust. Rate of climb will therefore remain unchanged. d. Banking will reduce the climb gradient, so more power will be available to increase rate of climb.

CLIMB 21. Select the appropriate words to complete the following statement. Fuel flow in a piston engine aircraft is proportional to .......... whilst that in a jet powered aircraft is directly proportional to ......... Thrust available from a jet engine .................. with increasing airspeed whilst that of a piston engine ................

1. 2. 3. 4. 5.

Is approximately constant Thrust Reduces rapidly Power Increases approximately linearly

60


CLlMB 22. An aircraft weighing 200000 Ibf has a maximum excess power What will be its maximum rate of climb? a. b. c. d.

ailable of 1500 Thrust Horse Power.

512 fpm. 753 fpm. 248 fpm. 358 fpm.

CLlMB 23. An aircraft weighing 50000 Ibf requires a thrust of 20000 Ibf when What will be its maximum climb gradient at this speed assuming it! Ibf and the total drag force does not change during the climb?

(ingstraight and level at 250 Kts IAS. laximum thrust at this speed is 40000

CLlMB 24. When does the second segment climb begin? a. b. c. d.

At the start of flap retraction. ,, to V,. At the start of acceleration from ,V At the end of landing gear retraction. At the start of acceleration from V, to flap retraction speec

CLlMB 25. When does the third segment begin? a. b. c. d.

When the flaps are fully retracted. 400 ft gross height minimum. ,, to V,. At the start of acceleration from ,V At the start of acceleration to flap retraction speed.

CLIMB26. In a constant mach descent the pitch angle will? a. Decrease. b. Remain constant. c. Increase. d" Increase then decrease.

CLlMB 27. An aircraft weighing 50000 Ibf requires a thrust of 20000 Ibf when If it were to climb to 40000 ft what would be its power available in a. b. c. d.

114 113 112 115

ing straight and level at 250 Kts IAS. ~mparisonto that at ISA msl?


61

CLIMB 28. What would be the effect of sweeping back the wings of a variable geometry aircraft in gliding flight? a. b. c, d.

Profile drag would reduce unless glide angle was reduced. Induced drag would increase unless glide angle was increased. Glide range and rate of descent would be increased. Glide range and rate of descent would be decreased.

CLIMB 29. How is maximum rate of emergency descent achieved in a high speed jet aircraft? a. Pitch down, retract flaps, spoilers and undercarriage and increase thrust, b. Deploy spoilers, reduce thrust to idle, use pitch to maintain speed within limits c. Deploy high lift device, retract undercarriage and spoilers. Increase thrust to maximum, pitch 45 degrees nose down. d. Thrust to idle, increase pitch to stalling angle, maintain nose up attitude to hold aircraft in the stall.

CI-IMB 30. What do points A and B represent on the whole aircraft polar diagram right? a. b. c. d.

Best L;D ratio Minimum drag Best L:D ratio Stalling

Minimum drag. Minimum power Stalling. Best L:D ratio.

CI-IMB 31. Why must there be an acceptable approach climb gradient? a. b. c. d.

To maintain control in turbulence. To maintain altitude on the approach. To retain speed stability on the approach. To ensure adequate go-around performance.

CLIMB 32. An aircraft weighing 50000 Ibf is able to achieve a maximum rate of climb of 1500 ft / min when climbing at 250 Kts TAS. What will be the maximum all up weight at which it can achieve a 5% climb gradient at this speed assuming its power available and total drag do not change during the climb?

62


CLIMB 33. Increased weight reduces the rate of climb and climb gradient but? a. b. c. d.

Vx increases and V, decreases. Vx and Vy increase. Vx and Vy decrease. Vx decreases and Vy increases.

CLIMB 34. Best climb gradient is achieved by flying at approximately? a. b. c. d.

1.1 V., 1.2 V., Best C, : CD2. Best C, : C,.

CLIMB 35. Absolute ceiling occurs when? a. b. c. d.

ROC is 300 Wmin. ROC is 750 Wmin. For a jet ROC is 100 Wmin and 500 Wmin for a propeller airc aft. ROC is zero for both jets and propeller aircraft.

CLIMB 36. What airspeed will produces the greatest glide endurance? a. ,V, b. V ,, C. .v , d. V ,,

which is higher than V .,, which is lower than V .,, or ,V,

depending on engine type.

CLIMB 37. What factors determine maximum glide range? a. b. c. d.

Wind, C,:C, ratio and weight. Wind, C,:C, ratio and height. Wind, Speed and weight. Wind, C,:C, and altitude.

CLlMB 38. What are,,V , a. b. c. d.

and?,V ,,

Maximum and minimum speeds for gliding, Minimum and maximum speeds for gliding. Speed for shallowest glide path and speed for lowest rate of jescent. Speed for maximum glide range and speed for minimum glic ? range.


CLlMB 39. How do VMGA and VMDR compare? a. b. d. e.

VMGA is greater than,,V , . VMGA is less than VMDR . VMGA is the same as,,V, . Depends on aspect ratio.

CLlMB 40. Which of the following is equal to lift in a steady climb or descent? a. b, c. d.

W Cos angle of climb or descent. W Sin angle of climb or descent. W Tan angle of climb or descent. None of the above.

CLlMB 41. What proportion of thrust is employed in supporting the weight of an aircraft in a steady climb? a. b. c. d.

W Sin angle of climb. W Cos angle of climb. W Tan angle of climb. None of the above.

CLlMB 42. What provides maximum glide range? a. b. c. d.

Strong headwind. Light headwind. Strong tailwind. Light tailwind.

CLlMB 43. What would give maximum glide range in a headwind? a. b. c. d.

Flying faster than V .,, Flying slower than V .,, Flying slower than V .,, Flying faster than V .,,

CLlMB 44. What would give maximum glide range in a tailwind? a. b. c. d.

Flying faster than V .,, Flying slower than .V ,, Flying faster than V .,, Flying slower than V.,

63

64


CLlMB 45. Which of the points on the CL:C, polar would give maximum glide range?

CLlMB 46. In a steady climb? a. b. c. d.

Lift is less than weight and thrust is less than drag Lift is less than drag and weight is less than thrust. Weight is less than lift and drag is more than thrust. Weight is more than lifl and drag is less than thrust.

CLlMB 47. What is the speed for rninimum sink rate'?

a. b. C. d.

VMsw V .,, VMP. V .,,

CI-IMB 48. What flap position would give maximum glide range?

CLlMB 49. What speed gives best angle of climb in a jet aircraft?

CLlMB 50. Which of the following occur at V ?,, 1. 2. 3. 4. 5.

L:D Max. Max jet endurance. Max prop range. Max jet climb angle. Max glide range all types.

1


a. b. c. d.

I,2, 3. 2, 3, 4. 3, 4, 5. All of the above.

CLIMB 51. What will be the effect of an increase in weight?

I. V ,, will increase. 2. Glide range will decrease. 3. Glide angle will increase. 4. Glide range and angle win be unaffected. a. b. c. d.

I,2, 3. 2 , 3, 4. Non,e of the above. 1,4.

CLIMB 52. In what direction does lift act in a steady climb? a. b. c. d.

Upwards. Vertically. At right angles to the Right path. None of the above.

CLIMB 53. What is the absolute ceiling of an aircraft? 1. 2. 3. 4.

The altitude where the low speed and high speed stall lines cross. The altitude at which power required is equal to power available. The altitude at which thrust available is equal to drag. The altitude at which rate of climb is zero.

a. b. c. d.

1, 2, 3. 2, 3, 4. None of the above. All of the above.

CLIMB 54. Increasing aircraft weight.. ........glide speed and.. .....rate of descent? a. b. c. d.


Decreases. Increases. Decreases. increases.

65

66

1


CLlMB 55. To descend at constant glide angle and IAS, the pitch attitude must.. .........? a. b. c. d.

Increase. Decrease. lncrease then decrease. Remain constant.

CLlMB 56. For a constant mach number descent, gradient must.. ....? a. b. c. d.

Decrease. Increase. lncrease then decrease. Remain constant.

CLlMB 57. For a constant IAS descent, gradient must.. ....? a. b. c. d.

Decrease. Increase. lncrease then decrease. Remain constant.

CLlMB 58. A headwind in a constant mach nurnber clirr~bwill ............ angle of a. b. c. d.

Decrease. Increase. Not affect. lncrease or decrease depending on mach number.

CLlMB 59. For maximum glide range, TAS must .....? a. b. c. d.

Decrease in a headwind. Decrease in a tailwind. lncrease in a tailwind. Remain constant.

CLlMB 60. To descend at constant IAS the a. b. c. d.

pitch down, Pitch down, Pitch up, pitch up,

............must be.. ........7. Increased. Maintained constant. Increased. Decrease.


CLIMB 61. Best climb angle is achieved at ............? a. .V ,, b. .V ,, c. v,. d. V.,

CLIMB 62. Maximum rate of climb occurs at ...........speed? a. b. c. d.

Maximum excess thrust. Minimum drag. Minimum thrust. Maximum excess power.

CLIMB 63. Maximum glide range is achieved at ............ ? a. C,4: C, b. C,3: C, C. C,: C, d. C,2 : C,

Max. Max. Max. Max.

CLIMB 64. A steady climb requires ...... thrust and a. b. c. d.

Less Less More 'More

..... power compared with constant speed level flight?

less. more. more. Less.

CLIMB 65. Increasing altitude at constant weight and mach number requires? a. b. c. d.

lncreased TAS. lncreased IAS. lncreased A of A. Constant A of A.

CLIMB 66. The correct procedure for an emergency descent is? a. b. c. d.

Maximise drag, minimise thrust, push the nose hard down. Minimise drag, maximise thrust, push the nose hard down. Maximise drag, minimise thrust, control speed with pitch attitude. Maximise drag, minimise thrust, bank steeply.

67

68


CLlMB 67. What is the correct equation?

CLlMB 68. Rate of climb is .......... by a headwind? a. Increased. b. Decreased. G. Not affected. d. lncreased or decreased depending on load factor.

CLIMB 69. Angle of climb is .......... by a headwind? a. b. c. d.

Increased. Decreased. Not affected. lncreased or decreased depending on load factor.

CLlMB 70. Climb gradient is ........ by a headwind? a. b. c. d.


CLlMB 71. Power required is .......... as altitude increases in a steady climb? a. b. c. d.


CLlMB 72. When climbing to cruise altitude with a headwind? a. b. c. d.

Climb time is decreased. Ground distance is decreased. Climb time is increased. Ground distance is increased.


69

CLlMB 73. If excess power is 25000 ft Ibflmin and aircraft weight is 10000 Ibf what will be the maximum rate of climb? a. b. c. d.

2.5 Wmin. 25 Wmin. 250 Wmin. 2500 Wmin.

CLlMB 74. If maximum thrust is 25000 Ibf, drag is 15000 Ibf and weight is 10000 Ibf what will be the maximum climb gradient? a. b. c. d.

100%. 50%. 70%. Infinite ( or vertical).

CLlMB 75. If at 250 Kts, excess power is 50,000,000 ft Ibflmin, and weight is 10000 Ibf, what is maximum angle of climb?

CLlMB 76. What might the points C and D represent on the whole aircraft polar Diagram right? 3

a. b. c. d.

Best L;D ratio Jet aircraft V, Best L:D ratio Prop aircraft V,

Minimum drag. Prop aircraft V., Stalling. Jet aircraft V.,

CLlMB 77. What might points C and D represent on the whole aircraft polar Diagram right? CL a. b. c. d.

Prop aircraft V, Prop aircraft V, VMD. Prop aircraft V,

Prop aircraft V.,

v~ Prop aircraft V.,

v ~ ~ -

70


CLlMB 78. Increasing weight will ........ glide speed and ........ maximum glide a. b. c. d.

Increase Increase Decrease Decrease

idurance?


CLlMB 79. Increasing weight will a. b. c. d.

......... ,V,,


and .......... rate of descent for b ,t glide range?


CLlMB 80. Deploying landing flap will a. b. c. d.


........ V,, and .......... maximum angle c :limb?


CLlMB 81. Use of reheat in a climb will ........ maximum angle of climb, and ... a. b. c. d.


... maximum rate of climb?


CLlMB 82. If use of reheat doubles thrust available this will a. b. c. e.

'

.......... the maxim1

1 angle of

Half. Double. More than double. Less than double.

CLIMB 83. If TAS is 200 Kts and rate of climb is 1000 Wmin, what is the climb

adient?

climb?


71

CLlMB 84. If climb gradient is 15% and TAS is 250 Kts, what is the rate of climb? a. b. c. d.

1750 ftlmin. 2750 ftlmin. 3750 ftlmin. 4750 ftlmin.

CLlMB 85. For maximum rate of descent use ....thrust, .......... Drag, and ...........to maintain airspeed within limits? a. b. c. d.

Maximum Maximum Minimum Minimum

Minimum Minimum Maximum Maximum

Spoilers. Attitude. Spoilers. Attitude.

CLlMB 86. In a constant IAS climb ....... might be inadvertently exceeded? a. .,V, b. V ., c. v,. d. M .,,

CLlMB 87. In a constant TAS climb ........ might be inadvertently exceeded? a. .V ,, b. V ., c. v,. d. .M ,,

CLlMB 88. In a constant mach climb ....... might be inadvertently exceeded? a. b. c. d.

.,V, V ., v,. .M ,,

CLlMB 89. In a constant mach descent ....... might be inadvertently exceeded? a. b. c. d.

.,V, V ., v,. .M ,,

72

1


CLIMB 90. When climbing at constant XpV2 IAS will a. b. c. d.

.......3.

Increase. Decrease. Remain constant. Increase up to 36000 ft then remain constant.

CLIMB 91. If at 40000 feet altitude air density is Xi of its sea level value, how c I best climb TAS compare with best climb IAS? a. b. c. d.

The same. Double. Quadruple. Less than.

CLIMB 92. If maximum thrust is 10000 Ibf, drag is 500 Ibf and weight is 5000 climb at the absolute ceiling?

f, what will be the maximum rate of

a. 100 Wmin. b. 500 Wmin. C. zero. d. More than 500 Wmin.

CLIMB 93. How does V, compare with V, at the absolute ceiling? a. b. c. d.

The same. More than. Less than. More or less than depending on aircraft type.

CLIMB 94. How does the angle of climb of a propeller aircraft vary with increa: rg airspeed? a. b. c. d.

Decreases. Increases. Remains constant. lncreases up to V, then decreases.

CLIMB 95. How does maximum angle of climb for a jet aircraft vary with incre: ing airspeed? a. b. c. d.

Decreases. Increases. Remains constant. lncreases up to V, then decreases.


73

CLlMB 96. How does maximum rate of climb for a jet aircraft vary with increasing airspeed? a. b. c. d.

Decreases. lncreases up to ,V, then decreases. Remains constant. lncreases up to 1.3VM, then decreases.

CLlMB 97. If weight is 15000 Ibf, maximum thrust is 25000 Ibf and drag is 5000 Ibf, what will be the maximum climb angle?

CLlMB 98. If excess power is 350 THP when TAS is 250 Kts and weight is 10000 Ibf, what will be the maximum rate of climb at that speed? a. b. c. d.

1155 ftlmiri. 2155 ftlmin. 3155 filmin. 4155 filmin.

CLlMB 99. If the weight of the aircraft in question 98 is doubled how will this affect maximum rate of climb? a. b. c. d.

Doubleit. Half it. Increase it. Decrease it by 4.1%.

CLlMB 100. If airspeed is maintained constant as aircraft weight reduces due to fuel use the aircraft will be in a ......? a. b. c. d.

'Constant rate climb. Cruise climb. Zoom climb. Constant altitude flight.

CLlMB 101. When climbing to cruise altitude with a headwind? a. b. c. d.

Climb rate is decreased. Climb rate is increased. Ground speed is decreased. Ground speed is increased.

74


CLlMB 102. An aircraft of mass 120000 Kg is at a pressure altitude of 10000 ft, vith an OAT of + I 3OC.If its IAS is 90 Kts and its climb gradient is 3.5%, which of the options below is clc ;est to its ROC? a. b. c. d.

245 Wmin. 360 Wmin. 945 Wmin. 300 Wmin.

CLlMB 103. When descending from FL400 to FL360, then from FL360 to F1 200 at constant mach number, the descent gradient will? a. b. c. d.

lncrease all the way. lncrease then decrease. lncrease then remain constant. Constant all the way.

CLlMB 104. When descending from FL360 to FL260 at constant mach number, ' en from FL260 to FL 100 at constant CAS, the descent gradient will? a. b. c. d.

lncrease all the way. lncrease then decrease. lncrease then remain constant. Cor~stantall the way.

CLlMB 105. A small twin prop aircraft is climbing from a screen height of 50 ft 2 a gradient of 10%. By how much will it clear an obstacle 850 m above field elevatio at a distance of 10000 m from the screen?

CLlMB 106. The climb schedule for an aircraft is 300 Kts 1 mach 0.84. At the cr ssover altitude? a. If the aircraft is at the correct CAS it will automatically be a1 b. The must be accelerated or decelerated to the correct mac c. If the CAS is correct and the mach number is too low that is too high the aircraft must be decelerated to the correct mac d. It the aircraft is at the correct CAS it will always be necess number.

he correct mach number. number. cceptable, but if the mach number is I number. ry to decelerate to the correct mach


75

CLIMB 107. The climb schedule for an aircraft is 300 Kts / mach 0.84. When climbing from the crossover altitude? a. b. c. d.

Climb gradient will increase. The must be accelerated or decelerated to the correct mach number to maintain constant ROC. Climb gradient will decrease. It the aircraft is at the correct CAS it will always be necessary to decelerate to the correct ROC.

CLIMB 108. The climb schedule for an aircraft is 300 Kts / mach 0.84. When climbing from the crossover altitude? a. b. c. d.

Pitch attitude must be reduced to maintain constant gradient. Pitch attitude must be increased to maintain constant gradient. Pitch attitude must be increased to increase gradient. Pitch must be increased to maintain mach number but gradient will increase.

CLIMB 109. An aircraft is conducting certification glide test flights. For the first test its mass is 100000 Kg and it achieves a maximum glide range of 100 nm from an altitude of 36000 ft. If it then climbs back to 36000 ft to repeat the test, its best glide speed will be ..... and its glide range will be.. ..... compared to the first test?

. a. Higher b. Higher c. Lower d. Lower

greater. less. The same. greater.

CLIMB 110. An aircraft is conducting certification glide test flights. For the first test its mass is 100000 Kg and it achieves a maximum glide range of 100 nm from an altitude of 36000 ft. If it then climbs back to 36000 ft to repeat the test, its best glide L:D ratio will be ..... and its glide angle will be.. ..... compared to the first test? a. b. c. d.

Higher The same Lower The same

less. greater The same. the same.

CLIMB 111. An aircraft is conducting certification glide test flights. For the first test its mass is 100000 Kg and it achieves a maximum glide range of 100 rim from an altitude of 36000 ft. If it then climbs back to 36000 ft to repeat the test? a. It will need to fly faster to achieve the same glide angle and ROD as in the first test. b. It will need to fly faster to achieve the same glide angle but at a higher ROD than in the first test. c. It will need to fly slower to achieve the same glide angle, but will have a lower ROD and glide range than in the first test. d. It will need to fly slower to achieve the same glide angle and glide range, but will have a lower ROD than in the first test.

76


CI-IMB 112. When conducting a type B noise abatement procedure ........ must be maintained up to ......? a. b. C. d.

V, V, V, + 10-20 Kts V, + 10-20 Kts

1500 ft. 1000 ft. 1500 ft. 1000 ft.

CLIMB 113. Decreasing nose down pitch angle in a glide will? a. b. c. . d.

lncrease range and endurance. Decrease range and endurance. lncrease range and decrease endurance. Decrease range and increase endurance.

CLIMB 114. What would reduce glide range by the greatest degree?

I I

a. b. c. d.

Tailwind. Weight increase. Weight decrease. Gear deployment.

CLIMB 115. What would increase glide range b) 'he greatest degree? a. b. c. d.

Tailwind. Weight increase. Weight decrease. Flap deployment.

CLIMB 116. If ROC = 250 Wmin and TAS = 150 Kts, which of the following be$ describes climb gradient in still air? (Assume IKt = 100 fpm)

CLIMB 117. As a flight progresses, how do best ROC and speed for best ROC pary? l

a. b. c. d.

lncrease lncrease Decrease decrease

decrease. increase. decrease. increase.

Climbing and Descelldillg

77

CLlMB 118. When climbing at a constant mach number in the troposphere what is the effect on CAS and TAS? a. b. c. d.

Decrease Decrease Increase Increase


CLlMB 119. What is the effect of increased weight on best ROC and speed for best ROC? a. b. c. d.

Increase Increase Decrease decrease


CLlMB 120. When gliding at constant mach number pitch angle and gradient will? a. b. c. d.

Increase Decrease Increase then decrease Decrease then remain constant.

increase. decrease. increase. Decrease.

CLIMB 121. When conducting a type A noise abatement climb procedure ........ must be maintained up to ........? a. V, + 10 to 20 Kts b. V, c. v, d. V, + 10 to20 kts

1500 ft. 1500 ft. 1000 ft, 4000 ft.

CLlMB 122. The first segment of a type B noise abatement climb procedure is from ..... to ..... with flaps and thrust at ....... and gear ... . ..? a. Screen height b. VLOF c. End ofTORR d. Screen height

1000 fi 1000 ft 1500 ft 1500 ft

take-off setting climb setting take-off setting climb setting

UP. UPdown. UP.

CLlMB 123. When would a type A noise abatement climb be used? a. b. c. d.

For all take-offs after 2300 For all take-offs between 2300 and 0700 the next day. When the noise sensitive area is very close to the airport. When the noise sensitive area is some distance from the airport.

78


CLIMB 124. When would a type 6 noise abatement climb be used? a. b. c. d.

For all take-offs after 2300 For all take-offs between 2300 and 0700 the next day. When the noise sensitive area is very close to the airport. When the noise sensitive area is some distance from the airport.

CLIMB 125. A noise abatement climb procedure comprises of ......segments, the first of which starts at ........, and the third of which ends at ........7 a. 3 b. 3

screen height screen height

C. 2 d. 2

v~~~ End of the TORR

cruise altitude. 3000 ft. 3000 ft. 1500 ft. I

CLIMB 126. The second segment of a type 6 noise abatement procedure take-off commences at ...... from where the aircraft is ............., then climbed to ....... f, whilst being ......? a. b. c. d.

1000ft Screen height 1000 ft 1500ft

accelerated to V4 accelerated to VzF accelerated to VzF accelerated to V,

3000 ft 1500 ft 3000 ft 2000 ft

CLIMB 127.

I

,

accelerating to V4 + 10 kts. accelerated to VzF+ 10 Kts. accelerated to VzF+ 10 Kts. accelerated to V.,

I

The second segment of a type A noise abatement procedure take-off commences at ...... from where the aircraft ....... to ........ ? a. b. c. d.

1000ft Screen height 1500 ft 1500ft

climbs at climb power climbs at take-off power climbs at climb power climbs at max continuous power

3000 ft. 1500 ft. 3000 ft. , 2000 ft.

'

I I

CLIMB 128. An aircraft climbing to cruising altitude after take-off has a thrust td weight ratio of 1:3. What will be its climb gradient if its lift : drag ratio at climb speed is 20:lli It may tJe assumed that lift = weight in this climb.


79

CLIMB 129. If an aircraft climbs to altitude at a constant IAS considerably lower than M ?,,

a. b. c. d.

Drag will be constant but climb gradient will decrease. Drag and climb gradient will decrease. Drag will increase causing climb gradient to decrease. Both drag and climb gradient will remain constant.

CLIMB 130. An aircraft climbing to cruising altitude after take-off has a thrust to weight ratio of 1:5. What will be its climb gradient if its lift : drag ratio at climb speed is 25:1? It may be assumed that lift equals weight in this question.

CLIMB 131. Descending at constant mach number might cause? a. b. c. d.

MMoto be exceeded. VMoto be exceeded. Shock stall. Tuck under.

CLIMB 132. Climbing at constant IAS might cause? a. b. c. d.

MMoto be exceeded. VMoto be exceeded. Low speed stall. TAS to reduce.

CLIMB 133. Increasing IAS in a climb will ......... the altitude at which the mach limit is reached? a. b. c. d.

Decrease. Increase. Not affect. lncrease or decrease depending on temperature.

CLIMB 134. When descending at constant IAS a headwind will? a. b. c. d.

lncrease distance covered over ground. Decrease descent time. lncrease descent time. Decrease distance covered over ground.

80


CLlMB 135. When descending at constant TAS a headwind will? a. b. c. d.

Decrease gradient. Decrease descent time. lncrease descent time. lncrease gradient.

CLlMB 136. not be affected by .........?

V,,wil a. b. c. d.

Weight. Altitude. Power setting. Humidity.

CLlMB 137. When descending at constant mach number below 35000 feet angle of attack must? a. b. c. d.

Increase. Decrease. Remain constant. lncrease or decrease depending on temperature.

CLlMB 138. When descending at constant IAS below 35000 feet angle of attack must? a. b. c. d.


CLIMB 139. When descending at constant TAS below 35000 feet, angle of a t t a ~ kmust? a. b. b. c.


CLlMB 140. When climbing at constant Mach number above the tropopause IAS will ..... and TAS will ..... ? a. b. c. d.

Decrease, Decrease, Increase, Increase,

Remain constant. Decrease. Remain constant. Increase.

I

I


CLIMB 141. In a constant mach number climb in the troposphere true airspeed? a. b. c. d.

Increases then decreases. Decreases. Remains constant. Increases.

CLIMB 142. In a constant mach number climb true airspeed? a. b. c. d.

Increases then remains constant. Decreases then remains constant. Decreases then decreases less slowly. Decreases then decreases more slowly.

CLIMB 143. In a constant TAS climb in the troposphere? a. b. c. d.

IAS increases. IAS decreases. IAS increases then remains constant. IAS decreases then remains constant.

CLIMB 144. In a constant IAS climb in the troposphere? a. b. c. d.

TAS increases. TAS decreases. TAS increases then remains constant. TAS decreases then remains constant.

CLIMB 145. In a constant TAS climb? a. b. c. d.

IAS increases. IAS decreases. IAS increases then increases more slowly. IAS decreases then decreases more quickly.

CLIMB 146. In a constant IAS climb in the troposphere? a. b. c. d.

TAS increases and mach number decreases. TAS increases and mach number increases. TAS decreases and mach number increases. TAS decreases and mach number decreases.

81

82


CLIMB 147. In a constant mach number descent there is a danger that? a. b. c. d.

MMowill be exceeded. M ,,,, will be exceeded. VMowill be exceeded. V, will be exceeded.

CLlMB 148. Climb gradient in still air is closest to? a. b. c. d.

Height gained divide by distance moved through the air. Distance moved through the air divided by height gained. TAS divided by ROC. ROC divided by CAS.

CLlMB 149. As altitude increases? a. b. c. d.

V, V, V, V,

and V, both increase. and V, both decrease. decreases and V, remains constant. decreases and V, remains constant.

CLlMB 150. What happens as a propeller aircraft climbs to its absolute ceiling? a. b. c. d.

V, VMax,VMin,and V, converge on V., VMin,VMax, VMpand VMDconverge on V., VMin,VMaxand V, converge on V., The CAS values equating to V, and V, both decrease.

CLlMB 151. What happens to the CAS values of V, and VMDas a propeller aircraft climbs to its absolute ceiling? a. b. c. d.

Increase. Decrease. Remain constant. Increase and decrease respectively.

CLlMB 152. Climb gradient is closest to? a. b. c. d.

Height gained divide by distance moved through the air. Distance moved through the air divided by height gained. TAS divided by ROC. ROC divided by ground speed.

I

I


83

CLlMB 153. Climb gradient is proportional to? a. b. c. d.

(T-D)/W. (T+D)/W. TASIW. Mach numberw.

CLlMB 154. If the climb schedule for an aircraft is changed from 27510.81 to 30010.81, how will the crossover altitude be affected? a. b. c. d.

Increase. Decrease. No change because it depends only on CAS. No change because it depends only on Mach number.

CLlMB 155. The climb schedule for an aircraft is 28010.81. When climbing above the tropopause TAS will? a. b. c. d.

Increase. Decrease. Remain constant. Decrease then increase at very high altitude

CLlMB 156. If the climb schedule for an aircraft is 28010.75 what will happen to the crossover altitude if temperature increases? a. b. c. d.

Increase. Decrease. Remain constant because TAS is not affected by temperature. Remain constant because mach number is not affected by temperature.

CLlMB 157. When gliding the speed giving minimum rate of descent will be?

CLlMB 158. When climbing the mach number for low speed buffet will? a. b. c. d.

Increase. Decrease. Remain constant because low speed buffet is not caused by mach number. Remain constant because mach number is not affected by altitude.

84


CLIMB 159. What happens to climb gradient as an aircraft climbs through the crossover altitude? a. b. c. d.

Increases. Decreases. Remains constant. Increases or decreases depending on mass.

CLIMB 160. What happens to speed as an aircraft climbs through its crossover altitude? a. b. c. d.

It changes from constant TAS to constant mach. It changes from constant CAS to constant TAS. It changes from constant CAS to constant mach. It changes from constant mach to constant CAS.

CLIMB 161. Why must the speed of an aircraft change when it climbs through its crossover altitude? a. b. c. d.

,, To avoid exceeding.V To avoid exceeding M .,, .,, To avoid exceeding M To avoid low speed stall.

POWER REQUIRED AND POWER AVAILABLE

/

POWER I. If TAS is increased from 300 Kts to 400 Kts with no change in altitude, configuration, or weight, by what percentage will power required change? a. b. c. d.

Decrease by 135%. Decrease by 35%. lncrease by 135%. lncrease by 235%.

POWER 2. What is the relationship between pewer required and TAS, as an aircraft accelerates above V ?,, a. b. c. d.

Power required increases in direct proportion to TAS. Power required increases in inverse proportion to TAS. Power required increases in proportion to (TAS)3. Power required increases in proportion to (TAS)2.

POWER 3. At what % of its stalling speed must a jet aircraft fly to achieve maximum endurance for a given fuel load?

POWER 4. For a piston aircraft at constant weight, angle of attack, and configuration, what will be the effect of increasing altitude? a. b. c. d.

lncreased power and TAS will be required. lncreased power will be required but at the same TAS. Lower power will be required at the same TAS. The same power will be required but at an increased TAS.

POWER 5. For a piston aircraft at a constant altitude, angle of attack, and configuration, what will be the effect of increasing weight? a. b. c. d.

More power will be required but at the same TAS. More power will be required but at a higher TAS. The same power will be required but at a higher TAS. More power will be required but at a lower TAS.

86

Power required and Power available

POWER 6. Compared with still air, when flying for maximum range into a headwind, speed should be? a. b. c. d.

Faster. The same. Slower. Depends on weight.

POWER 7. What flight condition requires least power at a given IAS? a. b. c. d.

Zero flap high altitude. Zero flap low altitude. 30° flap low altitude. 30° flap high altitude.

POWER 8. Maximum propeller aircraft range occurs at? a. .V ,, b. .V ,, c. v,. d. V.,

'

POWER 9. Power required for a given IAS at 40000 feet altitude, is .......... times that required at ISA MSL?

POWER 10. Power required is proportional to? a. b. c. d.

TAS. IAS. TAS2. TAS3.

POWER 11. Power required equals? a. b. c. d.

Drag x IAS. Drag x CAS. Drag x EAS. Drag xTAS.


87

POWER 12. As altitude increases, power available from a piston or turbo-prop ........,whilst that of a turbo-jet a. b. c. d.


..........7.

Increases. Decreases. Decreases. Increases.

POWER 13. As altitude increases, the power required curve moves ...., whilst the power available curve moves.. ......? a. b. c. d.

Up and right, Up and left, Down and left, Down and right,

Down. UP. UP. Down.

POWER 14. As altitude increases jet aircraft excess power? a. b. c. d.

Increases at a constant rate. Decreases at a constant rate. Remains constant. Decreases at a rate that decreases with increasing altitude.

POWER 15. Doubling IAS at a given altitude multiplies power required by?

POWER 16. Minimum power required speed is? a. b. c. d.

Less than V,. More than .V ,, Less than .V ,, V, of propeller aircraft.

POWER 17. Maximum range for a jet aircraft occurs at? a. b. c. d.

Less than V,. More than .V ,, Less than .V ,, .V ,,

88


POWER 18. Maximum excess power IAS .......... with increasing altitude? a. Increases. b. Decreases. c. Remains constant. d. lncreases then decreases.

POWER 19. Maximum excess power TAS ............ with increasing altitude? a. b. c. d.

Increases. Decreases. Remains constant. lncreases then decreases.

POWER 20. At the absolute ceiling, excess power? a. b. c. d.

Is zero. Is maximum. Varies with engine type. Varies with TAS.

POWER 21. At the absolute ceiling, V ,, a. b. c. d.

and V ?,,

Are the same. Both increase. Both decrease. Diverge.

POWER 22. The power available and power required curves? a. b. c. d.

Never meet. Never cross. Are parallel at the absolute ceiling. Cross at the absolute ceiling.

POWER 23. The power available and power required curves? a. b. c. d.

Cross at maximum and minimum speeds. Never cross. Are always parallel. Are never parallel.


POWER 24. In gliding flight? a. b. c. d.

The power required curves are irrelevant. The power available curves are irrelevant. No power is consumed. Power consumption is always at a minimum.

POWER 25. The best ratio of TAS to power required occurs at? a. .,V, b. .V ,, c. v,. d. V.,

POWER 26. Maximum jet endurance occurs at? ,, a. .V b. .V ,, C. v,. d. V.,

POWER 27. Decreasing weight moves the power required curve? a. b. c. d.

Down and right. Down and left. Up and right. Up and left.

POWER 28. ,V,

is point .... on the diagram right?

POWER 29. Power required is? a, b. c. d.

Drag x IAS. DragxTAS. Thrust required x TAS. Thrust required x IAS.

89

90


POWER 30. Excess power 1 Weight = ? a. b. c. d.

The sine of the maximum angle of climb. The cosine of the maximum angle of climb. The sine of the maximum angle of descent. The maximum rate of climb.

POWER 31. Moving C of G forward will? a. b. c. d.

Decrease range. lncreases range. Decrease range only if C of G is forward of C of P. Decrease range only if C of G is aft of C of P.

POWER 32. If the aft C of G limit is aft of the C of P and the forward limit if forward of the C of P, then moving the C of G from its aft limit to its forward limit will? a. b. c. d.

Decrease range: lncreases range then decrease range. Increase range. Decrease range then increase range.

POWER 33. Range will be maximum when? a. b. c. d.

C of G is forward of C of P. CofGisaftofCofP. C of G and C of P coincide. ,, When flying at .V

POWER 34. An aircraft of weight of 120000 Ibf can achieve a climb gradient of 2.5% using maximum climb power. At what weight will it achieve a gradient of 3% if the change in excess power is ignored?

POWER 35. An aircraft of weight 120000 Ibf can achieve a climb gradient of 2.5% using maximum climb power. What gradient will it achieve at a weight of 150000 Ibf if the change in excess power is ignored?


91

POWER 36. If weight is increased by 25% when flying at ,V,

power required will increase by?

POWER 37. Thrust equals drag when? a. Descending at constant IAS. 1;. Flying in level flight at constant IAS. c. Climbing at constant TAS. d. None of the above.

POWER 38. Maintaining altitude after an increase in mass without changing angle of attack requires? a. b. c. d.

lncreased power and airspeed. Decreased power and airspeed. Increased power and decreased airspeed. Decreased power and increased airspeed.

POWER 39. Two identical turbojet aircraft flying at the same airspeed and altitude have the same SFC. Aircraft A weighs 200000 Kg and consumes fuel at a rate of 5000 Kglhr. If aircraft B weighs 250000 Kg what will be its fuel consumption? a. b. c. d.

6250 Kglhr. 7812 Kglhr. 8250 Kglhr. 9812 Kglhr.

POWER 40. VIM, for a turboprop is? a. b. c. d.

Lower than V., Lowest fuel consumption speed. Higher than V., Lower than V,.

POWER 41. VIM, for a turboprop is? a. b. c. d.

Higher than V., V., Higher than V., Lower than V,.

92


POWER 42. If service ceiling is 12000 ft at a weight of 50000 Kg, at a weight of 75000 Kg it will be? a. b. c. d.

Higher. The same. Lower. Higher or lower depending on engine type.

POWER 43. Increasing altitude causes the power required curve to? a. b. c. d.

Move to the left and upwards. Move to the left and downwards. lblove to the right and upwards. I'Wove to the right and downwards.

POWER 44. Decelerating at the back of the drag curve? a. b. c. d.

lncreases drag and power required. Decreases drag and power required. lncreases drag and decreases power required. Decreases drag and increases power required.

POWER 45 A propeller aircraft is inherently ....... speed stable than a jet because its thrust airspeed? a. b. c. d.

More More Less More

...... with

increasing

increases. decreases. increases. decreases.

POWER 46. This diagram represents? a. b. c. d.

Drag. CL:CD ratio. Power required. Power available.

,V,

EAS


Drag. CL:CD ratio. Power required. Power available.

LID Max

EAS

Power requimd and Power available

93

POWER 48. V ,,

is most probably at point .. on the diagram?

POWER 49. Which point on the diagram best represents V, for a propeller aircraft?

POWER 50. Which point on the diagram best represents V, for a propeller aircraft?

POWER 51. As speed decreases below V ?,, a. Drag and power required decrease. b. Drag and power required increase. c. Drag decreases and power required increases. d. Drag increases and power required decreases.

POWER 52. As speed increases above V ?,, a. b. c. d.

Drag and power required decrease. Power required increases, whilst drag decreases then increases. Drag decreases and power required increases. Drag increases and power required decreases.

POWER 53. As speed changes from C, Max to ?V ,, a. b. c. d.

Both drag and power required decrease then increases at the same rate. Both drag and power required decrease then increase but at different rates. Drag increases and power required decreases. Drag decreases and power required increases.

EAS

94


POWER 54. If air density at 40000 feet is '/4 of that at sea level, how will power required change when climbing at constant IAS from MSL to 40000 feet? a. b. c. d.

Remain constant.. lncrease by a factor of 2. lncrease by a factor of 4. lncrease by a factor of 8.

POWER 55. If air density at 40000 feet is '/4 of that at sea level, how will power required change when climbing at constant TAS from MSL to 40000 feet? a. b. c. d.

Remain constant. .,, Decrease provided IAS remains above V Decrease provided IAS remains below.V ,, Decrease provided IAS remains above V ,,

POWER 56. If power lever setting is maintained at a constant value when climbing at V? , a. b. c. d.

Gradient and ROC will increzse. Gradient and ROC will decrease. Gradient will increase and ROC will decrease. Gradient will decrease and ROC will increase.

POWER 57. If power lever setting is maintained at a constant value when climbing at V? , a. b. c. d.

Gradient and ROC will increase. Gradient and ROC will decrease. Gradient will increase and ROC will decrease. Gradient will decrease and ROC will increase.


Variation of power available with increasing weight. Variation of power available with increasing altitude. Variation of power required with increasing weight. Variation of power required with increasing altitude.


Variation of power available with increasing weight. Variation of power available with increasing altitude. Variation of power required with decreasing weight. Variation of power required with increasing altitude.


95

POWER 60. This diagram represents? a. Variation of power available with increasing weight. b, Variation of power available with decreasing altitude. c. Variation of power required with increasing weight. d. Variation of power required with decreasing altitude.

I

EAS

POWER 61. Jet power available at high altitude is indicated by line ... on this diagram? Power

POWER 62. Propeller power available at high altitude is indicated by line ... on this diagram?

POWER 63. Jet power available at low altitude is indicated by line ... on this diagram?

POWER 64. Propeller power available at low altitude is indicated by line ... on this diagram?

e

96


POWER 65. Propeller thrust available at high altitude is indicated by the line .. on this diagram? Thrust

POWER 66. Propeller thrust available at low altitude is indicated by the line .. on this diagram? Thrust

POWER 67. Jet thrust available at high altitude is indicated by the line .. on this diagram?

. Thrust

POWER 68. Jet thrust available at low altitude is indicated by the line .. on this diagram?

Thrust

EAS

POWER 69. If weight is increased by 50% at V ,, power required at the same speed will increase by?


97

POWER 70. If weight is increased by 50% at a speed considerably higher than V ,, a. b. c. d.

,

\

power required will be increased by?

42.5%. More than 62.5%. Less than 62.5%. 62.5%.

POWER 71. If weight is increased by 50% at a speed considerably lower than V ,, a. b. c. d.

power required will?

lncrease by more than 62.5%. lncrease by less than 62.5%. Increase by more than 62.5% only if speed is lower than V ., Decrease by more than 62.5% only if speed is lower than V .,

\

POWER 72. ..

A multi engine aircraft is flying at its absolute ceiling when it suffers a single engine failure. In order to continue flying at that altitude it must? a. b. c. d.

Reduce speed. lncrease speed. Reduce weight. lncrease power setting.

POWER 73. A multi engine aircraft is flying at its absolute ceiling when it suffers a single engine failure. In order to continue flying at the same speed it must? a. b. c. d.

Climb. Descend. Accelerate. Decelerate.

POWER 74. An aircraft weighing 50000 Ibf is flying at its absolute ceiling when the pilot activates the reheat system. If reheat produces an additional 5000 Ibf of thrust it will enable the aircraft to? a. b. c. d.

Climb at a gradient of 10%. Climb at an angle of approximately 5.7 degrees. Climb at a rate of 1000 feet per minute. Climb at a gradient of 10% and at an angie of approximately 5.7 degrees.

POWER 75. As altitude increases the power required to fly at any given IAS? a. b. c. d.

lncreases because the drag force increases with altitude. Decreases because the drag force decreases wit altitude. lncreases because the drag force remains constant while TAS increases with altitude. Decreases because the drag force remains constant while TAS decreases with altitude.

98


POWER 76. For a turbojet the minimum power speed will be ..... than the minimum drag speed? a. b. c. d.

Higher. Lower. The same. Higher in a climb but lower in a descent.

POWER 77. The tangent to the power required curve gives ...... for a jet aircraft?

POWER 78. What is the main benefit gained from employing the step-climb technique? a. b. c. d.

lncreased range. lncreased endurance. lncreased speed. Avoidance of conflict with other traffic.

POWER 79. How does C of G movement affect SFC? a. b. c. d.

Forward movement will increase SFC only when C of G is already ahead of C of P. Forward movement will decrease SFC only when C of G is already forward of C of P. Forward movement will always increase SFC. Forward movement will never decrease SFC.

POWER 80. How does C of G movement within authorised limits affect SFC? a. b. c. d.

Forward movement will increase SFC only when C of G is already ahead of C of P. Forward movement will decrease SFC only when C of G is already forward of C of P. Forward movement will always increase SFC. Forward movement will never decrease SFC.

POWER 81. Flying into a headwind will? a. Increase range. b. Decrease range. C. Increase endurance. d. Decrease endurance.


99

POWER 82. When flying into a headwind the range of a jet aircraft can be maxirnised by?

%&.

\.

\

a. b. c. d.

Decreasing speed compared with that required in still air. Increasing speed compared with that required in still air. Maintaining the same speed as in still air. Decreasing altitude.

POWER 83. What will be the effect of flying at a cost index greater than zero? a. b. c. d.

Mach number will be greater than that for best distance flown per Kg of fuel. Mach number lower than that giving best distance per Kg of fuel. Improved climb performance. Degraded climb performance.

POWER 84. Maximum piston engine endurance occurs at? a. b. c. d.

Speed for best angle of climb V., Approximately the speed for best rate of climb V., Speed for maximum lift coefficient V,,g. Speed for minimum drag V,,,.

POWER-85. A jet aircraft fitted with old engines (0.06 kglN1h) gives an output of 14 Kglh. A jet fitted with new engines (0.035 kglN1h) gives an output of? a. b. c. d.

10.7 kglh. 11.7 kglh. 8.17 kglh. 14 kglh.

POWER 86. Moving C of g aft will? a. b. c. d.

Decrease range. lncrease range. lncrease stalling speed. lncrease longitudinal stability.

POWER 87. Which of the following is true of specific air range (SAR)? a. b. c. d.

SAR = IASIFF. SAR = TASIFF. SAR = GSIFF. SAR = MachIFF.


100

POWER 88. Long range cruise speed is? a. b. c. d.

Higher than max range cruise speed and gives approximately 99% of max SAR. Lower than max range cruise speed. Unaffected by mass: The same as max range cruise speed.

CURVES

CURVES 1. What is curve A in the diagram at the right? a. b. c. d.

lnduced drag. Total drag. Profile drag Form drag

CURVES 2. What is curve B in the diagram at the right? a. b. c. d.


CURVES 3. What iscurve C in the diagram at the right? a. b. c. d.


D

q EAS

CURVES 4. What is point D in the diagram at the right? a. b. c. d.

Minimum induced power. Minimum total drag. Minimum profile drag Minimum power required

D

"

f)

EAS

CURVES 5. What is the airspeed at point D in the diagram?

CURVES 6. What caused the change from curves 1 to 2? a. b. c. d.

Flap deployment. weight -increase. Weight decrease. Gear deployment. Curves EAS

Curves 2

CURVES 7. What caused the change from curves 1 to 2? a. b. c. d.

Flap deployment. Weight increase. Weight decrease. Gear deployment. EAS

CURVES 8. V ,,

is at point ..... on the diagram at the right?

Drag

A6

C D

EAS

CURVES 9. The greatest ratio of TAS : thrust required occurs at point .. ..?

Drag

AB

C D

EAS

CURVES 10. Propeller aircraft best endurance occurs at

.....?

A B

C D

EAS

A B

C D

EAS

A B C

D E

CURVES 11. Propeller aircraft best range occurs at

.....?

CURVES 12. Propeller aircraft V, might occur at .....?

CURVES 13. Propeller aircraft Vy might occur at .....?

CURVES 14. Jet aircraft Vy might occur at .....?

Drag

EAS

104

Curves

CURVES '15. Jet aircraft V, might occur at

.....? Drag

A B C D E

EAS

A B C D E

EAS

D E

EAS

CURVES 16. Jet aircraft best endurance occur at .....? Drag

CURVES 17. Jet aircraft best range occurs closest to point .....? Drag

A BC

CURVES 18. Curve A is proportional to?

CURVES 19. Curve B is proportional to? Drag

CURVES 20. Curve C is proportional to?

CURVES 21. What does the diagram at the right represent? a. b. c. d.

Variation of drag with altitude. Variation of power required with altitude. Variation of power available with altitude. Variation of thr~lstwith altitude.

I

TAS


Variation of power required with decreasing weight. Variation of power required with increasing weight. Variation of power required with decreasing altitude. Variation of power required with increasing altitude.


Variation of power required with Variation of power required with Variation of power required with Variation of power required with

decreasing weight. increasing weight. decreasing altitude. increasing altitude. EAS

CURVES 24. What does the diagram on the right represent? a. b. c. d.

Variation of drag against EAS increasing with altitude. Variation of drag against TAS increasing with altitude. Variation of power available increasing with altitude. Variation of power available Decreasing altitude.


Power available and power required for a propeller aircraft. Power available and power required for a jet aircraft. Thrust available and drag for a propeller aircraft. Thrust available and drag for a jet airciaft.


Power available and power required for a propeller aircraft. Power available and power required for a jet aircraft. Thrust available and drag for a propeiler aircraft. Thrust available and drag for a jet aircraft.


Power available and power required for a propeller aircraft. Power available and pcwer required for a jet aircraft. Thrust available and drag for a propeller aircraft. Thrust available and drag for a jet aircraft. I

EAS


Power available and power required for a propeller aircraft. Power available and power required for a jet aircraft. Thrust available and drag for a propeller aircraft. Thrust available and drag for a jet aircraft.

CURVES 29. What effect does the diagram on the right represent? a. b. c. d.

That of a weight increase. That of a weight decrease. That of an altitude increase. That of an altitude decrease.

I

TAS

Curves

107

CURVES 30. What effect does the diagram on the right represent? a. b. c. d.

That of a weight increase. That of a weight decrease. That of an altitude increase. That of an altitude decrease. Drag TAS

CURVES 31. 2

What effect does the diagram on the right represent? a. b. c. d.

That of a weight increase. That of a weight decrease. That of an altitude increase. That of an altitude decrease.

TAS


Aerodynamic ceiling. Maximum operating altitude. Absolute ceiling. Stabilising altitude.

EAS

CURVES 33. What do points A and B on the diagram at the right represent? a. b. c. d.

V, Prop and .V ,, V, Prop and .V ,, Vy Prop and .V ,, Vy Prop and .V ,,

CURVES 34. What do points A and B on the diagram at the right represent? a. b. c. d.

V, Jet and .V ,, V, Jet and V .,, V, Jet and .V ,, Vy Jet and .V ,,

$ :

EAS

:

.

:

j

:

1

A

108

Curves


Aerodynamic ceiling. Maximum operating altitude. Absolute ceiling. Stabilising altitude.

CURVES 36. What effect might the diagram on the right represent? a. b. c. d.

Single piston engine failure. Single jet engine failure. lncrease in altitude. lncrease in weight.

TAS

CURVES 37. What might have caused the change from point 1 to point 2 in the diagram at the right? a. b. c. d.

Decrease in weight. lncrease in weight. Decrease in altitude. lncrease in altitude.

CURVES 38. What does the change from point 1 to point 2 in the diagram at the right represent? a. b. c. d.

Decrease in minimum flying speed due to weight increase. lncrease in minimum flying speed due to weight increase. Decrease in minimum flying speed due to altitude increase. lncrease in minimum flying speed due to altitude decrease.

CURVES 39. The point providing best angle of climb in the diagram at the right is ...?

:

. j.

2 I

EAS

Curves

109

CURVES 40. The point providing best angle of climb in the diagram at the right is ...? a. b. c. d.

.....

A. B, C. D.

. .

AB

CURVES 41. The point providing best rate of climb in the diagram at the right is a. b. c. d.

.-.

...?

A. B. C. D.

.

.

..

.

C

D

P

D B

C D

CURVES 42. What point in the'diagram at the right would give best L:D ratio?

CURVES 43. What point in the diagram at the right would give minimum sink rate in a glide?

CURVES 44. What point in the diagram at the right might give best prop endurance? a. b. c. d.

A. B. C. D.

CD

110

Curves

CURVES 45. What point in the diagram at the right might give best prop range?

CURVES 46. What point in the diagram at the right might give best glide endurance? a. b. c. d.

A. B. C. D.

CURVES 47. What point in the diagram at the right might give best glide range? a. b. c. d.

A. B. C. D.

CURVES 48. What point in the diagram at the right might give minimum glide angle?

a. b. c. d.

A. B. C. D.

CURVES 49. What point in the dibgram at the right might give lowest flight speed? a. b. c. d.

A.

6. C. D.

A

co

Curves

111

CURVES 50. What point in the diagram at the right might give highest flight speed? a. b. c. d.

A. B.

C. D.

CD

CURVES 50. In what part of the diagram at the right might be best jet endurance? a. b. c. d.

Between A and B. Between B and C. C. D.

c

o

p -

CD

CURVES 51. In what part of the diagram at the right might be best jet angle of ciimb speed? a. b. c. d.

Between A and B. Between B and C. C. D.

c

D

p -

CD

CURVES 52. In what part of the diagram at the right might be best jet rate of clirnb speed? CD

a. b. c. d.

Between A and B. Between B and C. Between C and D. Below D.

CURVES 53. In what part of the diagram at the right might be best jet cruising speed? a. Between A and B. b. Between B and C. c. Between C and D. d. Below D.

CD

11'2

Curves

CURVES 54. Point D on the diagram at the right might be? a. .V ,, b. VMP. "XProp. d. ",Jet. C.

CURVES 55. Point E on the diagram at the right might be? a. .V ,, b. VMP. C.

Vxprop.

d.

J,'e't

CURVES 56. Point C on the diagram at the right might be?

,

a. .,v,

C

b. ",Jet.

B

CURVES 57. Point A on the diagram at the right might be? a. V,. b. V*. C. VMP. d.

",Jet.

CURVES 58. Point 6 on the diagram at the right might be .... if C is VYProp? a. .V ,, b. .V ,, C.

'*Jet.

d.

",Jet.

Curves

113

CURVES 59. Point B on the diagram at the right might be

... if C is VMp?

a. .V ,, b.

"XProp.

C.

"Yprop.

m d

",Jet.

CURVES 60. Point D on the diagram at the right might be? a. Prop best endurance speed. b. Prop best range speed. C. VMP. d.

J,e'.t

II

CURVES 61. Jet power available at high altitude is indicated by line ... on this diagram? a. A. b. B. c. C.

v

Power

CURVES 62. Propeller power available at high altitude is indicated by line ... on this diagram?

EAS

CURVES 63. Jet power available at low altitude is indicated by line a. b. c. d.

A. B. C. D.

... on this diagram? Povve:

EAS

CURVES 64. Propeller power available at low altitude is indicated by line ... on this diagram? a. b. c. d.

A.

Power

0. C.

D.

w EAS

CURVES 65. Propeller thrust available at high altitude is indicated by the line .. on this diagram? Thrust

EAS

CURVES 66. Propeller thrust available at low altitude is indicated by the line .. on this diagram?

Thrust

CURVES 67. Jet thrust available at high altitude is indicated by the line .. on this diagram?

CURVES 68. Jet thrust avaiiable at low altitude is indicated by the line .. on this diagram?

I

Curves

CURVES 69. What might be the cause of the change from take-off flight path I t o flight path 2 in the diagram at the right? a. b. c. d.

lncreased power setting. Decreased flap angle. lncreased weight. Decreased weight.

CURVES 70. What might be the cause of the changefrom take-off flight path 1 to flight path 2 in the diagram at the right? a. b. c. d.

lncreased flap angle. Decreased Rap angle. lncreased weight. Decreased weight.

BRP

CURVES 71. What might be the cause of the changefrom take-off flight path 1 to flight path 2 in the diagram at the right? a. b. c. d.

lncreased flap angle. Decreased flap angle. lncreased weight. Decreased weight.

BRP

CURVES 72. What might be the cause of the change from take-off flight path 1 to flight path 2 in the diagram at the right? a. b. c. d.

lncreased temperature. Decreased temperature. Decreased flap angle. Decreased weight.

BRP

115

Curves

116

CURVES 73. What might be the cause of the change from take-off flight path 1 to flight path 2 in the diagram at the right? a. b, c. d,

lncreased temperature. Decreased temperature. lncreased flap angle. lncreased weight.

CURVES 74. What might be the cause of the change from take-off flight path 1 to flight path 2 in the diagram at the right? a, b. c. d.

lncreased altitude. Decreased temperature. Decreased flap angle. Decreased weight.

BRP


Early rotation. Decreased temperature. Decreased flap angle. Decreased weight.

BRP


Late rotation. Decreased temperature. Decreased flap angle. Decreased weight.

BRP


lncreased density altitude. Decreased density altitude. Decreased pitch attitude. lncreased pitch attitude.


lncreased down slope. Decreased down slope. Decreased pitch attitude. lncreased pitch attitude.

BRP


lncreased down slope. lncreased up slope. Decreased density altitude. lncreased density altitude.

CURVES 80.

-

What might be the cause of the change from take-off flight path 1 to flight path 2 in the diagram at the right? a. b. c. d.

lncreased tailwind. Decreased tailwind. Decreased pitch attitude. lncreased pitch attitude.

BRP

2

Curves

118


Increased headwind. Decreased headwind. Decreased pitch attitude. lncreased pitch attitude.

BRP

-CURVES 82. What might be the cause of the change from take-off flight path 1 to flight path 2 in the diagram at the right? a. b. c. d.

Contaminated runway. Decreased tailwind. lncreased density altitude. lncreased weight.

c BRP

CLIRVES 83. What might be the cause of the change from take-off Right path Ito Right path 2 in the diagram at the right? a. b. c. d.

Decreased tailwind. lncreased headwind. lncreased field elevation. Decreased weight.


Contaminated runway. Decreased headwind. lncreased density altitude. lncreased weight.

BRP

Curves

CURVES 85. What might be the cause of the change from take-off flight path 1 to flight path 2 in the diagram at the right? a. e. f. g.

Anti-skid system failure. lncreased tailwind. lncreased field elevation. lncreased weight.

BRP

119

SPEEDS

SPEED 1. If V, at 60000 Kg is 250 Kts what will it be at 70000 Kg? a. b. c. d.

250 Kts. 260 Kts. 270 Kts. 280 Kts.

SPEED 2. If flaps up V, at 60000 Kg is 250 Kts what will it be at 70000 Kg with flaps down if the change in the C,:a curve due to flap deployment is ignored. a. b. c. d.

230 Kts. 240 Kts. 250 Kts. 260 Kts.

SPEED 3. If weight is increased by 25% by what % will V, increase?

SPEED 4. Descending at constant mach number might cause? a. b. c. d.

M ,, to be exceeded. V, to be exceeded. Shock stall. Tuck under.

SPEED 5. Climbing at constant IAS might cause? a. b. c. d.

M ,, to be exceeded. V, to be exceeded. Low speed stall. TAS to reduce.

Speeds

SPEED 6. The margin between V, and V, must be? a. b. c. d,

1.2 V., 1.3 V., Sufficient to avoid unintentional over-speeds. More than 50 Kts.

SPEED 7. lncreasing flap setting from 10 degrees to 30 degrees will ... ... V,? a. b. c. d.

Increase. Decrease unless limited by .V ,, Decrease or increase depending on altitude. Decrease or increase depending on C of G position.

SPEED 8. At const.ant altitude the buffet boundaries will converge with? a. b. c. d.

Forward movement of C of G. Aft movement of C of G. lncreasing temperature. Decreasing load factor.

SPEED 9. At constant altitude the buffet boundaries will converge with? a. b. c. d.

Rearward movement of C of P. Rearward movement of C of G. lncreasing temperature. lncreasing bank angle.

SPEED 10. lncreasing IAS in a climb will . . . ... . .. the altitude at which the mach limit is reached? a. b. c. d.

Decrease. Increase. Not affect. Increase or decrease depending on temperature.

SPEED 11. ,,V,

will not be affected by .........? a. b. c. d.

Weight. Altitude. Power setting. Humidity.

121

122

Speeds

SPEED 12. As altitude increases at constant weight and load factor, high speed buffet margin .......... and low speed buffet margin .......? a. b. c. d.

Increases, Increases, Decreases, Decreases.

Decreases. Increases. Increases. Decreases.

SPEED 13. When climbing at constant Mach number above the tropopause IAS will ..... and TAS will .....? a. b. c. d.

Decrease, Decrease, Increase, Increase,

Remain constant. Decrease. Remain constant. Increase.

SPEED 14. V, is ......... than V, and, V, is ....... than V,? a. b. c. d.

Less, Less, More, More,

Less. More. More. Less.

SPEED 15. V, is ......... than V, and ........... than VMcG? a. b. c. d.



SPEED 16. VMCA is .........than VRwhich is .............. than VMcG? a. b. c. d.


SPEED 17. The correct equation is? a. VMu< VMcA< V,. b. V, < VMcA < ,m ' "i. C. VR < < L' o,. d. Vzmin < VMcA '~VMU.


Speeds

123

SPEED 18. In a constant mach number climb in the troposphere true airspeed? a. b. c. d.

lncreases then dec~eases. Decreases. Remains constant. Increases.

SPEED 19. In a constant mach number climb true airspeed? a. b. c. d.

Increases then remains constant. Decreases then remains constant. Decreases then decreases less slowly. Decreases then decreases more slowly.

SPEED 20. V, is? a. The take-off safety speed. b. The speed attained in the fourth segment. c. The lowest speed at which directional control can be maintained following an engine failure after ta ke-off. d. The minimum lift off speed with one engine out.

SPEED 21. The minimum and maximum values of V, are?

SPEED 22. In a constant TAS climb in the troposphere? a. b. c. d.

IAS increases. IAS decreases. IAS increases then remains constant. IAS decreases then remains constant.

SPEED 23. In a constant IAS climb in the troposphere? a. b. c. d.

TAS increases. TAS decreases. TAS increases then remains constant. TAS decreases then remains constant.

124

Speeds

SPEED 24. In a constant TAS climb? a. b. c. d.

IAS increases. IAS decreases. IAS increases then increases more slowly. IAS decreases then decreases more quickly.

SPEED 25. In a constant IAS climb in the troposphere? a. b. c. d.

TAS increases and mach number decreases. TAS increases and mach number increases. TAS decreases and mach number increases. TAS decreases and mach number decreases.

SPEED 26. At low altitude? a. b. c. d.

The limiting variable is .V ,, The limiting variable is M .,, The limiting variable is V., The limiting variable is .M ,,,

SPEED 27. At high altitude? a. b. c. d.

,, The limiting variable is .V The limiting variable is M .,, The limiting variable is V., The limiting variable is .M ,,,

SPEED 28. In a constant mach number descent there is a danger that? a. b. c. d.

M ,, will be exceeded. M ,,,, will be exceeded. ,V, will be exceeded. V, will be exceeded.

SPEED 29. V, must be not less than . .... . for a Class A aircraft? a. V, plus 10%. , plus 5%. b. ,V c. ,,v , plus 10%. d. ,,V, plus 15%.

Speeds

SPEED 30. Increasing aircraft mass by 10% .. ....? a. b. c. d.

lncreases VM, lncreases VM, Decreases ,V, lncreases V ,,

by 5% but does not change parasite drag. by 5% but does not change induced drag. and induced drag by 5%. and induced drag by 5%.

SPEED 31. Decreasing aircraft mass by 20% ........ VM, by . .....% but ...........? a. b. c. d.

Decreases Decreases Increases Increases

10 20 10 20

does does does does

not affect profile drag. not affect induced drag. not affect profile drag. not affect induced drag.

SPEED 32. The correct sequence of velocities is?

SPEED 33. The relationship between Vx and V, is? a. b. c. d.

Vx is always less than V., Vx is always greater than V., V, is always less than or equal to V., Vx can never equal V.,

SPEED 34. The correct sequence is?

SPEED 35. As altitude increases? a. b. c. d.

V, V, V, V,

and V, both increase. and V, both decrease. decreases and V, remains constant. decreases and Vx remains constant.

125

126

Speeds

SPEED 36. ,,V ,

will be lowest with? a. b. c. d.

High ambient temperature humidity and pressure altitude. Low ambient temperature, humidity and pressure altitude. High ambient temperature and low humidity and pressure altitude. Low ambient temperature and high humidity and pressure altitude.

SPEED 37. At what speed must a jet aircraft and a propeller aircraft ,fly to maintain altitude in straight and level flight at the absolute ceiling?

SPEED 38. What happens as a propeller aircraft climbs to its absolute ceiling? a. b. C. d.

VMin,and V, converge on V., V, VMax, VMin,VMax, VMPand VMDconverge on V., VMi,, VMax and V, converge on V., The CAS values equating to V, and V, both decrease.

SPEED 39. What happens to the CAS values of V, and VMDas a propeller aircraft climbs to its absolute ceiling? a. b. c. d.

Increase. Decrease. Remain constant. Increase and decrease respectively.

SPEED 40. With an increase in weight? a. b. c. d.

V, V, V, V,

increases and V, increases and V, decreases and V, decreases and V,

decreases. increases. increases. decreases.

SPEED 41. What will limit the maximum value of V,? a. VRw b. V,. c. VR. d- "LO.,

Speeds

SPEED 42. V, is? a. b. c. d.

Take-off safety speed. Threshold speed. Take-off decision speed. Minimum control speed in the take-of configuration.

SPEED 43. lncreasing density will ...V,, .....V, a. b. c. d.

Increase Decrease Increase Decrease

.....VLoF?

increase decrease decrease increase

increase. decrease. decrease. decrease.

SPEED 44.

.....VO i F?

lncreasing altitude will ...V,, .....V, a. b. c. d.




SPEED 45. lncreasing weight will ...V,, .....V, a. b. c. d.


.....VLoF?



SPEED 46. lncreasing headwind will ...V,, .....V, a. b. c. d.


increase decrease not affect increase

.....V?,, increase. decrease. not affect. decrease.

SPEED 47. lncreasing flap angle will ...V,, .....V, a. b. C. d.



.....VLo,? increase. decrease. decrease. decrease.

127

Speeds

128

SPEED 48. lncreasing air temperature will ...V,, .....V, a. b. c. d.



.....VLoF? increase. decrease. not affect. decrease.

SPEED 49. lncreasing runway up slope will ...V,, .....V, a. b. C. d.



.....?V ,,, increase. decrease. not affect. decrease.

SPEED 50. lncreasing altitude will ...V, .....V, .....V? , a. b. c. d.


increase not affect decrease increase

increase. decrease. not affect. decrease.

SPEED 51. lncreasing temperature will ...V, :....V, .....VY? a. b. c. d.



increase. not affect. not affect. decrease.

SPEED 52. lncreasing weight will ...V, a. b. c. d.


.....Vxl .....V?, increase not affect decrease increase


SPEED 53. lncreasing runway up slope will ...V, .....V, .....V? , a. b. c. d.

Increase Nat affect Increase Decrease



Speeds

SPEED 54. lncreasing headwind will ...V, .....Vx, .....V? , a. b. C. d.

Increase Not affect Increase Decrease



SPEED 55. lncreasing flap angle will ...V, .....Vx, I ....V? , a, b. c, d.

increase Decrease Increase Decrease



SPEED 56. lncreasing temperature will ...VMcG,.....VMcA,.....V,? a. b. c. d.


increase not affect decrease decrease

increase. decrease. not affect. increase.

SPEED 57. lncreasing altitude will ...VMcG,.....VMcA,.....V,? a. b. c. d.


increase not affect decrease decrease

increase. decrease. not affect. increase.

SPEED 58. lncreasing weight will .. .VMcG,.....VMcA,.....V,? a. b. c. d.

Increase Not affect Not affect Decrease

increase not affect not affect decrease

increase. decrease the minimum value of. increase the minimum value of. increase.

SPEED 59. lncreasing runway up slope will ...VMcG,..... ,,,V, a. b. c. d.



.....V,?

increase. decrease. increase. increase.

129

130

Speeds

SPEED 60. lncreasing headwind will ...VMcG,.....VMcA,.....V,? a. b. c. d.



increase. decrease. increase. increase.

SPEED 61. lncreasing flap angle will ... ,,,V, a. b. c. d.


..... ,,,V ,


.....V,? increase. decrease minimum value of. increase minimum value of. increase.

SPEED 62. VATfor a class A aircraft is?

a. b. c. d.

lndependent of weight. lndependent of flap angle. ,, at screen height. 1.23 V 1.05 V,, at screen height.

SPEED 63. The landing climb speed is ? a. b. c. d.

1.08 V,, for 4 engine Class A aircraft. 5 V,, for twin engine Class A aircraft 1.I 1.25 V,, for 4 engine Class A aircraft. 1.25 VREFfor Class B aircraft.


1. I 3 V,, for 4 engine Class A aircraft. ,, for twin engine Class A aircraft 1. I 3 V 1.I 3 V,, for 4 engine Class A aircraft. 1. I 3 VREF for Class B aircraft.


1.25 V,, for 4 engine Class A zircraft. 1.25 V,, for twin engine Class A aircraft 1.2 V,, for 4 engine Class A aircraft. VREFfor all Class B aircraft.

Speeds


1.I 3 VsRfor 4 engine Class B aircraft. 1. I 3 VsRfor three engine Class A aircraft 1. I 3 VsRfor 4 engine Class A aircraft. 3 ,V ,, for Class B aircraft. 1.I

SPEED 67. The one engine inoperative approach climb speed is? a. b. c. d.

Not greater than 1.2 V ,, for 4 engine Class A aircraft. Not greater than 1.4 VsRfor all other Class A aircraft 1.25 VsRfor all Class B aircraft. . 1.2 ,V ,, for all Class B aircraft.

SPEED 68. The approach climb speed is? a. b. c. d.

Not greater than 1.2 V, for 4 engine Class A aircraft. Not greater than 1.25 V, for all Class B aircraft Not greater than 1.5 V, for all Class B aircraft. ,, for all Class B aircraft. 1.2 ,V

SPEED 69. The minimum acceptable value of V, is? a. b. c. d.

1.25 VsRfor 2 and 3 engine turbo-prop aircraft. 1.08 VsRfor 4 engine turbo-prop aircraft. 1. I VS for 4 engine jet aircraft. 1.15VMC.


1.25 V ,, for 2 and 3 engine turbo-prop aircraft. ,, for 4 engine turbo-prop aircraft. 1.25 V 1.08 VsRfor 4 engine jet aircraft. 1.15 VMC.


1.25 VsRfor 2 and 3 engine turbo-prop aircraft. 1.25 VsRfor 4 engine turbo-prop aircraft. 1. I V ,, for 4 engine jet aircraft. 1.1.V ,,

131

Speeds

132


V,. ,, 1.5 .V That required to reach V, at 50 ft. ,, That which causes VLoFto be 1.15 .V


V,. 1.05 .V ,, That required to reach V, at 50 ft. .,, That which causes VLoFto be 1.15 V


V,. 1.15VMc. That required to reach V, at 35 ft. ,, That which causes VLoFto be 1.15 .V


V,. 1.15VMc. That required to reach V, at 50 ft. ,, That which causes VLoFto be I. I .V


V,. 1.15 .V ,, That required to reach V, at 50 ft. ,, That which causes VLo, to be 1.05 V

withone engine out.

SPEED 77. What is a typical value of V,/Dmax? a. b. c. d.

Approximately 2.1Vs for jet aircraft. 94% of the I g stall speed. Approximately 1.5 V.,, Approximately 1.41Vs, of turbo-propeller aircraft

Speeds


Approximately 2.5 V, for jet aircraft. 94% of the l g stall speed. Approximately 1.32 V.,, Approximately 1.5 V, of piston propeller aircraft.


Approximately 2.5 V, for jet aircraft. 94% of the l g stall speed. .,, Approximately 1.5 V Approximately 1.7 V, of piston propeller aircraft.

SPEED 80. What is V ?, a. b. c. d.

At least 2 Kts or 2% above the stick pusher activation speed. The speed providing maximum angle of climb. The speed providing maximum ROC. At least 90% of the l g stall speed.

SPEED 81. What is V ?, a. b. c. d.

The minimum level flight CAS. The speed providing maximum angle of climb. The speed providing maximum ROC. At least the l g stall speed.


1.25 VS for 2 and 3 engine turbo-prop aircraft. 1.25 VS for 4 engine turbo-prop aircraft. 1.IVS for 4 engine jet aircraft. 1.1 VMC.

SPEED 83. For a class A aircraft V, must be not less than?

133

I34

Speeds

SPEED 84. For a class A aircraft V, with one engine inoperative, must be not less than? a. b. c. d.

110% of.,,V , 105% of VMcG. I15% of.,,V, That which causes ,V ,,

to be 105% of .V ,,

SPEED 85. An aircraft is flying at VIM,. By what % must its IAS reduce to continue flying at V ,, by 1O%?

as its weight decreases

SPEED 86. An aircraft is flying at VIM,. By what % must its IAS reduce to continue flying at V ,, by 20%?

as its weight decreases

SPEED 87. The value of,,V , a. b. c. d.

will not be affected by?

lncreased thrust. lncreased temperature. lncreased weight. lncreased altitude.

SPEED 88. For landing a Class A aircraft the speed should be? a. b. c. d.

Not less than Not less than Not less than Not less than

1.23 ,V ,, at a 35 ft screen height. , at a 50 ft screen height. 1.2,,V 1.3 ,V ,, at a 50 ft screen height. 1.2 ,V ,, at a 35 ft screen height.

SPEED 89. What is?,,V ,, a. b. c. d.

The highest decision speed at which it is possible to stop within the ASDA. The highest decision speed at which it is possible to stop within the ASDR. The lowest decision speed at which it is possible to continue a take-off within the TODA. The highest decision speed at which it is possible to continue a take-off within the TODR.

Speeds

SPEED 90. What is ?V ,, a. b. c. d.

The highest decision speed at which it is possible to stop within the ASDA. The highest decision speed at which it is possible to stop within the ASDR. The lowest decision speed at which it is possible to continue a take-off within the TODA. The highest decision speed at which it is possible to continue a take-off within the TODR.

SPEED 91. What is V,? a. b. c. d.

The take-off safety speed. The speed that must be achieved by screen height when all engines are operating. The speed that must be achieved by screen height. The steady take-off climb speed.

SPEED 92. What is ?,V ,, a. b. c. d.

The take-off safety speed. 'The speed that must be achieved by screen height when all engines are operating. The speed that must be achieved by screen height. The final take-off speed.

SPEED 93. What is the value for V,? a. b. c. d.

Never more than V2 plus 10 Kts. Never less than V2 plus 10 Kts. Never less than 1.2 VMcA. Never less than V,.

SPEED 94. What is the value for?,V ,, a. b. c. d.

Never more than V, plus 10 Kts. Never more than V .,, Never less than 1.I 8V ., Never more than V,.

SPEED 95. What is the value for V,? a. b. c. d.

Never more than V2 plus 10 Kts. Never less than V, plus 10 Kts. Never less than 1.3 VMcA. Never less than 1;3 V .,,

135

Speeds

136

SPEED 96. What is VAT,? a. b. c. d.

Target threshold speed with no engines operating. Target threshold speed with no engines failed. Target threshold speed with one engine operating. Target threshold speed with one engine failed.

SPEED 97. What is VAT,? a. b. c. d.

Target threshold speed with no engines operating. Target threshold speed with no engines failed. Target threshold speed with one engine operating. Target threshold speed with one engine failed.

SPEED 98. What is the minimum value of VREF?

SPEED 99. What is V ,, a. b. c. d.

for a Class A aircraft

m value of VLoFunder any conditions. The m i n i m ~ ~possible The maximum possible value of VLoFunder any conditions. The GAS at which the nose wheel first leaves the runway. The GAS at which the main wheels first leave the runway.

SPEED 100. What effect does increasing altitude have on ?,V ,, a. b. c. d.

lncrease. Decrease. No effect. lncrease or decrease depending on mass.

SPEED 101. What limits V ?,, a. b. c. d.

1. I 3 VsRfor 2 engine turbo-prop aircraft. 1.IV, for 3 engine turbo-prop aircraft. 1.25 V, for 4 engine turbo-prop aircraft. ,, 1.2 .V

Speeds

SPEED 102. What limits V? , a. b. c. d.

1.IV ,, for 2 engine turbo-prop aircraft. 3V ,, for 3 engine turbo-prop aircraft. 1.I 1.25 V, for 4 engine turbo-prop aircraft. ,, 1.2 .V


1.IV, for 2 engine turbo-prop aircraft. 1.25 V, for 3 engine turbo-prop aircraft. ,, for 4 engine turbo-prop aircraft. 1.08 V ,, 1.2 .V


1.25 V, for 2 engine turbo-prop aircraft. 1.3 V, for 3 engine turbo-prop aircraft. 1. I V, for 4 engine turbo-prop aircraft. 1.1 .v,

SPEED 105. Which of the following best defines ?V ,,, a. b. c. d.

1.2 V,. Not less than 1.23.,V ,, 1.1 VMCL. 1.05VMc,.

SPEED 106. ,V ,,

will be increased by? a. b. c. d.

lncreased landing mass. lncreased take-off mass. lncreased engine power output. lncreased flap angle.

SPEED 107. A 10% increase in landing mass will a. b. c. d.

Increase Decrease Increase Increase

5%. 10%. 15%. 25%.

........... VREFby ....... ?

137

138

Speeds

SPEED 108. An increase in .flap angle from 40 degrees to 45 degrees will ........ ? V ,,, a. b. c. d.

lncrease by 5%. Decrease. Decrease by 5%. Increase.

SPEED 109. Increasing pressure altitude will ..... ... ?V ,,, a. b. c. d.

Increase. Decrease. Not affect. lncrease or decrease depending on mass.

SPEED 110. If pressure altitude is increased V, and V, will? a. b. c. d.

Not be affected. Move further apart. Move closer together. lncrease by the same amoant.

SPEED 111. If pressure altitude i s increased V, and V, will? a. b. c. d.

Both increase. Both decrease. lncrease and decrease. Decrease and increase.

SPEED 112. V, must be greater than? a. b. c. d.

V, and 1.05.,,V, V, and 1.05.,V ,, V, and 1.05.,,V, ,V ,, and.,,V,

SPEED 113. Complete the following statement. The ....... must be long enough to enable an aircraft to accelerate to . Then decelerate to a stop without going off the end of the ........ ? a. b. c. d.

ASDR ASDA ASDA ASDA

the speed at V, plus 2 seconds the speed at V, plus 2 seconds the speed at V, the speed at V, plus 2 seconds

stopway. clearway. runway. stopway.

Speeds

SPEED 114. Which of the following might cause V, to be limited by VMcA? a. b. c. d.

Large flap angles. High mass. Low ambient pressure. Low ambient temperatures.

SPEED 115. Which of the following would be the first limit on increasing V,? a. VMc,. b. c. v,. d. V,.


Small flap angles. Low mass. Low ambient pressure. Low ambient temperatures.

SPEED 117. Which of the following is the maximum speed for raising or lowering the landing gear? a. V,. b. V,. C. v., d. V.,


Small flap angles. High mass. High ambient pressure. Low ambient temperatures.

SPEED 119. Which of the following is the maximum speed for flight with the landing gear deployed? a. V,. b. V,. C.

v.,

d. V.,

139

140

Speeds

SPEED 120. Which of the following might cause V2 to be limited by VMcA? a. b. c. d.

Small flap angles. High mass. Low ambient pressure. Low density altitude.

SPEED 121. If weight is increased by 25% by what % will V, increase?

SPEED 122. Increasing flap setting from 10 degrees to 30 degrees will ...... V",? a. b. c. d.

Increase. Decrease unless limited by VMcA. Decrease or increase depending on altitude. Decrease or increase depending on C of G position.

SPEED 123. V, is ......... than V, and ........... than VMc,? a. b. c. d.



SPEED 124. VMcAis .........than VRwhich is .............. than VMc,? a. b. c. d.

Less, Less, More, Iklore,

SPEED 125. The correct equation is?


Speeds

141

SPEED 126. V2 is? a. The take-off safety speed. b. The speed attained in the fourth segment. c. The lowest speed at which directional control can be maintained following an engine failure after take-off. d. The minimum lift off speed with one engine out.

SPEED 127. The minimum and maximum values of V, are?

SPEED 128. VR must be not less than ...... for a Class A aircraft? a. V, plus 10%. b. VMcplus 5%. c. v,, plus 10%. d. VMc, plus 15%.

SPEED 129. Increasing aircraft mass by 10% ......? a. b. c. d.

lncreases ,V, by 5% but does not change parasite drag. lncreases VMDby 5% but does not change induced drag. Decreases VMDand induced drag by 5%. lncreases ,V, and induced drag by 5%.

SPEED 130. Decreasing aircraft mass by 20% ........ ,V, a. b. c. d.

Decreases Decreases Increases Increases

10 20 10 20

by ...... % but ...... .....?

does not affect profile drag. does not affect induced drag. does not affect profile drag. does not affect induced drag.

SPEED 131. The correct sequence of velocities is?

142

Speeds

SPEED 132. The relationship between V, and V, is? a. b. c. d.

V, V, V, V,

is always less than V., is always greater than V., is always less than or equal to V., can never equal V.,

SPEED 133. The correct sequence is?

SPEED 134. VMc, will be lowest with? a. b. c. d.

High ambient temperature humidity and pressure altitude. Low ambient temperature, humidity and pressure altitude. High ambient temperature and low humidity and pressure altitude. Low ambient temperature and high humidity and pressure altitude.

SPEED 135. V, is? a. b. c. d.

Take-off safety speed. Threshold speed. Take-off decision speed. Minimum control speed in the take-of configuration.

SPEED 136. VMcAis the minimum control speed airborne. Which of the following is correct? a. Vs < VMcA < "M , .,i b- VR < VMcA < 'Lo,. C. M ",,i < VMcA < 'Mu. d. VMU< VMCA < V,.

SPEED 137. V, is the speed at which the critical engine is assumed to fail during take-off. Which of the following is correct? a. VMc, < VE, < V,. b' V2MIN < E' F < M' U. c. 1.05VMcA < V, < V,. d. 1.05VMcA< V, < VR.

Speeds

143

SPEED 138. A twin piston engine aircraft with a constant speed propeller has a V ,, 100 Kts. Which of the following is correct? a. b. c. d.

Vy = 130 Kts. Vy = 100 Kts. Vy is less than 100 Kts. 100Kts
SPEED 139. What is V,? a. b. c. d.

The take-off (35 ft) climb speed. The take-off decision speed. The speed at which it is assumed the critical engine will fail. The speed at which the aircraft leaves the ground.

SPEED 140. If, V ,

is ignored what is?,,V ,

c. v,. d. V,.

SPEED 141. The minimum acceptable margin between VREF and,,V,

is?

SPEED 142. The minimum acceptable margin between V, and VMcAis?

SPEED 143. The minimum value for V, is?

of 130 Kts and a holding speed of

144

Speeds

SPEED 144. ,V,

is? a. Minimum speed at which an aircraft can lift-off and continue a take-off safely. b. Minimum speed at which to rotate into the lift-off attitude. c. The maximum speed at which an aircraft can leave the ground without exceeding tyre speed limits. d. The minimum speed for flight with the flaps up.

SPEED 145. The initial climb-out speed with all engines operating is? a. V., b. V2.5. . c. v,. d. V, + 10 Kts.

TAKE-OFF

TAKE-OFF 1. Flight planning has been completed for a take-off using 10 degree flap deflection. The climb limited TOM is 4550 Kg, and the Field length limited TOM is 5650 Kg. If the aircraft is loaded to a mass of 5000 Kg, which of the following statements is correct? a. b. c. d.

15 degree flap deflection will increase both limits. 15 degree flap deflection will increase climb limit and decrease field length limit. 5 degree flap deflection will decrease both limits. 5 degree flap deflection will increase climb limit and decrease field length limit.

TAKE-OFF 2. The end of the first segment is? a. b. c. d.

At 400 feet gross height. At flap retraction altitude. At gear up altitude. Above the obstacle dcmain at flap retraction height.

TAKE-OFF 3. Take-off performance is best with? a. b. c. d.

Low altitude, low temperature, low humidity. High altitude, high temperature, high humidity. Low altitude, low temperature, high humidity. High altitude, low temperature, low humidity.

TAKE-OFF 4. Why does the take-off climb chart (table 4.5) in the CAP 698 not take account of headwinds and tailwinds? a. b. c. d.

Because it is already factorised into the calculations. Because headwinds and tailwinds do not affect climb gradient. Because the chart relztes only to still air gradients. Because headwinds and tailwinds do not affect ROC.

TAKE-OFF 5. E IT what speed is,,V, a. b. c. d.

referenced?

V, for air speed. ,V ,, For ground speed. V, for ground speed. V, for ground speed.

TAKE-OFF 6. To what speed is the limiting tyre speed referenced ? a. b. c. d.

,V ,, for ground speed. V, for air speed. VATfor ground speeds. V, for ground speeds.

TAKE-OFF 7. A contaminated runway will? a. b. c. d.

lncrease TORR and decrease TODR. lncrease TODR more than TORR. lncrease TORR more than TODR. lncrease TORR but not affect TODR.

TAKE-OFF 8. lncreasing ambient pressure? a. b. c. d.

lncrease take-off run and decrease climb gradient. lncrease take-off run and climb gradient. Decrease take-off run and climb gradient. Decrease take-off run and increase climb gradient.

TAKE-OFF 9. lncreasing pressure altitude will? a. b. c. d.

lncrease take-off run and climb gradient. lncrease take-off run and decrease climb gradient. Decrease take-off run and climb gradient. Decrease take-off run and increase climb gradient.

TAKE-OFF 10. If TOM is field limited? a. b. c. d.

,,V, is too high. Mass equals field limited mass but is greater than climb limited mass. Mass equals climb limited mass. Mass equals field limited mass.

TAKE-OFF 11. If an engine fails after V,? a. b. c, 6.

Continue or abort depending on which engine has failed. Continue or abort depending on mass. Abort. Continue.

TAKE-OFF 12. Take-off performance calculations have been made based on on field limited mass at 30" flap angle. If 25O is inadvertently set? a. b. c. d.

The calculated V, will be too low. The aircraft will overrun the stopway if take-off is abandoned at V,. V, will be found to be too low. V, will be found to be too high.

TAKE-OFF 13. A field is said to be balanced when? a. b. c. d.

EMDA = TODA. TODA= T O W . EMDA=TORA. EMDA = 1.33 TODA

TAKE-OFF 14. The take-off distance is? a. b. c. d.

From BRP to screen height. From BRP to end of clearway. From BRP to end of stopway. From BRP to 400 feet height.

TAKE-OFF 15. Increasing upward runway slope will? a. b. c. d.

lncrease TORR and TODR. lncrease TORR and TODA. Decrease TORR and TODR. Increases TODR and EMDA.

TAKE-OFF 16. Increasing flap angle from 10" to 20" will? a. b. C. d.

Decrease V, and VLoF. lncrease V, and VLoF. Increase V, and decrease VLoF. Decrease V, and increase VLoF.

TAKE-OFF 17. Decreasing flap angle from 15O to 5O will? a. b. C. d.

Decrease V, and VLoF. Increase V, and VLoF. lncrease V, and decrease VLoF. Decrease V, and increase VLo,.

TAKE-OFF 18. Increasing flap angle beyond the normal take-off setting will? a. b. c. d.

Decrease VLoFand TODR. lncrease VLoFand TODR. Decrease VLoFand increase TODR. lncrease VLoFand decrease TODR.

TAKE-OFF 19. A headwind will ........TORR and ..........TODR? a. b. c. d.



TAKE-OFF 20. A headwind will.. ......V, and ..........? V ,,, a. b. c. d.

increase Not affect Decrease Not affect

Not affect. Decrease. Decrease. Not affect.

TAKE-OFF 21. A tailwind will ........TORR and ..........TODR? a. b. c. d.

Increase Not affect Decrease Not affect

Increase. Decrease. Decrease. Not affect.

TAKE-OFF 22. A tailwind will.. ......V, and .-. .......VLoF? a. b. c. d.

Increase Not affect Decrease Not affect

Not affect. Decrease. Decrease. Not affect.

TAKE-OFF 23. For take-off from a contaminated runway it is necessary to? a. b. c. d.

lncrease V, and V, by a safety factor of 1.3. Increase V, and V, by a safety factor of 1.I. Use the standard V, and V,. Increase V,, V, and V,.

TAKE-OFF 24. The value of V, depends upon? a. b. c. d.

Mass, air density, and flap angle. Mass, air temperature, and runway slope. C of G position, mass, and runway slope. Runway condition, runway slope and aircraft mass.

TAKE-OFF 25. The value of V, depends upon? a. b. c. d.

Mass, air density, and runway condition. Mass, air temperature, and flap angle. C of G position, mass, and runway slope. Runway condition, runway slope and aircraft mass.

TAKE-OFF 26. Flight planning has been completed for a take-off using 10 degree flap deflection. The climb limited TOM is 4500 Kg, and the Field length limited TOM is 5000 Kg. If the aircraft is loaded to a mass of 5000 Kg, which of the following statements is correct? a. Using 5 degrees of flap might increase the climb limited mass sufficiently to enable take-off to be made at the new mass. b. Using 20 degrees of flap might increase the climb limited mass sufficiently to enable take-off to be made at the new mass. c. Using 5 degrees of flap will increase the climb limited mass but will decrease the field length limited mass. d. Using 20 degrees of flap will increase the climb limited mass but will decrease the field length limited mass.

TAKE-OFF 27. Flight planning has been completed for a take-off using 10 degree flap deflection. The climb limited TOM is 5000 Kg, and the Field length limited TOM is 4500 Kg. If the aircraft is loaded to a mass of 5000 Kg, which of the following statements is correct? a. Using 5 degrees of flap might increase the climb limited mass sufficiently to enable take-off to be made at the new mass. b. Using 20 degrees of flap might increase the climb limited mass sufficiently to enable take-off to be made at the new mass. c. Using 5 degrees of flap will decrease the climb limited mass but will increase the field length limited mass. d. Using 20 degrees of flap will decrease the climb limited mass but will increase the field length limited mass.

TAKE-OFF 28. Flight planning has been completed for an early morning take-off using 10 degree flap at a mass of 4300 Kg. The climb limited TOM is 5000 Kg, the Field length limited TOM is 4500 Kg. The take-off is then delayed due to traffic problems. Which of the following statements is correct? a. b. c. d.

Take-off perfqrmance will be degraded if OAT increases. Take-off performance will be improved if OAT increases. Take-off performance will be unaffected by changes in OAT. Decreases in take-off performance due to increasing OAT can be overcome by reducing flap deflection.

TAKE-OFF 29. Operating from a slush contaminated runway will require ........compared to operating from the same runway in dry conditions? a. b. c. d.

Unchanged V,, increased V, decreased weight. Decreased V,, decreased V, decreased weight. Increased V,, increased V, unchanged weight. Decreased V,, unchanged V, decreased weight.

TAKE-OFF 3Q. An MRJT aircraft at mass 52000 Kg is to take-off from a runway contaminated with 6 mm of standing water. Take-off performance calculations indicate a dry runway field limited take-off mass of 52000 Kg. The effect of the contamination will be to? a. b. c. d.

lncrease V, and V, but not affect mass. Decrease V, mass, and V,. lncrease V, and decrease V, and mass. Decrease V, but not affect V,or mass.

TAKE-OFF 31. A defective anti-skid system will? a. b. c. d.

Reduce take-off performance. Reduce landing performance. Reduce both take-of and landing performance. Not affect take-off nor landing performance.

TAKE-OFF 32. An aircraft is cleared for take-off using either 5, 10, or 15 degrees of flap, the normal setting being 10 degrees. Take-off using the 5 degree setting might be appropriate for? a. b. c. d.

Long runway, low pressure altitude and close obstacles. Long runway, high pressure altitude and close obstacles. Short runway, low pressure altitude and distant obstacles. Short runway, high pressure altitude and distant obstacles.

TAKE-OFF 33. An aircraft is cleared for take-off using either 5, 10, or 15 degrees of flap, the normal setting being 10 degrees. Use of the 15 degree setting will ..... compared to the standard setting? a. b. c. d.

Decrease take-off run, TODR and obstacle clearance. lncrease take-off run, TODR and obstacle.clearance. lncrease take-off run and decrease TODR and obstacle clearance. Decrease take-off run and increase TODR and obstacle clearance.

TAKE-OFF 34. If atmospheric pressure increases whilst other factors remain unchanged, take-off distance required will ......and climb performance will ......?

a. b. c. d.


increase. decrease. decrease. increase.

TAKE-OFF 35. Decreased ambient temperature causes? a. b. c. d.

Decreased VMBE masslimit. Increased climb limited mass limit. Decreased field length limited mass limit. Decreased climb limited mass limit.

TAKE-OFF 36. Increasing flap setting from 5 degrees to 15 degrees will? a. b. c. d.

IncreaseV,. Decrease V,. lncrease V, if not limited by.,V ,, Decrease V, if not limited by.,V ,,

TAKE-OFF 37. When taking-off in high temperatures or at high pressure altitudes? a. b. c. d.

Tyre speeds become more limiting. Tyre speeds become less limiting. Tyre speeds are not affected. Tyre speeds may become more or less limiting depending on runway condition.

TAKE-OFF 38. The 2 second time interval between V , a. b. c. d.

and V, is?

Not required for jet engines. To allow the propeller to automatically feather. To allow the engine to run down so that no thrust is produced. To allow for the failure to be detected.

TAKE-OFF 39. From commencement of the take-off run to screen height is? a. b. c. d.

Gross TODR. Net TODR. Gross TOR. Measured TOD.

TAKE-OFF 40. What is TORR? a. b. c. d.

The distance between the BRP and the lift off point. ,, is achieved. The calculated distance between BRP and the point at which ,V The distance between BRP and the point at which screen height IS achieved. The calculated distance between BRP and the point at which screen weight is achieved.

TAKE-OFF 41. What is the TODR? a. b. c. d.

The distance between the BRP and the lift off point. The calculated distance between BRP and the point at which ,V ,, is achieved. The distance between BRP and the point at which screen height is achieved. The calculated distance between BRP and the point at which screen weight is achieved.

TAKE-OFF 42. The minimum acceptable value for V, is? ,, a. 1.1 .V b. 1.1 Vs for 4 engine turbo-props and jets with provision for significant reduction in one-engineinoperative stalling speed. c. 1.25 Vs for 2 and 3 engine turbo-props and jets without provision for significant reduction in oneengine-inoperative stalling speeds. d. VMcG.

TAKE-OFF 43. The minimum acceptable value for V, is? a. 1.15VMc. b. 1.2 Vs for 4 engine turbo-props and jets with provision for significant reduction in one-engineinoperative stalling speed. c. 1.25 Vs for 2 and 3 engine turbo-props and jets without provision for significant reduction in oneengine-inoperative stalling speeds. d. VMcG'

TAKE-OFF 44. The minimum acceptable value for V, is? a. 1.15VMC. b. 1.1'5 Vs for 4 engine turbo-props and jets with provision for significant reduction in one-engineinoperative stalling speed. c. 1. I 5 Vs for 2 and 3 engine turbo-props and jets without provision for significant reduction in oneengine-inoperative stalling speeds. d. VMcG'

TAKE-OFF 45. The minimum value of V, is? a. V,. b. 110% VMC. c. The speed required to attain V, the end of the TORA. d. VMcG.

a •

I

TAKE-OFF 46.

The minimum value of V, is? a. l.lVl. b. 105% .V ,, c. The speed required to attain \I, by the end of the clearway. d. VMcG.

TAKE-OFF 47. The minimum value of V, is? a. l.lVl. b. 115% VMc. c. The speed required to attain V, by screen height. d. VMcG.

TAKE-OFF 48. ATC require a class A aircraft to conduct a 20° banked turn as soon as possible after lift-off. What is the minimum allowable height for the commencement of this turn and by what margin must the aircraft clear obstacles during the turn? a. b. c. d.

35 ft 50 ft 400 ft

400ft

35 ft. 50 ft. 35 ft. 50 ft.

TAKE-OFF 49. When taking-off from a contaminated runway? a. b. c. d.

Rotate at the normal V., lncrease V, by 10%. lncrease V, by 10 Kts. Decrease V, by 10% or 10 Kts, whichever is greater.

TAKE-OFF 50. An error is made carrying out the take-off performance calculations, resulting in the calculated V, figure of 160 Kts, being 10 kts too high. Which of the following statements is true? a. b. c. d.

If an engine fails at 155 Kts and the take-off is abandoned the aircraft will overrun. If an engine fails at 155 Kts and the take-off is continued the aircraft will not achieve screen height. If an engine fails at 150 kts and the take-off is abandoned the aircraft will overrun. Engine failures at any speed up to the calculated V, can be dealt with safely by continuing the ta ke-off.

TAKE-OFF 51. An error is made carrying out the take-off performance calculations, resulting in the calculated figure of 140 Kts, being 10 kts too low. Which of the following statements is true? a. b. c. d.

If an engine fails at 145 Kts and the take-off is abandoned the aircraft will overrun. If an engine fails at 155 Kts and the take-off is continued the aircraft will not achieve screen height. If an engine fails at 150 kts and the take-off is abandoned the aircraft will overrun. Engine failures at any speed up to the calculated V, can be dealt with safely by abandoning the take-off.

TAKE-OFF 52. An error is made carrying out the take-off performance calculations, resulting in the calculated figure of 160 Kts, being 10 kts too high Which of the following statements is true? a. b. c. d.

If an engine fails at 145 Kts and the take-off is abandoned the aircraft will overrun. If an engine fails at 155 Kts and the take-off is continued the aircraft will not achieve screen height. If an engine fails at 150 kts and the take-off is abandoned the aircraft will overrun. Engine failures at any speed above the calculated V, can be dealt with safely by continuing the take-off.

TAKE-OFF 53. What proportion of a headwind should be taken into account when calculating take-off performance for a single engine propeller aircraft of 3000 Kg mass, when operating from a dry paved runway?

TAKE-OFF 54. What proportion of a tailwind should be taken into account when calculating take-off performance for a single engine propeller aircraft of 3000 Kg mass, when operating from a dry paved runway?

TAKE-OFF 55. An aerodrome has a runway of 4000 m and the first obstacle is 1500 meters from the end of the take-off run. What is the maximum take-off distance that may be assumed to be available.

TAKE-OFF 56. Defective anti-skid systems? a. b. c. d.

Decrease maximum landing mass. Decrease field limited TOW. Decrease both field limited TOW and maximum landing mass. lncrease maximum TOW but decrease maximum landing mass.

TAKE-OFF 57. Defective anti-skid systems?

1. 2. 3. 4.

lncrease LDR. lncrease TORA. lncrease ASDR. lncrease TORR.

a. b. c. d.

1, 2 and 3. 2 and 4. 1 and 3. 1,2and4.

TAKE-OFF 58. TODR is? a. b. c. d.

The distance between BRP and achieving VLoF. 'The distance between BRP and achieving .V ,, The calculated distance between BRP and the achievement of screen height. The calculated distance between BRP and the achievement of VLoF.

TAKE-OFF 59. TORR is? a. b. c. d.

,,, The distance between BRP and achieving .V The distance between BRP and achieving V .,, The calculated distance between BRP and screen height. The calculated distance between BRP and the achievement of VLoF.

TAKE-OFF 60. Minimum acceptable obstacle clearance for class A aircraft in the obstacle domain is.... increasing to ...... in turns above 400 ft using bank angles of ..... ?

TAKE-OFF 61. Minimum acceptable all engines operating gradient for a twin piston aircraft at screen height in the takeoff is?

TAKE-OFF 62. A light piston twin climbs at a gradientof 10% from the screen height of 15m (50 ft). There is an obstacle 900 m high, 3280 m from the screen. By how much will the aircraft clear the obstacle? a. b. c. d.

100 m. 115m. 85m. It does not clear the obstacle.

TAKE-OFF 63. During certification testing of a twin engine jet aircraft the true take-off distances were 1547 m with all engines operating, and 1720 m with the critical engine inoperative at V,. What distance would be used for certification purposes?

TAKE-OFF 64. If an aircraft takes-off using the wet V, instead of the dry V, how will TODR and climb performance be affected? a. b. c. d.

Increased Increased Decreased Decreased

improved. degraded. degraded. improved.

TAKE-OFF 65. Can take-off be carried out if V, is less than V,? a. b. c. d.

No. Yes. , is decreased to equal V,. Only if,,V Only if V, is increased to equal.,,V ,

TAKE-OFF 66. V2 is? a. Take-off safety speed. b. Fourth segment commencement speed. c. The lowest speed at which control can be maintained following a single engine failure during takeoff. d. The speed at which an aircraft lifts off the ground.

TAKE-OFF 67. Which of the following is true of the increased V2 procedure? a. b. c. d.

It increases the maximum TOW. It decreases the maximum TOW. It increases TODR and climb performance at any given TOW. It can be used to increase climb performance only if the field length limited mass is less than the climb limited mass.

TAKE-OFF 68. Wind component at take-off does not affect? a. b. c. d.

Climb gradient. Obstacle limited TOW. Field length limited TOW. Rate of climb.

TAKE-OFF 69. TODA is? a. b. c. d.

Declared runway length. Declared runway plus stopway length. Declared runway length plus clearway. Declared runway plus stopway plus clearway.

TAKE-OFF 70. The first segment ends? a. b. c. d.

When the gear is fully retracted. When the flaps are fully retracted. At 1500 ft above runway level. At 400 ft above runway level.

TAKE-OFF 71. Can a stopway be included in take-off run required (TORR) calculations? a. Yes. b. No. c. Only if the clearway is longer than the stopway. 6. Only if the clearway is shorter than the stopway.

TAKE-OFF 72. Reduced thrust take-off should not be used? a. b. c. d.

Temperature is more than ISA + 7%. When it is dark. When the runway is wet. When anti-skid system is unserviceable.

TAKE-OFF 73. Can anti skid systems be used for take-off and landing? a. b. c. d.

Yes. No. Take-off only. Landing only.

TAKE-OFF 74. An increase in OAT will .. .. Field length limited TOW and .. ... climb limited TOW? a. b. c. d.



TAKE-OFF 75. What will happen to ASDA if anti-skid is ~~nserviceable? a. b. c. d.

Nothing. Increase. Decrease. lncrease or decrease depending on slope.

TAKE-OFF 76. The distance to be maintained between aircraft is greatest when? a. b. c. d.

Taking-off heavy. Landing heavy. Taking-off light. Landing light.


Temperature is less than ISA. When it is raining. At night. When the runway is icy or very slippery.


Temperature is less than ISA. When thrust reverse systems are unserviceable. At night. At pressure altitudes above 3500 ft amsl. .

TAKE-OFF 79. Variable thrust take-off procedures should not be used? a. b. c. d.

Temperature is greater than ISA. When increased V, procedure is being used. At night. At pressure altitudes above 5000 ft amsl.

TAKE-OFF 80. Assumed temperature take-off procedures should not be used? a. b. c. d.

Temperature is greater than ISA. At night. When the PMC is switched off or unserviceable. At pressure altitudes above 10000 ft amsl.

TAKE-OFF 81. Assumed temperature take-off procedures should not be used? a. b. c. d.

On upward sloping runways. At night. When the runway is contaminated. When the runway is wet.

TAKE-OFF 82. The maximum thrust reduction that may be used when conducting a reduced thrust take-off is?

TAKE-OFF 83. The assumed temperature take-off procedure can be used only when? a. b. c. d.

The runway is dry. The available distance greatly exceeds that required. The field length limited mass is too low. The climb limited mass is too low.

TAKE-OFF 84. The purpose of the reduced thrust take-off procedure is to? a. b. c. d.

Conserve engine life. Reduce TORR. Reduce fuel consumption. Reduce TODR.

TAKE-OFF 85. ,,V,

is most likely to be a limiting factor when? a. There is a tailwind. b. There is a headwind. c. Pressure altitude is low. d. Air temperature is low.

TAKE-OFF 86. ,,V,

is most likely to be a limiting factor when? a. b. c. d.

There is a wet runway. There is a headwind. Pressure altitude is high. Air temperature is low.

TAKE-OFF 87. ,,V,

is most likely to be a limiting factor when? a. b. c. d.

There is a wet runway. There i s a headwind. Pressure altitude is low. Air temperature is high.

TAKE-OFF 88. Tyre speed is likely to be greatest when? a. b. c. d.

QFE is low. QFE is high. QNH is high. OAT is low.

TAKE-OFF 89. During certification test flying of a one hundred and twenty seat, twin jet aircraft, the distance from BRP to ,V ,, is 1500 m with all engines operating, and 2000 m with one engine out. Which of the following figures is most likely to be the one used for TORR certification purposes?

TAKE-OFF 90. During certification test flying of a one hundred and twenty seat, twin jet aircraft, the distance from BRP to VLoFon a wet runway is 1500 m with all engines operating, and 2000 m with one engine out. Which of the following figures is most likely to be the one used for TORR certification purposes?

TAKE-OFF 91. During certification test flying of a one hundred and twenty seat, twin jet aircraft, the distance from BRP to Screen height from a dry runway is 1500 m with all engines operating, and 2000 m with one engine out. Which of the following figures is most likely to be the one used for TODR certification purposes?

TAKE-OFF 92. During certification test flying of a one hundred and twenty seat, twin jet aircraft, the distance from BRP to Screen height from a contaminated runway is 1500 m with all engines operating, and 2000 m with one engine out. Which of the following figures is most likely to be the one used for TODR certification purposes?

TAKE-OFF 93. If the mass of an aircraft was greater than that calculated, which of the following speeds would be found to be greater than expected during the take-off? ab. c. d.

L'o., V., v,. V,.

TAKE-OFF 94. Which of the flowing conditions is best suited to the use of low flap angles? a. High field elevation, high temperature, long runway, distant obstacles. b. High field elevation, low temperature, short runway, no obstacles. c. Low field elevation, high temperature, long runway, close obstacles. d. Low field elevation, low temperature, short runway, no obstacles.

TAKE-OFF 95. A depth of 5 mm of standing water will ....... field length limited mass, compared with a dry runway? a. b. c. d.

Decrease. Increase. Not change. Increase or decrease depending on aircraft type.

TAKE-OFF 96. What is the takeoff run required for a class B aircraft, with one engine inoperative? a. b. c. d.

Distance from BRP to .V ,,, Distance from BRP to V,. Distance from BRP to rr~idpoint between Vr and 35 ft screen height. Distance between BRP and mid point between VLoFand 35 ft screen height.

TAKE-OFF 97. What term is used for the distance from BRP to screen height? a. b. c. d.

Gross TODR. Net TODR. Gross TOD. Measured TOD.

TAKE-OFF 98. What is the measured TOD? a. b. c. d.

BRP to screen height. and screen height. BRP to mid point between VLOF BRP to first obstacle. VLoFto screen height.

TAKE-OFF 99. An aircraft is prepared for take-off from runway 09 which has a 1% downward slope. The take-off mass is ,,, If the effects of winds are ignored, what will happen if operations are switched to runway limited by .V 27, immediately before take-off? a. b. c. d.

,, will be exceeded. Mass must be reduced or ,V ,V ,, will be decreased so mass must be reduced. ,,V, will not change but mass need not be changed because it is known to be within limits. has increased. Mass may be increased because,,V,

TAKE-OFF 188. During a take-off, where is the reference point zero located? a. b. c. d.

Where the aircraft reaches screen height. Where the aircraft reaches V,. Where the aircraft reaches.,V , Where the aircraft reaches V,.

TAKE-OFF 101. If conditions are such that V, is equal to , , ,V standard setting for take-off? a. b. c. d.

what will happen to V, if Rap angle is increased beyond the

Decrease. Remain unchanged. Increase. Decrease if not limited by.,,V,

TAKE-OFF 102. When taking -off from a runway with 5 mm of standing water, field limited mass will ..... and V, will Compared to a damp runway? a. b. c. d.


....


TAKE-OFF 103. Which of the following is most likely to require a reduced V, value? a. b. c. d.

Inoperative anti-skid system. Reduced thrust take-off. Noise abatement take-off. Take-off with close obstacles.

TAKE-OFF 104. An aircraft is prepared for take-off from runway 09 which has a 1% upward slope. The take-off mass is limited by.,V ,, If the effects of winds are ignored, what will happen if operations are switched to runway 27, immediately before take-off? a. b. c. d.

will be exceeded. Mass must be reduced or,,V, ,,V, will be decreased so mass must be reduced. ,,V, will not change but mass need not be changed because it is known to be within limits. Mass may be increased because,,V, has increased.

TAKE-OFF 105. An aircraft has been prepared for take-off and all take-off performance calculations have been completed. What will happen if it is then decided to load one additional passenger? a. Performance calculation must be reworked to ensure safe operation at the increased mass. b. Performance calculations need not be reworked because the increase in mass will be insignificant. c. Performance calculations need not be reworked provided extra fuel is burned at the ramp prior to take-off, to reduce mass to its original value. d. Taking on an extra passenger under these circumstance is illegal so this situation could not arise.

TAKE-OFF 106. Which of the following factors might limit take-off mass? a. b. c. d.

Wheel rotation speed. Tyre energy absorption capability. Wheel energy absorption capability. Aquaplaning speed.

TAKE-OFF 107. What is the gross flight path? a. b. c. d.

The flight path of an aircraft in as-new condition flown by an experienced test pilot. The flight path of an average aircraft flown by an average pilot. The flight path of a poorly maintained aircraft flown by an inexperienced pilot. The actual flight path of a particular aircraft under consideration.

TAKE-OFF 108. What is the purpose of using a net flight path? a. b. c. d.

To ensure that all obstacles can be cleared safely. To ensure optimum performance. To minimise noise nuisance to people on the ground. To maximise fuel economy.

TAKE-OFF 109. For an airbus A320 aircraft the net flight path is equal to the gross flight path reduced by . . .. ?

TAKE-OFF 110. For a Boeing 747 aircraft the net flight path is equal to the gross flight path reduced by .... ?

TAKE-OFF 111. For 3 engine class A aircraft the net flight path is equal to the gross flight path reduced by

.... ?

TAKE-OFF 112. The NFP of any given segment will be? a. b. c. d.

Less than the gross. More than the gross. More or less than gross depending on headwinds and tailwinds. Gross.

TAKE-OFF 113. The minimum acceptable GFP in the first segment for a twin engine class A aircraft is 0%. What is its minimum acceptable NFP?

TAKE-OFF 114. ,,V , is the minimum speed at which it is possible to maintain control on the ground following a failure of the critical engine using? a. b. c. d.

Primary flying controls only. Primary flying controls and nose wheel steering. All means available. Nose wheel steering only.

TAKE-OFF 115. What is the minimum acceptable climb gradient for a ciass B aircraft taking-off below a 1000 ft cloud base.

TAKE-OFF 116. Flight planning has been completed for a take-off using 10 degrees of flap. The climb limited TOM is 4670 Kg, and the field length limited TOM is 5070 Kg. If the aircraft is loaded to 5070 Kg, which of the following is true. a. The take-off cannot be carried out with 10 degrees of flap but might be possible with 20 degrees. b. The take-off cannot be carried out at this mass and no change of flap setting will overcome this problem

c. 15 degrees of flap might enable the take-off to be made at this mass. d. The mass must be reduced unless the flap setting is reduced.

TAKE-OFF 117. Compared to still air conditions, a headwind will? a. b. c. d.

lncrease TODR and decrease climb gradient. lncrease TODR and climb gradient. Decrease TODR and increase climb gradient. Decrease TODR and climb gradient.

TAKE-OFF 118. Flight planning has been completed for a take-off using 10 degrees of flap. The climb limited TOM is 4670 Kg, and the field length limited TOM is 5070 Kg. If the aircraft is loaded to 5070 Kg, which of the following is true. a. b. c. d.

Increasing flap angle will enable take-off at the new mass. Decreasing flap setting will enable take-off at the new mass. A tailwind might enable take-off at the new mass. A headwind might enable take-off at the new mass.

TAKE-OFF 119. lncreasing take-off flap setting will ......... TODR and a. b. c. d.


.

.........climb limited TOM.

decrease. increase. increase. decrease.

TAKE-OFF 120. Take-off performance has been calculated for a maximum TOM from a balanced field. The take-off is delayed for three hours during which QFE decreases by 60 hPa, and ambient temperature decreases by 3.96" C. How will take-off performance be affected? ,, might be exceeded. a. ,V b. To, might be unchanged because the reducing pressure will be off-set by the reducing temperature. c. To, will be reduced because there will be less drag in thinner air. d. Pressure altitude will decrease so TODR wilt decrease.

TAKE-OFF 121. Runway slope has certain effects on take-off performance? a. b. c. d.

Downhill slope increases V., Downhill slope decreases V,. Downhill slope increases V,. Downhill slope decreases V,.

TAKE-OFF 122. Reduced thrust (flex thrust) must not be used when? a. b. c. d.

It is dark. OAT is above ISA+ 10. Anti-skid is unserviceable. Runway is wet.

TAKE-OFF 123. During certification test flying a 4 engine turbojet on a dry runway, the measured take-off distance to screen height is 3050 m with one engine failed at V,, and 2555 m with all engines operating. The TODR adopted for certification will be? a. b. c. d.

3513 m. 2938 rn. 2555 rn. 3050 rn.

TAKE-OFF 124. Which of the foltowing combinations of take-off conditions is likely to result in the worst climb performance? a. b. c. d.

Sea level, ISA + 5, 10 degree flap setting. Sea level, ISA + 10, 10 degree flap setting. 5000 ft field elevation, ISA + 5, 10 degree flap setting. 5000 ft field elevation, ISA + 10, 30 degree flap setting.

TAKE-OFF 125. How are ASDR and TODR affected if flaps are set at 10 degrees instead of 30 degrees for take-off? a. b. c. d.

lncrease both ASDR and TODR. lncrease ASDR and decrease TODR. Decrease both ASDR and TODR. Decrease ASDR and increase TODR.

TAKE-OFF 126. Using 10 degrees of flap instead of 30 degrees might improve .... for take-off? a. b. c. d.

TOW vs Obstacle clearance requirements (depending on obstacle location). TOW vs Climb requirements. Both a and b. TOW vs Runway length requirements.

TAKE-OFF 127. Will slush affect the TOW vs Runway length requirements? a. b. c. d.

No because this is already taken into account in the performance charts. No provided no engines fail. Yes. TOW for any given runway length may be decreased considerably. Yes. Reduced surface friction will enable faster acceleration so TOW for any given runway length can be increased.

TAKE-OFF 128. How does a downhill slope affect take-off performance? a. b. c. d.

V, V, V, V,

is unchanged. is increased. is decreased. is increased.

TAKE-OFF 129. How are ASDR and TODR affected if V, is increased? a. b. c. d.

Increase both ASDR and TODR. lncrease ASDR and decrease TODR. Decrease both ASDR and TODR. Decrease ASDR and increase TODR.

EN-ROUTE

EN-ROUTE 1. A commercial jet aircraft will fly? a. b. c. d.

Always above its optimum altitude. Always below its optimum altitude in order to avoid mach buffet. Always at its optimum altitude to achieve optimum economy. At its optimum altitude for only part of each flight because this altitude increases as the flight progresses.

EN-ROUTE 2. A 5% reduction in mass will produce a ......% decrease in jet aircraft fuel consumption if cruising speed and altitude remain constant.

EN-ROUTE 3. Why is there a delay of several seconds between engine failure and commencement of drift down? a. b. c. d.

This is the time required for the crew to become aware of the failure. This is the time taken for the aircraft to decelerate to drift down speed. This is the time required to carry out the calculations. This is the time required to complete post failure drills.

EN-ROUTE 4. An aircraft is flying at V,,, when it jettisons fuel equivalent to 10% of its total mass. What will be the effect on fuel consumption if its speed remains unchanged? a. b. c. d.

9% decrease. 1g0/0 decrease. 29% decrease. 39% decrease.

EN-ROUTE 5. An aircraft is flying at 1.5V, when it jettisons fuel equivalent to 10% of its total mass. What will be the effect on fuel consumption if its speed and altitude remain unchanged? a. b. c. d.

Less than 19% decrease. 19% decrease. More than 19% decrease. No change.

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EN-ROUTE 6. An aircraft is flying at 0.9VMDwhen it jettisons fuel equivalent to 10% of its total mass. What will be the effect on fuel consumption if its speed remains unchanged? a. b. c. d.

Less than 9% decrease. 19% decrease. More than 9% decrease. No change.

EN-ROUTE 7. Two identical turbojets are cruising at the same speed, SFC and altitude. If jet A weighs 150000 Kg and consumes fuel at 5000 Kglhr, what will be the consumption rate for jet B which weighs 200000 Kg? a. b. c. d.


EN-ROUTE 8. Two identical turbojets are cruising at V ,, at the same SFC and altitude. If jet A weighs 175000 Kg and consumes fuel at 2000 Kglhr, what will be the consumption rate for jet B which weighs 150000 Kg? a. b. c. d.


EN-ROUTE 9. The absolute ceiling is? a. b. c. d.

The altitude at which maximum rate of climb is 100 fpm. The altitude at which maximum rate of climb is 500 fpm. The altitude at which maximum rate of climb is zero. The altitude at which maximum angle of clirr~bis less than 1.5%.

EN-ROUTE 10. The maximum range speed in a turbojet aircraft is?

EN-ROUTE 11. If service ceiling is 5000 m at a mass of 3000 Kg, it will be a. b. c. d.

Higher. Lower. Unchanged. Higher or lower depending on speed.

..... at a mass of 3500 Kg?

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EN-ROUTE 12. The drift down procedure is used on a twin turbojet when? a. b. c. d.

An engine fails at any altitude above the single engine ceiling. An engine fails at any altitude. In the event of sudden depressurisation or structural failure. To avoid bad weather.

EN-ROUTE 13. Optimum turbojet cruise altitude is that at which?

-

a. b. c. d.

is greatest. M ,, is greatest. SAR is maximum. Power required is equal to power available so no power is wasted.

V,

EN-ROUTE 14. Optimum turbojet cruising altitude is used? a. b. c. d.

Never. Rarely. Constantly in the cruise. The majority of the time in the cruise.

EN-ROUTE 15. In figure 4.25 of CAP 698, the drift down curve for 70000 Kg mass starts immediately whereas that for 35000 Kg does not. Why is this? a. The heavier aircraft has more inertia so it loses speed more slowly after engine failure. b. The lighter aircraft has less inertia so it loses speed more slowly after engine failure. c. The heavier aircraft produces more induced drag so it slows down more quickly after engine failure. d. The lighter aircraft is at a less efficient angle of attack.

EN-ROLITE 16. Increasing mass will.. ....drift down rate and ...... stabilising altitude? a. b. c. d.



EN-ROUTE 17. Icing conditions will a. b. c. d.


..... the drift down rate and ...... the stabilising altitude? Increase. Decrease. Increase. Decrease.

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EN-ROUTE 18. Switching off the air conditioning packs below 1700 feet will ...... the drift down rate and ...... the stabilising altitude? a. b. c. d.



EN-ROUTE 19. A tailwind will a. b. c. d.

...... drift down rate and .......drift down distance?



EN-ROUTE 20. A headwind will ...... drift down rate and .......drift down distance? a. b. c. d.



EN-ROUTE 21. A headwind will a. b. c. d.

...... drift down rate and .......drift down gradient?



EN-ROUTE 22. Increasing density altitude will ...... drift own rate? a. b. c. d.

Increase. Decrease. IVot affect. Depends on temperature.

EN-ROUTE 23. Why are the curves in the net level-off graph (CAP 698 figure 4.23) broi en at 1700 ft? a. Because the AIC system must be used above this altitude. b. Because the N C system cannot be used below this altitude. c. Because the N C system must be used above this altitude but would not be used following engine failure at lower altitudes. d. Because the aircraft depicted is more aerodynamically efficient at low altitudes.

-

.

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EN-ROUTE 24. Maximum continuous N, is typically ........ by increasing pressure altitude and ..... by increasing air temperature? a. b. c. d.

Increased Increased Decreased Decreased

Increased. Not affected. Decreased. Increased.

EN-ROUTE 25. The positive limit load factor of a JAR certificated passenger aircraft is ......... flaps up and .......flaps down?

EN-ROUTE 26. At VA? a. b. c. d.

The aircraft will ,suffer permanent damage. The aircraft cannot be damaged by control inputs. The aircraft cannot be stalled before exceeding the limiting load factor. Manoeuvring is not possible.

EN-ROUTE 27. If load factor is increased beyond...... at maximum operating altitude it will cause ...........7. a. b. c. d.

Ig, 1.3g, 1.5g, 2.5g,

mach tuck under. buffeting. clear air turbulence. overstressing of the structure.

EN-ROUTE 28. VAis? a. b. c. d.

Minimum speed at which maximum nose up control input is permitted. Maximum speed at which full nose up control input is permitted. Minimum control speed. Maximum control speed.

EN-ROUTE 29. Which of the following would increase VAfor a given aircraft structure? a. b. c. d.

lncreased aspect ratio. lncreased camber. lncreased wing area. lncreased sweep back.

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EN-ROUTE 30. Which of the following would improve ride quality in turbulence? a. b. c. d.

lncreased aspect ratio. lncreased camber. lncreased wing area. lncreased sweep back.

EN-ROUTE 31. VAis? a. b. c. d.

The The The The

minimum speed at which it possible to attain limiting positive load factor. maximum speed at which it is possible to attain limiting positive load factor recommended rough air flying speed. maximum operating speed.

EN-ROUTE 32. Deploying flaps in turbulence will? a. b. c. d.

Reduce stalling hazard but increase overload hazard. Reduce stalling hazard and overloading hazard. lncrease stalling hazard but decrease overloading hazard. lncrease stalling hazard and overloading hazard.

EN-ROUTE 33. Aircraft manoeuvrability.. ...... at high altitude because.. .........? a. b. c. d.

Increases, Increases, Decreases, Decrease,

Stability decreases. TAS increases. Buffet margins converge. Buffet margins diverge.

EN-ROUTE 34. .............. aircraft are least affected by turbulence? a. b. c. d.

Dihedral winged. Anhedral winged. Swept winged. Straight winged.

EN-ROUTE 35. The left boundary of the V-n and gust envelopes is? a. b. c. d.

High speed stall. Low speed stall. Mach tuck under. VAand V, respectively.

En-Route

EN-ROUTE 36. The right boundary of the V-n and gust envelopes is? a. Vc. b. V., MC,. d. V, and,,V ,

C.

respectively.

EN-ROUTE 37. Exceeding maximum operating altitl~deis likely to cause? a. b. c. d.

Structural damage due to excessive load factor. High speed buffet. Shock stall. Tuck under.

EN-ROUTE 38. Buffeting must not occur between

..........and ...........3.

a. ,V , and M ,,. b. Vslg and.,V, C. Vs,g and .M ,, d. 1.2Vslg and M V /,.,

EN-ROUTE 39. At the aerodynamic ceiling? a. b. c. d.

Load factors above 1.3 will cause high speed buffet. Load factor above 1 will cause high speed buffet. Load factors above 2.5 will cause high speed buffet. Any increase in load factor above 1 or speed change will cause buffet.

EN-ROUTE 40. At the absolute ceiling? a. b. c. d.

Load factors above 1.3 will cause loss of height. Load factor above 1 will cause loss of height. Load factors above 2.5 will cause loss of height. Any manoeuvres will cause loss of height.

EN-ROUTE 41. At the intersections of the power available and power required curves? a. b. c. d.

Constant speed level flight is possible. Accelerating level flight is possible. Steady speed climbing flight is possible. Accelerating climbing flight is possible.

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EM-ROUTE 42. If fuel consumption is defined in Kglnm? a. b. c. d.

Forward movement of C of G decreases fuel consumption if C of G is ahead of C of P. C of G position has no effect on fuel consumption. Forward movement of C of G increases fuel consumption if C of G is ahead of C of P. Aft movement of C of G reduces fuel consumption if C of G is aft of C of P.

EM-ROUTE 43. In level flight? a. b. c. d.

Maximum speed occurs when power required equals maximum power available. Minimum speed occurs when power required equals minimum power available. Maximum speed occurs when power required equals power available. Maximum speed occurs when maximum excess power is available.

EN-ROUTE 44. Flying at maximum range speed in a jet aircraft the angle of attack will be? a. b. c. d.

. , ,C , That giving Higher than that giving best L:D ratio. That giving optimum excess power available. Lower than that giving best L:D ratio.

EM-ROUTE 45. Changing speed from V ,, a. b. c. d.

to ?V ,,

Decreases drag and increases power required. lncreases drag and power required. Decreases drag and power required. lncreases drag and decreases power required.

EN-ROUTE 46. What is indicated in the buffet boundary chart? a. b. c. d.

Relationship between aircraft altitude, mass and low and high speed stall mach numbers. ,,, Relationship between aircraft altitude, mass and .M Relationship between aircraft altiiude, mass and high and low speed stall CAS. Relationship between aircraft altitude, mass and low and high speed stall IAS.

EM-ROUTE 47. When flying close to the maximum operating altitude? a. b. c. d.

Manoeuvrability is improved. Manoeuvrability is degraded. Manoeuvring is not possible. Only one flight speed is possible.

En-Route

EN-ROUTE 48. When flying close to the maximum operating attitude stability is? a. b. c. d.

Improved. Degraded. Not possible. Negative.

EN-ROUTE 49. Flying at maximum range speed in a propeller driven aircraft the angle of attack will be? a. b. c. d.

That giving best L:D ratio. Higher than that giving best L:D ratio. That giving optimum excess power available. Lower than that giving best L:D ratio.

EN-RQUTE 50. Which of the following occur when a propeller aircraft is at its absolute ceiling? a. b. c. d.

Maximum thrust is equal to drag and maximum power available is equal to power required. Power available is equal to power required but maximum thrust available is more than drag. Maximum thrust available is equal to drag but power available is more than power required. Thrust, drag, power available and power required are all equal.

EN-ROUTE 51. Flying at maximum endurance speed in a propeller driven aircraft the angle of attack will be? a. b. c. d.


EN-ROUTE 52. Flying at maximum endurance speed in a jet aircraft the angle of attack will be? a. b. c. d.


EN-ROUTE 53. A commercial jet will fly? a. b. c. d.

Always below optimum altitude to avoid mach buffet. Always at optimum altitude to achieve maximum economy. At optimum altitude when possible but this increases as the flight progresses. At the optimum altitude when possible but this decreases as the flight progresses.

177

178

En-Route

EN-ROUTE 54. A twin engine jet aircraft has six passenger seats and 5000 Kg maximum TOM. What is its minimum allowable EN-ROUTE obstacle clearance when drifting down to an alternate airfield following a single engine failure? a. b. c. d.

2000 ft. 1500 ft. 1000 ft. 2500 ft.

EN-ROUTE 55. What does the buffet boundary chart indicate? a. b. c. d.

M ,, at various altitudes and masses. ,V , at various altitudes and masses. Low speed stall and high speed stall boundaries for various masses and altitudes. Low speed and high speed buffet onset boundaries for various masses and altitudes.

EN-ROUTE 56. At the absolute ceiling for a jet aircraft? a. b. c. d.

Maximum ROC is zero. Flight is possible only at .V ,, Maximum ROC is 500 fpm. Flight is possible only at M .,,

EN-ROUTE 57. The most efficient EN-ROUTE flight profile is? a. b. c. d.

Step climb at constant N,. Cruise climb at constant N,. Step climb at increasing N,. Cruise climb at increasing N,.

EN-ROUTE 58. If it is believed that an aircraft drift down to an EN-ROUTE diversion, following single engine failure will not clear an obstacle, the correct procedures is to? a. b. c. d.

Jettison fuel irr~mediatelyto reduce drift down rate. Increase speed to reduce drift down rate. Select a different diversion. Assess fuel requirements then jettison fuel as soon as possible.

EN-ROUTE 59. When flying a jet at optimum altitude? a. b. c. d.

SFC remains constant. SFC increases as the flight progresses. SFC decrease as the flight progresses. Variation in SFC depends upon altitude.

En-Route

179

EN-ROUTE 60. High altitude cruise range in a twin jet aircraft will ...... when flying with one engines failed? a. b. c. d.

Be unchanged. Increase. Decrease. Depends on engine type.

EN-ROUTE 61. How does the long range cruise speed vary? a. b. c. d.

LRC TAS decreases with increasing altitude. LRC IAS increases with increasing altitude. LRC mach number decreases with increasing altitude. LRC mach number decreases with decreasing altitude.

EN-ROUTE 62. Which of the following is the correct procedure during drift down following engine failure? a. b. c. d.

Discard the flight manual and descend at V, + 10 Kts to the destination. Descend in the approach configuration. Begin fuel jettison immediately, ensuring sufficient reserves at the destination. Do not commence fuel jettison until en-route obstacles have all been cleared.

EN-ROUTE 63. What effect does an increase in mass have on glide range? a. b. c. d.

Increase. Decrease. Not affect. Decreases both range and ROD.

EN-ROUTE 64. What is the stabilising altitude? a. 'The altitude at which aircraft must temporarily cruise during a climb to stabilise temperatures and pressures. b. The altitude at which rate of descent becomes zero in a drift down following engine failure. c. The altitude at which an aircraft cruises temporarily between each step of a step climb. d. The altitude at which an aircraft cruises temporarily between each step of a step descent.

EN-ROUTE 65. What does a buffet boundary chart indicate? a. b. c. d.

M ,, for various masses. ,V , for various masses. Speeds for low speed buffet and high speed buffet onset at various masses and attitudes. Mach numbers for low speed stall and high speed stall at various masses and altitudes.

En-Route

180

EN-ROUTE 66. Maximum operating altitude for a jet aircraft is normally limited by? a. b. c. d.

Power and thrust available. High and low speed buffet boundaries. Jimitations of cabin pressurisation systems. Certificated maximum altitudes.

EN-ROUTE 67. Drift down procedure is employed? a. Following engine failure at any altitude above that at which reduced power can maintain level flight. b. For visual approaches only. c. Following explosive decompression of pressurisation failure. d. For ILS approaches only.

EN-ROUTE 68. How does optimum cruise altitude for a jet aircraft vary during flight? a. b. c. d.

Increase as weight decreases. Decreases as weight increases. Decreases as weight decreases. Remains constant.

EN-ROUTE 69. Optimum cruise altitude for a jet transport aircraft? a. b. c. d.

Cannot be used as to do so would cause ATC congestion. Is always used throughout cruise flight. Is never used by commercial aircraft as to do so would increase fuel costs. Is used whenever ATC considerations permit.

EN-ROUTE 70. The most efficient en-route profile is? a. b. c. d.

Cruise climb. Step climb. Constant speed. Constant altitude.

EN-ROUTE 71. Commercial jets always cruise? a. b. c. d.

At their most efficient altitude and speed. At 2000 ft above or below their optimum altitude, but at their most efficient speed. As close to their most efficient altitude and speed as ATC considerations will permit. At their maximum altitude and speed whenever possible.

En-Route

181,

EN-ROUTE 72. Flying a jet aircraft at maximum speed requires? a. b. c. d.

Maximum thrust and minimum drag. Minimum thrust and maximum drag. Maximum thrust and maximum drag. Best lift : drag ratio and maximum thrust.

EN-ROUTE 73. At constant altitude the buffet boundaries will converge with? a. b. c. d.

Forward movement of C of G. Aft movement of C of G. lncreasing temperature. Decreasing load factor.

EN-ROUTE 74. At constant altitude the buffet boundaries will converge with? a. Rearward movement of C of P. b. Rearward movement of C of G. c. lncreasing temperature. d. lncreasing bank angle.

EN-ROUTE 75. As altitude increases at constant weight and load factor, high speed buffet margin speed buffet margin . ......? a. b. c. d.


Decreases. Increases. Decreases. Increases.

EN-ROUTE 76. At low altitude? a. b. c. d.

The limiting variable is.,V, The lirniting variable is .M ,, The limiting variable is V .,, The limiting variable is.,M ,,

EN-ROUTE 77. At high altitude? a. b. c. d.

, The limiting variable is.,V The limiting variable is .M ,, The limiting variable is V ., The limiting variable is.,M ,,

.......... and low

182

En-Route

EN-ROUTE 78. If mach number and weight remain constant, increasing altitude requires? a. b. c. d.

Lower angle of attack. Higher angle of attack. Lower pitch attitude. Lower C,.

EN-ROUTE 79. At what speed must a jet aircraft and a propeller aircraft fly to maintain altitude in straight and level flight at the absolute ceiling? a. b. C.

VM,

Vxprop.

VMD

VMD-

VMP

VMP-

'XJet

M ' D.

EN-ROUTE 80. At any given altitude the buffet margins will? a. Increase with increasing weight. b. Decrease with increasing weight. c. The low speed margin will increase and the high speed margin will decrease with increasing weight. d. The low speed margin will decrease and the high speed margin will increase with increasing weight.

EN-ROUTE 81. Flight at maximum range speed in a jet aircraft requires? a. b. c. d.

Best L:D ratio. Angle of attack greater than that for best L:D ratio. Angle of attack less than that for best L:D ratio. Maximum angle of attack.

EN-ROUTE 82. For a jet aircraft to cruise at constant angle of attack and constant altitude? a. b. c. d.

Thrust must be gradually decreased to reduce IAS. Thrust must be gradually increased to maintain constant IAS. Thrust must be gradually decreased to maintain constant IAS. Drag will remain constant so thrust must remain constant.

EN-ROUTE 83. Following an engine failure an aircraft is drifting down at 140 Kts TAS with a gradient of 5%. What is its ROD if 1 Kts = 100 ft/min? a. b. c. d.

7 fpm. 70 fpm. 700fpm. 7000 fpm.

En-Route

183

EN-ROUTE 84. Following an engine failure an aircraft is drifting down at 140 Kts TAS and 5% gradient. If it encounters a headwind of 20 Kts, what will its ground speed be? a. b. c. d.

109.8 Kts. 119.8 Kts. 129.8 Kts. 139.8 Kts.

EN-ROUTE 85. Following an engine failure an aircraft is drifting down at 140 Kts TAS and 5% gradient. If it encounters a headwind of 20 Kts, what will its new gradient be?

EN-ROUTE 86. What rate of climb is available in a class B aircraft operating at its maximum cruise pressure altitude with all engine operating? a. b. c. d.

300 fpm. 400 fpm. 500 fpm. 600 fpm.

EN-ROUTE 87. The minimum acceptable vertical separation for class B aircraft in the accountability area is? a. b. c. d.

40 ft. 50 ft. 60 ft. 75 ft.

EN-ROUTE 88. Increasing the number of engines will? a. b. c. d.

lncrease drift down rate. Decrease drift down rate. Not affect drift down rate. lncrease or decrease drift down rate depending on mass.

EN-ROUTE 89. Dumping fuel following engine failure at high altitude will ..... ROD, and a. b. c. d.


decrease. increase. increase. decrease.

...... stabilising altitude?

En-Route

184

EN-ROUTE 90. What is the semi-width of the en-route obstacle accountability area for a class A aircraft fitted with navigation equipment unable to meet a containment criteria of 95%? a. b. c. d.

5 nm. 10 nm. 15 nm. 20 nm.

EN-ROUTE 91. By what vertical distance must the drift down path of a class A aircraft with one engine inoperative, clear obstacles en-route to the intended landing site? a. b. c. d.

50 ft. 500 ft. 1000 ft. 2000 ft.

EN-ROUTE 92. For a class A aircraft with one engine inoperative? a. b. c. d.

GFP must be positive 1000 ft above the assumed landing site. NFP must be positive 1000 ft above the assumed landing site. GFP must be 1.5% 1000 ft above the assumed landing site. NFP must be positive 1500 ft above the assumed landing site.

EN-ROUTE 93. For a class A aircraft with one engine inoperative? a. b. c. d.

GFP must be positive 1000 ft above all terrain and obstructions in the accountability area. NFP must be positive I000 ft above all terrain and obstacles in the accountability area. GFP must be 1.5% 1000 ft above the assumed landing site. NFP must be positive 1500 ft above all terrain and obstacles in the accountability area.

EN-ROUTE 94. The absolute~ceilingis the altitude at which the maximum attainable rate of climb is? a. b. c. d.

0 fpm. 50 fpm. 100 fpm. 500 fpm.

EN-ROUTE 95. The maximum operating altitude for a pressurised aircraft is? a. b. c. d.

The highest altitude for normal operations. Only applicable to 4 engine aircraft. The absolute ceiling. Dependent upon outside air temperature.

En-Route

185

EN-ROLITE 96. A 4 engine aircraft is operating at a mass for which the 3 engine service ceiling is FL 260. It suffers a single engine failure when cruising at FL400. If the remaining engines are immediately throttled up to maximum continuous power, which of the following statement will be true of the flight between FL400 and FL260? a. b. c. d.

Net flight path must always be at least 2000 ft above all terrain. Net gradient must be positive or better. Net gradient is less than gross gradient. Gross gradient is equal to net gradient.

EN-ROUTE 97. An aircraft is cruising at high altitude when it encounters as strong headwind. How will this affect the speed required to achieve maximum range? a. b. c. d.

TAS must be increased. TAS must be decreased. TAS is not affected. TAS must be increased by an amount equal to the headwind.

EN-ROUTE 98. When descending at constant mach number in the stratosphere? a. b. c. d.

TAS increases. TAS decreases. TAS remains constant. IAS increases.

EN-ROLITE 99. When cruising at constant mach number and constant FL in the stratosphere? a. b. c. d.

C, must increase. C, must decrease. C, must remain constant. SAR will increase.

EN-ROUTE 100. Comparing the ranges achievable using (1) Constant altitude, (2) Step climb, and (3) cruise climb techniques?

-EN-ROUTE 101. If weight decreases when cruising at constant altitude and C,? a. b. c. d.

Thrust and IAS required will decrease. Thrust and IAS required will increase. Thrust required will increases and IAS decrease. Thrust required will remain unchanged but IAS will decrease.

En-Route

186

EN-ROUTE 102. At the maximum operating altitude? a. b. c. d.

Load factors above 1.3 will cause high speed buffet. Load factor above 1 will cause high speed buffet. Load factors above 2.5 will cause high speed buffet. Any manoeuvres will cause high speed buffet.

LANDING

LAND I. If a pilot elects to use 45 degree flap instead of the normal 35 degrees for landing what change in landing can be expected? a. LDR decreases and go-around performance is improved. b. LDR increases and go- around performance is degraded. c. LDR is decreased and go-around performance is degraded. d. LDR is increased and go-around performance is improved.

LAND 2. LDR will be increased by? a. b. c. d.

High mass, high density, uphill runway slope. High mass, low density, downhill runway slope. Low mass, low density, uphill runway slope. Low mass, high density, downhill runway slope.

LAND 3. A slope contaminated with 5 mm of dry snow will .... landing distance compared to a dry runway? a. b. c. d.

Decrease. Increase. Be unchanged. Increase or decrease depending on flap setting.

LAND 4. Landing distance available at .the destination field .... compared to that at the departure field? a. b. c. d.

Must be more. May be less. Must be the same. Must be more or less depending on aircraft type. %

LAND 5. - --.

By what % must LDR change for a Class A jet aircraft operating to a wet field compared to the same field In dry conditions.

188

Landing

LAND 6. When operating to a dry runway the landing distance used by a Class A turbojet must not exceed ... . .. % of the LDA?

LAND 7. When operating to a dry runway the LDA for a Class A jet must be at least ..... of the landing distance used?

LAND 8. When operating to a dry runway a Class A turbo-prop aircraft requires a .... % safety margin of LDA over the landing distance used.

LAND 9. When operating to a dry runway the LDA for a Class A turbo-prop aircraft must be at least ..... of the landing distance used?

LAND 10. When operating to a dry runway a Class A turbo-prop aircraft requires a .....% safety margin of LDA landing distance used?

-.

Landing

189

LAND 11. When landing on a grass runway? a. b. c. d.

LDR will decrease due to increased rolling resistance through the grass. LDR will increase due to reduced friction between the tyres and the surface. LDR will be unaffected unless the grass is wet or very long. LDR will increase due to aquaplaning.

LAND 12. If the still air approach is 150 kts at a gradient of -5%, what will the gradient be if TAS, configuration and attitude are unchanged, when landing in a 20 Kts tailwind?

LAND 13. Increasing ambient temperature will ..... LDR and a. Decrease b. Decrease C. Increase d. Increase

....... WAT limit?


LAND 14. Maximum landing weight is most limited by? a. b. c. d.

Grass runway and tailwind. Grass runway and headwind. Concrete runway and tailwind. Concrete runway and headwind.

LAND 15. Decreasing ambient temperature will ..... LDR and ....... WAT limit? a. b. c. d.



LAND 16. When landing with all engines operating the speed at screen height should be?

190

Landing

LAND 17. If a pilot elects to use 35 degrees of flap instead of the standard 25 degrees when landing? a. b. c. d.

LDR is increased and go-around performance improved. LDR is increased and go-around performance degraded. LDR is decreased and go-aroundperformance improved. LDR is decreased and go-around performance degraded.

LAND 18. Landing on 0.5 mm of wet snow will? a. b. c. d.

Increase LDR. Decrease LDR. Not affect LDR. decrease LDA.

LAND 19. By what must the landing distance be factorised for a class A turbo-prop aircraft or: a wet runway? a. b. c. d.

0.51. 0.61 .. 0.7. 0.8

LAND 20. What technique should be employed when approaching a wet runway where there is a danger of aquaplaning? a. Land as normal then apply full reverse thrust and brakes upon reaching VA. b. Use a shallow approach to minimise impact at touchdown then apply full reverse and brakes as soon as possible. c. Carry out a positive touch down then apply full reverse and brakes as soon as possible. d. Use a shallow approach to minimise impact at touch down the apply full reveres and brakes'only when speed is less than VA.

LAND 21. What landing distance must be available for landing a turbo-prop aircraft at an alternative destination compared to at a main destination? a. b. c. d.

Identical. More. Less. More or less depending on pilot experience.

LAND 22. 5 mm of snow will ....... landing distance required? a. b. c. d.

Decrease. Increase. Depends on pressure altitude. Not affect.

Landing

191

LAND 23. For a class A turbojet aircraft landing on a wet runway, the landing distance available must be factorised by?

LAND 24. By what % must calculated landing distance be increased for a turbojet aircraft landing on a wet runway?

LAND 25. Which of the following does not depend upon the strength of a tailwind? a. b. c. d.

V,,,calculations. Landing climb limit mass calculations. Field length limited mass calculations. Approach gradient.

LAND 26. What extra % of the dry LDR must be used as the wet LDR for a turbo jet aircraft?

LAND 27. If the LDA for a turbojet planning to land on a on a given runway in dry conditions is 4000 m, what will LDA be if the runway becomes wet before the landing takes place.

LAND 28. If stalling speed is 120 Kts, at what speed should it cross the threshold on approach to landing? a. b. c. d.

100 Kts. 146 Kts. 156 Kts. 166 Kts.

192

Landing

LAND 29. When operating to a dry runway a Class A jet aircraft requires a .....% safety margin of LDA over LDR.

LAND 30. What is the purpose of the approach climb limit? a. b. c. d.

To clear obstacles on the go-around. To ensure maximum descent gradient is not exceeded. To ensure obstacles are cleared in the approach. To ensure that at least the minimum acceptable climb gradient is possible with one engine out in a go-around.

LAND 31. An aircraft is established on the approach to landing when it encounters a headwind? What must be done to maintain constant descent gradient? a. b. c. d.

Increase ground speed. Increase TAS. Decrease TAS. Decrease ground speed.

LAND 32. What is the minimum acceptable all engines operating landing climb gradient for a class B aircraft?

LAND 33. What is the minimum acceptable one engine inoperative landing climb gradient for a class B aircraft?

LAND 34. For a class B aircraft landing on a dry concrete runway the LD must not exceed ...... of the LDA?

Landing

193

LAND 35. For a class B aircraft landing on a dry concrete runway the minimum acceptable LDA is the landing distance used plus a safety factor of ....... ?

LAND 36. For a class B aircraft landing on a dry hard surface the minimum acceptable LDA is ... ... times the LD?

LAND 37. For a class B aircraft landing on dry grass the LD must not exceed .. .... of the LDA?

LAND 38. For a class B aircraft landing on dry grass the minimum acceptable LDA is the landing distance used plus a safety factor of . ......?

LAND 39. For a class B aircraft landing on dry grass the minimum acceptable LDA is ...... times the LD?

LAND 40. By what additional factor must the LDR for a class B aircraft be multiplied to calculate LDR for wet hard surfaces?

194

Landing

LAND 41. For a class B aircraft landing on wet grass the LD must not exceed

...... of the LDA?

LAND 42. For a class B aircraft landing on wet grass the minimum acceptable LDA is the landing distance used plus a safety factor of .......?

LAND 43. For a class B aircraft landing on wet grass the minimum acceptable LDA is ...... times the LD?

.LAND 44. For a class B aircraft landing on a wet concrete runway the LD must not exceed

...... of the LDA?

LAND 45. For a class B aircraft landing on a wet concrete runway the minimum acceptable LDA is the landing distance used plus a safety factor of .... ...?

LAND 46. For a class B aircraft landing on a wet hard surface the minimum acceptable LDA is ...... times the LD?

Landing

195

LAND 47. If a grass surface becomes wet this will ........ causing ...............? a. b. c. d.

Decrease friction between tyres and surface, Decrease the efficiency of the wheel brakes, Increase rolling resistance, Decrease rolling resistance,

LDR to increase. LDR to decrease. LDR to decrease. LDR to increase.

LAND 48. If the thrust reverse system is unserviceable it will cause? a. b. c. d.

LDR to increase and maximum landing mass to decrease. LDR and maximum landing mass to increase. LDR to decrease and maximum landing mass to increase. ,, to increase. LDR to decrease and ,V

LAND 49. The speed for landing a class A aircraft is? a. b. c. d.

Not less than 1.3VSat a 50 ft screen height. Not less than 1.3 V, at a 35 ft screen height. , at a 50 ft screen height. Not less than,,V at a 35 ft screen height. Not less than, ,V

LAND 50. What minimum acceptable climb gradient is used when calculating WAT limited landing mass for a class A twin engine aircraft? a. b. c. d.

2.0% net with one engine inoperative. 2.1% gross with one engine inoperative. 2.4% gross with one engine inoperative. 2.7% net with two engines inoperative.

LAND 51. What minimum acceptable climb gradient is used when calculating WAT limited landing mass for a class A three engine aircraft? a. b. c. d.

2.0% net with one engine inoperative. 2.1 % gross with one engine inoperative. 2.4% gross with one engine inoperative. 2.7% net with two engines inoperative.

LAND 52. What minimum acceptable climb gradient is used when calculating WAT limited landing mass for a class A four engine aircraft? a. b. c. d.

2.0% net with one engine inoperative. 2.1 % net with one engine inoperative. 2.4% gross with two engine inoperative. 2.7% gross with one engines inoperative.

196

Landing

LAND 53. What minimum acceptable climb gradient is used when calculating WAT limited landing mass for a class A twin engine aircraft conducting an instrument approach with a decision height of 150 ft? a. b. c. d.

2.0% net. 2.1% net. 2.4% gross. 2.5% gross.

LAND 54. An Airbus A320 is to conduct an instrument approach to a high altitude runaway with very high ambient temperature. Performancecalculations indicate that at its minimum practicable landing mass (after dumping fuel to minimum acceptable reserves) its maximum attainable climb gradient following engine failure will be 2.4%. Which of the following is true. a. b. c. d.

It cannot carry out an instrument approach. It cannot land at this field. The instrument landing can be carried out provided decision height is at least 200 ft. The landing must be abandoned if an engine fails during the approach.

LAND 55. If LDR is 4000 m, ,V ,, is 110 Kts and VATis increased to 130 Kts, to allow for turbulent conditions. What would be the new LDR?

LAND 56. What descent path is considered to be a steep approach for a Class A aircraft? a. b. c. d.

3.5' or greater. 4.5' or greater. 5.5' or greater. 6.5O or greater.

LAND 57. An aircraft is programmed to fly to an airfield with a single runway on which there is predicted to be a 15 Kt tailwind at the time of its arrival. Performance calculations show that although the aircraft can comply with the still air requirement, it cannot do so with the predicted tailwind. Which of the following is true? a. The flight cannot take place until the'tailwind has subsided to enable full compliance. b. The flight can be planned provided at least two suitable diversion fields enabling full compliance are available. c. The flight can take place provided at least one suitable diversion field enabling full compliance is available. d. The flight can take place because only the still air requirements are mandatory.

Landing

197

LAND 58. Which of the following best defines the measured landing distance? a. LDA x a safety factor of 1.3. b. LDA x a safety factor of 1.45. c. Horizontal distance between attaining screen height and the point at which the aircraft comes to a full stop when landing on a hard dry surface. d. Horizontal distance between attaining a screen height to the point at which the aircraft comes to a full stop on whatever surface it is landing on.

LAND 59. What LDR is to be used for a class B aircraft using a wet grass surface? a. b. c. d.

Landing distance x Landing distance x Landing distance x Landing distance x

a factor of a factor of a factor of a factor of

1.75. 1.97. 1.45. 1.64.

LAND 60. What is a declared safe area and when would it be required? a. The clearway. It is used for normal take-offs and landings. b. The clearway. It is used for steep approaches. c. An area free of obstructions located immediately before the threshold, It is required for short landings. d. An area free of obstructions immediately before the threshold. It is used for all take-offs and landings.

SECTION 3 SUMMARY OF ANSWERS

Atmospheres Airfields

Lift and drag Flaps Climbing and descending Power required and power available Curves Speeds Take-off En-route Landing

SUMMARY OF ANSWERS

ATMOS 1. c. A-TMOS 2. a. A-TMOS 3. c. A-TMOS 4. d. ATMOS 5. b. A-TMOS 6. d. A-TMOS 7. a. ATMOS 8. a. ATMOS 9. c. ATMOS 10. b. ATMOS 11. a. ATMOS 12. b. ATMOS 13. a. ATMOS 14. d. ATMOS 15. b. ATMOS 16. a. ATMOS 17. c. ATMOS 18. a. ATMOS 19. b A-TMOS 20. a. ATMOS 21. d. A-TMOS 22. a. A-TMOS 23. c. ATMOS 24. c. ATMOS 25. a. ATMOS 26. c. ATMOS 27. d. ATMOS 28. d. ATMOS 29. d. ATMOS 30. a. ATMOS 31. a. ATMOS 32. b. ATMOS 33. d. ATMOS 34. a. A-TMOS 35. d. ATMOS 36. b. ATMOS 37. d. ATMOS 38. d. ATMOS 39. a. ATMOS 40. b. ATMOS 41. c.

ATMOS 42. c. ATMOS 43. b, ATMOS 44, a. ATMOS 45. c. ATMOS 46. c. ATMOS 47, b. ATMOS 48. c. ATMOS 49, a. ATMOS 50. b. ATMOS 51. b. ATMOS 52. a. ATMOS 53. c. ATMOS 54, d. ATMOS 55. c. ATMOS 56. a. ATMOS 57. b. ATMOS 58. a. ATMOS 59. b. ATMOS 60. c. FIELDS 1. b. FIELDS 2. c. FIELDS 3. a. FIELDS 4. d. FIELDS 5. c. FIELDS 6. a. FIELDS 7. c. FIELDS 8. a. FIELDS 9. a. FIELDS 10, c. FIELDS 11. c. FIELDS 12, c. FIELDS 13. b. FIELDS 14. c. FIELDS 15. b. FIELDS 16. b. FIELDS 17. a. FIELDS 18. a. FIELDS 19. a. FIELDS 20. a. FIELDS 21. d.

FIELDS 22. b. FIELDS 23. a. FIELDS 24. c. FIELDS 25. c. FIELDS 26. a. FIELDS 27. b. FIELDS 28. c. FIELDS 29. b. FIELDS 30. d. FIELDS 31. a. FIELDS 32. d. FIELDS 33. a. FIELDS 34. b. FIELDS 35. b. FIELDS 36. d. FIELDS 37. a. FIELDS 38. a. FIELDS 39. a. FIELDS 40. b. FIELDS 41. d. FIELDS 42. d. FIELDS 43. c. FIELDS 44. a. FIELDS 45. c. FIELDS 46. c. FIELDS 47. a. FIELDS 48. c. FIELDS 49. a. FIELDS 50. c. FIELDS 51. 5. FIELDS 52. a. FIELDS 53. d. FIELDS 54. a. FIELDS 55. a. FIELDS 56. b. FIELDS 57. a. FIELDS 58. c. FIELDS 59. b. FIELDS 60. c. FIELDS 61. a. FIELDS 62. b.

FIELDS 63. b. FIELDS 64. c. FIELDS 65. c. - FIELDS 66. b. FIELDS 67. d. FIELDS 68, c. FIELDS 69. c. FIELDS 70. c. FIELDS 71. d. FIELDS 72. a. FIELDS 73. a. FIELDS 74. c. FIELDS 75. c. FIELDS 76. a. FIELDS 77. c. FIELDS 78. d. FIELDS 79. a. FIELDS 80. b. FIELDS 81. c. FIELDS 82. b. FIELDS 83. c. FIELDS 84. a. FIELDS 85. b. FIELDS 86. a. FIELDS 87. b.

202

Summary of Answers

FLAP 1. c. FLAP 2. a. FLAP 3. a. FLAP 4. c. FLAP 5. c. FLAP 6. a. FLAP 7. d. FLAP 8. b. FLAP 9. a. FLAP 10. b. FLAP 11. a. FLAP 12. c. FLAP 13. d. FLAP 14. d. FLAP 15. c. FLAP 16. b. FLAP 17. a. FLAP 18. a. FLAP 19. b. FLAP 20. d. FLAP 21. a. FLAP 22. a. FLAP 23. c. FLAP 24. b. FLAP 25. b. FLAP 26. c. CLlMB 1. b. CLlNlB 2. c. CLlMB 3. c. CLlMB 4. a. CLlMB 5. a. CLlMB 6. b. CLlMB 7. d.

CLIMB 8. c. CLlMB 9. d. CLlMB 10. b. CLlMB 11. c. CLlMB 12. a. CLlMB 13. c. CLlMB 14. d. CLlMB 15. c. CLlMB 16. a. CLIMB 17. d. CLlMB 18. d. CLlMB 19. c. CLlMB 20. b. CLIMB 21. c. CLlMB 22. c. CLlMB 23. c. CLlMB 24. c. CLlMB 25. b. CLlMB 26. c. CLlMB 27. a. CLlMB 28. b. CLlMB 29. b. CLlMB 30. d. CLIMB 31. d. CLIMB 32. c. CLlMB 33. b. CLIMB 34. d. CI-IMB 35. d. CLIMB 36. b. CLlMB 37. b. CLIMB 38. c. CLlMB 39. a. CLlMB 40. a. CLlMB 41. a. CLlMB 42. c. CLIMB 43. a. CLlMB 44. 0. CLlMB 45. d. CLlMB 46. d. CLIMB 47. c. CLIMB 48. b. CLlMB 49. b. CLlMB 50. d. CLlMB 51. d. CLlMB 52. c. CLlMB 53. b. CLlMB 54. b. CLlMB 55. d.

CLlMB 56. b. CLlMB 57. d. CLlMB 58. b. CLlMB 59. b. CLlMB 60. b. CLlMB 61. c. CLlMB 62. d. CLlMB 63. c. CLlMB 64. c. CLlMB 65. c. CLlMB 66. c. CLIMB 67. c. CLlMB 68. c. CLlMB 69. a. CLlMB 70. a. CLlMB 71. a. CLlMB 72. b. CLlMB 73. a. CLlMB 74. d. CLlMB 75. b. CLlMB 76. d. CLlMB 77. a. CLIMB. 78. b. CLIMB. 79. a. CLIMB. 80. c. CLIMB. 81. a. CLIMB. 82. c. CLIMB. 83. a. CLIMB. 84. c. CLIMB. 85. d. CLIMB. 86. d. CLIMB. 87. c. CLIMB. 88. c. CLIMB. 89. a. CLIMB. 90. c. CLIMB. 91. b. CLIMB. 92. c. CLIMB. 93. a. CLIMB. 94. a. CLIMB. 95. d. CLIMB. 96. d. CLIMB. 97. d. CLIMB. 98. a. CLIMB. 99. b. CLIMB. 100. b. CLIMB. 101. c. CLIMB. 102. b. CLIMB. 103. a.

CLINIB. 104. c. CLIMB. 105. c. CLIMB. 106. a. CLIMB. 107. a. CLINIB. 108. d. CLIMB. 109. c. CLIMB. 110. d. CLIMB. 111. d. CLIMB. 112. d. CLIMB. 113. d. CLIMB. 114. d. CLIMB. 115. a. CLIMB. 116. b. CLIMB. 117. a. CLIMB. 118. a. CLIMB. 119. d. CLIMB. 120. a. CLIMB. 121. a. CLIMB. 122. a. CLIMB. 123. d. CLIMB. 124. c. CLIMB. 125. a. CI-IMB. 126. c. CI-IMB. 127. c. CLIMB. 128. c. CLIMB. 129. a. CI-IMB. 130. b. CI-IMB. 131. b. CLIMB. 132. a. CI-IMB. 133. a. CI-IMB. 134. d. CLIMB. 135. d. CLIMB. 136. a. CI-IMB. 137. b. CLIMB. 138. c. CLIMB. 139. b. CLIMB. 140. a. CLIMB. 141. b. CLIMB. 142. b. CLIMB. 143. b. CLIMB. 144. a. CLIMB. 145. d. CLIMB. 146. b. CLIMB. 147. c. CLIMB. 148. a. CLIMB. 149. d. CLIMB. 150. c. CLIMB. 151. c.

CLIMB. 152. d. CLIMB. 153. a. CLINIB. 154. b. CLINIB. 155. c. CLIMB. 156. a. CLIMB. 157. a. CLIMB. 158. a. CLIMB. 159. a. CLINIB. 160. c. CLIMB. 161. c. POW. 1. c, POW. 2. c. POW. 3. b. POW. 4. a. POW. 5. b. POW. 6. a. POW. 7. b. POW. 8. a. POW. 9. a. POW. 10. d. POW. 11. d. POW. 12. c. POW. 13. a. POW. 14. d. POW. 15. c. POW. 16. c. POW. 17. b. POW. 18. b. POW. 19. a. POW. 20. a. POW. 21. a. POW. 22. c. POW. 23. a. POW. 24. b. POW. 25. b. POW. 26. a. POW. 27. b. POW. 28. c. POW. 29. c. POW. 30. d. POW. 31. c. POW. 32. b. POW. 33. c. POW. 34. a. POW. 35. a. POW. 36. b. POW. 37. b.

Summary of Answers

POW. 38. a. POW. 39. b. POW. 40. b. POW. 41. a. POW. 42. c. POW. 43. c. POW. 44. a. POW. 45. b. POW. 46. c. POW. 47. c. POW. 48. a. POW. 49. a. POW. 50. c. POW. 51. b. POW. 52. b. POW. 53. b. POW. 54. b. POW. 55. d. POW. 56. b. POW. 57. b. POW. 58. b. POW. 59. d. POW. 60. c. POW. 61. c. POW. 62. d. POW. 63. a. POW. 64. b. POW. 65. d. POW. 66. c. POW. 67. b. POW. 68. a. POW. 69. d. POW. 70. c. POW. 71. a. POW. 72. c. POW. 73. b. POW. 74. d. POW. 75. c. POW. 76. b. POW. 77. b. POW. 78. a. POW. 79. a. POW. 80. c. POW. 81. b. POW. 82. b. POW. 83. a. POW. 84. b. POW. 85. c.

POW. 86. b. POW. 87. b. POW. 88. a. CURVES. 1. a. CURVES. 2. b. CURVES. 3. c. CLIRVES. 4. b. CURVES. 5. b. CURVES. 6. b. CLIRVES. 7. d. CLIRVES. 8. a. CURVES. 9. d. CURVES. 10. a. CURVES. 11. c. CLIRVES. 12. a. CURVES. 13. c. CURVES. 14. e. CURVES. 15. d. CURVES. 16. d. CURVES. 17. d. CURVES. 18. c. CURVES. 19. a. CURVES. 20. a. CURVES. 21. a. CURVES. 22. b. CURVES. 23. d. CURVES. 24. b. CURVES. 25. c. CURVES. 26. a. CURVES. 27. d. CURVES. 28. b. CURVES. 29. c. CURVES. 30.d. CURVES. 31. c. CURVES. 32. c. CLIRVES. 33. b. CLIRVES. 34. b. CLIRVES. 35. c. CURVES. 36. b. CURVES. 37. a. CURVES. 38. b. CURVES. 39. a. CURVES. 40. c. CURVES. 41. d. CLIRVES. 42. c. CURVES. 43. b. CURVES. 44. b.

CURVES. 45. c. CURVES. 46. b. CURVES. 47. c. CURVES. 48. c. CURVES. 49. a. CURVES. 50. d. CURVES. 51. c. CURVES. 52. d. CURVES. 53. d. CURVES. 54. a. CURVES. 55. d. CURVES. 56. d. CURVES. 57. a. CURVES. 58. b. CURVES. 59. b. CURVES. 60. b. CURVES. 61. c. CURVES. 62. d. CURVES. 63. a. CURVES. 64. b. CURVES. 65. d. CURVES. 66. c. CURVES. 67. b. CURVES. 68. a. CURVES. 69. c. CURVES. 70. a. CURVES. 71. b. CURVES. 72. a. CURVES. 73. b. CURVES. 74. a. CURVES. 75. a. CURVES. 76. a. CURVES. 77. b. CURVES. 78. a. CURVES. 79. b. CURVES. 80. b. CURVES. 81. a. CURVES. 82. a. CURVES. 83. c. CURVES. 84. a. CURVES. 85. a. SPEED. 1. c. SPEED. 2. b. SPEED. 3. b. SPEED. 4. b. SPEED. 5. a. SPEED. 6. c.

SPEED. 7. b. SPEED. 8. a. SPEED. 9. d. SPEED. 10. a. SPEED. 11. a. SPEED. 12. d. SPEED. 13. a. SPEED. 14. d. SPEED. 15. b. SPEED. 16. b. SPEED. 17. b. SPEED. 18. b. SPEED. 19. b. SPEED. 20. a. SPEED. 21. a. SPEED. 22. b. SPEED. 23. a. SPEED. 24. d. SPEED. 25. b. SPEED. 26. a. SPEED. 27. b. SPEED. 28. c. SPEED. 29. b. SPEED. 30. a. SPEED. 31. a. SPEED. 32. a. SPEED. 33. c. SPEED. 34. b. SPEED. 35. d. SPEED. 36. a. SPEED. 37. a. SPEED. 38. c. SPEED. 39. c. SPEED. 40. b. SPEED. 41. c. SPEED. 42. a. SPEED. 43. b. SPEED. 44. a. SPEED. 45. a. SPEED. 46. c. SPEED. 47. b. SPEED. 48. a. SPEED. 49. c. SPEED. 50. b. SPEED. 51. b. SPEED. 52. a. SPEED. 53. b. SPEED. 54. b.

203

SPEED. 55. b. SPEED. 56. d. SPEED. 57. d. SPEED. 58. c. SPEED. 59. c. SPEED. 60. c. SPEED. 61. b. SPEED. 62. c. SPEED. 63. a. SPEED. 64. b. SPEED. 65. d. SPEED. 66. b. SPEED. 67. b. SPEED. 68. c. SPEED. 69. b. SPEED. 70. c. SPEED. 71. d. SPEED. 72. a. SPEED. 73. b. SPEED. 74. c. SPEED. 75. d. SPEED. 76. d. SPEED. 77. a. SPEED. 78. c. SPEED. 79. d. SPEED. 80. a. SPEED. 81. d. SPEED. 82. d. SPEED. 83. c. SPEED. 84. d. SPEED. 85. a. SPEED. 86. b. SPEED. 87. c. SPEED. 88. a. SPEED. 89. a. SPEED. 90. c. SPEED. 91. b. SPEED. 92. d. SPEED. 93. b. SPEED. 94. c. SPEED. 95. d. SPEED. 96. b. SPEED. 97. d. SPEED. 98. a. SPEED. 99. a. SPEED. 100. b. SPEED. 101. a. SPEED. 102. b.

204

Summary of Answers

SPEED. 103. c. SPEED. 104. d. SPEED. 105. b. SPEED. 106. a. SPEED. 107. a. SPEED. 108. d. SPEED. 109. c. SPEED. 110. c. SPEED. 111. a. SPEED. 112. a. SPEED. 113. d. SPEED. 114. a. SPEED. 115. c. SPEED. 116. b. SPEED. 117. c. SPEED. 128. c. SPEED. 119. d. SPEED. 120. d. SPEED. 121. b. SPEED. 122. b. SPEED. 123. b. SPEED. 124. b. SPEED. 125. b. SPEED. 126, a. SPEED. 127. a. SPEED. 128. b. SPEED. 129. a. SPEED. 130. a. SPEED. 131. a. SPEED. 132. c. SPEED. 133. b. SPEED. 134. a. SPEED. 135. c. SPEED. 136. a. SPEED. 137. a. SPEED. 738. d. SPEED. 139. a. SPEED. 140. c. SPEED. 141. a. SPEED. 142. d. SPEED. 143. c. SPEED. 144. a. SPEED. 145. d. TAKE-OFF. I.d. TAKE-OFF. 2. C. TAKE-OFF. 3. a. TAKE-OFF. 4. C.

TAKE-OFF. 5. C. TAKE-OFF. 53. b. TAKE-OFF. 6. a. TAKE-OFF. 54. d. TAKE-OFF. 7. C. TAKE-OFF. 55. b. TAKE-OFF. 8. d. TAKE-OFF. 56. C. TAKE-OFF. 57. C. TAKE-OFF. 9. b. TAKE-OFF. 10. d. TAKE-OFF. 58. C. TAKE-OFF. 11. d. TAKE-OFF. 59. d. TAKE-OFF. 12. b. TAKE-OFF. 60, C. TAKE-OFF, 13. a. 1 TAKE-OFF. 61. a. TAKE-OFF. 62. d. TAKE-OFF. 14. a. TAKE-OFF. 15. a. TAKE-OFF. 63. d. TAKE-OFF. 16. a. I TAKE-OFF. 64. b. TAKE-OFF. 17. b. TAKE-OFF. 65. d. TAKE-OFF. 18. C. TAKE-OFF. 66. a. TAKE-OFF. 19. C. TAKE-OFF. 67. C. TAKE-OFF. 20. d. TAKE-OFF. 68. d. TAKE-OFF. 21. a. TAKE-OFF. 69. C. TAKE-OFF. 22. d. TAKE-OFF. 70. a. TAKE-OFF. 23. C. TAKE-OFF. 71. b. TAKE-OFF. 24. a. TAKE-OFF. 72. d. TAKE-OFF. 73. a. TAKE-OFF. 25. b. TAKE-OFF. 26. C. TAKE-OFF. 74. C. TAKE-OFF. 27. d. TAKE-OFF. 75. a. TAKE-OFF. 28. a. TAKE-OFF. 76. a. TAKE-OFF. 29. b. TAKE-OFF. 77. d. TAKE-OFF. 30. b. TAKE-OFF. 78. b. TAKE-OFF. 31. C. TAKE-OFF. 79. b. TAKE-OFF. 32. b. TAKE-OFF. 80. C. TAKE-OFF. 33. a. TAKE-OFF. 81. C. TAKE-OFF. 82. C. TAKE-OFF. 34. d. TAKE-OFF. 83 b. TAKE-OFF. 35. b. TAKE-OFF. 36. d. TAKE-OFF. 84. a. TAKE-OFF. 85. a. TAKE-OFF. 37. a. TAKE-OFF. 86. C. TAKE-OFF. 38. d. TAKE-OFF. 39. d. TAKE-OFF. 87. d. TAKE-OFF. 40. b. TAKE-OFF. 88. a. TAKE-OFF. 89. d. TAKE-OFF. 41. d. TAKE-OFF. 42. a. TAKE-OFF. 90. b. TAKE-OFF. 43. b. TAKE-OFF. 91. C. TAKE-OFF. 44. C. I TAKE-OFF. 92. d. TAKE-OFF. 45. a. TAKE-OFF. 93. a. TAKE-OFF. 46. b. TAKE-OFF. 94. C. TAKE-OFF. 47. C. TAKE-OFF. 95. a. TAKE-OFF. 48. d. TAKE-OFF. 96. d. TAKE-OFF. 49. a. TAKE-OFF. 97. d. TAKE-OFF. 98. a.. TAKE-OFF. 50. a. TAKE-OFF. 51. d. TAKE-OFFS9. d. TAKE-OFF. 52. d. TAKE-OFF. 100. a.

'

,

TAKE-OFF. 101. b. TAKE-OFF. 102. C. TAKE-OFF. 103. a. TAKE-OFF. 104. b. TAKE-OFF. 105. C. TAKE-OFF. 106. a. TAKE-OFF. 107. b. TAKE-OFF. 108. a. TAKE-OFF. 109. a. TAKE-OFF. 110. d. TAKE-OFF. 111. b. TAKE-OFF. 112. a. TAKE-OFF. 113. a. TAKE-OFF. 114. a. TAKE-OFF. 115. a. TAKE-OFF. 116. b. TAKE-OFF. 117. C. TAKE-OFF. 118. d. TAKE-OFF. 119, d. TAKE-OFF. 120. a. TAKE-OFF. 121. d. TAKE-OFF. 122. C. TAKE-OFF. 123. d. TAKE-OFF. 124. d. TAKE-OFF. 125. a. TAKE-OFF. 4 26. C. TAKE-OFF. 127. C. TAKE-OFF. 128. C. TAKE-OFF. 129. a. EN-ROUTE. 1. d. EN-ROUTE. 2. a. EN-ROUTE. 3. b. EN-ROUTE. 4. a. EN-ROUTE. 5. a. EN-ROUTE. 6. C. EN-ROUTE. 7. b. EN-ROUTE. 8. b. EN-ROUTE. 9. C. EN-ROUTE. 10. C. EN-ROUTE. 11. b. EN-ROUTE. 12. a. EN-ROUTE. 13. C. EN-ROUTE. 14. d. EN-ROUTE. 15. C. EN-ROUTE. 16. b. EN-ROUTE. 17. b. EN-ROUTE. 18. C.

I

I

1

EN-ROUTE. 19. C. EN-ROUTE. 20. b. EN-ROUTE. 21. C. EN-ROUTE. 22. a. EN-ROUTE. 23. C. EN-ROUTE. 24. a. EN-ROUTE. 25. b. EN-ROUTE. 26. b. EN-ROUTE. 27. b. EN-ROUTE. 28. b. EN-ROUTE. 29. d. EN-ROUTE. 30. d. EN-ROUTE. 31. a. EN-ROUTE. 32. d. EN-ROUTE. 33. C. EN-ROUTE. 34. c. EN-ROUTE. 35. b. EN-ROUTE. 36, b. EN-ROUTE. 37. b. EN-ROUTE. 38. d. EN-ROUTE. 39. d. EN-ROUTE. 40. d. EN-ROUTE. 41. a. EN-ROUTE. 42. C. EN-ROUTE. 43. C. EN-ROUTE. 44. d. EN-ROLITE. 45. a. EN-ROUTE. 46. a. EN-ROUTE. 47. b. EN-ROUTE. 48. b. EN-ROUTE. 49. a. EN-ROUTE. 50. a. EN-ROUTE. 51. b. EN-ROUTE. 52. a. EIV-ROUTE. 53. C. EN-ROUTE. 54. a. EN-ROUTE. 55. d. EIV-ROUTE. 56. C. EN-ROUTE. 57. b. EN-ROUTE. 58. d. EN-ROUTE. 59. C. EN-ROUTE. 60. C. EN-ROUTE. 61. a. EN-ROUTE. 62. C. EN-ROUTE. 63. C. EN-ROUTE. 64. b. EN-ROUTE. 65. C. EN-ROUTE. 66. d.

Summary of Answers

EN-ROUTE. 67. a. EN-ROUTE. 68. a. EN-ROUTE. 69. d. EN-ROUTE. 70. a. EN-ROUTE. 71. C. EN-ROUTE. 72. C. EN-ROUTE. 73. a. EN-ROUTE. 74. d. EN-ROUTE. 75. C. EN-ROUTE. 76. a. EN-ROUTE. 77. b. EN-ROUTE. 78. b. EN-ROUTE. 79. a. EN-ROUTE. 80. b. EN-ROUTE. 81. C. EN-ROUTE. 82. C. EN-ROUTE. 83. C. EN-ROUTE. 84. b. EN-ROUTE. 85. d. EN-ROUTE. 86. a.

EN-ROUTE. 87. b. EN-ROUTE. 88. b. EN-ROUTE. 89. b. EN-ROUTE. 90. b. EN-ROUTE. 91. d. EN-ROUTE. 92. d. EN-ROUTE. 93. b. EN-ROUTE. 94. a. EN-ROUTE. 95. a. EN-ROUTE. 96. a. EN-ROUTE. 97. a. EN-ROUTE. 98. C. EN-ROUTE. 99. b. EN-ROLITE. 100. b. EN-ROUTE. 101. a. EN-ROLITE. 102. a. LAND. 1. c. LAND. 2. b. LAND. 3. b.

LAND. 4. b. LAND. 5. a. LAND. 6. c. LAND. 7. a. LAND. 8. c. LAND. 9. a. LAND. 10. d. LAND. 11. b. LAND. 12. c. LAND. 13. c. LAND. 14. a. LAND. 15. b. LAND. 16. b. LAND. 17. d. LAND. 18. c. LAND. 19. b. LAND. 20. c. LAND. 21. a. LAND. 22. b. LAND. 23. c.

LAND. 24. a. LAND. 25. b. LAND. 26. a. LAND. 27. b. LAND. 28. c. LAND. 29. b. LAND. 30. d. LAND. 31. b. LAND. 32. d. LAND. 33. c. LAND. 34. d. LAND. 35. b. LAND. 36. d. LAND. 37. c. LAND. 38. b. LAND. 39. a. LAND. 40. b. LAND. 41. b. LAND. 42. a. LAND. 43. d.

LAND. 44. c. LAND, 45. a. LAND. 46. c. LAND. 47. a. LAND. 48. a. LAND. 49. a. LAND. 50. b. LAND. 51. c. LAND. 52. d. LAND. 53. d. LAND. 54. c. LAND. 55. a. LAND. 56. b. LAND. 57.c. LAND. 58. c. LAND. 59. b. LAND. 60. c.

205

SECTION 4 ANSWERS AND EXPLANATIONS

Atmospheres Airfields Lift and drag Flaps Climbing and descending Power required and power available Curves Speeds Ta ke-off

Landing

ATMOSPHERES

ATMOS 1.6. This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a pressure altitude of 30000 ft and a mach number of 0.84, the TAT in the standard atmosphere is -12.

ATMOS 2. a. This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a pressure altitude of 30000 ft and a mach number of 0.84, the TAT in the standard atmosphere is -12. But the indicated TAT is -10. The difference between these two figures is a measure of the temperature deviation in that particular atmosphere. This is found by subtracting the ISA TAT from the indicated TAT. This is -10 -(-12), which gives a temperature deviation of 92. The atmosphere used in this question is therefore 2 degrees warmer than the international standard atmosphere.

ATMOS 3.6. This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a pressure altitude of 40000 ft and an indicated TAT of-29, the Mach number in the standard atmosphere would be 0.8.

ATMOS 4. d. This question can be answered using figure 4.3 on page 51 of CAP 698. From this it can be seen that at a mach number of 0.88 and an ISA TAT of -4, the pressure altitude in the standard atmosphere must be 28000 ft.

ATMOS 5. b. This question must be addressed in two stages. First, the temperature deviation must be found from figure 4.3 on page 51 of CAP 698. This temperature deviation must then be used in the standard equation to calculate density altitude. From figure 4.3, with a pressure altitude of 30000 ft, and a mach number of 0.82, the ISA TAT must be 14. But the indicated TAT is -10. The temperature deviation is equal to the indicated TAT minus the ISA TAT, which in this case is -10 -(-14), which is +4. This temperature deviation of +4 can now be used to calculate density altitude using the following equation: Density alt = Pressure alt + (118 x temperature deviation). So density altitude = 30000 ft + (118 x 4) = 30472 ft which is option b.

ATMOS 6. d. This question can be answered using figure 4.2 on page 50 of the CAP 698. From this it can be seen that if QNH is 999 hPa, a correction of +400 ft must be added to the elevation to give the pressure altitude. This gives a pressure altitude of 25400 ft at an elevation of 25000 ft.

ATMOS 7. a. This question can be answered using figure 4.2 on page 50 of CAP 698. The question states that at an elevation of 22000 ft the pressure altitude is 22800 ft. This means that a correction of +800 ft has been

210

Atmospheres

added to the elevation. From figure 4.2, it can be seen that a correction of +800 ft equates to a QNH of between 983 and 986 hPa. Option a, 985 hPa is therefore the most accurate.

ATMOS 8. a. This question can be answered using figure 4.1 on page 49 of the CAP 698. A wind blowing from 330 approaches runway 04 from the left at an angle of 70 degrees. So a wind of 30 Kts is blowing from the left at an angle of 70 degrees. From figure 4.1 this resolves to give a headwind component of approximately 10 Kts and a crosswind from the left of approximately 28 Kts.

ATMOS 9. c. This question can be answered using figure 4.1 on page 49 of the CAP 698. A wind blowing from 330 approaches runway 09 from the left at an angle of 120 degrees. Its will therefore comprise of components of a tailwind and crosswind from the left. From figure 4.1 it can be seen that the maximum permissible crosswind component of 10 Kts produced by a wind approaching at 120 degrees, equates to a wind of approximately 11.5 degrees. The most accurate option in this question is therefore option c, 12 kts.

ATMOS 10. b. This question can be answered usirlg figure 4.1 on page 49 of the CAP 698. A wind blowing from 270 approaches runway 29 from the left at an angle of 20 degrees. Its will therefore comprise of components of a headwind and crosswind from ,the left. From figure 4.1 it can be seen that a wind of 30 Kts approaching from an angle of 20 degrees gives a headwind of approximately 28 Kts and a crosswind of approximately 10.5 Kts from the left.

ATMOS 11. a. A QNH of 900 mb (or hPa) is outside of the range covered in figure 4.2 on page 50 of the CAP 698, so the problem must be solved using the standard equation: Pressure Altitude = Elevation + (30 x (1013 - QNH)) So in this question Pressure altitude = 4000 ft + (30 x (1013 - 900)) Which resolves to give Pressure altitude = 7390 ft.

ATMOS 12. b. This problem can be solved using the standard equation: Pressure altitude = 30 x (1013 - QFE) So in this question Pressure altitude = 30 x (1013 - 1020) Which resolves to give Pressure altitude = - 210 ft.

ATMOS 13. a. This problem can be solved using the standard equation: Pressure altitude = Elevation + (30 x (1013 - QNH)) Rearranging this gives Elevation = Pressure altitude - (30 x (1013 - QNH)) Which resolves to give Elevation = 3700 - 390 which = 3310 ft.

ATMOS 14. d. When calculating take-off and landlag performance prior to the commencement of a flight the actual temperatures at the take-off site are known, whereas only forecast temperatures are available for the destination. Take-off performance calculations are therefore based on actual temperatures, whilst landing calculations are based on forecast or predicted temperatures.

Atmospheres

21 1

ATMOS 15. b. In the International Standard Atmosphere (ISA) the temperature at mean sea level is 15OC and the lapse rate is 1.98OC per 1000 ft increase in altitude. At a pressure altitude of 5000 ft amsl, the temperature should therefore be 15OC - (5 x 1.98j. This means that the temperature at 5000 ft amsl pressure altitude should be 5.1°C. The temperature deviation is the actual temperature minus the ISA standard temperature. The temperature deviation in this case is therefore +lO°C - 5.1°C that is +4.g°C

ATMOS 16. a. The density of the air at any point in the atmosphere is dependent upon air pressure, temperature and humidity. The density altitude at any given point in the atmosphere at any given time, is the altitude at which the prevailing density would occur in the international standard atmosphere. It can be calculated using the equation: Density altitude = Pressure altitude plus (118 x the temperature deviation)

ATMOS 17. c. The pressure altitude at any point can be found by setting 1013.2 mp (hPa) on the altimeter subscale.

ATMOS 18. a. Density altitude is the altitude at which the prevailing air density would occur in the international standard atmosphere (ISA). It can be calculated from pressure altitude and temperature using the standard equation: Density altitude = Pressure altitude + (118 x temperature deviation) The sea level temperature in the ISA is 15% and the standard temperature lapse rate is 1.98OC per 1000 ft. The temperature in the ISA at a pressure altitude of 5000 ft should therefore ,be 15OC - (5 x 1.9g°C). The temperature deviation is the actual temperature minus the ISA standard temperature, which in this case is (25OC ) - (15OC -(5 x 1.98OC). Inserting these figures into the original equation gives: Density altitude = 5000 + 118(25 - (15 - (5 x 1.98))) = 7348.2 ft.

ATMOS 19. b. QFE is the air pressure at the field elevation. This problem can be so!ved using the standard equation: Pressure altitude = 30 x (1013 - QFE) = 30 x (1013 -1 022) That is the pressure altitude is -270 ft amsl.

ATMOS 20. a. This problem can be solved using the two standard equations: Pressure altitude = Elevation + (30 x (1013 - QNH)) And Pressure altitude = (30 x (1013 - QFE)) Combining the two equations gives: Elevation + (30 x (1013 -QNH)) = (30 x (1013 - QFE)) Inserting the data provided in the question gives: 4500 + (30 x (1013 - 1000)) = (30 x ('I013 -QFE))

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This simplifies to give 4890 = (30 x (1013 - QFE)) This can be rearranged to give:

Which is -850 = - QFE So QFE = 850 hPa.

ATMOS 21. d. Pressure altitude is the altitude at which the existing air pressure would occur in the international standard atmosphere. It is the altimeter indication when 1013.25 hPa is set on the altimeter sub-scale.

ATMOS 22. a. Density is a measure of how tightiy packed the molecules of a material are, or how much mass of a material can be held in a given volume of space. lncreasing altitude causes air pressure to decrease. Decreasing air pressure causes air to expand such that it is less tightly packed into a given volume. This decreases density. lncreasing temperature causes air to expand, reducing the mass that can be contained within a given volume. This reduces density. Water vapour is less dense than air so increasing humidity causes density to decrease. Air density is therefore decreased by increasing humidity, increasing altitude and increasing temperature.

ATMOS 23. c. QNH is the ambient air pressure at mean sea level. QFE is the ambient air pressure at the airfield elevation. The two are related in that any increase in one results in an equal increase in the other. Option d is therefore incorrect in that an increase in QNH would increase QFE. The elevation of an airfield is its vertical distance above mean sea level. Elevation does not change with varying atmospheric condition. Option c is therefore correct.

ATMOS 24. c. QNH is the ambient air pressure at mean sea level. QFE is the ambient air pressure at the airfield elevation. The two are related in that any increase in one results in an equal increase in the other. Option d is therefore incorrect in that an increase in QNH would increase QFE. Option b is incorrect because a change in QFE must cause a change in QNH. The elevation of an airfield is its vertical distance above mean sea level. Elevation does not change with varying atmospheric condition. Option a is therefore incorrect. Option c, increase QNH, is therefore correct.

ATMOS 25. a. EAS is the Equivalent Airspeed. The movement of an aircraft through the air causes the air pressure close to it to increase. At low speeds this increased pressure is not significant but as speed increases the magnitude of the pressure rise increases. At high speed the pressure rise is sufficient to compress the air slightly, thereby increasing the dynamic pressure. This process is called adiabatic compression. If left uncorrected this would cause the airspeed indicator to become progressively less accurate, over indicating as speed increases. The EAS is the Calibrated Airspeed corrected for the effects of this adiabatic compression of the air.

ATMOS 26. c. An airspeed indicator gives an indication proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressuie, so climbing at constant IAS or CAS means

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climbing at constant dynamic pressure. But dynamic pressure is equal to %pV2,where p is air density and V is TAS. So when climbing at constant IAS, the value of TAS must increase to compensate for decreasing density, in order to maintain a constant dynamic pressure and airspeed indication. TAS therefore increases as altitude increases in a constant IAS climb.

ATMOS 27. d. Mach number is the speed of an aircraft as a fraction of the local speed of sound. Mach 1 for example means a speed equal to the local speed of sound, whereas mach 0.5 means a speed only half of the local speed of sound. The speed of sound in the atmosphere is not constant but is proportional to the square root of absolute temperature (LSS = 38.94 TM Absolute temperature ). This means that if air temperature decreases, the local speed of sound also decreases. But if the local speed of sound decreases, then the TAS equating to any given mach number also decreases. Decreasing temperature therefore causes the TAS at any given mach number to decrease. As pressure altitude increases up to the tropopause at 36000 ft, the air temperature decreases at a rate of approximately 1.98 degrees per 1000 ft. Above 36000 ft the temperature remains constant. This means that as pressure altitude increases up to 36000 ft, the local speed of sound decreases, then remains constant at higher altitudes. So when climbing at constant Mach number TAS will decrease up to 36000 ft, then remain constant.

ATMOS 28. d. As pressure altitude increases up to the tropopause at 36000 ft, the air temperature decreases at a rate of approximately 1.98 degrees per 1300 ft. Above 36000 ft the temperature remains constant. So as altitude increases the temperature decreases up to 36000 ft then remains constant above this altitude.

ATMOS 29. d. An airspeed indicator gives an indication (IAS) proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so descending at constant IAS means descending at constant dynamic pressure. Airspeed indicators are susceptible to individual instrument errors and errors due to minor inaccuracies in the pressure sensing systems. The Calibrated Airspeed (CAS) is the IAS corrected for these minor errors. So descending at constant IAS or CAS means descending at constant dynamic pressure.

ATMOS 30. a. Density altitude is the altitude at which the prevailing air density would occur in the International Standard Atmosphere (ISA). As altitude increases, the air pressure decreases allowing the air to expand. This expansion causes density to decrease, so density decreases with increasing altitude. This means that anything which causes the density to decrease causes the density altitude to increase. Increasing air terr~peraturealso causes the air to expand, thereby reducing its density and increasing the density altitude. So increasing temperature decreases density but increases density altitude.

ATMOS 31. a. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determifled by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so as the LSS decreases, the Mach number at any given TAS increases. So when descending through an inversion at any given TAS, the temperature and LSS decrease, causing the mach number to increase. This is the opposite to what happens in a normal atmosphere. So option a is correct and options b and c are incorrect.

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As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to increase. So option d is incorrect.

ATMOS 32. b. In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so as the LSS increases, the Mach number at any given TAS decreases. So when climbing through an inversion at any given TAS, the temperature and LSS increase, causing the mach number to decrease. This the opposite to what happens in a normal atmosphere. So option b is correct and options a and c incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the CAS at any given TAS to decrease. So when climbing at constant TAS the CAS decreases. So option d is incorrect.

ATMOS 33. d. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS, the TAS decreases. So option d is correct and a is incorrect. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any given TAS, the terr~peratureand LSS decrease, causing the mach number to increase. But when descending through an inversion at constant CAS, the overall effect on Mach number is a balance between decreasing TAS and decreasing LSS. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, decreasing TAS is therefore the only one that is definitely true.

ATMOS 34. a. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS, the TAS increases. So option a is correct and d incorrect. In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any given TAS, the temperature and LSS increase, causing the mach number to decrease. But when climbing through an inversion at constant CAS, the overall effect on Mach number is a balance

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between increasing TAS and increasing LSS. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option a, increasing TAS is therefore the only one that is definitely true.

ATMOS 35. d. In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any given mach number, the temperature and LSS increase, causing the TAS to increase. So option d is correct and options b and c, incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant TAS the CAS decreases. But when climbing through an inversion at constant mach number, the overall effect on CAS is a balance between increasingTAS due to increasing LSS and decreasing CAS :TAS ratio due to decreasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, increasing TAS is therefore the only one that is definitely true.

ATMOS 36. b. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any given mach number, the temperature and LSS decrease, causing the TAS to decrease. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant TAS the CAS increases. But when descending through an inversion at constant mach number, the overall effect on CAS is a balance between decreasing TAS due to decreasing LSS and increasing CAS :TAS ratio due to increasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option b, decreasing TAS is therefore the only one that is definitely true.

ATMOS 37. d. In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through

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an inversion at any given mach number, the temperature and LSS increase, causing the TAS to increase. Option d is therefore correct and option c incorrect. Also because temperature and LSS both increase when climbing in an inversion, option b is incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an inversion or in a normal atmosphere, the presswe decreases. This causes the TAS at any given CAS to increase. So when climbing at constant TAS the CAS decreases. But when climbing through an inversion at constant mach number, the overall effect on CAS is a balance between increasing TAS due to increasing LSS and decreasing CAS :TAS ratio due to decreasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, increasing TAS is therefore the only one that is definitely true.

ATMOS 38. d. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any given mach number, the temperature and LSS decrease, causing the TAS to decrease. Option d is therefore correct and option c incorrect. Also because temperature and LSS both decrease when descending in an inversion, option b is incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant TAS the CAS decreases. But when descending through an inversion at constant mach number, the overall effect on CAS is a balance between decreasing TAS due to decreasing LSS and increasing CAS :TAS ratio due to increasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on CAS without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option d, decreasing TAS is therefore the only one that is definitely true.

A'rMOs 39. a. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbirlg, whether in an inversion or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS the TAS increases. So option a is correct and options b and c are incorrect. In an inversion the normal relationship between temperature and altitude is reversed, so climbing causes temperature to increase. The local speed of sound (LSS) is determined by temperature alone, so increasing temperature causes the local speed of sound to increase. Mach number represents TAS as a fraction of LSS, so if the LSS increases, the mach number at any given TAS decreases. So when climbing through an inversion at any TAS, the temperature and LSS increase, causing the mach number to decrease. But when climbing through an inversion at constant CAS, the overall effect on mach number is a balance between decreasing mach number due to increasing LSS and decreasing CAS :TAS ratio due to decreasing pressure.The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore

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possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option a, increasing TAS is therefore the only one that is definitely true.

ATMOS 40. b. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an inversion or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS the TAS decreases. So option b is correct and options a and c are incorrect. In an inversion the normal relationship between temperature and altitude is reversed, so descending causes temperature to decrease. The local speed of sound (LSS) is determined by temperature alone, so decreasing temperature causes the local speed of sound to decrease. Mach number represents TAS as a fraction of LSS, so if the LSS decreases, the mach number at any given TAS increases. So when descending through an inversion at any TAS, the temperature and LSS decrease, causing the mach number to increase. But when descending through an inversion at constant CAS, the overall effect on Mach number is a balance between increasing Mach number due to decreasing LSS and increasing CAS :TAS ratio due to increasing pressure. The overall effect will depend upon the magnitude of the temperature lapse rate. It is not therefore possible to predict the effect on mach number without first knowing the temperature lapse rate in the inversion. Of the options offered in this question, option b, decreasing TAS is therefore the only one that is definitely true.

ATMOS 41 c. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach nurnber represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So option c is correct and options a and b are incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isotherinal layer or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to increase. So when descending at constant TAS the CAS increases. So option d is incorrect.

ATMOS 42. c. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal Layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So option c is correct and options a and b are incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to decrease. So when climbing at constant TAS the CAS decreases. So option d is incorrect.

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ATMOS 43. b As altitude changes, the main factor controlli~gthe ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS the TAS decreases. So option d is incorrect. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. But when descending at constant CAS the TAS decreases so the Mach number decreases. Option b is therefore correct and option a and c incorrect.

ATMOS 44. a. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS the TAS increases. So option d is incorrect. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. But when climbing at constant CAS the TAS increases so the Mach number increases. Option a is therefore correct and option b and c incorrect.

ATMOS 45. c. An isothermal layer is one in which the terr~peraturedoes not vary with altitude. The local speed of sound is determined by temperature alone, so in an isohermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach nbmber at any given TAS is constant. So when climbing at constant Mach number in an isothermal the TAS remains constant. Option c is therefore correct and option a and b incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the CAS at any given TAS to decrease. So when climbing at constant Mach number and constant TAS in an isothermal, CAS decreases. So option d is incorrect.

ATMOS 46. c. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when descending at constant mach number in an isothermal the TAS remains constant. Option c is therefore correct and option a and b incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the CAS at any given TAS to increase. So when descending at constant Mach number and constant TAS in an isothermal, CAS increases. So option d is incorrect.

ATMOS 47. b. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach

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number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when climbing at constant Mach number in an isothermal the TAS remains constant. Option d is therefore incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing , whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the CAS at any given TAS to decrease. So when climbing at constant Mach number and constant TAS in an isothermal, CAS decreases. So option b is correct and a and c incorrect.

ATMOS 48. c. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when descending at constant CAS in an isothermal the LSS remains constant. Option c is therefore correct and options a and b incorrect. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending , whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS in an isothermal, TAS decreases. So option d is incorrect.

ATMOS 49. a. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermal layer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when climbing at constant CAS in an isothermal the LSS remains constant. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When climbing, whether in an isothermal layer or in a normal atmosphere, the pressure decreases. This causes the TAS at any given CAS to increase. So when climbing at constant CAS in an isothermal, TAS increases. But LSS is constant in an isothermal, so increasing TAS causes Mach number to increase. So option a is correct and option b, c and d, are incorrect.

ATMOS 50. b. As altitude changes, the main factor controlling the ratio of TAS : CAS is the ambient air pressure. When descending, whether in an isothermal layer or in a normal atmosphere, the pressure increases. This causes the TAS at any given CAS to decrease. So when descending at constant CAS in an isothermal, TAS decreases. So option b is correct and option a and c incorrect. An isothermal layer is one in which the temperature does not vary with altitude. The local speed of sound is determined by temperature alone, so in an isothermallayer the LSS is constant at all altitudes. Mach number represents the TAS as a fraction of the LSS, so in an isothermal layer the Mach number at any given TAS is constant. So when descending at constant CAS in an isothermal the decreasing TAS causes Mach number to decrease. option d is therefore incorrect.

ATMOS 51. b. This problem can be solved using the following equation: Pressure altitude = Elevation + (30 x (1013 - QNH))

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Inserting the figures provided in the question gives: Pressure altitude = 4000 ft +(30 x (1013 - 1025)) Which is 4000 ft + -360 ft

= 3640 ft.

ATMOS 52. a. In the international standard atmosphere (ISA) the relationship between altitude, temperature, pressure and density are fixed. As altitude decreases in the ISA, both the air pressure and temperature increase. The increasing pressure tends to compress the air increasing its density, whilst the increasing temperature tends to expand the air, decreasing its density. The effects of the pressure change are however the dominant factor so as pressure and temperature increase in the ISA, so does the air density.

ATMOS 53. c. In the lnternational Standard Atmosphere (ISA) the relationship between altitude, temperature, pressure and density are fixed. As altitude decreases in the ISA both the air pressure and temperature increase. The increasing pressure tends to compress the air thereby increasing its density, whilst the increasing temperature tends to expand the air thereby decreasing its density. The effect of the pressure change is however the dominant factor so as pressure and temperature increase in the ISA, so does the air density. But in a non-standard atmosphere these relationships are not fixed. Meteorological conditions can cause temperature and pressure to vary such that density might increase or decrease, depending upon the relative magnitudes of the variation in each. If the increase in pressure is of a greater proportion than that of temperature then the density will increase. Conversely, if the increase in temperature is of a greater proportion than that of pressure, the density will decrease.

ATMOS 54. d. As atmospheric pressure increases, the air is compressed causing its density to increase. As air temperature increases, the air expands causing its density to decrease. In the ISA where both pressure and temperature increase together, the pressure change is the dominant factor, so density increases. If however pressure remains constant whilst temperature increase, then the air will expand and air density will decrease. Calibrated Airspeed (CAS) is proportional to dynamic pressure (%pV2),where p is air density and V is the True Airspeed (TAS). This means that any given value of %pV2will occur at the same CAS at all altitudes. If air density decreases, the TAS equating to any given CAS must increase, so that the increasing V2 balances the decreasing p, such that %pV2remains constant. This in turn means that if density decreases, then the ratio of CAS:TAS decreases. So if pressure remains constant whilst temperature increases the density will decrease, causing the CAS:TAS ratio to decrease.

N M O S 55. c. In the International Standard Atmosphere (ISA) the relationships between altitude, temperature, pressure and density are fixed. As altitude increases in the ISA, both the air pressure and temperature decrease. The decreasing pressure tends to expand the air thereby decreasing its density, whilst the decreasing temperature tends to compress the air thereby increasing its density. The effect of thgpressure change is however the dominant factor, so as altitude increases, the pressure, temperature and density all decrease at a fixed rate. The pressure at any altitude and the temperature at any altitude are therefore fixed. In the real atmosphere the rates of change of pressure and temperature with changes in altitude are not fixed. It is therefore possible that the pressure and density at any given altitude do not conform to the ISA predictions.

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Pressure altitude is the altitude at which the prevailing pressure at any point in a real atmosphere would occur in the ISA. It is the altitude indicated on a barometric altimeter when 1013.25 hPa is set on the subscale. Elevations in the ISA are vertical distances above msl. Elevation in the ISA is therefore the same as pressure altitude. The density altitude is the pressure altitude at which the prevailing density would occur in the ISA. So density altitude could also be said to be the elevation at which the prevailing density would occur in the ISA. Statements 1 and 3 in this question are therefore true.

ATMOS 56. a. In the International Standard Atmosphere (ISA), the relationships between altitude, temperature, pressure and density are fixed. As altitude increases in the ISA, both the air pressure and temperature decrease. The decreasing pressure tends to expand the air thereby decreasing its density, whilst the decreasing temperature tends to compress the air thereby increasing its density. The effect of the pressure change is however the dominant factor, so as altitude increases, the air density decreases. Aircraft performance is directly proportional to power available. Power available is directly proportional to the mass flow of air passing through the engines. As air density decreases the mass flow of air passing through the engines decreases, causing power available and aircraft performance to decrease. This is the primary factor responsible for the reduction in aircraft performance as altitude increases so option a is the most appropriate in this question. A secondary factor is that as altitude increases, the lift and drag forces at any given IAS remain constant but the TAS increases. But power required is equal to TAS multiplied by drag, so as altitude increases, the power required for any given flight condition increases. This increasing power requirement, coupled with decreasing power available increases the rate of reduction in performance.

ATMOS 57. b. In the ISA the relationships between altitude, pressure, temperature and density are fixed such that the density at any given pressure altitude is always the same. In the real atmosphere the rates of change of pressure and temperature with changes in altitude are not fixed. It is therefore possible that the density at any given altitude does not conform to the ISA predictions. Pressure altitude is the altitude at which the prevailing pressure at any point in a real atmosphere would occur in the ISA. It is the altitude indicated on a barometric altimeter when 1013.25 hPa is set on the subscale. The density at any given pressure altitude is not however constant in a real atmosphere. The density altitude is the pressure altitude at which the prevailing density would occur in the ISA. Option b is therefore correct. Options a and care incorrect because they use the term "conditions" but do not state what these conditions are (temperature, pressure, density, humidity). This option is therefore too loosely worded to be accurate. Option d is also incorrect because dividing actual density by density lapse rate will not give the density altitude. It would be more accurate to say that: Density alt = (ambient density - ISA msl density) 1 density lapse rate.

ATMOS 58. a. This problem can be solved using the following equation: Density alt = pressure alt +(118.6 x temperature deviation)

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Where temp deviation = (ambient temp -(I 5 -(I .98 x pressure alt I 1000 ft))) degrees. At a pressure altitude of 10000 ft and an ambient temperature of 12O C, Temp deviation = (12 -(I 5 -(I .98 x 10))) which is 12 + 4.8 or 16.8 Inserting this value into the original equation gives: Density altitude = 10000 ft + (I18.6 x 16.8) which is 11992.4. The closest option is therefore option a, 11993 ft. It should be noted that in some examinations a less accurate of Density alt = Pressure alt + (120 x ternp deviation) is sometimes used.

ATMOS 59. b. This problem can be solved using the following equation: Density alt = pressure alt +(I 18.6 x temperature deviation) 5 -(1.98 x pressure alt I 1000 ft))) degrees. Where temp deviation = (ambient temp -(I At a pressure altitude of 0 ft and an ambient temperature of

lo0 C,

Temp deviation = (10 -(I5 -(I .98 x 0))) which is -5 Inserting this value into the original equation gives: Density altitude = 0 ft + (118.6 x -5) which is -593 ft. It should be noted that in some examinations a less accurate of Density alt = Pressure alt + (120 x temp deviation) is sometimes used.

ATMOS 60. c. This problem can be solved using the following equatio~s: Pressure altitude = Elevation + (30 x (1013 - QNH)) lnsertiqg the values given in the question gives: Pressure altitude = 1000 ft + (30 x (1013 - 1025)) which is 640 ft. It should be noted that the problem can also be solved using the following equation: Pressure altitude = 30 x (1013 - QFE) Which gives pressure altitude = 30 x (1013 -991.67) = 639.9 ft.

AIRFIELD

FlELD 1. b. CAP 168 Section 3, paragraph 9.3 states that the maximum allowable length of the clearway is 50% of the TORA.

FlELD 2. c. CAP 168 Section 3, paragraph 8 defines the stopway. This includes a requirement that the stopway must be capable of sustaining the weight of an aircraft without causing structural damage to the aircraft. The stopway therefore ends at the point where its structural strength falls below that level.

FlELD 3. a. The definition of the stopway in CAP 168 includes the requirement that its slope should be the same as the runway and should not change by more than 0.5% per 30 m.

FIELD 4. d. The definition of the stopway in CAP 168 includes a requirement that its width be no less than that of the runway it serves.

FlELD 5. c. CAP 168 Chapter 3, Appendix 3C paragraph 2.1.3 defines a balanced field as one in which the length of the clearway is equal to that of the stopway, or no clearway or stopway exist. Under these circumstances the ASDA (sometimes called the EMDA) is equal to the TODA.

FlELD 6. a. CAP 168 Chapter 3, paragraph 13 (c), defines the TODA as being equal to the TORA plus the clearway. Although the clearway ends at the first obstacle, its maximum length is 50% of the TORA. The maximum possible length of the TODA is therefore 150% of the TORA.

FIELD 7. c. CAP 168 Chapter 3, paragraph 13C, defines the TODA as being equal to the TORA plus the clearway. Although the clearway ends at the first obstacle, its maximum length is 50% of the TORA. The maximum possible length of the TODA is therefore 150% of the TORA so option b is incorrect. The EMDA is equal to the TORA plus the stopway, so the maximum length of the TODA must be less than 150% of the EMDA so option a is incorrect. The ASD is the accelerate stop distance required and this varies with aircraft weight and ambient conditions. Option d is therefore incorrect. Option c is therefore the most appropriate.

FIELD 8. a. CAP 168 defined the required width of the clearway as being not less than that of the runway strip for the visual runway, at the end of the TORA, expanding from the centreline linearly to a semi-width of 90 m. This equates to 150 m a i the end of the TORA, expanding to 180 m at the end of the clearway.

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FLIED 9. a. CAP 168 Chapter 3, paragraph 13C, defines the TODA as being equal to the TORA plus the clearway.

FIELD 10. c. CAP 168 Chapter 3, Appendix 3C paragraph 2.1.3 defines a balanced field as one in which the length of the clearway is equal to that of the stopway, or no clearway or stopway exist. Under these circumstances the ASDA (sometimes called the EMDA) is equal to the TODA. Option a and b in this question both satisfy this condition. A field is unbalanced when the clearway is longer than the stopway or when the stopway is longer than the clearway (option c). It should be noted that option d is highly improbable in that the stopway must be capable of supporting the weight of the aircraft whereas the clearway need not. Clearways are therefore rarely if ever harder than stopways.

FIELD 11. c. CAP 168 Chapter 3, Appendix 3C paragraph 2.1.3 defines a balanced field as one in which the length of the clearway is equal to that of the stopway, or no clearway or stopway exist. Under these circumstances the ASDA (sometimes called the EMDA) is equal to the TODA. Option a and b in this question both satisfy this condition. A field is unbalanced when the clearway is longer than the stopway (option c), or when the stopway is longer than the clearway. It should be noted that option d is highly improbable in that the stopway must be capable of supporting the weight of the aircraft whereas the clearway need not. Clearways are therefore rarely if ever harder than stopways.

FIELD 12. c. CAP 168 Chapter 3, Appendix 3C paragraph 2.1.3 defines a balanced field as one in which the length of the clearway is equal to that of the stopway, or no clearway or stopway exist. Under these circumstances the ASDA (sometimes called the EMDA) is equal to the TODA. Option a and b in this question both satisfy this condition. A field is unbalanced when there is a clearway but no (option c), or when the stopway is longer than the clearway. It should be noted that option d is highly improbable in that the stopway must be capable of supporting the weight of the aircraft whereas the clearway need not. Clearways are therefore rarely if ever harder than stopways.

FIELD 13. b. CAP 168 Chapter 3, Appendix 3C paragraph 2.1.3 defines a balanced field as one in which the length of the clearway is equal to that of the stopway, or no clearway or stopway exist. Under these circumstances the ASDA (sometimes called the EMDA) is equal to the TODA. A field with a stopway but no clearway does not satisfy this condition and hence is unbalanced (option b). Such fields are however permitted but are less operationally flexible in that they limit take-off mass to the field length limits and hence do not permit the full exploitation of the climb limited mass capability of the aircraft.

FIELD 14. c. CAP 168 Chapter 3, Appendix 3C paragraph 2.?.3 defines a balanced field as one in which the length of the clearway is equal to that of the stopway, or no clearway or stopway exist. Under these circumstances the ASDA (sometimes called the EMDA) is equal to the TODA. When operating from such a field the calculation of take-off performance for some aircraft types is simplified (option c).

FIELD 15. b. This problem can be resolved using the standard equation: Slope = (difference in threshold elevations 1 TORA) x 100%

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But both difference in elevation and TORA must be expressed in the same units so if elevation is in ft, then a TORA in meters must be multiplied by 3.28 ftlm. lnserting the data provided in this question gives: Slope = (150 ft - 200 ft) 1(1200 m x 3.28 ftlm) XI 00% Which resolves to give slope = -1.27%. That is a slope of 1.27% downwards.

FIELD 16. b. This problem can be resolved using the standard equation: Slope = (difference in threshold elevations IT O W ) x 100% But both difference in elevation and TORA must be expressed in the same units so if elevation is in ft, then a TORA in meters must be multiplied by 3.28 ftlm. This problem is compounded by the fact that the two runaways have different TORAs. This means that one runway (in this case 27) has a displaced threshold. In such cases the shorter of the two values of TORA must be used in calculating slope for both runways. lnserting the data provided in this question gives: 00% Slope = (275 ft - 310 ft) I (900 m x 3.28 ftlm) XI Which resolves to give slope = -1 .18% This is a slope of approximately 1.2% downwards.

FIELD 17. a. This problem can be resolved using the standard equation: Slope = (difference in threshold elevations I T O W ) x 100% But both difference in elevation and TORA must be expressed in the same units so if elevation is in ft, then a TORA in meters must be multiplied by 3.28 ftlm. lnserting the data provided in this question gives: Slope = (300 ft - 320 ft) 1 (4000 m x 3.28 ftlm) ~ 1 0 0 % . Which resolves to give Slope = -0.15% That is a slope of 0.15% downwards.

FIELD-18. a. JAR-OPS 1.480 (a) (2) defines a contaminated runway as one on which more than 25% of the runaway surface is covered with more than 3 mm depth of water, or by slush or snow to an equivalent depth of more than 3 mm.

FIELD 19. a. The term water equivalent depth is a means of comparing snow or slush with water. The water equivalent depth is equal to the actual depth of the snowlslush multiplied by its specific gravity. For example if the contaminant has a depth of 6 mm and a specific gravity of 0.5, then its water equivalent depth is 3 mm.

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FlELD 20. a. The definition of a contaminated runway given in JAR-OPS 1.280, lists three conditions, each of which qualify as contaminated. One of these conditions is snow which has been compressed into a solid mass which resists further compression and will hold together or break into lumps if picked up. This is compacted snow. No minimum thickness is specified for this form of contamination.

FlELD 21. d. JAR-OPS 1.480 (a) (3) defines a damp runway as one on which the surface is not dry, but when the moisture on it does not give a shiny appearance.

FlELD 22. b. JAR-OPS 1.475 (d) states that for performance purposes, a damp runway, other than damp grass, may be considered to be dry. A damp runway does not therefore affect performance except when it is a damp grass runway.

FlELD 23. a. JAR-OPS 1.480 (a) (4) states that a dry runway is one which is neither wet nor contaminated, and includes paved surfaces which have been prepared with grooves or a porous surface such that effectively dry braking effectiveness is retained when moisture is present.

FlELD 24. c. JAR-OPS 1.480 (10) defines a wet runway as one on which the surface is covered with water or equivalent, less than 3 mm deep, or when there is sufficient moisture to make it appear reflective.

FlELD 25. c. JAR-OPS 1.480 defines a flooded runway as one on which more than 50% of the surface is covered with more than 3 mm of water.

FlELD 26.a. UK AIP AGA 5, defines the reporting action required for runway contamination. This includes a requirement that snow or slush be measured at intervals of approximately 300 m along the runway, between 5 and 10 meters from the runway centre line.

FlELD 27. b. JAR-OPS 1.480 (9) states that the maximum allowable slope for European runways is + or - 2%. This is reduced to 1% in the UK.

FlELD 28. c. JAR-OPS 1.490 (c) (b) requires that published distances available be reduced by the amount taken by aircraft to line up with the runway for take-off. This is termed the runway alignment reduction.

FlELD 29. b. CAP 168, chapter 3, paragraph 9, defines the clearway as an area free of obstructions above an upward slope of 1.25% from the end of the TORA.

FlELD 30. d. JAR-OPS 1.480 (a) (1) defines the ASDA as the length of the TORA plus the associated stopway. This is also sometimes referred to as the EMDA or EDA.

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227

FlELD 31. a. JAR-OPS 1.480 (a) (7),defines the TODA as the distance from the point on the surface at which an aircraft can commence its take-off run, to the nearest obstacle in the direction of take-off. The TORA is from the start of the take-off run to the end of the runway and the clearway is from this point to the first obstacle. The TODA is therefore the TORA plus the clearway. This definition is also included in ICAO Annex 14.

FlELD 32. d. JAR-OPS 1.490 (c) (b) requires that published distances available be reduced by the amount taken by aircraft to line up with the runway. This is termed the runway alignment reduction. The TODA and TORA, are related to the distance available up to the point of main wheel lift off so the alignment reduction is taken from this point on the aircraft. But the ASDA is concerned with the distance available to stop the aircraft without its nose wheel running off the stopway. Runway alignment reduction is therefore greater for the ASDA than for the TORA or TODA.

FlELD 33. a. JAR-OPS 1.490 (c) (b) requires that published distances available be reduced by the amount taken by aircraft to line up with the runway. This is termed the runway alignment reduction. The TODA and TORA, are related to the distance available up to the point of main wheel lift off so the alignment reduction is taken from this point on the aircraft. But the ASDA is concerned with the distance available to stop the aircraft without its nose wheel running off the stopway. The ASDA is therefore taken to commence at the nose wheel.

FlELD 34. b. JAR-OPS 1.490 (c) (b) requires that published distances available be reduced by the amount taken by aircraft to line up with the runway. This is termed the rur;way alignment reduction. The TODA and TORA, are related to the distance available up to the point of main wheel lift off so the alignment reduction is taken from this point on the aircraft. But the ASDA is concerned with the distance available to stop the aircraft without its nose wheel running off the stopway. The TODA is therefore taken to commence at the main wheels.

FlELD 35. b. JAR-OPS 1.490 (c) (b) requires that published distances available be reduced by the amount taken by aircraft to line up with the runway. This is termed the runway alignment reduction. The TODA and TORA, are related to the distance available up to the point of main wheel lift off so the alignment reduction is taken from this point on the aircraft. But the ASDA is concerned with the distance available to stop the aircraft without its nose wheel running off the stopway. The TORA is therefore taken to commence at the main wheels.

FlELD 36. d. ICAO Annex 14 defines the TODA as the length of the TORA plus the associated clearway. ICAO Annex 6 page 44, defines the TORA as the length of the declared runway. Combining these two definitions shows that the TODA is equal to the length of the runway plus its associated clearway.

FlELD 37. a. The field length limited TOM is the maximum mass at which an aircraft can safely take-off using a given runway length. Any factor which decreases engine power will increase the length of the take-off run

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Airfields

required at any given mass. Decreasing engine power therefore reduces field length limited TOM. The power output of any engine is proportional to the mass of air passing through it. This mass of air is in turn proportionalto local air density. lncreasing ambient ternperature decreases air density and hence decreases engine power and field length limited TOM.

FlELD 38. a. Following initial construction, the load bearing strength of concrete increases over a period of months, before reaching its maximum value. The ability of a runway to sustain the operation of high mass aircraft, therefore increases during the first year following its construction. The compatibility between aircraft and runways is controlled through international agreements using a system of classification numbers. The Aircraft Classification Number (ACN) is a measure of the load imposed by a given aircraft type. The Pavement Classification Number (PCN) is a measure of the load bearing capability of a surface. In ordgr to maintain the integrity of newly constructed concrete runways, the ACN must not be greater than the PCN during the first 12 months following runway construction.

FlELD 39. a. The ASDA is the distance available in which an aircraft can accelerate to V, and decelerate to a stop in the event of an aborted take-off. The length of the ASDA is equal to the length of the runway plus the associated stopway. lncreasing take-off mass (TOM) reduces acceleration rates and hence increases the distance required to accelerate and stop. An increase in stopway length will therefore increase the maximum allowable TOM. V, is the speed from which it is possible to either continue the take-off safely or abort, without runniug off the end of the stopway. The longer the ASDA, the greater the maximum value of V,. lncreasing the stopway length will therefore increase the maximum TOM and also increase the maximum V,.

FlELD 40. b. ICAO Annex 14 defines the TODA as the length of the TORA plus the associated clearway. ICAO Annex 6 page 44, defines the TORA as the length of the declared runway. Combining these two definitions shows that the TODA is equal to the declared length of the runway plus its associated clearway.

FlELD 41. d. ICAO Annex 14 defines the TODA as the length of the TORA plus the associated clearway. ICAO Annex 6 page 44, defines the TORA as the length of the declared runway. Combining these two definitions shows that the TODA is usually equal to the declared length of the runway plus its associated clearway. In some case however the geometry of the intersection between the taxiway and runway is such that some of the runway must be used to align the aircraft for take-off. This reduces the TODA. The magnitude of this reduction depends upon the geometry and size of individual aircraft types. The TODA is therefore sometimes dependent upon aircraft type. The reduction is not however directly related to aircraft mass.

FlELD 42. d. The TODA is equal to the runway length plus the clearway length. The maximum allowable length of the clearway for a class A aircraft is 50% of the runway length. The maximum TODA length is therefore 150% of the runway length. In this case a runway length of 4000 m gives a maximum TODA of 6000 m.

FlELD 43. c. The TODA is equal to the runway length plus the clearway length, where the clearway is from the end of the runway to the first obstacle. The maximum allowable length of the clearway for a class A aircraft is 50% of the runway length. The maximum TODA length is therefore 150% of the runway length. In this

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case a runway length of 4000 m plus a 1000 m clearway gives a 5000 m TODA, which is less than 150°h of the TORA.

FlELD 44. a. The maximum permissible length of the clearway is 50% of the TORA so option c is incorrect. The TODA is the TORA plus the clearway, so the maximum clearway (50% of TORA) cannot be more than 113 of the TODA. Options b and d are therefore incorrect. The clearway is an area beyond the end of the TORA, clear of obstacles that might be struck by an aircraft after take-off. Although its maximum length is 50% of the TORA, its length will be reduced if a closer obstacle exists. Option a is therefore the most appropriate.

FlELD 45. c. The clearway is an area beyond the end of the TORA, clear of obstacles that might be struck by an aircraft after take-off. It starts at the end of the runway, which is also the end of the TORA. This point is also the beginning of the stopway, so option c is the most appropriate.

FlELD 46. c. A field is unbalanced on three occasions. These are when the stopway and clearway lengths are not equal, or when there is a stopway but no clearway or when there is a clearway but no stopway. Options b and d are incorrect because they describe a balanced field condition. Option a is incorrect because the use of the word "only" excludes two of the other possible unbalanced conditions.

FlELD 47. a. The TODA is equal to the runway length or TORA, plus the clearway. The maximum permissible length of the clearway is 50% of that of the TORA. The TODA is therefore never more than 50% greater than the TORA.

FlELD 48. c. The accelerate step distance or ASDA is equal to the length of the TORA plus the stopway. The purpose of adding a stopway is therefore to increase the ASDA. The stopway plays no part in determining the lengths of the TORA or TODA so options a and b are incorrect. Option d describes the purpose of the clearway. Option c is therefore the most appropriate.

FlELD 49. a. Area A to B is the runway. Area B to C is the stopway. Area B to D is the clearway. The TORA is the distance between the commencement of the take-off run to the point at which VLoFmust be achieved. The stopway is used only in the case of aborted take-off so VLoFmust be achieved by point 6. The distance A to B is therefore the TORA. The accelerate stop distance or ASDA is the TORA plus the stopway. This is between points A and C. The take-off distance or TODA is the TORA plus the clearway. This is between points A and D. The area between points A and B is therefore the TORA (option a).

FlELD 50. c. Area A to B is the runway. Area B to C is the stopway. Area B to D is the clearway. The TORA is the distance between the commencement of the take-off run to the point at which VLoFmust be achieved. The stopway is used only in the case of aborted take-off so VLo, must be achieved by point B. The distance A to B is therefore the TORA. The accelerate stop distance or ASDA is the TORA plus the stopway. This is between points A and C. The take-off distance available or TODA is the TORA plus the clearway. This is between points A and D. The area between points A and C is therefore the ASDA (option c).

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Airfields

FIELD 51. b. Area A to B is the runway. Area B to C is the stopway. Area B to D is the clearway. The TORA is the distance between the commencement of the take-off run to the point at which VLo, must be achieved. The stopway is used only in the case of aborted take-off so VLo,must be achieved by point B. The distance A to B is therefore the TORA. The accelerate stop distance or ASDA is the T O W plus the stopway. This is between points A and C. The take-off distance or TODA is the TORA plus the clearway. This is between points A and D. The area between points A and D is therefore the TODA (option b).

FIELD 52. a. The stopway is an area of prepared surface beyond the end of the runway that is capable of supporting the weight of an aircraft. It provides an additional length of stopping distance to be used in the event of an aborted take-off. The stopway is area B to C in the diagram.

FIELD 53. d. The clearway is an area free from obstacles, beyond the end of the T O W . Its pLlrpose is to ensure that aircraft do not strike obstacles during the initial climb away from the ground after lift-off. It is marked as area B to D in the diagram.

FIELD 54. a. The clearway is an area free from obstacles, beyond the end of the TORA. Its purpose is to ensure that aircraft do not strike obstacles during the initial climb away from the ground after lift-off. It is marked as area B to D in the diagram. The maximum permitted length of a cleanvay is 50% of the TORA. In this case the TORA is the area from A to B, so the maximum length to area B to D is 50% of area A to B.

FIELD 55. a. The gradient or slope of a runway is the difference in elevation between its end, divided by its length. It is normally expressed as a percentage where: Slope = (Elevation difference 1 TORA) x 100%. The diagram indicates north as being at the top of the diagram so runway 27 runs from right to left. This ends of this runway are at points A and B, so the elevation difference between its ends is 50 ft down or 50 ft. The TORA of the runway is length A to B which is 2000 m. This can be converted into ft by multiplying by 3.28 ft/m. This gives a length of 6560 ft. The slope is therefore -50 ft 1 5650 ft x 100% Which is 0.76% down slope. ,

FIELD 56. b. The situation referred to in this question is illustrated below. ....

BRP

.......................-....

............... Distance required

1500 ft

3000 rn

I

I

.....

2000 ft

......

".

Reference zero ms~ r

v

The height required is the elevation of the obstacle (2000 ft), minus the elevation of reference zero. It should be noted that elevations are always measured above msl. The elevation of reference zero is the elevation of the BRP + (slope x TODR) So reference zero elevation = 1500 ft + (-0.02 x 3000 m x 3.28 ft/m) Which is 1500 - 196.8 ft = 1303.2 ft. The height of the obstacle above reference zero is the elevation of the obstacle minus that of reference zero. Which is 2000 ft - 1303.2 ft = 696.8 ft.

FIELD 57. a. The situation referred to in this question is illustrated below. -...r..

Distance required

"".

It "

""

The height required is the elevation of the obstacle (2000 ft), minus the elevation of reference zero. It should be noted that elevations are always measured above msl. The elevation of reference zero is the elevation of the BRP + (slope x TODR) So reference zero elevation = 1500 ft + (0.02 x 3000 rn x 3.28 ft/m) Which is 1500 + 196.8 ft = 1696.8 ft. The height of the obstacle above reference zero is the elevation of the obstacle minus that of reference zero. Which is 2000 ft - 1696.8 ft = 303.2 ft.

FIELD 58. c. From question FIELD 57, the height of the obstacle above reference zero is 303.2 ft. The obstacle is 1000 m from the end of the TODR. The gradient required in this question is illustrated below. .........

Reference zero

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The gradient of a flight path = (height change / horizontal distance) x 100%. So in this case the gradient = ( +303.2 ft / (1000 m x 3.28 Wm)) x 100%. Which is +9.25%.

FlELD 59. b. From question FlELD 56, the height of the obstacle above reference zero is 696.8 ft. The obstacle is 10000 m from the end of the TODR. The gradient required in this question is illustrated below.

Reference zero

rnsl

The gradient of a flight path = (height change / horizontal distance) x 100%. So in this case the gradient = ( +696.8 ft / (10000 m x 3.28 Wm)) x 100%. Which is +2.11%.

FlELD 60. c. The TODR is the required distance between the commencement of the take-off to achieving screen height. The screen is therefore located at the end of the TODR. The length of the TODR varies with aircraft mass and ambient conditions, so the screen is not located at any fixed point on the airfield. Options a, b and d are therefore incorrect. For Class A aircraft the screen height is 35 ft.

FlELD 61. a. CAP 168 Section 3 defines the requirements for a stopway. This includes a requirement that the minimum stopway semi-width is that of its associated runway. This applies to all stopways, regardless of runway width.

FlELD 62. b. CAP 168 section 3 defines the requirements of a stopway. They do not include any specific maximum length, so option a is incorrect. The TORA is the distance from the point on the aerodrome at which an aircraft can commence its take-off run, to the end of the runway. In effect this usually means the entire runway, although it might be reduced by any runway alignment area. Option c is therefore incorrect. The TODA is the TORA plus the clearway. The TODA is therefore almost always greater than the runway length, so option d is incorrect. CAP 168 Section 3 and ICAO Annex 14 define the requirements of a clearway. These include a maximum permissible length of 50% of the associated runway. Option b, the clearway, is therefore correct.

FlELD 63. b. CAP 168 and ICAO Annex 14 define the clearway as an area at ihe end of the TORA which is free of objects which may cause a hazard to aircraft in flight. The clearway therefore starts at the end of the TORA, so option b is correct. Although the TORA usually ends at the end of the runway (option c) this is not always the

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case. If for example some form of maintenance work were being carried out at the end of a runway, the declared T O W would need to be reduced. In this case the clearway would not start at the end of the runway, but would still start at the end of the TORA. The TODA includes the clearway so option a is incorrect. Both the stopway and the clearway start at the same point at the end of the TORA, option d is incorrect.

FlELD 64. c. CAP 168 and ICAO Annex 14 define the stopway. These documents include the fact that its length is limited by a reduction in load bearing strength, a ditch or depression, deterioration in friction characteristics, and any obstacle likely to cause structural damage to an aircraft if the take-off is abandoned. These documents do not include any specific maximum measurement. The definition for a clearway does however include option c, 50% of the TORA. Option c is therefore correct.

FlELD 65. c. The Aircraft Classification Number (ACN) of an aircraft represents the load it will apply to a runway surface. The Pavement Classification Number (PCN) represents the load bearing capability of a runway surface. ICAO Annex 14 (Amendment 35), defines the system of load classification numbers, and the conditions under which ACN may exceed PCN. This document includes a requirement that operations of aircraft with an ACN value more than 50% greater than the PCN of the runway, may be conducted for emergency only. An ACN 50°/0 more than the PCN, means an ACN that is more than 150% of the PCN. Option c is therefore correct.

FlELD 66. b. The T O W is the take-off run available. If there is no runway alignment reduction, the TORA is equal to the runway length. If CD in the diagram is the stopway, then BC is probably the runway and CE is the clearway. If this is the case, then the TORA is distance BC.

FlELD 67. d. The TODA is the distance from the point at which the take-off run can be commenced, to the first obstacle. Unless there is a runway alignment reduction, the TODA is the runway plus the stopway. If BC in the diagram is the runway, then CD is probably the stopway and CE the clearway. If this is the case, the TODA is the distance BE.

FlELD 68. c. The ASDA is the accelerate stop distance. This is the distance available over which an aircraft can accelerate to V, then stop safely after an aborted take-off. The ASDA is the length of aerodrome surface that is capable of supporting the weight of an aircraft without damaging the aircraft. The ASDA is the runway plus any stopway that is available. If CE is the clearway then CD is probably the stopway and BC the runway. If this is the case the ASDA is distance BD.

FlELD 69. c. The EMDA is the emergency stopping distance. It is an alternative name sometimes used for the ASDA. The ASDA is the accelerate stop distance. This is the distance available over which an aircraft can accelerate to V, then stop safely after an aborted take-off. The ASDA is the length of aerodrome surface that is capable of supporting the weight of an aircraft without damaging the aircraft. The ASDA is the runway plus any stopway that is available. If CE is the clearway then CD is probably the stopway and BC the runway. If this is the case then the ASDA is distance BD.

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FlELD 70. c. The runway and stopway must both be capable of supporting the weight of any aircraft using them, without damage to the aircraft. The clearway need not be capable of supporting the weight if an aircraft because lift-off must be achieved before the end of the stopway. If BC in the diagram is the runway then CD is the stopway and CE is the clearway. If this is the case then the area that must be capable of supporting the weight of the aircraft is BD.

FlELD 71. d. Although both the runway and the stopway must be capable of supporting the weight of the aircraft without damage, the stopway never sustains the impact loads caused by the landing. Also, the stopway is called upon to carry the weight of the aircraft only in the event of an aborted take-off, whereas the runway must do so for every take-off and landing. The stopway can therefore be of a lower load bearing capability than the runway. The minimum acceptable level is 30% of that of the runway.

FlELD 72. a. A mu-meter is one of the devices employed to test the coefficient of friction of runways. The lower the coefficient of friction, the less effective will be the braking action of aircraft using the runway. When the mu-meter reading is more than 0.4 it is only necessary that the pilots be given a single reading for the whole runway. When however the reading falls below 0.4, separate readings must be reported for each 11 3 of the runway length.

FlELD 73. a. Dry snow has a specific gravity of less than 0.35 so 10 mm of dry snow has a water equivalent depth of only 3.5 mm. The minimum depth of water for which corrections to take-off calculations are necessary is 3 mm. For this reason, dry snow of less than 10 mm depth need not be taken into account.

FlELD 74. c. CAA AIC 6111999 (Pink 195) paragraph 2.1 . I . specifies that the maximum allowable depth of wet snow for take-off and landing operations is 15 mm.

FlELD 75. c. Reference Zero (RZ) is an imaginary horizontal plain passing through a point directly below the aircraft at the end of the TODR. It is sometimes used as a point of reference for various points on the take-off flight path.

FlELD 76. a. The relationship between T O M , TODA, ASDA, the stopway and the clearway, is illustrated below.

..

: .

.

.. .

9 4

I

I

:

RUNWAY

*STOPWAY

ASDA

Y

TODA

From this it can be seen that the TORA is the TODA minus the clearway.

H

I

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235

FlELD 77. c. The surface must be capable of supporting the weight of an aircraft at all points between the BRP and the latest acceptable stopping point in the event of a rejected take-off. This distance is the ASDA and includes both the runway and the stopway. The clearway need not be capable of supporting the weight of an aircraft. But both the stopway and the clearway commence at the end of the TORA so the first part of the clearway is also the stopway. This means that the area which need not be capable of supporting the weight of an aircraft is the part of the clearway beyond the end of the stopway. This is the clearway minus the stopway.

FlELD 78. d. An unbalanced field is one in which the TODA is not equal to the ASDA. Of the options offered in this question only option d, satisfies this condition.

FlELD 79. a. Vp is the aquaplaning speed. It is proportional to tyre pressure and contaminant specific gravity and is lower for landirrg than for take-off, such that for landing: Vp (in Kts) = 7.7dp 1SG of contaminant, where p is the tyre pressure in psi

The SG of water is 1 so for a wet runway Vp = 7.74150 / 1 Which gives a value of Vp = 94.3 kts.

FlELD 80. b.

.

.

Vp is the aquaplaning speed. It is proportional to tyre pressure and contaminant specific gravity and is lower for landing than for take-off, such that for landing: Vp (in Kts) = 7.7dp ISG of contaminant, where p is the tyre pressure in psi

So in this case Vp = 7.74150 10.5 Which gives a value of Vp = 188.6 kts.

FlELD 81. c. Vp is the aquaplaning speed. It is proportional to tyre pressure and contaminant specific gravity and is lower for landing than for take-off, such that for landing: Vp (in Kts) = 94p ISG of contaminant, where p is the tyre pressure in psi

The SG of water is 1 so for a wet runway Vp = 94150 I1 Which gives a value of Vp = 110.2 kts.

FlELD 82. b. Vp is the aquaplaning speed. It is proportional to tyre pressure and contaminant specific gravity and is lower for landing than for take-off, such that for landing: Vp (in Kts) = 9dp ISG of contaminant, where p is the tyre pressure in psi

So in this case Vp = 94150 / 0.5 Which gives a value of Vp = 220.5 kts.

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Airfields

FlELD 83. c. Vp is the aquaplaning speed. It is proportional to tyre pressure and contaminant specific gravity. At touchdown the wheels must spin up very rapidly to match ground speed so high sliding forces are applied between the runway and the tyre. In the case of a rejected take-off, the tyres are already running a speed equal to ground speed when the brakes are applied. The sliding forces are therefore initially much lower in take-off than in landing. For this reason the aquaplaning speed is lower in landing than in take-off.

FlELD 84. a. Measured performance is the average performance achieved by an aircraft or group of aircraft of a given type, when tested using an acceptable testing method in specified conditions. Of the options offered in this question, option a, is closest to compliance with this definition.

FlELD 85. b. JAR OPS 1.480 defines a flooded runway as one on which more than 50 % of the surface is covered with water more than 3 mm deep.

FlELD 86. a. The stopway is an area of prepared surface, capable of supporting the weight of an aircraft, beyond the end of the runway. Its purpose is to provide an additional space in which an aircraft can be brought to a halt following a rejected take-off. The stopway therefore increases the ASDA. This increasedASDA means that a greater distance is available to accelerate to V, then decelerate to a halt. This in turn enables an aircraft at any given mass to operate at a greater V,. The overall purpose of a stopway is therefore to increase both the maximum allowable V, and the ASDA.

FlELD 87. b. The clearway is an obstacle free area beyond the end of the runway. Its purpose is to enable aircraft to carry out the initial climb to screen height without being endangered by obstacles. The clearway is never required to support the weight of an aircraft and hence may be over land or water. The take-off distance available (TODA) is equal to the runway plus the clearway, so the provision of a clearway increases the TODA. The take-off distance required (TODR) is from the commencement of the take-off run to the point at which screen height is attained. V, is the take-off decision speed and is the maximum speed at which, in the event of an engine failure, it is possible for the take-off to be completed or rejected safely. If V, is reduced the distance required to attain screen height will increase. So by increasing the distance available from the start of the take-off run to the nearest obstacle, a clearway enables a lower value of V, to be employed. The purpose of a clearway is therefore to increase TODA and hence decrease the minimum value of V,.

LIFT AND DRAG

LD 1. a. lnduced drag = CDl% pV2S where: CDl is the coefficient of induced drag. p is the local air density. V is the TAS of the relative airflow. S is the wing area. So Dl is proportional to all of the variables in this equation. IAS is derived from X pV2 so as factors such as ambient temperature or altitude alter local air density these changes will affect induced drag to the same degree. lnduced drag is therefore more closely proportional to IAS than to TAS. High airspeeds tend to increase air density above the local value, increasing both the induced drag and IAS, so the more accurate EAS is employed to take account of these effects. lncreasing speeds also enable the required lift to be generated at lower angles of attack. This reduces wingtip vortices and hence induced drag as the square of speed increases. lnduced drag is therefore inversely proportional to EAS2.

LD 2. b. lnduced drag is produced as a direct result of the generation of lift. lncreasing the lift force using any given combination of wing, airspeed, and local atmosphere, requires an increase in angle of attack. This increases the pressure differences between the lower and upper surfaces of the wing and hence increases wing tip vortices which are the causeof induced drag. lnduced drag is therefore directly proportionalto lift. lncreasing airspeed enables the required lift to be generated at lower angles of attack, reducing both wingtip vortices and induced drag. lnduced drag is therefore inversely proportional to airspeed. Combining the two proportionalities gives the result that induced drag is proportional to LN. Although options a, b and c are all correct, option b is the most accurate.

LD 3. b. L:D ratio is the ratio of lift to drag. In unaccelerated straight and level flight, lift is equal to weight and drag is equal to thrust. In this case L:D ratio = 50000:25000 Which is 2:l

LD 4. b. Load factor is equal to lift divided by weight, so an increase in load factor at any given weight results in an increase in lift. lncreasing lift at any given TAS increases induced drag and hence thrust required. Power required is equal to thrust multiplied by TAS so an increase in load factor at any given TAS, will result in a proportional increase in power required.

LD 5. a. An increase in weight at any given load factor and airspeed requires an equal increase in lift. This is achieved by increasing angle of attack, which results in stronger wingtip vortices. These vortices are the principal cause of induced drag (Dl), but have no significant effect on profile drag (D,) within the

238

Lift and Drag

normal angle of attack range. Total drag DTotalis the sum of D, and D, so increasing weight also increase D~otal'

LD 6. b. lncreasing weight requires a corresponding increase in lift. This results in an increase in induced drag at any given airspeed. Induced drag is proportional to 1/(EAS2) the magnitude of the increase diminishes with increasing airspeed. This means that the increase in induced drag is most prominent at the low end of the speed range. lncreasing weight has no significant effect on profile drag. ,V , is the speed at which Dl and D, are equal so the increase in Dl in the low speed range causes,V , to occur at a higher speed. Aircraft are speed stable at all speeds above ,V , so the increase in,V , reduces the speed range over which the aircraft is speed stable. An increase in weight therefore increases,V , and decreases speed stability.

LD 7. c. Profile drag (D,) is proportional to V and induced drag (Dl) is proportional to IN2. ,V , is the speed at which Dp and Dl are equal. Lowering of landing gear increases D, but has no significant effect on Dl. Because Dp is proportional to V2 the increase is greatest at the high end of the speed range. The increase , to occur at a lower airspeed. Aircraft are in D, therefore causes the crossover of the two curves at ,V speed stable at all speeds above VMDso any reduction in,V , increases the speed range over which the , and increases speed aircraft is speed stable. The lowering of landing gear therefore decreases ,V stability.

LD 8. c. Dl is proportional to 1/(EAS2) and D, is proportional to EAS2. EAS is IAS corrected for the effects of compressibility. These effects are small at low speeds so because thisquestion stipulates a constant low IAS it can be assumed that EAS also remains constant. Both Dl and D, will therefore remain unchanged. Had the question not stipulated a low IAS then the effects of compressibility WOULDneed tcrbe taken into account. Climbing at constant IAS results in an increase in TAS so climbing through a large altitude range at a constant high IAS would result in the aircraft attaining speeds at which compressibility effects became significant causing EAS to increase. Under these circumstance Dl would reduce and Dp would increase with increasing altitude.

LD 9. b. Although deployment of trailing edge flaps increases both Dl and Dp, the effect on each is not constant at all angles. Small angles of deflection produce a greater increase in Dl than in D, and a small increase in DTota,.,V , is the speed at. which Dl is equai to Dp so increasing D, by a greater proportion than D, will Aircraft are speed stable at all speeds above,V , so increasing,V , will decrease cause an increase in.,V, the proportion of the speed range over which an aircraft is speed stable. Deployment of trailing edge ,flaps to the 10 degree setting would therefore increase both Dl and Dp and decrease speed stability.

LD 10. a. Although deployment of trailing edge flaps increases both Dl and Dp, the effect on each is not constant at all angles. Large angles of deflection produce a smaller increase in Dl than in D, and a large increase in DTOtaI. ,V , is the speed at which Dl is equal to D, so increasing Dl by a smaller proportion than D, will Aircraft are speed stable at all speeds above,,V , so decreasing ,V , will cause a decrease in .,V, increase the proportion of the speed range over which an aircraft is speed stable. Deployment of trailing edge flaps to the 40 degree setting would therefore increase Dl, D, and speed stability.

Lift and Drag

239

LD 11. b. Drag = C, %pV2SWhere

C, is the coefficient of total drag p is the local air density V is the TAS S is the wings area Water vapour is less dense than dry air so increasing humidity reduces the local air density. If all other factors in the drag equation remain unchanged then total drag will decrease.

LD 12. b. Best L:D ratio is achieved when flying at VMD.At this speed D, is equal to Dl so the ratio of Dp:Dl is 1:l.

LD 13. b. VMDis the speed at which the total drag is at a minimum value. At any given weight and load factor the lift required remains constant regardless of speed so the greatest L:D ratio occurs when drag is minimum. VMDis therefore the speed at which drag is at a minimum and the L:D ratio is greatest. It is also the speed at which jet aircraft achieve best endurance, propeller aircraft maximum range and all aircraft maximum glide range.

LD 14. b. Drag is the force which opposes the motion of an aircraft through the air so except when the aircraft is climbing or descending vertically drag cannot be parallel to lift or weight. Although it is commonly assumed that drag acts.paralle1 to the relative airflow it is more useful to consider it to act parallel but opposite to the direction of flight. Strictly speaking drag only acts parallel to the relative airflow when the aircraft is flying at its zero lift angle of attack. Under these circumstances there is no downwash so the relative airflow is also parallel but opposite to the direction of flight.

LD 15. a. At zero lift angle of attack the wing generates no lift and hence no induced drag. Under these circumstances the only force generated is profile drag so the total reaction is equal to Dp.

LD 16. a. At zero lift angle of attack the wing produces no lift and hence no induced drag or downwash. The total reaction is therefore the profile drag acting parallel but opposite to the direction of flight.

LD 17. a. Dl is proportional to IN2and hence decreases with increasing airspeed. Dp is proportional to V2 and so increases with increased airspeed. VMDis the speed at which D, and Dp are equal so at all lower speeds Dl is greater than D., ,,,D , is the sum of Dl and D, at all speeds so D, can never equal,,,D , and D, can do so only at the zero lift angle of attack where Dl is zero.

LD 18. b. Dl is proportional to IN2so it increases as speed decreases. Dp is proportional to V2 SO it decrease as speed decreases. CLMaxoccurs at the stall and VMo is the maximum speed permitted under any circumstances. Changing speed from VMoto CLMax therefore represents a reduction in speed from maximum to minimum values. This reduction in speed will cause D, to increase and Dp to decrease.

LD 19. c. An aircraft may be subject to five types of drag. These are friction, form, interference, induced and shock.

240

Lift and Drag

lnduced drag is caused by the generation of lift whilst shock drag is the result of flight at transonic or supersonic speeds. Profile drag is caused by the moticn of the aircraft through the air and is not connected with the generation of lift or any specific speed range. At low speeds profile drag is the sum of friction, form and interference drags. At higher speeds shock drag makes a further contribution to it.

LD 20. d. Options a, b, and c are inadequate for the following reasons. A cambered aerofoil and an aerofoil at a positive angle will generate lift only when subject to a moving airflow. A high speed airflow will generate lift only when passing over an aerofoil possessing the correct combination of camber and angle of attack. Option d represents the generation of lift as explained by Newton's second and third laws of motion. That is to say that a force is generated when a mass of matter is accelerated (F=MA) and the lift force generated is the equal and opposite reaction to that which accelerates the mass. In this case the pressure differences above and below the wing exert a force on the airflow causing it to accelerate downwards. Lift is the equal an opposite reaction to this acceleration and is equal to the mass of air multiplied by the downward acceleration given to it. Option d is therefore the most accurate.

LD 21. c. In order to create lift a wing must accelerate airflow downwards by generating pressure differences above and below its surfaces. By leaking from the lower to the upper surfaces the pressure differences produce rotating airflows or vortices at the wing tips. The mass of air flowing upward in the part of the vortices outboard of the tips is precisely equal to the mass of air flowing downwards inboard of the tips. The vector sum of these flows is therefore zero so the vortices do not generate any additional lift. The energy used to drive the vortex flow is extracted from the aircraft and this represents additional drag. This additional drag caused by the rotational airflow in the wingtip vortices is lift induced drag or D,.

LD 22. c. In order to create lift a wing must accelerate airflow downwards by generating pressure differences above and below its surfaces. By leaking from the lower to the upper surfaces the pressure differences produce rotating airflows or vortices at the wing tips. The mass of air flowing upward in the parts of the vortices outboard of the tips is precisely equal to the mass of air flowing downwards in the parts inboard of the tips. The vector sum of these flows is therefore zero so the vortices do not generate any additional lift. Inboard of the tips the downward flow of the air in the vortices reduces the effective angle of attack of the wing and so reduces the C, and hence lift. In order to maintain the required lift force the wing must be pitched up to a higher angle in order to restcre the original angle of attack. Although this process restores the lift force, it also tilts the total reaction further aft such that it includes a greater drag component. This additional drag due to downwash of airflow over the trailing edge caused by wingtip vortices is lift induced drag or D,.

LD 23. a. lnduced drag is caused by the creation of lift. Lift is created by the downward acceleration of air due to the pressure differences above and beiow the wing. These pressure differences leak around the wingtips as vortices causing upwash outboard of the wingtips and dowfiwash at the training edges. The upwash is equal to the downwash so no additional lift is produced by the vortices but the downwash over the trailing edge reduces angle of attack and hence lift. To restore lift it is necessary to increase pitching angle and this tilts the total reaction aft increasing the drag component. This additional component is induced drag. To create induced drag a wing must produce lift and a cambered wing will produce lift at zero angle of attack. Also a symmetrical (zero camber) wing will also produce lift and induced-drag when at a non-zero angle of attack. Pitching angle alone does not create camber or angle of attack and hence cannot create lift or induced drag. Of the options provided only option a includes all of the essential elements of induced drag.

Lift and Drag

241

LD 24. a. ,V , is the speed at which total drag is at its minimum value for a given weight and configuration. Its location on the Drag:EAS curve is indicated by the EAS at which the induces drag and profile drag components are equal, and total drag is the sum of the two. lnduced drag is therefore 50% of total drag when flying at.,V,

LD 25. c. Power required = Drag x TAS and Drag = CD%pV2Swhere V = TAS So power required = C,'/zpVS

xV

This means that power required is proportional to V2 x V which is V3 So power required is proportional to V3 In accelerating from 300 Kts to 400 Kts TAS is increased to 133% of its previous value. So power required is increased to (1.33)3times its previous value. So power required is 2.35 time sit previous value which is an increase of 135%

LD 26. b. lncreasing weight requires an increase in lift to maintain any given flight profile. lncreasing lift creates additional induced drag but has little effect on profile drag. Total drag is the sum of induced and profile so as weight increases total drag increases. Minimum drag is the total drag when flying at VMDso increasing weight also increases rr~inimumdrag.

LD 27. b. Drag is equal to C,X pV2S where

C, is the coefficient of drag p is the local air density V is the true speed of the free stream airflow S is the surface area

C, , V and S are not affected by any change in air density. Drag is therefore directly proportional to air density so if density is reduced by a factor of 4, drag will decrease by the same factor.

LD 28. d. Drag is equal to C,X pV2Sand indicated airspeed is proportional to l/z pV2If air density is reduced by half with no change in C, V2 or S, then both drag and indicated airspeed will also reduce by half. If however indicated airspeed is maintained constant while density reduces by half then V2must be doubled such that l/z pV2remains constant. Doubling of V will affect drag in the same manner so if density is halved while indicated airspeed remains constant then V2 will double and drag will also remain constant.

LD 29. c. lnduced drag D,, is proportional to IN2and the coefficient of induced drag CDl, is proportional to 1N4. In this case V is multiplied by 2, so Dl is multiplied '/z2which is %, and C, is multiplied by X4which is 1/16.

LD 30. c. Weight is the force generated by the earth's gravity acting upon the mass of an aircraft. It always acts directly downwards toward the centre of the earth. Although lift acts in the opposite direction to weight in

242

Lift and Drag

straight and level flight, this is not the case when an aircraft climbs, descends, or turns. Option b is therefore correct only in straight and level flight. Similarly lift always acts at right angles to the flight path but weight does so only in straight and level flight. Rate of turn has no effect on the direction of the weight force.

LD 31. b. It is convenient to consider two possible definitions of the direction of the lift force dependirlg on what aspect of flight is being considered. In straight and level Right lift opposes weight and because weight acts vertically downwards lift is said to act vertically upwards. Although this condition is true only in straight and level flight it satisfies a more generally applicable definition that lift acts at right angles to the flight path. This definition is used when considering manoeuvres such as climbing, descending, gliding, banking and turning. It is not however generally applied when considering the aerodynamic processes affecting the generation of lift. Under these circumstances it is commonly accepted that lift acts at right angles to the relative airflow. Because the generation of lift creates downwash the relative airflow direction is not exactly the same as the reciprocal of the flight path. In effect this means that the lift force is angled backwards slightly. The direction of the thrust line depends only on aircraft structural considerations and has no direct relationship to the direction of the lift force. Of the options offered in this question, a, b and c are all correct to some extent. Option b however is the one specified in JAR 25.

LD 32. b. Lift in generated by the downward acceleration of air. 'This downward acceleration is produced by pressure differences created above and below the wings, which are in turn generated by the acceleration of airflow around the curved surfaces of the wings.. At all normal angles of attack the static air pressure both above and below the wing is lower than ambient. The pressure drop above the wing is however the greater of the two. Lift is the vector sum of the upward and downward forces exerted by these pressure differences. It is only at very high angles of attack that static pressure above ambient levels is exerted on the lower surface of the wing. The majority of lift is therefore produced by low pressure above the wings.

LD 33. c. At low positive angles of attack the acceleration of airflow over the lower surface of a wing causes the air velocity to increase. This in turn causes its static pressure and temperature to decrease. The static pressure below a wing at low angles of attack is therefore lower than ambient.

LD 34. d. IMinimum glide gradient gives maximum glide range. Glide range = L:D ratio multiplied by height so for a given height, maximum range and minimum gradient are achieved when flying a maximum L:D ratio. That is to say when C, I C, is at a maximum.

LD 35. b. As altitude increases air pressure decreases causing the air to expand. The overall effect of this process is that in a standard atmosphere air density at 40000 feet is approximately % of its sea level value. IAS is proportional to %pV2 so climbing at constant IAS causes TAS to increase. In a standard atmosphere the rate of change of TAS is such that at 40000 feet IAS is approximately half of TAS. Profile drag is proportional to EAS2 and at low speeds EAS is approximately equal to IAS. Profile drag is therefore approximately proportional to IAS2. In this case IAS is multiplied by % so profile drag will be multiplied by (%)2. Profile drag will therefore be multiplied by %.

Lift and Drag

243

LD 36. d. Induced drag is caused by the generation of lift so raising or lower the undercarriage does not affect it in any way. Undercarriage position does however affect the frontal area and streamlining of the aircraft so raising it will reduce profile drag. Because profile drag is proportional to EAS2, the drag reduction is most significant at high speeds. This means that VMDwhere induced drag and profile drag are equal, occurs at a higher speed. An aircraft is speed stable at all speeds above VMDSO increasingVMDreduces the proportion of the speed range over which it is speed stable. Raising the undercarriage therefore reduces profile drag and makes an aircraft less speed stable.

LD 37. b. In order to generate lift a wing alters static air pressures such that pressure above the wing is lower than that below it. The vector sum of the forces generated by these pressure differences is the total reaction and lift is a part of this total reaction. These pressure differences leak around the wing tips such that air flows from the comparatively high pressure area below the wing to the lower pressure area above. This causes spanwise ,flow from root to tip below the wing and from tip to root above it. Options c and d are incorrect because in that they specify the wrong directions for these spanwise airflows. Although pressure differences do exist in front of and behind the wing, these contribute to drag but do not affect wingtip vortices.

LD 38. a. Aspect ratio is the ratio of wingspan to mean chord length so a high aspect ratio wing has a greater wingspan and shorter chord than one of lower aspect ratio and similar wing area. At any given airspeed a high aspect ratio wing will therefore affect a greater mass of air than its low aspect ratio equivalent. Lift is the (Newton's third law) reaction to the downward acceleration of air and is equal to the mass of air accelerated multiplied by the downward acceleration given to it. To produce any given lift force the larger mass of air affected by a high aspect ratio wing will require a smaller acceleration. The magnitude of the downward acceleration is proportional to the pressure difference multiplied by the time during which it acts on the air. Because of its longer span and shorter chord length a high aspect ratio wing produces the same lift force by giving a larger mass of air a smaller acceleration. Wingtip vortices are caused by the pressure differences leaking around the wingtips. The strength of these vortices is proportional to the pressure difference multiplied by the time during which they act on the air. 'The shorter chord length of a high aspect ratio wing reduces the time during which the pressure differences act thereby reducing vortex strength.

LD 39. d. D, is caused by wing tip vortices due to the leakage of pressure differences from the lower to the upper wing surfaces around the wing tips. Although the magnitude of these pressure differences is constant for a given lift force, increased airspeed reduces the time during which they act upon any given mass of air. This reduces the strength of the wing tip vortices and hence the induced drag. D, therefore reduces with increasing airspeed such that it is proportional to 1/V2. Because profile drag is caused by friction and pressure forces generated by the flow of air over the surface of the aircraft, it increases with the square of airspeed. Dp is therefore proportional to V2.

LD 40. c. Below VMDthe major component of total drag is induced drag which is proportional to 1/V2. Although profile drag increases with V2, this is a very small part of total drag at low speeds. V2 is lower than VMDso total drag is therefore approximately proportional to the inverse of the square of speed when accelerating to V.,

244

Lift and Drag

LD 41. b. Because this question does not state which drag is to be considered, both induced and profile must be considered. Although this is a basic drag question it is complicated by the fact that decreasing IAS by a factor of 5 means multiplying it by 115. D, is proportional to V2 SO decreasing IAS by a factor of five will multiply D, by (1/52)which is 1/25. D, is proportionalto 1n/2 so this will be multiplied by (1/(1/5)2),which is 25.

LD 42. b. Because this question does not state which drag is to be considered, both induced and profile must be considered. Although this is a basic drag question it is complicated by the fact that decreasing IAS by a factor of 5 means multiplying it by 115. D, is proportionalto IN2so this will be multiplied by (1/(1/5)2),which is 25. D, is proportional to V2 SO decreasing IAS by a factor of five will multiply D, by (1h2) which is 1/25.

LD 43. b. Drag = CD%pV2S

where

CDis the coefficient of total drag p is the air density V is the free stream TAS S is the surface area

If all other factors remain unchanged, the magnitude of the total drag force will vary directly with air density.

LD 44. b. Because power required is drag multiplied by TAS, VMPis always lower than VMD.Also because fuel flow in a propeller aircraft is proportional to power, the lowest fuel flow and hence maximum endurance for , Fuel flow in a jet aircraft is proportional to thrust and best range requires the such aircraft occurs at.,V , The VMDfor any given combination best ratio of low fuel flow :TAS. This occurs at a speed higher than.,V of weight, configuration and load factor, is the speed at which the total drag force is at a minimum. This mean that for any given weight VMDproduces the lowest drag force and hence the greatest ratio of lift to drag. That is to say VMDprovides the best L:D ratio.

LD 45. b. Drag = CD%pV2S

Where

C, is the coefficient of total drag p is the air density V is the free stream TAS S is the surface area

Decreasing pressure causes air to expand thereby reducing its density. If all other factors remain unchanged this reduction in air density will cause a reduction in total drag.

LD 46. b. Flap retraction at constant IAS reduces the magnitude of the lift force produced. Induced drag is directly proportional to the lift force so flap retraction also decreases the magnitude of induced drag. Retraction also decreases wing camber and angle of attack, producing a small decrease in profile drag. The combined effect of these processes is a reduction in total drag.

LD 47. d. Lift = C, %pV2 and up to the stalling angle CL increases linearly with angle of attack. In straight and level flight lift equals weight so to remain straight and level as speed increases, CLmust be reducedto compensate

Lift and Drag

245

for increasing V2 in order to maintain a constant lift force. This means that low angles of attack are employed at high speeds. Total drag is made up of two co-mponents, profile drag and lift induced drag. Profile drag is proportional to V2 whilst lift induced drag is proportional to I N 2 ,where V is the equivalent air speed (EAS). This means that as speed increases, profile drag increases and induced drag decreases. At the high speeds associated with low angles of attack, profile drag is large and induced drag is small. Shock induced drag is significant only in the transonic and supersonic speed ranges. The term vortex drag is a non-standard term sometimes used when referring to lift induced drag.

LD 48. a. Lift = CL %pV2and up to the stalling angle CL increases linearly with angle of attack. In straight and level flight lift equals weight, so to remain straight and level as speed increases, C, must be reduced to compensate for increasing V, in order to maintain a constant lift force. This means that low angles of attack are employed at high speeds. Total drag is made up of two components, profile drag and induced drag. Profile drag is proportional to V2, whilst lift induced drag is proportional to I N 2 , where V is the equivalent air speed (EAS). This means that as speed increases profile drag increases and induced drag decreases. At the low speeds associated with high angles of attack profile drag is small and induced drag is large. Shock induced drag is significant only in the transonic and supersonic speed ranges. The term vortex drag is a non-standard term sometimes used when referring to lift induced drag.

LD 49. d. The principal cause of induced drag is the downward deflection of airflow caused by wing tip vortices. Although the downward acceleration of air would normally be expected to produce an upward lift force, the wing tip vortices produce an equal and opposite upward flow of air outboard of the tips. The overall effect of these upward and downward flows of air is not an increase in lift but a downward deflection of the relative airflow passing over the wing. This decreases effective angle of attack causing a reduction in the magnitude of the total reaction, whilst simultaneously tilting it in a more rearward direction. These changes cause a reduction in lift and an increase in drag. Although the lost lift can be recovered by increasing the pitching angle of the aircraft to restore angle of attack, this process tilts the total reaction still further rearwards causing a further increase in drag. The additional drag produced by this sequence of events is called lift induced drag.

LD 50.

b.

lnduced drag is caused by wing tip vortices which are in turn caused by the pressure differences generated above and below a wing in order to create lift. Although increasing angles of attack below the stalling angle increase both lift and induced drag, the relationship reverses beyond the stalling angle with both lift and lift induced drag rapidly decreasing. It cannot therefore be said that induced drag is caused by angle of attack. lnduced drag is proportional to I/(EAS), so option c is also incorrect. Increasing aspect ratio reduces the intensity of wingtip vortices and hence induced drag. lnduced drag is not therefore caused by aspect ratio. For any given corr~binationof aspect ratio and airspeed, both wingtip vortices and lift induced drag increase with increasing lift. lnduced drag is proportional to the square of the lift force as indicated in the following formula. C, = kC,2/7cA

where

A is the aspect ratio.

C, is the coefficient of induced drag. C, is the coefficient of lift. K is a factor accounting for wing plan form.

246

Lift and Drag

LD 51. d. Induced drag is caused by the downward acceleration of air behind the wing. This deflects the relative airflow downwards reducing effective angle of attack and hence lift. In order to restore the original lift value the pitch attitude and hence angle of attack must be increased. Although this restores the lost lift, it also angles the total reaction in a more rearward direction, increasing the magnitude of the drag force. The principal cause of the downwash is wing tip vortices caused by air migrating around the wingtips from the lower to upper surfaces. Upwash ahead of the wing tends to increase effective angle of attack and so to some extent offsets the effects of downwash. Tip tanks produce a small reduction in induced drag by reducing the intensity of the wingtip vortices. High tailplanes reduce fin tip vortices so reducing the induced drag produced by the fin.

LD 52. c. Lowering the undercarriage increases profile drag. Undercarriage position does not alter aircraft weight however and hence has no effect on lift induced drag. Because of its position below the thrust line the extra drag produced by the undercarriage produces a nose down pitching moment.

LD 53. d. Because this question does not refer to any specific type of drag it is necessary to consider the effects on both lift induced and profile drag. Lift induced drag is proportional to so doubling IAS multiples induced drag by 1/(2)2. That is the new induced drag is % of its original magnitude. None of the options satisfies this condition. Profile drag is proportional to V2 SO doubling IAS will multiply profile drag by 22. That is the new profile drag will be 4 times its original magnitude. Option d specifies this result.

lw,

LD 54. b. This question does not specific any drag type so both lift induced and profile must be considered. Profile drag is proportional to V2 SO tripling IAS will multiply profile drag by 32which is 9. This result is not specified in any of the options. Lift induced drag is proportional to 1 w so tripling IAS will multiply induced drag by 11 32. That is the new induced drag will be 119 times its previous value. Option b specifies this result.

LD 55. a. C, = kC,2/nA

C, = coefficient of lift. k is a coefficient relating to plan form. A is aspect ratio. So C, is proportional to C,2 1/A 1/k Although C, is directly proportional to angle of attack up to the stalling angle, this relationship does not apply at greater angles. Of the options provided in this question option a, is the most accurate. where

LD 56. c. The effect of increasing wing camber is to move the lift curve up and to the left without altering its gradient. This causes an increased C, at any given pre-stall angle of attack, a reduced stalling Angle and an increased C, max.

LD 57. d. The effect of increasing wing camber is to move the lift curve up and to the left without altering its gradient. This causes an increased C,- at any given pre-stall angle of attack, a reduced stalling angle and an increased C, max. The increased C, max, reduces the stalling speed for any given weight.

Lift and Drag

247

LD 58. c. lncreasing air temperature reduces air density. Lift = C,%pVS where p is air density. Lift is therefore proportional to air density. The reduction in air density caused by increasing temperature will therefore reduce the lift force produced at any given combination of C, V and S.

LD 59. b. Because water vapour is lighter than dry air, increasing humidity will reduce ambient air density. Lift = CL%pV2S, so reducing air density will reduce lift for any combination of C, V and S. In order to maintain constant lift force in any given combination of V and S, an increase in humidity will therefore require a higher C,. Humidity has no effect on C, at any given angle of attack or C, max for any given wing section. Because of the reduction in air density, increased humidity will also increase stalling speed and decrease profile drag.

Because water vapour is lighter than dry air, increasing humidity will reduce ambient air density. Lift = = C:hpV2 S so reducing air density will reduce lift for any combination of C, V and S. In order to maintain constant lift force in any given combination of V and S, an increase in humidity will therefore require a higher C,. Humidity has no effect on C, at any given angle of attack or C, , ,,for any given wing section. Because of the reduction in air density, increased humidity will also increase stalling speed and decrease profile drag.

LD 61. a. In a steady climb lift = weight multiplied by the cosine of the angle of climb, so if changes in weight are ignored a constant angle of climb requires a constant lift force. IAS is proportional to %pV2 where p is air density and V is TAS. This means that when climbing at constant IAS, %pV2 must also be constant. Air density reduces with altitude, so when climbing at constant tAS, the reduction in density must be matched by an increase in TAS. This means that for a constant TAS climb, IAS and hence WpV2 reduces. But lift is equal to CLXpV2so to maintain constant lift when climbing at constant TAS, C, must be increased to match the reducing value of WpV2. Angle of attack must therefore be increased to increase C,.

LD 62. a. Lift is created by pressure differences above and below the wing, which cause air to be accelerated downwards. Leakage of these pressure differences causes air to migrate from the lower to upper surfaces around the wingtips. This migration produces wingtip vortices. Induced drag occurs because these wingtip vortices cause the relative airflow to be deflected downwards behind the wing, tilting the total reaction to a more rearward inclination. This reduces lift and increases drag. lncreasing C, increases the magnitude of these pressure differences and so increases induced drag. lncreasing aspect ratio increases wing span and reduces wing chord so at any given airspeed the mass of air affected by the wing is increased and the time during which it is caused to leak around the wing tips is reduced. Increasing aspect ratio therefore reduces the magnitude of the wingtip vortices and hence reduces induced drag. Statement 2 is incorrect because increasing sweep back angle decreases aspect ratio and so increases induced drag. Statement 3 is incorrect because increasing EAS increases the mass of air affected by the wing thereby reducing the downward acceleration required for a given lift force. This in turn reduces the time during which the air is driven into the wing tip vortices. So increasing EAS reduces wingtip vortices and induced drag. lncreasing C, increases the lift force generated at any given airspeed and hence increases the pressure differences between the upper and lower surfaces. lncreasing C, therefore increases wingtip vortex strength and induced drag. Statements 1 and 4 are therefore correct as indicated in option a.

248

Lift and Drag

LD 63. a. D, is the sum of friction drag, form drag and interference drag. Within the normal operating range of ,, is approximately constant. D, = CD&pV2S, so for angles of attack these do not vary significantly so C ,, V and S, D, is proportional to V2. Statement 1 is therefore correct. Although any given combination of ,C it is true that C, is proportional to I N 4 , increasing EAS increases the mass of air affected by the wing and so reduces the time during which it is affected by the wing. This reduces wingtip vortex strength and hence induced drag. Because of these effects, induced drag is proportional to IN2so statement 2 is incorrect. An airspeed indicator produces an output that is directly proportional to WpV2 so for a given MpV2 it will give a constant IAS. As altitude increases p decreases, so to maintain a constant value of %pV2 at any given IAS, V must increase. But V is TAS so this means that the TAS corresponding to any given IAS so both are proportional increases with altitude. Induced drag = CD,%pV2Sand profile drag is C,%pV% to IAS, rather than to TAS. But EAS is simply IAS corrected for pressure sensing errors and compressibility errors, so both types of drag are proportional to EAS. Statements 1 and 4 are therefore correct.

LD 64. b. At V, L = C, MaxWpVs2S and at any speed V, L = CL%pV2S so L is proportional to C, V ,. S2 So C, .,VS2

and to CLV2

.,, .,,

is proportional to C,V

So at any speed V, C, / C,

= Vs2 / V2

Which means that at any speed V, C, = (C,

.,,

So when V = 1.3Vs, then C, = (C,),,. That is C, = C,

) Vs2 / V2

1 / 1.32

/ 1.69 which simplifies to give C, = 0.59 C,

This means that at 1.3Vs, the C, is 59% of C, .,.

.,,

Note: All questions of this form can be solved using the following equation: At any speed V, C, = (C,),,.

Vs2 / V2

LD 65. a. Lift = CLWpV2Swhere p is air density and V is TAS. lncreasing altitude causes p to decrease. To maintain constant lift force at constant TAS as altitude increases it is therefore necessary to increase C,. But C, is proportional to CL2SO increasing CL will increase C, and hence induced drag. lncreasing altitude at constant TAS therefore increases induced drag.

LD 66. b. Induced drag is caused by wingtip vortices which are in turn caused by the leakage of air pressure from the lower surface to the upper surface of a wing around the wingtips. This leakage takes the form of wingtip vortices. The strength of these vortices and hence the magnitude of induced drag is proportional to the pressure difference between the upper and lower surfaces. That is to say, increasing pressure differences increases both the vortex strength and the induced drag force. Lift is generated by accelerating a mass of air downwards and the magnitude of the lift force generated equal to the mass of air affected multiplied by the downward acceleration given to it. In straight and level .flightlift must equal weight so increasing weight requires either a greater mass of air or a greater downward acceleration rate.

Lift and Drag

249

The mass of air affected by the wing is proportional to its aspect ratio and airspeed, so increasing either will reduce the downward acceleration required for any given lift force. Induced drag will therefore reduce with increasing aspect ratio and/or increasing airspeed. For any given combination of aspect ratio and airspeed, increasing lift requires an increase in pressure difference between the upper and lower wing surfaces. This increased pressure difference increases wingtip vortices and induced drag. lnduced drag is therefore increased by, increasing aircraft weight, decreasing airspeed and decreasing aspect ratio.

LD 67. c. Load factor = lift / weight so for any given weight, increasing load factor requires an increased lift force. At any given airspeed this is achieved by increasing angle of attack, which in turn increases C, and CDl. Within the normal operating range of angles of attack ,C , and hence D, do not change. lncreasing load factor does not therefore affect D, but increases Dl. Because Dl is proportional to IN2the increase is most significant at low speeds so the total drag curve is pushed upward and right towards the higher , Bkcause any aircraft is speed range. This increases drag at any given speed and also increases.,V , the increase in,V , caused by the increased load factor will reduce the speed stable at speeds above,,V proportion of the speed range over which it is speed stable. This means that speed stability is decreased. , and will decrease speed stability. lncreasing load factor will therefore not affect D, will increase,,V

LD 68. d. As the speed of an aircraft decreases it is necessary to increase angle of attack in order to increase C, to and this occurs at V., As speed is reduce maintain a constant lift force. The maximum value of C, is C, further beyond the stall both C, and the lift force decrease rapidly as indicated in

,,,

the diagram. At the stall the airflow,,,C , separates from the upper surfaces of Decrease in C, the wing producing a highly turbulent wake which causes a large increase in C,

in CD

profile drag. At 0.9 V, the C, is

.: .

..: .. ..__

therefore less than C,

,,,and CDis C,

greater than at 1.IV, when the wing has not stalled.

LD 69. c. C, = kCL2/ 7rA Rearranging this equation gives C, is proportional to d~,,.

..: .

.. .:

FLAPS

FLAPS 1. c. Deployment of trailing edge flaps increases wing camber and angle of incidence. This has the effect of increasing C, which enables the required lift to be produced at a lower airspeed. The overall effect is an increase in C, and a decrease in V, allowing the aircraft to take-off and land at lower speeds.

FLAPS 2. a. Deployment of trailing edge flaps increases wing camber, thereby increasing the coefficient of lift at any angle of attack. This increased coefficient of lift enables the wing to generate the required lift force at a lower airspeed. The higher camber also means that the maximum coefficient of lift (C, and wing stall, will be achieved at a lower angle of attack. The overall effects of trailiqg edge Flap deflection are therefore reductions in landing speed, take-off speed, and stalling angle. The greater the flap deflection angle, the greater will be the reduction in stalling angle.

FLAPS 3. a. Although deployment of trailing edge flaps increases C, it also increases C., In most cases the overall effect of deployment is a reduction in L : D ratio, so the best L :D ratio is usually achieved with the flaps in the up (zero degree) position.

FLAPS 4. c. The coeficient of induced drag (C,,) is proportional to CL2.Because flap deployment always increases the coefficient of lift, it also increases the coefficient of induced drag. Induced drag will therefore always be increased by flap deployment.

FLAPS 5. c. Although trailing edge flap deployment increases C, it also increases C., The effect of these changes on the L : D ratio depends upon the relative magnitudes of the changes in C, and C., The majority of the benefit in terms of increasing C, is produced during the first few degrees of deployment, whilst the greatest proportion of the increase in C, occurs during the last few degrees. Deploying trailing edge flaps to the 45O position will therefore reduce the L:D ratio the most. Deployment of slats produces very little additional drag and hence always improves the L:D ratio.

FLAPS 6. a. The principal purpose of deploying flaps is to enable an aircraft to fly at lower airspeeds during landing and take-off. During deployment they increase the camber, angle of incidence, and angle of attack of the wings, thereby increasing C,.

FLAPS 7. d. Flap deployment increases the C, at any given angle of attack, including CLM, at the stall, but decreases the stalling angle of attack. The increased C, enables more lift to be generated at any given airspeed, thereby making it possible for the aircraft to fly at lower speeds without stalling. Flap deployment therefore decreases stalling speed and stalling angle, but increases CLM .,

Flaps

251

FLAPS 8. b. JAR 25 definesVFoas the maximum speed at which the flaps may be extended or retracted. It is commonly termed maximum flap operating speed.

FLAPS 9. a. JAR 25 defines V ,,

as the maximum speed at which an aircrah may be flown with its flaps extended.

FLAPS 10. b. Deployment of slats produces a convergent duct or slot between the leading edge ofthewing and the slatAir passing through this slot is accelerated, increasing the energy of the boundav layer passing over the upper surface. This energising of the upper surface boundary layer increases its ability to resist separation and so increases the stalling angle of attack. If an aircraft is already flying close to its stalling angle when slats are prematurely retracted, the decrease in stalling angle might cause it to stall.

FLAPS 11. a. Deployment of flaps increases the camber , angle of incidence and angle of attack of the wings, and so increases both C, and C., The greatest increase in C, is however produced during the first few degrees of deployment, while the greatest increase in C, occurs during the last few degrees.

FLAPS 12. c. Deployment of trailing edge flaps prior to landing increases the camber, angle of incidence and angle of attack of the wings. This increases C, at any given angle of attack and so reduces tlie landing speed and nose up attitude required for landing. If the flaps fail to deploy for landing then the angle of attack at any given attitude will be reduced, so a higher nose up attitude and airspeed will be required. Because of the higher airspeed, the landing roll will also increase.

FLAPS 13. d. When leading edge slats are deployed they energise the boundary layer passing over the top of the wing, thereby increasing both C, and C., The increase in C, is however much greater than the increase in C, so slat deployment increases the L:D ratio. Deployment of trailing edge flaps also increases both CLand C, but the effect on C, is much greater than that caused by slats. Trailing edge flap deployment therefore tends to decrease L:D ratio. If trailing edge flaps are retracted whilst leaving slats deployed, both lift and drag will decrease, but the L:D ratio will increase due to the removal of the high drag penalty imposed by the trailing edge flaps.

FLAPS 14. d. Deployment of slats produces a convergent duct or slot betweenthe leading edge of the wing and the slat. Air passing through this slot is accelerated, increasing the energy of the boundary layer passing over the upper surface. This energising of the upper surface boundary layer increases its ability to resist separation and so increases the stalling angle of attack. If an aircraft is already flying close to its stalling angle when slats are prematurely retracted, the decrease in stalling angle might cause it to stall.

FLAPS 15. c. Deployment of trailing edge flaps increases the camber, angle of incidence, and angle of attack of the wings. The increased camber increases the tendency of the boundary layer to separate at any given angle of attack, so flap deployment decreases the stalling angle of attack. But because flap deployment also increases C,, it increases the lift force that can be generated at any given airspeed and so decreases stalling speed. Trailing edge flap deployment therefore decreases stalling speed and stalling angle of attack.

252

Flaps

FLAPS 16. b. Spoilers are hinged panel fitted to the upper surfaces of the wings of some aircraft. When deployed they protrude into the airflow, causing the downstream airflow to separate from the wing. This destroys the lift force that would otherwise be created by that part of the wing and also increases drag. When employed as lift dumpers, spoilers deploy in the landing run immediately after touch-down. The overall effect of spoiler deployment is therefore a decrease in the L:D ratio.

FLAPS 17. a. Profile drag (Dp) is proportional to the frontal area of a wing and induced drag (Dl) is proportional to CL2. Deployment of flaps increases wing camber, angle of incidence, and angle of attack, thereby increasing both CL and the frontal area of the wing. This increases both D, and Dl. The majority of the increase in lift and induced drag occurs during the first few degrees of deployment, whilst the majority of the increase in profile drag occurs during the last few degrees.

FLAPS 18. a. Flap deployment increases both the lift and drag forces generated by an aircraft. The relative magnitudes of the increases in lift and drag vary with deployment angle, such that most of the extra lift is generated by the first few degrees of deployment, whilst most of the extra drag is generated by the last few degrees. During the take-off run an aircraft must accelerate to flight speed as quickly as possible, so the requirement is for a ,flap angle giving maximum extra lift but minimum extra drag. Conversely, in the landing run, additional drag is needed to slow the aircraft after touch down. The normal configuration for landing is therefore slats and flaps fully deployed.

FLAPS 19. b. Flap deployment increases both the lift and drag forces generated by an aircraft. The relative magnitudes of the increases in lift and drag vary with deployment angle, such that most of the extra lift is generated by the first few degrees of deployment, whilst most of the extra drag is generated by the last few degrees. During the landing run the flaps are fully deployed in order to maximise drag and hence decrease the landing run distance.

FLAPS 20. d. Flap deployment increases both the lift and drag forces generated by an aircraft. The relative magnitudes of the increases in lift and drag vary with deployment angle, such that most of the extra lift is generated by the first few degrees of deployment, whilst most of the extra drag is generated by the last few degrees. During the take-off run an aircraft must accelerate to flight speed as quickly as possible, so the requirement is for a flap angle giving maximum extra lift but minimum extra drag. Deploying flaps to small angles will therefore minimise the take-off run. If however flaps are fully deployed during the takeoff, the extra drag will reduce the rate of acceleration and hence increase the take-off distance required. Flap deployment can therefore either increase or decrease the take-off run depending on the angle of deployment.

FLAPS 21. a. In addition to increasing the lift produced by a wing, deployment of trailing edge flaps also increases the magnitude of the drag farce. Power required to maintain an aircraft in straight and level flight is equal to the drag force multiplied by the TAS. Because flap deployment increases drag, it also increases the power required.

Flaps

253

FLAPS 22. a. Profile drag (D,) is proportional to the frontal area of a wing and induced drag ( Dl) is proportional to C:. Deployment of flaps increases wing camber, angle of incidence, and angle of attack, thereby increasing both C, and the frontal area of the wing. This increases both D, and Dl. The majority of the increase in lift and indl~ceddrag occurs during the first few degrees of deployment, whilst the majority of the increase in profile drag occurs during the last few degrees. The use of small angles of flap deflection therefore increases Dl more than D.,

FLAPS 23. c. JAR 25 defines VF0 as the maximum speed at which the flaps of an aircraft can be operated. This means that it is the maximum speed at which flaps can be extended or retracted.

FLAPS 24. b. Flap deployment increases both lift and drag at any given flight speed. Whilst the increased lift decreases the take-off speed, the increased drag decreases the rate of acceleration. The greatest increase in lift occurs during the first half of the deployment cycle, whilst the greatest increase in drag occurs during the second half. For take-off the flaps are set a low angle of deflection in order to maximise the ratio of increased lift to increased drag. If however the flaps are fully deployed for take-off the decreased acceleration due to increased drag, will outweigh the decreased take-off speed due to increased lift. The overall effect will therefore be a longer take-off distance.

FLAPS 25. b.

.

Flap deployment increases both lift and drag at any given flight speed. Whilst the increased lift decreases the take-off speed, the increased drag decreases the rate of acceleration. The greatest increase in lift occurs during the first half of the deployment cycle, whilst the greatest increase in drag occurs during the second half. If full flap is employed in landing, the higher lift at lower airspeeds will enable a steeper approach to be used.

FLAPS 26. c. The stalling angle of a wing depends upon a number of factors including surface condition, camber, thickness to chord ratio and plan form. Typical figures are 12 degrees for wings with plain flaps deployed and 14 degrees with split flaps deployed.

CLIMBING AND DESCENDING

CLlMB 1. b. Maximum endurance requires minimum fuel consumption rate. Fuel consumption in a piston engine is In a jet proportional to power output, so maximum endurance is achieved at minimum power speed (V,). aircraft fuel consumption is proportional to thrust, which in straight and level flight is equal to drag. So for a jet aircraft maximum endurance is achieved at minimum drag speed (V),.

CLlMB 2. c. Fuel flow in a piston engine aircraft is proportional to power, whilst that in a jet-powered aircraft is directly proportional to thrust. Thrust output of a jet engine is approximately constant with increasing airspeed, whilst that of a piston engine reduces rapidly.

CLlMB 3. c. When the engines of an aircraft fail in flight, the aircraft possesses a limited store of potential and kinetic energy. Throughout the glide this energy is dissipated in overcoming drag. For maximum range it is necessary to achieve the best compromise between drag and TAS. This is achieved when flying at the best L:D ratio, which occurs at.,V , Although deployment of flaps increases C, it also increases C, and reduces L:D ratio. Maximum glide range is therefore achieved by fling with flaps up at.,V,

CLlMB 4. a. The sine of the angle of CLlMB is equal to excess thrust divided by aircraft weight. Thrust is approximately constant in a jet aircraft, so best excess thrust and hence best angle of climb is achieved at minimum drag speed (V),. In a propeller aircraft thrust reduces rapidly v~ithspeed, so best excess thrust and angle of climb are achieved at the minimum safe flying speed. Best angle of climb is used to clear obstacles after take-off.

CLlMB 5. a. This problem can be solved using the equation, % gradient = ((thrust - drag)/mass) x 100.

In the initial state this gives 5 = (( thrust - drag) I200000 Kg) x 100 Which can be rearranged to give. (thrust - drag) = (200000 x 5) 1 100 And usirrg the same equation in the new condition: (thrust - drag) = (new mass x 4) / 100 Assuming thrust and drag do not change, these two equations are equal so (200000 x 5) 1100 = (new mass x 4) I100 This simplifies to give new mass = (200000 x 5) 14 which is 250000 So the aircraft can achieve a gradient of 4% at a amass of 250000 Kg.


255

CLIMB 6. b. The low speed stall limits the minimum speed at which a subsonic aircraft can fly, whilst its maximum speed is limited by high-speed shock stall, as it approaches the local speed of sound. As altitude increases air density decreases, causing a similar decrease in dynamic pressure (%pV2) at any given TAS. Both airspeed indication (IAS) and lift are derived from dynamic pressure, so as altitude increases, reducing dynamic pressure affects both stalling speed and IAS to the same degree. Indicated stalling speed therefore remains constant with changing altitude. Increasing altitude does however'alter the ratio of IAS to TAS such that for a given IAS, the corresponding TAS increases. This means that the low speed limit, (the stalling TAS) increases with altitude. The local speed of sound is 38.94 times the square root of the absolute air temperature. So as altitude increases, temperature and the local speed of sound both decrease. A subsonic aircraft will suffer shock stall when TAS reaches some critical fraction of the local speed of sound (Merit), so as the speed of sound reduces so does the maximum speed of an aircraft. The overall effect is that the gap between maximum and minimum speed reduces with increasing altitude.

CLIMB 7. d. Thrust is equal to the mass of air affected by the propulsion system multiplied by the acceleration given to that mass. This acceleration is equal to the speed at which the air leaves the system minus the TAS of the aircraft. Because of the low speed of its propeller wash, the thrust of a propeller reduces rapidly as TAS increases. In the case of a jet, because of its high velocity jet stream, thrust reduces much less slowly and is approximately constant over a wide speed range. Thrust horsepower is equal to thrust multiplied by TAS so for a propeller aircraft, thrust horsepower increases rapidly at low speeds, before the reducing thrust causes it to decrease rapidly as TAS increases. For a jet with its approximately constant thrust, the thrust horsepower increases approximately linearly with TAS.

CLIMB 8. c. Maximum glide range = (L:D ratio) x ( height) To calculate range in nautical miles the height must also be in nautical miles. Using the conversion factor 1 nm = 6000 feet, 36000 feet = 6 nm. C, = 0.45 and C, = 0.0225 and L:D ratio = C,:C,

ratio

So L:D ratio = 0.45 1 0.225 = 20 This gives Range = L:D ratio x Height = 20 x 6 = 120-nautical miles. So maximum glide range = 120 nm.

CLIMB 9. d. Maximum glide range is equal to the L:D ratio multiplied by the height and is not affected by changes in (), at which best L:D ratio is achieved. weight. Reducing weight does however reduce the speed V For range in nautical miles the height must also be in nautical miles. Using the conversion factors, 1 nm = 6000 feet, and 36000 feet = 6 nm. C, = 0.45 and C, = 0.0225 and L:D ratio = C,:C,

ratio

So L:D ratio = 0.45 10.225 = 20 This gives Range = L:D ratio x Height = 20 x 6 = 120 nm So maximum glide range = 120 nm which is not affected by weight changes.

256


CLlMB 10. b. Rate of CLlMB (ROC) = excess power 1 weight. As altitude increases, power available decreases and power required increases. Increasing altitude therefore reduces both excess power and ROC. The service ceilings of propeller and jet aircraft are the altitudes at which maximum rates of climb are 100 fpm arid 500 fpm respectively.

CLlMB 11. c. Rate of CLlMB equals excess thrust power divided by aircraft weight, so for any given weight, the maximum rate of clirr~boccurs at the speed at which excess power is maximised. The thrust horsepower output of a propeller aircraft increases initially at low speeds, before reducing rapidly with increasing airspeed. That of a jet increases approximately linearly with increasing speed. Power required for all aircraft is equal to drag multiplied by TAS. This decreases as speed approaches minimum power speed, then increases rapidly with increasing speed. For a propeller aircraft the maximum value of excess power and hence rate of climb, occurs at a comparatively low airspeed between,V , and .V ,, Because of its increasing power output, the best rate of CLlMB speed for a jet aircraft occurs at a much higher speed between ,V , and the never exceed speed (V,).

CLlMB 12. a. The absolute ceiling of an aircraft is the altitude at which the rate of CLlMB is zero. Rate of climb is excess thrust power divided by aircraft weight, so when rate of climb is zero excess power is also zero. The absolute ceiling is therefore the altitude at which the power available curve is just tangential to the power required curve.

CLIMB 13. c. In gliding flight range through still air is equal to the L:D ratio divided by the height of the aircraft at the start of the glide. The Tangent of the glide angle is the rate of descent divided by the ground speed. Although gliding into a headwind does not alter the range through the air, the air itself is moving in the opposite direction to the aircraft. Range over the ground is therefore decreased. Because the ground speed is also reduced whilst the rate of descent remains constant, the Tangent of the glide angle increases. Between zero and ninety degrees an increasing tangent equates to an increasing angle, so the glide angle increases in a headwind. The overall result is that a headwind reduces glide range and increases glide angle. An alternative way of describing this situation is that because the headwind opposes the forward motion of the aircraft, the pilot must minimise the time during which the aircraft is affected by it, in order to maximise glide range. To do this he must increase glide speed by pushing the nose further down. This' increases glide angle but reduces range.

CLlMB 14. d. In gliding flight range through still air is equal to the L:D ratio divided by the height of the aircraft at the start of the glide. The Tangent of the glide angle is the rate of descent divided by the ground speed. Although gliding out of a tailwind does not alter the range through the air, the air itself is moving in the same direction as the aircraft. Range over the ground is therefore increased. The tailwind will not however alter glide endurance or rate of descent.

CLlMB 15. c. Although flap deployment increases the C, at any given angle of attack and hence increases lift at any speed, it also increases C, and drag. In most cases the increase in C, is less than the increase in C, so the L : D ratio is decreased. The increased C, enables the aircraft to fly at lower speeds, so both take-off


257

and landing speeds are reduced. But the reduced L : D ratio reduces excess thrust and so reduces the maximum angle of climb.

CLlMB 16. a. In order to maximise endurance it is necessary to minimise the rate at which fuel is consumed. In a piston engine fuel consumption is proportional to power output. So minimum fuel flow and best endurance occur at the minimum power speed, .V ,, But in a jet engine fuel flow is proportional to thrust. So minimum fuel , flow and maximum endurance are achieved at the speed requiring minimuin thrust. This is.,V

CLlMB 17. d. Power available from any air-breathing engine is proportional to the mass flow of air passing through it. As altitude increases, air density decreases, so power available decreases. Power required is equal to drag multiplied by TAS. As altitude increases the drag at any given IAS remains constant but the TAS increases. This means that the power required to fly at any given IAS increases with altitude. Combining the above two factors means that as altitude increases power available decreases while power required increases, causing excess power to decrease.

CLlMB 18. d. Maximum rate of CLlMB is equal to excess power divided by aircraft weight. Power required is equal to drag multiplied by TAS. As altitude increases the drag at any given IAS remains constant but the TAS increases. This means that the power required to fly at any given IAS increases with altitude such that the power required : TAS curve moves upwards and to the right, Power available from any air-breathing engine is proportionalto the mass flow of air passing through it. As altitude increases,air density decreases, so power available decreases causing the power available : TAS curve to move downwards. The overall effect of these changes is that as altitude increases the EAS at which best rate of CLlMB is achieved decreases whilst the corresponding TAS increases. Because of the different shapes of the power available curves, the effect on TAS is greater for piston engines than for jets.

CLlMB 19. c. When gliding for best range it is necessary to maximise the ground distance covered. When flying into a head wind, this is achieved by increasing speed to minimise the time during which the aircraft must oppose the headwind. When gliding out of a tailwind a reduced speed extends the time during which the aircraft benefits from the tailwind.

CLlMB 20. b. When an aircraft is banked its lift vector is tilted, thereby reducing the vertical component of lift. In order to maintain height it is necessary to increase angle of attack to increase the total lift force. This constitutes an increase in load factor, and in addition to increasing lift, the drag force is also increased. This increased drag force means that more thrust is required to balance drag and this constitutes an increase in power required. This decreases the excess power available. But rate of CLlMB is proportional to excess power . so banking also reduces rate of climb.

CLlMB 21. c. A piston engine produces brake horsepower, which is converted into thrust by mean of a propeller. The fuel flow in a piston engine is proportional to power output. A jet engine produces thrust directly and an increase in fuel flow will increase thrust. Fuel flow in a jet is therefore directly proportional to thrust. As TAS increases, the thrust produced by a jet engine decreases slightly, levels of at about 250 Kts, then returns' to its original value by about 500 Kts. The overall effect is that jet thrust is approximately constant with

258


increasing speed. The thrust of a propeller is equal to the mass of air passing through it multiplied by the acceleration imparted to that air. Because the propeller wash speed is quite low, the acceleration given to the air rapidly decreases as aircraft speed increases. This reduction in airflow acceleration causes thrust to decrease rapidly with increasing airspeed.

CLlMB 22. c. This problem can be solved using the following equation: Rate of CLlMB = Excess power (in Wlbflmin) / weight (in Ibf) In this case rate of CLlNlB = 1500 HP x 33000 ft Ibf/min/HP / 200000 Ibf Which simplifies to give rate of CLlMB = 247.5 ft / min The closest option to this is c, 248 Wmin.

CLlMB 23. c. The Sin of the best angle of CLlMB = ~ i c e s thrust s / Weight Excess thrust = thrust available - thrust required for straight and level flight In this case excess thrust = 40000 Ibf - 20000 Ibf = 20000 Ibf So for best angle of CLIMB Sinq = 20000 Ibf 150000 Ibf = 0.4 so best angle of climb = 23.578 degrees. This equates to a gradient of approximately 44%, so the closest option is c, 40%.

CLlMB 24. c. The second segment starts as soon as the landing gear becomes fully retracted.

CLlMB 25. b. The third segment ends when the aircraft reaches its minimum flap retraction height of 400 ft.

CLlMB 26. c. As'altitude decreases, the IAS equating to any given mach number increases. IAS must therefore increase to maintain constant mach number in a descent. Assuming an aircraft is descending at idle power this increase in IAS must be achieved by decreasing drag. But drag is both proportional to IAS and angle of attack, so to decrease drag it is necessary to decrease angle of attack. So angle of attack must decrease in a constant mach descent. This is achieved by pushing the nose of the aircraft down, thereby increasing its nose down pitch attitude.

CLlMB 27. a. lncreasing altitude causes air density and hence mass flow through the engine to reduce. Power output is directly proportional to mass flow so power available at 40000 f? from a typical jet engine will be approximately % that at ISA msl. Few if any piston engines can operate at this altitude.

CLlMB 28. b. Sweeping back the wings would reduce aspect ratio causing an increase in induced drag for any given lift value. Increased induced drag would reduce the glide endurance and range and hence increase the rate of descent. lncreasing the glide angle would reduce Angle of Attack and hence induced drag but


259

the higher speed attained would increase profile drag and rate of descent, again reducing glide endurance and range. Induced drag would therefore increase unless glide angle was increased to decrease lift.

CLlMB 29. b. An aircraft in flight possesses potential energy proportional to its altitude, and kinetic energy proportionai to its TAS. This energy is provided to the aircraft by the engines. Maximum rate of descent requires maximum dissipation of potential energy, whilst the need to remain within VMo, and MM0 requires that increases in TAS be limited. Maximum rate of descent is therefore achieved by reducing the supply of energy from the engines, whilst dissipating energy by increasing drag. This is most effectively achieved, by deploying spoilers, whilst reducing thrust to idle and using pitch to limit speed.

CLlMB 30. d. Point A indicates the maximum value of C, which occurs immediately prior to the stalling angle of attack. This is the best L:D ratio. Point B indicates the best ratio of C,:C,.

CLlMB 31. d. In the event of a rejected approach it must be possible for an aircraft to regain height and conduct a goaround safely. In order to achieve this an aircraft must have an acceptable approach CLlMB gradient.

CLIMB 32. c. Using the approximation 1 Kt = 100 ft 1min, the % gradient of the CLlMB path is equal to the Rate Of Climb (ROC) in ft min divided by the TAS in Kts Maximum ROC = Excess Power IWeight Excess Power = (Thrust - Drag) x TAS Which = (50000 Ibf x 1500 ft Imin) = 75000000 ft Ibf Imin % CLlMB gradient = (ROC ITAS) and ROC = (Excess Power IWeight) So % CLlMB gradient = (Excess Power IWeight) ITAS So to achieve a 5% CLlMB gradient we require 5% = (75000000 1Weight ) 1250 Transforming this equation gives Weight = 75000000 l ( 5 x 250) = 60000 The maximum weight at which the aircraft can achieve a 5% CLlMB gradient is therefore 60000 Ibf.

CLlMB 33. b. V, is the speed for best angle of CLlMB and \Iy is the speed for best rate of climb. Best angle of climb occurs when excess thrust is maximum. Best rate of climb occurs when excess power is maximum. Increasing weight requires increased lift to maintain any given flight path, and the creation of this extra lift increases induced drag. Because induced drag is proportional to I N 2 , the increase in drag is greatest at low speeds. The increases both total drag and minimum drag speed pushing the drag curve up and to the right on the D: speed curve. This also moves the point at which excess thrust is a maximum to the right, which represents an increase in V, . Also because power required is drag multiplied by TAS, moving the drag curve up and to the right moves the power-requiredcurve in the same direction. The point of minimum power required is therefore also moved to a higher speed, causing the speed for maximum excess power (V,) to increase. Both V, and Vy therefore increase with increasing aircraft weight.

CLIMB 34. d. Best CLIMB gradient for any given weight occurs when flying at the speed at which excess thrust is maximum. This speed depends upon the shapes and locations of the thrust available and thrust required -curves.Thrust required is proportional to total drag. This is high at the low speed stalling speed, decreasing gradually up to the minimum drag speed ,,V, then increasing approximately exponentially, as speed

260


increases further. For a propeller aircraft thrust available decreases rapidly with airspeed, so excess thrust is maximum at zero speed. In the case of a jet aircraft thrust is approximately constant with increasing speed, so excess thrust is maximum when drag is minimum at.,V, ,V , is also the speed at which the C,: C, ratio is best. Of the options available in this question, none are correct for a propeller aircraft, but option d is correct for a jet aircraft. Option d is therefore the most appropriate answer.

CLlMB 35. d. As altitude increases the power required for any flight condition increases, whilst the power available from any form of normally aspirated air breathing engine decreases. The absolute ceiling is the altitude at which the power available curve just touches the power required curve, and excess power is zero. Because rate of CLlMB is equal to excess power divided by weight, this is also the'altitude at which rate of clirr~bis zero.

CLlMB 36. b. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy consumption rate. But energy consumption rate is power required, so maximum glide endurance is achieved by flying at the minimum power speed. Power required is equal to drag multiplied by TAS and although,V , produces ,, which is lower the lowest drag force, it does not produce the lowest power required. This occurs at ,V than the minimum,V , for all aircraft.

CLlMB 37. b. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity and potential energy due to it height. Throughout the subsequent glide this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy consumption rate, which occurs at.,V , But maximum glide range requires the best ratio of energy consumed to distance flown. This occurs when flying at the best C, : C, ratio, which occurs at.,V , In still air the glide range is equal to the C, : C, ratio multiplied by the height at which the glide commences. When flying into a headwind the ground speed corresponding to any given airspeed is lower, and hence distance covered is reduced. Flying out of a tail wind has the opposite effect, increasing ground speed and hence range. The factors determining maximum glide range are therefore wind, C, : C, ratio and height.

CLlMB 38. c. Despite not being standard JAR defined terminology,, , V , and, ,V , are sometimes used in JAR ATPL examination questions. VMGA(Velocity Minimum Glide Angle), is the speed at which the decent angle is minimum., ,V , (Velocity IVlinimum Descent Rate), is the speed at which the descent rate is minimum. VMGAwill therefore give the shallowest glide angle and hence maximum glide range. In still air this is.,V , Because, ,V , gives the lowest decent rate it also gives the maximum glide endurance., ,V , in still air conditions is.,V,

CLlMB 39. a. Despite not being standard JAR defined terminology ,,V , and, , V , are sometimes used in JAR ATPL , (Velocity Minimum Glide Angle), is the speed at which the decent angle is examination questions., , V lowest., ,V , (Velocity Minimum Descent Rate), is the speed at which the descent rate is lowest. VMGAwill

Cljnbing and Descending

261

therefore give the shallowest glide angle and hence maximum glide range. In still air this is.V ,, Because , ,V , gives the lowest decent rate, it also gives the maximum glide endurance., ,V , in still air conditions ,, ,V , is greater than ,V , for all aircraft types so, , V , is greater than .,,V , is.V

CLIMB 40. a. For an aircraft in a steady climb, the thrust line is angled upwards, so a proportion of its weight is carried by engine thrust. This means that the lift force generated by the wings will be less than the weight. When in a steady descent at idle power, the drag line is angled upwards, such that part of the weight is carried by the drag. This again means that the lift generated by the wings will be less than the weight. The situation in a CI-IMB is illustrated on the previous page. From this it can be seen that the cosine of the angle of climb (f) is equal to the lift divided by the weight. This means that the lift is equal to the weight multiplied by the cosine of the angle of climb. The situation is similar in a descent where lift equals weight multiplied by the cosine of the angle of descent.

Weight

/cp Cosin cp = UW

So L = Cosine cp

CLIMB 41. a. For an aircraft in steady climb, the thrust line is angled upwards and so aproportion of its weight is carried by engine thrust. This means that the lift force generated by the wings will be less than the weight. Since the thrust must also oppose the drag force, it is only the thrust minus the drag that is available to support part of the weight.This portion of the thrust is called the excess thrust and as illustrated in the diagram, is equal to weight multiplied by the sin of the angle of CLIMB ($). Sin cp = Excess t h r u s w So Excess thrust = W Sin cp

CLIMB 42. c. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy , But maximum glide range requires the best ratio of energy consumed consumption rate, which occurs at.,V ' to distance flown. This occurs when flying at the best C, : C, ratio, which occurs at .V ,, In still air the glide range is equal to the C, : C, ratio multiplied by the height at which the glide commences. Glide range is also equal to ground speed multiplied by glide duration. When flying into a headwind the ground speed corresponding to any given airspeed is lower, and hence glide range is reduced. Flying out of a tail wind has the opposite effect, increasing ground speed and hence range. The factors determining maximum glide range are therefore wind, CL : C, ratio and height. For any given combination of CL : C, ratio and height, a strong tailwind will give the maximum range.

CLIMB 43. a. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished,

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the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy , But maximum glide range requires the best ratio of energy consumed consumption rate, which occurs at.,V to distance fown. This occurs when flying at the best C, : C, ratio, which occurs at.,V, In still air the glide range is equal to the CL : C, ratio multiplied by the height at which the glide commences. Glide range is also equal to ground speed multiplied by glide duration. When flying into a headwind the ground speed corresponding to any given airspeed is lower, and hence the range covered is reduced. In , will maximise range by minimising the time during such circumstance flying at a speed higher than ,V which the headwind is reducing ground speed.

CLlMB 44. b. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy But maximum glide range requires the best ratio of energy consumed consumption rate, which occurs at.,V, to distance flown. This occurs when flying at the best C, : C, ratio, which occurs at.,V, In still air the glide range is equal to the C, : C, ratio multiplied by the height at which the glide commences. Glide range is also equal to ground speed multiplied by glide duration. When flying out of a tailwind the ground speed corresponding to any given airspeed is higher, and hence the range covered is increased. In such circumstance flying at a speed lower than ,V , will maximise range by maximising the time during which the headwind is increasing ground speed.

CLlMB 45. d. Maximum glide range in still air is achieved by flying at,,V, at which speed the C, : C, ratio is maximum. , The diagram in this question is a whole aircraft C, : C, polar. The greatest C, : C, ratio and hence ,V occurs on such a diagram where a tangent drawn from the origin just touches the curve. This is point D in the diagram.

CLlMB 46. d. When an aircraft is in steady CLlMB the thrust line is angled upwards, so a proportion of its weight is carried by engine thrust. This means that the lift force generated by the wings will be less than the weight. But because the thrust must carry a proportion of the weight whilst also opposing the drag, thrust must be greater than drag. So in a steady CI-IMB weight is more than lift and drag is less than thrust.

CLlMB 47. c. In gliding flight minimum sink rate will produce maximum glide endurance. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance , will be achieved by flying at the speed requiring the lowest energy consumption rate. This occurs at.,V Although option a, ,V , might be assumed to mean the velocity for minimum sink rate, this is not a is the minimum drag speed, which would achieve the maximum range but recognised term. Option b,,V, not the maximum endurance. ,V, is an outdated term for the maximum allowable flight speed. This would produce neither maximum endurance nor maximum range.


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CLlMB 48. b. The sin of the maximum angle of CLlMB that can be achieved by an aircraft is equal to its excess thrust divided by its weight. Excess thrust in any flight condition is the maximum thrust attainable at that speed, minus the drag. In jet aircraft thrust remains almost constant at all speeds so maximum excess thrust and such that the drag force is at a minimum. Although hence maximum climb angle will be achieved at,,V, deployment of flaps increases lift, it also increases drag. At all but very small angles, the overall effect of flap deployment is decreased L : D ratio and hence decreased angle of climb. The best angle of climb will , with flaps at zero deflection. therefore be achieved by flying at ,V

CLlMB 49. b. The sine of the maximum angle of CLlMB that can be achieved by an aircraft is equal to its excess thrust divided by its weight. Excess thrust in any flight condition is the maximum thrust attainable at that speed, minus the drag. In jet aircraft thrust remains almost constant at all speeds, so maximum excess thrust and such that the drag force is at a minimum. hence maximum climb angle will be achieved at,,V,

CLlMB 50. d. ,V , for any given weight is the speed at which the drag force on an aircraft is at its minimum. This means that ,V , is also the speed at which the L :D ratio is maximum. Fuel flow in a jet aircraft is proportional to thrust, which in straight and level flight is equal to drag. So by flying straight and level at ,V ,, a jet aircraft will achieve minimum fuel consumption and hence maximum endurance. Fuel flow in a propeller aircraft is proportional to power output, so best propeller aircraft range is achieved by flying the speed at which that ratio of power required : TAS is greatest. This occurs at .,V, The sin of the maximum angle of CLlMB that can be achieved by an aircraft is equal to its excess thrust divided by its weight. Excess thrust in any flight condition is the maximum thrust attainable at that speed, minus the drag. In jet aircraft thrust remains almost constant at all speeds, so maximum excess thrust and hence maximum climb angle will be achieved at ,V ,, such that the drag force is at a minimum. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide range will be achieved by flying at the speed providing the best ratio of power required , for all aircraft types. to TAS. This occurs at ,V From the above analysis it can be seen that all of the conditions specified in this question occur at.,V,

CLlMB 51. d. ,V , is the minimum drag speed and is the speed at which induced drag equals profile drag. Increasing , to increase. aircraft weight increases induced drag, causing,V Although increasing weight increases,,V, it does Maximum glide range is achieved when flying at.,V, not affect the glide angle, nor glide range. The aircraft merely arrives at its landing point sooner.

CLlMB 52. c. JAR 25 defines lift is as that component of the total reaction that acts perpendicular to the flight path.

CLlMB 53. b. As altitude increases,the power required for any given flight condition increases, whilst the power avail~ble decreases. The absolute ceiling is the altitude at which the power available is equal to the power required.

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Power available is equal to thrust multiplied by TAS. Power required is equal to drag multiplied by TAS. So at the absolute ceiling where power available equals power required, thrust available equals drag. The maximum rate of CLlMB that can be achieved by an aircraft is equal to its excess power divided by its weight. So at the absolute ceiling where power available equals power required, the excess power and hence rate of climb are zero. The altitude at which the low speed and high speed stall lines cross is the aerodynamic ceiling.

CLlMB 54. b. VMDis the minimum drag speed and is the speed at which induced drag equals profile drag. Increasing aircraft weight increases induced drag, causing VMDto increase. Maximum glide range is achieved when flying at VMD.Although increasing weight increases VMD,it does not affect the glide angle, nor glide range. The aircraft merely flies down the same slope at a higher speed, and so arrives at its landing point sooner. lncreasing aircraft weight therefore increases both glide speed and rate of descent.

CLlMB 55. d. IAS is proportional to %pV2, lift is proportional to CL%pV2,and drag is proportional to CDWpV2.Also CL is proportional to angle of attack. So in a constant IAS gliding descent, if the angle of attack is kept constant then the C, lift and drag will all remain constant. But in a steady descent lift equals the weight multiplied by the cosine of the angle or gradient of descent. So in a constant IAS descent if the angle of attack is kept constant, the descent gradient will remain constant. Also if descent gradient is constant then constant angle of attack will be achieved by maintaining constant pitch attitude. The pitch attitude of an aircraft must therefore remain constant in a constant IAS descent.

CLlMB 56. b. The mach number of an aircraft in flight is the speed of the aircraft as a fraction of the local speed of sound. The speed of sound is proportional to absolute temperature, so as altitude decreases, increasing terr~peraturecauses the local speed of sound to increase. This means that in a constant mach descent the TAS and IAS of an aircraft will increase. IAS is proportional to %pV2, lift is proportional to CL%pV2,and drag is proportional to CD%pV2.Also CL is proportional to angle of attack. So in a constant Mach descent, if angle of attack remains constant as IAS increases, both lift and drag will increase. But increasing drag will reduce speed. Angle of attack must therefore be decreased in order to reduce drag and hence allow TAS and IAS to increase to maintain constant Mach number. But decreasing the angle of attack in order to decrease drag will also decrease lift. Because lift equals the weight multiplied by the cosine of the angle of descent, decreasing lift will increase the angle of descent. The descent gradient will therefore increase in a constant mach gliding descent.

CLlMB 57. d. IAS is proportional to %pV2, lift is proportional to CLWpV2,and drag is proportional to CD%pV2.Also CL is proportional to angle of attack. So in a constant IAS gliding descent, if the angle of attack is kept constant then the C, lift and drag will all remain constant. But in a steady descent lift equals the weight multiplied by the cosine of the angle or gradient of descent. So in a constant IAS descent if the angle of attack is kept constant the descent gradient will remain constant. Also if descent gradient is constant then constant angle of attack will be achieved by maintaining constant pitch attitude. The pitch attitude of an aircraft, angle of attack and descent gradient must therefore remain constant in a constant IAS gliding descent.


CLlMB 58. b. Although a headwind would not affect the Distance by which headwind pushes rate of CLlMB or airspeed of an aircraft, it the air backwards in given timewould cause the air in which the aircraft is flying, to be pushed backwards relative to the ground. This would cause the angle of CLlMB to increase as indicated in the diagram at the right. The fact that the aircraft was in a constant mach number-CLIMB would have no bearing on this situation.

265

Distance by which headwind pushes the air backwards in given

Actual CLlMB gradient (over ground) is increased CLlMB gradient through the air

Height gained in given time

CLlMB 59. b. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy consumption rate, which occurs at VM,. But maximum glide range requires the best ratio of energy consumed to distance flown. This occurs when flying at the best C, : C, ratio, which occurs at .V ,, In still air the glide range is equal to the C, : C, ratio muitiplied by the height at which the glide commences. When flying out of a tailwind the ground speed corresponding to any given airspeed is increased, and hence the range covered is increased. In such circumstances, flying at a speed lower than,V , will maximise range by maximising the time during which the headwind is increasing ground speed.

CLlMB 68. b. IAS is proportional to %pV2, lift is proportional to C,'/2pV2, and drag is proportional to CD%pV2.Also CL is proportional to angle of attack. So in a constant IAS gliding desceni, if the angle of attack is kept constant then the C, lift and drag will all remain constant. But in a steady descent lift equals the weight multiplied by the cosine of the angle or gradient of descent. So in a constant IAS descent if the angle of attack is kept constant the descent gradient will remain constant. Also if descent gradient is constant, then constant angle of attack will be achieved by maintaining constant pitch attitude. If an aircraft is perrr~ittedto descend when in a nose up attitude there is a considerable danger that the stalling angle will be exceeded. Gliding descents are therefore conducted in a nose down attitude. The nose down pitch attitude of an aircraft, angle of attack and descent gradient must therefore remain constant in a constant IAS gliding descent.

CLlMB 61. c. The sine of the maximum angle of CLlMB that can be achieved by an aircraft is equal to its excess thrust divided by its weight. Excess thrust in any flight condition is the maximum thrust attainable at that speed, minus the drag. The speed at which excess thrust and hence angle of climb is a maximum is called Vx. This term is common to both propeller and jet aircraft, although the corresponding airspeeds are very different. In jet aircraft thrust remains almost constant at all speeds, so maximum excess thrust and hence maximum climb angle will be achieved at,,V, such that the drag force is at a minimum. In the case of a propeller aircraft, thrust decreases rapidly with increasing airspeed, so Vx is very close to the minimum safe flying speed. Of the options offered in this question, ,V , is correct only for jets whereas Vx is correct for all aircraft.

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CLlMB 62. d: The maximum rate of CI-IMB that can be achieved by an aircraft is equal to its excess power divided by its weight. Excess power is equal to the power available minus the power required. In a jet aircraft power available increases linearly with TAS and maximum excess power and best rate of climb, occur at a speed somewhat higher than .V ,, In a propeller aircraft power available increase with airspeed, before again decreasing. The maximum excess power and hence best rate of climb in propeller aircraft occurs at a speed somewhat lower than .V ,, For all aircraft the speed at which rate of climb is maximum is that at ., which excess power is a maximum. This speed is called V

CLlMB 63. c. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide, this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy consumption rate, which occurs at .V ,, But maximum glide range requires the best ratio of energy consumed to distance flown. This occurs when flying at the maximum CL : C, ratio, which occurs at .V ,,

CLlMB 64. c. When an aircraft is in steady CLIMB the thrust line is angled upwards and so a proportion of its weight is carried by engine thrust. This means that the lift force generated by the wings will be less than the weight. When in a steady descent at idle power, the drag line is angled upwards, such that part of the weight is carried by the drag. This again means that the lift generated by the wings will be less than the weight. The situation in a CLlMB is illustrated at the right. - From this it can / 4 be seen that the cosine of the angle of CLlMB ($), is equal to Cosine Q = L M So L = W Cosine Q the lift divided by the weight. This means that the lift force in such a CLlMB equals the weight multiplied by the cosine of he angle of climb. But the cosine cf all nonzero angles is less than 1, so lift equals less than weight. Load factor is equal to lift divided by weight, so in a steady climb when lift is less than weight, the load factor must be less than 1. In considering this situation it should be noted that this result is due to the definition of load factor. The aircraft is still subject to l g and hence weighs the same as in level flight.

CLlMB 65. c. The mach number of an aircraft represents its TAS as a fraction of the local speed of sound (LSS) LSS is proportional to temperature, so as altitude increases, the TAS equating to any given mach number decreases. Increasing altitude also increases the ratio of IAS : TAS, so as altitude increases, the IAS at any given TAS decreases. This means that climbing at constant mach number causes IAS to decrease. But lift is proportional to IAS multiplied by CL and CL is proportional to angle of attack. So to maintain a CLlMB at constant mach number the angle of attack must gradually increase to compensate for the decreasing IAS.

CLlMB 66. c. An aircraft in flight possesses a store of kinetic energy proportional to its speed, and potential energy proportional to its height above ground. In an emergency descent it is necessary to reduce this store of energy so that the aircraft attains the desired lower altitude, as quickly as possible without exceeding its


267

maximum speed. If this is attempted by simply pushing the nose down, then the aircraft will accelerate into a dive in which potential energy is transformed into kinetic energy with little energy being lost. The overall effect of this manoeuvre will be that the aircraft will exceed its limiting speed or mach nurr~ber(VMoIMMo) If however the thrust levers are set to idle, this will prevent further energy being supplied to the aircraft. If the speed brakes, flaps and gear are deployed the increased drag will increase the rate at which energy is extracted from the aircraft. If the nose is then pushed forward the aircraft will enter a gliding dive in which speed can be controlled by use of pitch attitude.

CLlMB 67. c. V, is the take-off safety speed.,V , is the speed at which power required is minimum. VMDis the speed at which drag is minimum. V, is the speed at which the maximum angle of CLlMB can be achieved. V, is the speed at which the maximum rate of climb can be achieved. For a jet aircraft V, < VMp< VMD= V, < V ., For Option c is therefore correct for a jet aircraft. a propeller aircraft V, < V, < VMp< V, <.,V,

CLlMB 68. c. The maximum rate of CLlMB that an aircra'ft can achieve is equal to the excess power minus the power required at that speed. Neither excess power nor power required are affected by headwinds. Headwinds will therefore have no effect on the rate of climb. Distance by which headwind pushes

CLlMB 69. a. Although a headwind would not affect the rate of CI-IMB or air speed of an aircraft, it would cause the air in which the aircraft was flying, to be pushed backwards relative to the ground. This would cause the angle of CLlMB to increase as indicated in the diagram at the right.

Actual angle of CLIMB ( ground) is increased

/

Height gained in given time

Angle of CLlMB through the air Distance by which headwind pushes the air backwards in given time

CLlMB 70. a. Although a headwind would not affect the rate of CLlMB or airspeed of an aircraft, it would cause the air in which the aircraft is flying, to be pushed backwards relative to the ground. This would cause the CLlMB gradient to increase as indicated in the diagram at the right.

/

/

Actual angle of CLIMB (over / / ground) is increased Angle of CLIMB through the air Height gained in given time

CLlMB 71. a. Power required in straight and level flight is equal to drag multiplied by TAS. Both drag and IAS are proportional to %pV2 so at any given IAS the drag remains constant at all altitudes. But the ratio of TAS : IAS increases with increasing altitude. At 40000 feet ISA for example, the TAS is twice the IAS. This means that at any given IAS, as altitude increases power required increases, because the TAS by which drag is multiplied is increasing while the drag remains constant. The power required therefore increases with increasing altitude.

268

Climbing and Descending Distance by which headwind pushes the air backwards in given time

CLIMB 72. b. Although a headwind would not affect the rate of CLlMB or airspeed of an aircraft, it would cause the air in which the aircraft is flying, to be pushed backwards relative to the ground. This would cause the CLlMB gradient to increase as indicated in the diagram at the right. But by increasing CLlMB gradient, a headwind will decrease distance moved over the ground in a climb.

ground) is increased

through the air

CLIMB 73. a. This problem can be solved using the following equation: Rate of CLl NIB = Excess power Iweight So in this case rate of CLlMB = (25000 ft Ibflmin) I10000 Ibf Which is rate of CLlMB = 2.5 Wrrrin

CLIMB 74. d. This problem can be solved using the following equation: The Sine of the angle of CLIMB = Excess thrust Iweight Excess thrust = thrust - drag In this question excess thrust = 25000 Ibf - 15000 Ibf = 10000 Ibf So in this case the Sine of the angle of CLlMB = 10000 Ibf I10000 Ibf = 1 If the sine of the angle of CLlMB is 1 then the angle of climb is 90° This aircraft is therefore capable of climbing vertically.

CLlMB 75. b. This problem can be solved by first using the figures for excess power and TAS to drive excess thrust using the following equation: Excess power = excess thrust x TAS So excess thrust = excess power ITAS Both sides of the equation must be in the same units so using 1 Kt = 100 Wmin In this case (50000000 ft Ibflmin) l ( 2 5 0 Kts x 100 WminIKt) Which simplifies to give excess thrust = 2000 Ibf. This can then be used to calculate maximum angle of CLlMB using the following equation: Sine of the angle of CLlMB = excess thrust 1weight So in this case the sine of the angle of CLlMB = 2000 Ibf I10000 Ibf = 0.2 If the sine of the angle of CLINIB is 0.2 then the angle of climb is 11.5".


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CLlMB 76. d. The maximum value of C,,),C ,(, occurs at the stall speed (V,), so point A is the stall. The best L :D ratio occurs on a whole aircraft polar where a tangent drawn from the origin touches the curve. This is point B so point B is.,V, So point C is in the diagram provided in this question. But best L :D ratio occurs at,,V, Point D is at a speed higher than.,V , V, for a propeller aircraft occurs at more than V, but less than.,V, whilst V, for a jet occurs at a speed higher than .,V, Of the options a speed between V, and ,,V, provided in this question only option d, propeller aircraft V, and Jet aircraft V, satisfy the above criteria.

CLlMB 77. a. The maximum value of C,,),C ,(, occurs at the stall speed (V,), so point A is the stall. The best L :D ratio occurs on a whole aircraft polar where a tangent drawn from the origin touches the curve. This is point B in the diagram provided in this question. But best L :D ratio occurs at,,V , so point B is.,V, So point D i s more than V, but less than point C and point C is more than both V, and point D, but less than .,V,

V, for a propeller aircraft occurs at a speed between V, and.,V, Vx for a propeller aircraft occurs at a ., Of the options provided in this question only option a, propeller speed higher than V, but lower than V aircraft V, and propeller aircraft Vx satisfy the above criteria.

CLlMB 78. b. An aircraft in flight possesses kinetic energy proportional to its TAS, and potential energy proportional to its height. When the engines fail in flight this energy cannot be replenished, but is dissipated in doing work to push the aircraft forward against the drag force. The rate of energy dissipation is the power required. To achieve maximum glide endurance it is necessary to minimise the energy dissipation rate. This is achieved by flying at.,V, ,V , is the minimum power speed and is proportional to aircraft weight. Increasing aircraft and hence increases the glide speed required for best glide endurance. lncreasing weight increases,,V, weight also increases the power required-at all speeds including,,V, and hence decreases maximum glide endurance.

CLlMB 79. a. An aircraft in flight possesses kinetic energy proportional to its TAS, and potential energy proportional to its height. When the engines fail in flight this energy cannot be replenished, but is dissipated in doing work to push the aircraft forward against the drag force. The rate of energy dissipation is the power required. To achieve maximum glide range it is necessary to fly at the speed at which the ratio of power required to TAS is greatest. This is achieved by flying at .,V, ,V , is the minimum drag speed and is the speed at , to which induced drag equals profile drag. lncreasing aircraft weight increases induced drag, causing,V increase. Maximum glide range is achieved when flying at.,V , Although increasing weight increases ,,V, it does not affect the glide angle, nor glide range. The aircraft merely flies down the same slope at a higher speed. Its rate of descent is therefore increased and it arrives at its landing point sooner. Increasing glide speed and rate of descent. aircraft weight therefore increases,,V,

CLlMB 80. c. The sine of the maximum angle of CLIMB that an aircraft can achieve is equal to its excess thrust divided by its weight. At any given weight, the maximum angle of climb is achieved by flying at the speed providing the greatest excess thrust. This speed is called Vx. lncreasing aircraft weight increases the lift required and hence increases induced drag. But induced drag is greatest at low speeds, so the effect of increasing speed is to push the Drag : TAS curve upwards and to the right. This means that drag at any given speed is greater, and also that the speed providing minimum drag is increased. The increase in drag means that more thrust is required to overcome drag, so excess thrust is reduced. The pushing of the drag curve to the right means that 'Jx is increased. The overall effect of these changes is that increasing weight increases Vx and decreases the maximum angle of climb.

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CLlMB 81. a. The sine of the maximum angle of CLIMB that an aircraft can achieve is equal to excess thrust divided by weight. The maximum rate of climb is equal to excess power divided by weight. Activation of reheat increases both thrust and power available and hence increases both angle of climb and rate of climb.

CLlMB 82. c. The sine of the maximum angle of CLIMB that an aircraft can achieve is equal to excess thrust divided by weight. The maximum rate of climb is equal to excess power divided by weight. Activation of reheat increases both thrust and power available and hence increases both angle of climb and rate of climb. But if an aircraft is already flying at the speed at which a climb is to be conducted, then all of the extra thrust and power produced by activation of reheat will be excess thrust and excess power. That is to say, neither drag nor power required will increase, so all of the extra thrust and power is excess to that required to maintain airspeed. If activation of reheat doubles thrust and power available while thrust and power required remain constant, then excess power and thrust will be more than doubled. This means that the angle of climb and rate of climb will be more than doubled.

CLlMB 83. a. For low angles of CLlMB (up to about 15O) a reasonably acc~~rate estimate of climb gradient in still air can be calculated using the following equation: % gradient = Rate of CLlMB (in ftfmin) 1TAS (in Kts) % So in this case the % gradient = 1000 ftlmin 1200 Kts = 5% Notes:

1. The above equation is based on the assumption that I Kts = 100 ftfmin so the equation becomes: % gradient = ROC in ftlmin I((TAS in ftlmin) 1 loo)), Which is why the result is the % gradient. 2. This equation gives only an approximate solution, a more accurate figure would be provided using the following equation: % gradient = ROC in ftfmin 1 Ground speed in Kts.

CLlMB 84. c. For low angles of CLlMB (up to about 15") a reasonably accurate estimate of climb gradient in still air can be calculated using the following equation: % gradient = Rate of CLlMB (in Wmin) 1TAS (in Kts) % Rearranging this equation gives rate of CL!MB = TAS x % gradient. So in this case rate of CLlMB = 250 Kts x 15% Which is 3750 ftlmin Notes:

1. The above equation is based on the assumption that I Kts = 100 ftfmin so the equation becomes: % gradient = ROC in ftlmin 1 ((TAS in ftlmin) 1 100)). Which is why the result is the % gradient. 2. This equation gives only an approximate solution, a more accurate figure would be provided using the following equation: % gradient = ROC in Wmin 1 Ground speed in Kts.


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CLIMB 85. d. An aircraft in flight possesses a store of kinetic energy proportional to its speed, and potential energy proportional to its height above ground. In an emergency descent it is necessary to reduce this store of energy so that the aircraft attains the desired lower altitude as quickly as possible, without exceeding speed limits. If this is attempted by simply pushing the nose down, then the aircraft will accelerate into a dive in which potential energy is transformed into kinetic energy, with little energy being lost. The overall effect of this manoeuvre will be that the aircraft will exceed its limiting speed or mach number (VMo1 MMo). If however the thrust levers are set to idle, this will prevent further energy being supplied to the aircraft. If the speed brakes, flaps and gear are deployed, the increased drag will increase the rate at which energy is extracted from the aircraft. If the nose is then pushed forward, the aircraft will enter a gliding dive in which speed can be controlled by use of pitch attitude. Maximum rate of descent is therefore achieved by using minimum thrust, together with maximum drag, and varying pitch attitude to maintain airspeed within limits.

CLIMB 86. d. VMo is the maximum operating speed of an aircraft, expressed as CAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at low altitudes where air density is high, VMo is the limiting parameter. MMois the maximum operating speed of an aircraft, expressed as a mach number. If the aircraft is operated at higher mach numbers, the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. MMo is some specific fraction of the local speed of sound. But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the CAS equating to MModeceases as altitude increases. The overall effect of these factors is that VMo is the limiting speed at low altitude and MMois the limiting speed at high altitude. The altitude at which the TAS equating to VMo is equal to that equating to MMo is called the crossover altitude. When climbing at constant CAS or IAS, both TAS and Mach number increase. There is therefore a danger that MMowill be exceeded in a constant IAS CLIMB above the crossover altitude.

CLIMB 87. c. VMo is the maximum operating speed of an aircraft expressed as CAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at lo-w altitudes where air density is high, VMo is the limiting parameter. MMo is the maximum operating speed of an aircraft expressed as a mach number. If the aircraft is operated at higher mach numbers the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. MMois some specific fraction of the local speed of sound. But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the CAS equating to MModeceases as altitude increases. The overall effect of these factors is that VMo is the limiting speed at low altitude and MMois the limiting speed at high altitude. When climbing at constant TAS, both CAS and Mach number decrease so there is no danger that VMo or MMo will be exceeded. If however the CLIMB is continued to a sufficiently high altitude, the CAS might reduce to less than the low speed stalling speed V., The aircraft will then stall.

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CLlMB 88. c. The mach number of an aircraft is its TAS, expressed as a fraction of the local speed of sound. Mach 0.8 for example represents 80% of the local speed of sound. But the local speed of sound is proportional to air temperature, and hence decreases with increasing altitude, up to the tropopause at 36000 feet ISA. Above this altitude both temperature and the local speed of sound remain constant. This means that the TAS equating to any given mach number, decreases with increasing altitude up to 36000 feet ISA, and remains constant at higher altitudes. But the CAS equating to any given TAS decreases with increasing altitude throughout the entire atmosphere. This means that the CAS equating to any given mach number decreases with increasing altitude throughout the atmosphere. So if a constant mach CLlMB is continued to a sufficiently high altitude, or if the constant mach number is sufficiently low, the CAS might decrease to less than the low speed stalling speed V., The aircraft will then stall.

CLlMB 89. a. VMo is the maximum operating speed of an aircraft expressed as CAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportionalto air density, so at low altitudes where air density is high, VMois the limiting parameter. ,M , is the maximum operating speed of an aircraft expressed as a mach number. If the aircraft is operated at higher mach numbers the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. M, is some specific fraction of the local speed of sound. But the local speed of sound is proportional to air temperature and hence increases with decreasing altitude. This means that the CAS equating to ,M , increases as altitude decreases. The altitude at which the CAS equating to VMo is the same as that equating to MMois called the crossover altitude. The overall effect of these factors is that VMo is the limiting speed below the crossover altitude and MMois the limiting speed above the crossover altitude. When descending at constant mach number there is a danger that VMo will be exceeded below the crossover altitude.

CLlMB 90. c. lndicated air speed (IAS) is proportional to % p v , where p is air density and V is TAS. Whenever the airspeed indicator senses any given %pV2, it will always produce the same 1.4s indication regardless of altitude. When climbing at constant %pV2, the IAS w ~ ltherefore l remain constant.

CLlMB 91. b. lndicated air speed (IAS) is proportional to '/zpV2, where r is air density and V is the true airspeed (TAS). Whenever the airspeed indicator senses any given '/zp@, it will always produce the same IAS indication regardless of altitude. When climbing at constant % O F , the IAS will therefore remain constant. But as altitude increases, p decreases. So if IAS remains constant then V (TAS) must increase such that the increase in V2 is equal to the decrease in p, and '/zpV2 remains constant. At 40000 feet in the standard atmosphere, p is '/4 of its sea level value, so for any given IAS, V2 must be 4 times its sea level value. This means that at 40000 feet ISA, the best GLIIVB TAS is twice the value of the best climb IAS. P

CLlMB 92. c. The maximum rate of CLlNlB that an aircraft can achieve at any altitude is equal to its excess power divided by its weight. As altitude increases power available decreases and power required increases. This means that excess power and hence rate of climb decrease with increasing altitude. The absolute ceiling is the altitude at which power available is equal to power required for level flight and excess power and rate of climb are zero. The rate of climb of any aircraft at its absolute ceiling is therefore zero.


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CLIMB 93. a. The maximum rate of CLlMB that an aircraft can achieve at any altitude is equal to its excess power divided by its weight. As altitude increases power available decreases and power required increases. This means that excess power and hence rate of climb decrease with increasing altitude. The absolute ceiling is the altitude at which power available is eql;al to power required for level flight and excess power and rate of climb are zero. The rate of climb of any aircraft at its absolute ceiling is therefore zero. Also because power required and power available vary with airspeed, straight and level flight is possible at only one speed at the absolute ceiling. So at the absolute ceiling Vx and Vy are the identical.

CLIMB 94. a. The maximum angle of CLlMB that an aircraft can achieve at any given airspeed is proportional to excess thrust divided by weight. But the thrust of a propeller aircraft reduces rapidly with increasing airspeed, whilst drag decreases up to ,V , then increases at higher speeds. The shapes of the thrust and drag curves are such that a propeller aircraft achieves its maximum angle of climb at a speed close to its minimum safe flying speed. Excess thrust and angle of climb both decrease with increasing airspeed.

CLlMB 95. d. The maximum angle of CLlNlB that an aircraft can achieve at any given airspeed is proportional to excess thrust divided by weight. The thrust of a jet aircraft remains approximately constant at all speeds, whilst , then increases at higher speeds. Excess thrust and angle of climb both therefore drag decreases up to,V , then decrease with increasing airspeed. increase up to ,V

CLlMB 96. d. The maximum rate of CLlMB that an aircraft can achieve at any given speed is equal to its excess power divided by its weight. Power available is equal to thrust multiplied by TAS. Thrust from a jet aircraft remains approximately constant at all speeds so power available increases linearly with increasing TAS. But power then increases, required is equal to drag multiplied by TAS. Drag decreases with increasing speed up to ,V, so the power required curve is of a generally similar shape to the drag curve. The minimum power required , but at a lower speed termed .,V, does not however occur at,V Because of the shapes of the concave power required curve and the linearly increasing power available curve, maximum excess power and hence maximum rate of CLlMB occurs at a speed greater than.,V, This speed is termed Vy and is approximately 1.3VMD.The rate of climb of a jet aircraft therefore increases up to approximately 1.3VMDthen decrease with increasing speed.

CLlMB 97. d. This problem can be solved using the following equations: Excess thrust = thrust - drag And the sine of angle of CLlMB = Excess thrust 1 weight In this case excess thrust = 25000 Ibf - 5000 Ibf = 20000 Ibf. So the sine of angle of CLlMB = 20000 Ibf 1 15000 Ibf = 1.67 But the sines of angles cannot be greater than 1, which equates to an angle of 90°. This means that the aircraft can CLlMB at an angle of 9C0, which would require 15000 Ibf of excess thrust. In this condition it would still have 5000 Ibf of excess thrust in reserve to enable it to accelerate in the climb.

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CLlMB 98. a. This question is complicated by the fact that the excess power is expressed as thrust horsepower THP. This problem can be overcome by using the conversion factor 1 THP = 33000 ft lbf Imin. So 350 THP = 350 THP x 33000 ft lbf I min I THP = 11550000 ft Ibf Imin. The problem can now be solved using the following equation: Rate of CLlMB = excess power Iweight. So in this case rate of CLlMB = 11550000 ft lbf 1 min I10000 Ibf Which gives a rate of CI-IMB .of 1155 ftlmin.

CLlMB 99. b. Rate of CLlMB = excess power divided by weight. So if the weight of an aircraft is doubled and its excess power remains unchanged its rate of climb would be halved. This is however an underestimation of the decrease in rate of climb because the lift required to support the extra weight would also increase drag and hence decrease excess power. Doubling the weight of an aircraft would therefore decrease its rate of climb by more than half. This option is not however available in this question so option b (half) is the most accurate.

CLlMB 100. b. In straight and level flight the attitude of an aircraft is such that the lift equals the weight. As fuel is consumed during flight, the weight decreases, so either angle of attack or speed must be decreased in order to educe lift to match the reducing weight. If both angle of attack and airspeed remain constant as weight reduces, the lift will exceed the weight, causing the aircraft to gain altitude. This situation is termed a cruise climb.

CLlMB 101. c. The rate of CLIMB of an aircraft is equal to its excess power divided by its mass. A headwind will not affect either of these and hence will neither increase nor decrease rate of climb.

Distance by which headwind pushes the air backwards in given time

Although a headwind would not affect the rate of CLlMB or airspeed of an aircraft, it would cause the air in which the aircraft was flying, to be pushed backwards relative to the ground. This would cause the angle of CLIMB to increase as indicated in the diagram at the right. But changes in angle of CLIMB are not included in the options Angle of climb through given time i n this question. As indicated in the above diagram, a the air headwind would cause the mass of air through which the aircraft was flying, to be pushed backwards as the aircraft flies into it. This means that although the TAS would be unchanged, the ground speed would be reduced. Option c is therefore the most appropriate.

CLlMB 102. b. This problem can be solved using the equation: gradient - ROC ITAS. Which can be rearranged to give ROC = TAS x gradient. The question is complicated by the fact that information give includes pressure altitude, IAS and OAT but


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not TAS. In order to obtain a truly accurate answer it would be necessary to calculate TAS from the given values of pressure altitude and OAT, then use this value of TAS to calculate ROC. If however an initial estimate of TAS is made, assuming ISA msl conditions, this can be compared with the options to identify the most appropriate. In ISA rnsl conditions TAS = IAS. Using the above equation and the given IAS of 90 Kts, in place TAS gives: ROC = 90 Kts x 3.5%. giving an initial estimate of 315 fprn ROC. But the question specifies a pressure altitude of I0000 ft and an OAT of +13O C, so the effects of these conditions need to be investigated. In ISA conditions the OAT at 10000 ft would be approximately -5O C, so the conditions in this question indicate a higher than ISA temperature and a higher than msl pressure altitude. Both of these factors would decrease air density causing the TAS at any given IAS to increase. This means that the actual TAS would be greater than 90 Kts. This in turn means that the actual ROC would be greater than the figure of 315 fprn calculated above. Of the options offered in this question, option b (360 fpm) and c(945 fpm) satisfy these criteria. But to obtain a ROC of 945 fprn with a gradient of 3.5%, the TAS would need to be approximately 270 Kts. This is approximately three times greater than the 90 Kts IAS given in the question. Even at 40000 ft ISA, TAS is only twice IAS, so a figure of three times IAS is not possible at 10000 ft. The most appropriate option is therefore option b, 360 fpm.

CLIMB 103. a. Although this question does not specify gliding flight, it should always be assumed that such questions refer to gliding, unless powered flight is explicitly stated. This is because the ability to vary thrust enables descent gradient to be controlled independently of IAS. Under these circumstance there could be no single correct answer to this question. In steady gliding flight the IAS is such that the component of aircraft weight acting down the gradient balances the drag force acting up the gradient. In this condition the aircraft is in equilibrium at a steady IAS. This condition will persist provided the lift and drag forces do not change. But lift and drag at any given IAS are proportional to angle of attack, so to maintain a steady glide IAS, it is necessary to maintain a steady angle of attack. In steady gliding flight the tangent of the descent gradient is equal to the drag force divided by the lift. But drag divided by lift is the inverse of the 1ift:drag ratio. This means that if a gliding aircraft is flown such that its L:D ratio is constant, then the tangent of iis descent grzdient, and hence the descent gradient will also be constant. L:D ratio is proportional to angle of attack so to maintain constant L:D ratio and descent gradient, it is necessary to maintain constant angle of attack. But lift and drag are both proportional to angle of attack and IAS, so if an aircraft descends at constant IAS and constant angle of attack, it will also maintain constant lift, drag and descent gradient. But this question specifies a constant Mach number, rather than a constant IAS. To understand the implications of this it is necessary to examine the effects of decreasing altitude on the ratio of mach number to TAS, and the ratio of TAS:IAS. Above FL360 (36000 ft) in the ISA, the temperature is constant. This means that when descending above FL360, the temperature will remain constant. But mach number is the ratio of TAS: local speed of sound (LSS) and LSS is proportional to temperature. So when descending from FL400 to FL360 at constant mach number, the TAS remains constant. Bellow FL360, temperature and LSS increase with decreasing altitude. So descending from FL360 to

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FL200 at constant mach number causes TAS to increase. TAS is the true airspeed, whilst IAS is proportional to TAS2 x density. As altitude decreases density increases, so the TAS at any given IAS decreases as altitude decreases. This means that for any given TAS, the IAS increases as altitude decreases. So when descending at constant mach number, the TAS increases because of increasing temperature and the IAS increases because of increasing density. The rate of increase being greatest below FL360. But increasing IAS causes both lift and drag to increase, so if angle of attack is unchanged, the aircraft will slow down due to increased drag, and the gradient will decrease due to increased lift. In order to maintain constantpmachnumber it will therefore be necessary to reduce angle of attack in order to reduce drag and hence increase IAS. This will involve pushing the nose of the aircraft further down, and hence will increase the descent gradient. The overall effect of these processes is that descent gradient will increase throughout all stages of a constant mach gliding descent.

CLIMB 104. c. The first part of this question is the same as question 103 in that it involves a constant mach number. As explained above this will cause gradient to increase. But below FL260, the aircraft in this question maintains constant CAS. As explained in the first three paragraphs of the answer to question 103, this will produce a constant gradient. The most appropriate option in this question is therefore c, increasing then constant gradient.

CLIMB 105. c. The situation

illustrated in the

Screen height 50 ft (15 m) . A.

Obstacle

From the diagram it can be seen that the Tan of the gradient = height gained 1 1000 m The gradient = 10% so height gained = Tan 10% x 10000 m. To calculate what angle represents a 10% gradient it must be noted that gradient = height gained divide by distance flown up the gradient. But this ratio is also the sin of the angle , so a 10% gradient represents an angle for which the sin is 10% (or 0.1) This gives an angle of 5.739 degrees. So the height gained = Tan 5.739 degrees x 10000 m, which is 1005 m. But the gradient starts at the screen height of 50 ft which is 15 m, so the height of the aircraft at the obstacle is 1005 + 15 m which is 1020 m. The height of the obstacle is 850 m so the aircraft clears it by 1020 - 850 = 170 m.

CLIMB 106. a. The term CLIMB schedule refers to the rnacner in which aircraft speed is controlled for optimum efficiency in a climb. The two limiting speeds are the Velocity Maximum Operating (V), and the Mach Number Maximum Operating (M).,


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VMo is the maximum operating speed of an aircraft, expressed as CAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at low altitudes where air density is high, VMo is the limiting parameter. MMo is the maximum operating speed of an aircraft, expressed as a mach number. If the aircraft is operated at higher mach numbers, the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. MMo is some specific fraction of the local speed of sound. But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the CAS equating to MMo deceases as altitude increases. The overall effect of these factors is that VMo is the limiting speed at low altitude and MMo is the limiting speed at high altitude. The altitude at which the TAS equating to VMo is equal to that equating to MMo is called the crossover altitude. So if an aircraft is at the correct CAS upon reaching its crossover altitude, it will automatically be at the correct mach number.

CLlMB 307. a. The term CLlMB schedule refers to the manner in which aircraft speed is controlled for optimum efficiency and the Mach Number in a climb. The two limiting speeds are the Velocity Maximum Operating (V), Maximum Operating (FAMo).VMois limiting the parameter at iow altitudes and MMois the limiting altitude at high altitudes. The crossover altitude is the altitude at which the CAS equating to VMo is equal to that equating to MMo. Below the crossover an aircraft will operate at constant CAS. Above the crossover altitude it will operate at constant mach number. Crossover altitude varies with mass but for a modern jet aircraft is typically in the region of 26000 ft. This question therefore refers to flight above approximately 26000 ft. The mach number of an aircraft in flight is the speed of the aircraft as a fraction of the local speed of sound. The speed of sound is proportional to absolute temperature, so as altitude increases, decreasing temperature causes the local speed of sound to decrease. This means that in a constant mach CLlMB the TAS will decrease. But the ratio of TAS:CAS increases with altitude, so constant mach number and decreasing TAS causes IAS to decrease with increasing altitude. In commercial operations aircraft are climbed at a constant power setting, so to maintain constant mach number, the necessary reductions in TAS and IAS are achieved by increasing the angle of attack to increase drag. This increase in angle of attack also increases lift. But in a steady CLlMB the lift force equals the weight multiplied by the cosine of the climb gradient. So increasing angle of attack to maintain constant mach number, also increases the climb gradient. The climb gradient will therefore increase in a constant mach climb above the crossover altitude.

CLlMB 108. d. The term CLlMB schedule refers to the manner in which aircraft speed is controlled for optimum efficiency and the Mach Number in a climb. The two limiting speeds are the Velocity Maximum Operating (V), Maximum Operating (MMo)VMo is limiting the parameter at low altitudes and MMois the limiting altitude at high altitudes. The crossover altitude is the altitude at which the CAS equating to VMo is equal to that equating to MMo. Below the crossover an aircraft will operate at constant CAS. Above the crossover altitude it will operate at constant mach number. Crossover altitude varies with mass but for a modern jet aircraft is typically in the region of 26000 ft. This question therefore refers to flight above approximately 26000 ft. The mach number of an aircraft in flight is the speed of the aircraft as a fraction of the local speed of sound. The speed of sound is proportional to. absolute temperature, so as altitude increases, decreasing

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temperature causes the local speed of sound to decrease. This means that in a constant mach CLIMB the TAS will decrease. But the ratio of TAS : CAS increases with altitude, so constant mach number and decreasing TAS causes IAS to decrease with increasing altitude. In commercial operations aircraft are climbed at a constant power setting, so to maintain constant mach number, the necessary reductions in TAS and IAS are achieved by increasing the angle of attack to increase drag. This increase in angle of attack also increases lift. Bui in a steady CLIMB the lift force equals the weight multiplied by the cosine oT the climb gradient. So increasing angle of attack to maintain constant mach number, also increases the climb gradient.

CLIMB 109. c. The important points to be noted in answering this question are that glide speed is directly proportional to mass, but glide range is independent of mass. The glide range of an aircraft is equal to its L:D ratio multiplied by its height at the commencement of the glide. In order tomaximise glide range from any given altitude, the aircraft must be flown at its best L:D ratio. This is achieved by flying at minimum drag speed (VIM,). The value of the best L:D ratio attainable is a function aerofoil section, surface condition and plan form. It is not affected by aircraft mass. VIM, is however a function of mass, such that decreasing mass decreases ,,, but not the L:D ratio. the value of VIM,. This also decreases the drag value at ,V In this question a glide is completed and the aircraft is then climbed back to the same altitude and the test repeated. This will involve the burning of fuel so the mass of the aircraft will be less during the second test. This will reduce glide speed but will not change the L:D ratio or glide range.

CLIMB 110. d. The important points to be noted in answering this question are that glide speed is directly proportional to mass, but glide range is independent of mass. The glide range of an aircraft is equal to its L:D ratio multiplied by its height at the commencement of the glide. In order to maximise glide range from any given altitude, the aircraft must be flown at its best L:D ratio. his is achieved by flying at mir~imumdrag speed (VIMD). The value of the best L:D ratio attainable is a function aerofoil section, surface condition and plan form. It is not affected by aircraft mass. VIM, is however a function of mass, such that decreasing mass decreases the value of VIM,. This also decreases the drag vaiue at VIM,, but not the L:D ratio. The overall effect of this is that as mass decreases the aircraft must fly down the same gradient at a lower speed in order to achieve the same L:D ratio and glide range. In this question a glide is completed and the aircraft is then climbed back to the same altitude and the test repeated. This will involve the burning of fuel so the mass of the aircraft will be less during the second test. This will reduce glide speed but will not change the glide range or gradient.

CLIMB 111. d. The important points to be noted in answering this question are that glide speed is directly proportional to mass, but glide range is independent of mass. The glide range of an aircraft is equal to its L:D ratio multiplied by its height at the commencement of the glide. In order to maximise glide range from any given altitude, the aircraft must be flown at its best L:D ratio. This is achieved by flying at minimum drag speed (VIM,). The value of the best L:D ratio attainable is a function of aerofoil section, surface condition and plan form. It is not affected by aircraft mass. V ,,, is however a function of mass, such that decreasing mass decreases


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the value of V .,, This also decreases the drag value at V ,, but not the L:D ratio. The overall effect of this is that as mass decreases the aircraft must fly down the same gradient at a lower speed in order to achieve the same L:D ratio and glide range. Also the reduced glide speed down the same gradient, will decrease the rate of descent, and hence increase glide endurance. In this question a glide is completed and the aircraft is then climbed back to the same altitude and the test repeated. This will involve the burning of fuel, so the mass of the aircraft will be less during the second test. It will therefore need to fly slower to achieve the same glide angle and glide range, but will have'a lower ROD than in the first test.

CLIMB 112. d. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are cbse to the airport. In both types, the first segment is an initial CLIMB at V, plus 10 to 20 Kts, using take-off flap setting. The power setting is at take-off for type B and climb power for type A procedures. In the type A procedure this climb continues to 1500 ft, whereas in the type B procedure it is terminated at 1000 ft above the airport.

CLIMB 113. d. The wording of this question is typical of many used in JAR examinations, in that it requires a number of assumptions to be made. In this case, no mention is made concerning the initial condition of the aircraft. This is significant in that if it is initially at its optimum glide attitude, it will be flying at V ,, and will be achieving its best possible glide range, but not its best possible best glide endurance. Any change in attitude will decrease L:D ratio and glide range, but might iricrease or decrease glide endurance. This is because best glide range occurs at the minimum drag speed (V,,), whereas best glide endurance occurs at the minimum power required speed (V,,,). For all aircraft types V,, is lower than V ,, so any ,,, towards VIM, will decrease range but increase endurance. If action that decreases glide speed from V ,, when its nose is brought up to reduce its nose down attitude, both lift and drag an aircraft-is gliding at V will increase. This will cause speed to decrease from V ,, towards V ,, causing glide range to decrease and endurance to increase. It should be noted that the assumption of initial flight at V ,,, is essential in that if the aircraft had been flying at a higher speed, the effects would have been different. Reducing nose down attitude would take ,, and VIMP, thereby increasing both glide range and endurance. The overall the speed closer to both V effect would then be an increase in both range and endurance (option a).

CLIMB 114. d. Best glide range occurs at the minimum drag speed ( V,). Any factor that increases speed above or decreases speed below VIM, will decrease glide range. If an aircraft is flying at VIM, in a tailwind, it will be achieving maximum range through the air, whilst the air is moving in the same direction. This means that the glide range over the ground will be increased. Option a is therefore incorrect. Provided an aircraft is flown at V ,, it will be at its maximum L:D ratio and achieve its maximum still air glide range. If its weight is increased or decreased the value of VIM, will increase or decrease ,, it will achieve the same L:D ratio and glide accordingly, but provided the aircraft flies at its new V range. Changes in weight do not therefore affect glide range so options b and c are incorrect. If however the landing gear is deployed this will increase the drag and decrease the L:D ratio and glide range at all speeds.

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CLIMB 115. a. Best glide range occurs at the minimum drag speed ( VIM,). Any factor that increases speed above or ,,, in a tailwind, it will be decreases speed below VIM, will decrease glide range. If an aircraft is flying at V achieving maximum range through the air, whilst the air is moving in the same direction. This means that the glide range over the ground will be increased. Option a is therefore correct. Provided an aircraft is flown at ,V ,,, it will be at its maximum L:D ratio and achieve its maximum still air glide range. If its weight is increased or decreased the value of VIM, will increase or decrease accordingly, ,, it will achieve the same L:D ratio and glide range. Changes but provided the aircraft flies at its new V in weight do not therefore affect glide range so options b and c are incorrect. If the flaps are lowered this will increase both lift and drag, but will decrease the L:D ratio and glide range. Option a is therefore the only one that will increase glide range.

CLIMB 116. b. If it is assumed that 1 Kt = 100 fpm, then the CLIMB gradient in still air can be calculated using the following equation: % gradient = ROC in fpm / (TAS in Kts x 100 ipmlKt) x 100%

Using this equation would give % gradient = 2501150 = 1.66%. Which is approximately 1.67 %. A more accurate estimate is 1 Kt = 101.3 fpm. Using this figure would give: % gradient = ROC in fpm / (TAS in Kts x 101.3 fpm/Kt) x 100% = 1.65 %.

If no conversion factor for Kts - fpm is given, the calculation shouid be made using both 100 fpmlKt and 101.3 fpm/Kt, in order to find the best option.

CLIMB 117. a. The best rate of CLIMB that an aircraft can achieve is equal to its excess power divided by its mass. This means that if mass decreases while excess power remains constant or increases, then maximum rate of climb will increase. The speed at which the best rate of climb is achieved is termed V., This is the speed at which excess power (power available - power required) is maximum. Decreasingaircraft mass decreases both the power requited at any speed and the speed at which excess power is greatest. So decreasing mass decreases the speed for best rate of climb and increases the best rate of climb. As a flight progresses, fuel will be burned, thereby causing mass to decrease. This will increase rate of climb and decrease the speed for best rate of climb.

CLIMB 118. a. The troposphere is that part of the international standard atmosphere (ISA) below a pressure altitude of 36000 ft. As altitude increases throughout the troposphere, both air density and pressure decrease. So this question relates to flight in which both air density and temperature are decreasing as altitude increases. Calibrated Airspeed (CAS) is proportional to dynamic pressure ('/2pV2)where p is the air density and V is the True Airspeed (TAS), such that the CAS at any given dynamic pressure is constant at all altitudes. This means that as altitude increases at any given CAS, the decreasing density must be matched by an increasing TAS, such that '/2pV2 remains constant. Conversely when climbing at constant TAS the value of dynamic pressure and hence CAS must decrease.


But as altitude increases in the troposphere, air temperature also decreases. The local speed of sound is proportional to temperature, so as altitude increases, the decreasing temperature causes LSS to decrease. Mach number is the ratio of TAS:LSS . So as altitude increases, decreasing LSS causes the mach number at any given TAS to increase Altand the TAS at any given mach number to decrezse. This means that as altitude increases, an aircraft climbing at constant mach number, will experience decreases in both TAS and CAS. The overall effect of these changes is illustrated in the diagram at the right.

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t

Alt

Speed -+

CLIMB 119. d. The best rate of CLIMB that an aircraft can achieve is equal to its excess power divided by its weight. This means that if weight increases while excess power remains constant or decreases, then maximum rate of ., This is the speed climb will decrease. The speed at which the best rate of climb is achieved is termed V at which excess power (power available- power required) is maximum. Increasing aircraft weight increases both the power required at any speed and the speed at which excess power is greatest. So increasing weight decreases the best rate of climb and increases the speed for best rate of climb

CLIMB 120. a. This question refers to gliding descent at constant mach number. Its solution therefore requires an understanding of how such a descent would affect TAS and CAS, and how these would affect glide pitch attitude and gradient. As altitude decreases below 36000 ft ISA, both temperature and air densiiy increase. The local speed of sound (LSS) is prcportional to temperature, so as altitude decreases, the increasing temperature causes LSS to increase. Mach number is the ratio of TAS : LSS . So as altitude decreases in a constant mach glide, the increasing LSS causes the mach number at any given TAS to decrease. This means that the TAS at any given mach number increases, so in a constant mach glide the TAS increases as aititude decreases. Calibrated Airspeed (CAS) is proportional to dynamic pressure (%pV2)where p is the air density and V is the True Airspeed (TAS), such that the CAS at any given dynamic pressure is constant at all altitudes. This means that as altitude decreases at any given CAS, the increasing density must be matched by an decreasing TAS, such that XpV2 remains constant. Conversely when descending at constant TAS the value of dynamic pressure and hence CAS must increase. But in a constant mach glide the TAS is increasing so CAS must increase at an even greater rate. Both lift and drag are proportional to angle of attack and CAS, so if angle of attack remains constant in a constant mach glide, the increasing CAS will cause both lift and drag to increase. The increasing drag will cause CAS to decrease, which in turn will decrease mach number. This means that in order to maintain constant mach number it will be necessary to reduce angle of attack in order to reduce drag, so that both TAS and CAS increase. This reduction in angle of attack is achieved by increasing nose down pitch attitude. But the reduced angle of attack will also reduce lift. In a steady descent lift is equal to weight multiplied by the cosine of the gradient, so as lift decreases the gradient must increase. The overall effect of these factors is that in a constant mach glide both nose down pitch angle and gradient will increase.

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CI-IMB 121. a. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whi!st type B is for use when noise sensitive areas are close to the airport. In both types, the first segment is an initial CLlMB from screen height. It is conducted with gear up, at a speed of V, plus 10 to 20 Kts, using take-off flap setting. The power setting is at take-off for type E and climb power for type A procedures. In the type A procedure this climb continges to 1500 ft, whereas in the type B procedure it is terminated at 1000 ft above the airport.

CLIMB 122. a. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are close to the airport. In both types, the first segment is an initial CLlMB from screen height. It is conducted with gear up, at a speed of V, plus 10 to 20 Kts, using take-off flap setting. The power setting is at take-off for type B and climb power for type A procedures. In the type A procedure this climb continues to 1500 ft, whereas in the type B procedure it is terminated at 1000 ft above the airport.

CLlMB 123.d. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are close to the airport.

CLlMB 124. c. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are close to the airport.

CLIMB 125. a. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are close to the airport. In both types, the procedure comprises of three segments. In both the type a and B procedures the first segment commences at screen height and the third segment ends at cruising altitude.

CLlMB 126. c. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are close to the airport. In the type B procedure the first segment is an initial CLlMB from screen height to 1000 ft above airfield level. It is conducted with the gear up, at a speed of V, plus 10 to 20 Kts, using takeoff thrust and flap setting. The second segment commences at 1000 ft, from where the aircraft is accelerated to the minimum safe manoeuvring speed with zero flap (VzF).It is then climbed to 3000 ft whilst accelerating to VzF + 10 Kts. The third segment then commences at 3000 ft.


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CLIMB 127. c. The noise abatement procedures are intended to reduce noise nuisance to people living and working in the area surrounding airports. Two types are commonly employed. Type A is used for take-off where the noise sensitive area is some distance from the airport, whilst type B is for use when noise sensitive areas are close to the airport. In the type A procedure the first segment is an initial CLlMB from screen height to 1500 ft above airfield level. It is conducted with the gear up, at a speed of V, plus 10 to 20 Kts, using clinib thrust and flap setting. The second segment commences at 1500 ft, from where the aircraft is accelerated to the minimum safe manoeuvring speed with zero flap (V,), at climb power. It is then climbed to 3000 ft at the same speed and power setting. The third segment then commences at 3000 ft.

CLIMB 128. c. The problem can be solved using the following equation:

% gradient = ((Thrust - Drag)/Weight) x 100% T W ratio is 1:3 which means that T = 113 W L:D ratio = 20:l which means that D = 1/20 L But the questions specifies that L = W, so if D = 1/20 L, then D = 1/20 W Substituting these values in the original equation gives: %gradient = ((113 W - 1/20W)/ W ) x 100% This means that % gradient = ( (20160 W - 3/60 W) / W) x 100% Which simplifies to give % gradient = 17/60 x 100 % which is 28.33%. It should be noted that in a steady CLIMB or descent the lift is equal to the weight multiplied by the cosine of the gradient. This is less than the weight. At very small angles however the cosine of the angle is almost equal to the angle so the assumption that lift is equal to weight is often used in this type of question to simplify matters.

CLIMB 129. a. Indicated Airspeed (IAS) is proportional to dynamic pressure (WpV2)where p is the air density and V is the True Airspeed (TAS), such that the dynamic pressure at any given IAS is constant at all altitudes. So if an aircraft climbs at constant IAS it is climbing at constant WpV2. Both lift and drag are proportional to angle of attack and WpV2 , so if an aircraft climbs at constant IAS (and hence constant WpV2 ) if its angle of attack remains constant, its lift and drag would also remain constant. But in a steady CLlMB lift is equal to the cosine of the gradient, so climbing at constant IAS, angle of attack and lift, would also result in a constant climb gradient. Power required is equal to drag multiplied by TAS, and as altitude increases the TAS at any given IAS increases. This means that the power required to maintain constant IAS in a CLlMB increases with altitude. But as altitude increases the power available from all types of engine decreases. In order to maintain a constant IAS climb, the angle of attack must be gradually reduced, to reduce drag such that power required decreases to match the decreasing power available. But decreasing angle of attack also decreases lift, and lift is equal to weight multiplied by the cosine of the climb gradient. so as the angle of attack is reduced to maintain constant IAS, the reduction In lift also causes the gradient to decrease. So in a constant IAS climb both drag and climb gradient decrease. It should be noted that if continued to an altitude at which TAS approached,,M ,,, wave drag would decrease gradient at a greater rate.

the sudden increase in

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CLlMB 130. b. The problem can be solved using the following equation:

% gradient = ((Thrust -Drag)/Weight) x 100% T W ratio is 1.5 which means that T = 115 W L:D ratio = 25:l which means that D = 1/25 L But the questions specifies that L = W, so if D = 1/25 L, then D = 1/25 W Substituting these values in the original equation gives:

% gradient = ((115 W - 1/25 W) / W ) x 100°h This means that % gradient = ( (5125 W - 1/25 W) / W) x 100%

-

Which simplifies to give % gradient = 4/25 x 100 % which is 16% It should be noted that in a steady CLlMB or descent the lift is equal to the weight multiplied by the cosine of the gradient. This is less than the weight. At very small angles however the cosine of the angle is almost equal to the angle so the assumption that lift is equal to weight is often used in this type of question to simplify matters.

CLlMB 131. b. The maximum speed at which an aircraft may be routinely flown is limited by the ability of the structure to support the resulting aerodynamic loads and by the compressibility effects encountered when flying close to the local speed of sound. At low altitudes the calibrated airspeed producing limiting aerodynamic loads is the limiting value. This is called VMo. The mach number at which compressibility effects become the limiting value is termed MMo.As altitude is increased, the local speed of sound decreases. At the crossover , At higher altitudes VMo is greater than altitude, the speed equating to MMois equal to that equating to.,V MMoand at lower altitudes MMo is greater than.,V , When descending at constant mach number close to MMo,there is a danger that VMowill be exceeded. CLlMB 132. a. When climbing at constant IAS, both TAS and Mach number increase with increasing altitude. Also because the local speed of sound decreases with increasing temperature, the IAS and TAS equating to MMo will decrease with increasing altitude up to 36000 feet. If a constant IAS CLlMB is continued to a sufficiently high altitude ,M , will be exceeded.

CLlMB 133. a. The mach limit, McRITis the ratio of TAS:LSS at which the airflow at some point on the aircraft first reaches the local speed of sound. Because the local speed of sound decreases with temperature and the ratio of TAS:IAS increases with altitude, the IAS equating to McRITdecreases with increasing altitude. Climbing at constant IAS will therefore cause TAS to increase until at some altitude IAS equals McRIT.Increasing IAS in the CLIMB increases the rate at which TAS increases, thereby decreasing the altitude at which IAS equals MCRIT.

CLlMB 134. d. In descending flight in still air the TAS of an aircraft can be considered to be made up of two components. The vertical component of TAS is the rate of descent and the horizontal component is the ground speed. A headwind will reduce the ground speed component but will not affect the rate of descent component.


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The duration of the descent will therefore be unchanged, but because of the lower ground speed, the aircraft will cover less ground in any given descent.

CLIMB 135. d. In descending flight in still air the TAS of an aircraft can be considered to be made up of two components. The vertical corr~ponentof TAS is the rate of descent and the horizontal component is the ground speed. A headwind will reduce the ground speed component but will not affect the rate of descent component. The duration of the descent will therefore be unchanged, but because of the lower ground speed, the aircraft will cover less ground in any given descent. The combined effect of reduced ground speed and unaffected rate of descent, is to increase the descent gradient. This is illustrated below. Gradient without headwind Increased gradient with headwind

Rate of descent

Reduced ground speed with headwind Ground speed without headwind

CLIMB 136. a. , ,V , is the minimum calibrated airspeed at which it is possible to maintain control of an aircraft in flight, in the landing configuration, following the failure of a critical engine. It is the speed at which the control forces provided by full deflection of the flying controls is just sufficient to overcome the yawing and rolling moments caused by the asymmetric power effects of the engine failure. For any given control deflection the magnitude of the control forces generated depends upon airspeed, and air density. The magnitude of the moments generated by asymmetric power depend upon the thrust produced by the engine, which in turn depends upon air density and power setting. Because none of the above factors very with aircraft weight, variations in weight do not affect V,,-,.

CLIMB 137. b. As altitude decreases below 36000 feet, air temperature increases. The local speed of sound is proportional to the square of the absolute air temperature, so this also increases with decreasing altitude. When descending at constant mach number, the aircraft speed as a proportion of the local speed of sound will remain constant. Because the local speed of sound is increasing, the airspeed (IAS and TAS) will increase at constant mach number. Assuming idle power is used, this increase in airspeed can be achieved only by decreasing angle of attack to reduce the drag force acting on the aircraft. Angle of attack must therefore be reduced in order to carry out a constant mach gliding descent. It should be noted that if power were increased during the descent the angle of attack could be maintained at a constant setting. In questions of this form, it should be assumed that the aircraft is in gliding (zero thrust) descent unless otherwise stated.

CLIMB 138. c. When an aircraft is in a steady descent, the lift force is equal to the weight of the aircraft multiplied by the cosine of its angle of descent. In order to maintain a constant angle of descent it is therefore necessary to maintain a constant lift force. But lift equals C, WpV2S and IAS is constant at all altitudes for any given value of WpV2. This means that to maintain constant IAS an aircraft must maintain constant %pV2 . Assuming that the wing area (S) does not vary, the maintenance of constant lift at constant IAS requires constant C,. This in turn requires a constant angle of attack.

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CLlMB 139. b. When an aircraft is in a steady descent, the lift force is equal to the weight of the aircraft multiplied by the cosine of its angle of descent. In order to maintain a constant angle of descent it is therefore necessary to maintain a constant lift force. Lift equals C, %pV2Sand for any given value of %pV2the IAS is constant at all altitudes. But as altitude decreases the TAS : IAS ratio decreases such that a constant TAS descent results in a continuous increase in IAS and %pV2. This increase in IAS would increase drag, thereby causing the aircraft speed to decrease. This means that to maintain constant TAS, without varying thrust, the drag must be reduced by decreasing angle of attack. It should be noted that angle of attack would be reduced by pushing the nose down. This constitutes an increase in pitch attitude. So an increase in (nose down) pitch to decrease arrgle of attack is required to maintain constant TAS in a constant thrust descent.

CLlMB 140. a. The mach number of an aircraft is a measure of its true airspeed (TAS) as a fraction of the local speed of sound. A constant mach CLlMB therefore means that the aircraft's TAS as a proportion of the local speed of sound remains constant. But the local speed of sound is proportional to temperature, which remains constant above the tropopause. This means that a constant mach climb above the tropopause, causes TAS to remain constant. IAS is proportional to %pV2 such that any given value of %pV2will produce the same value of IAS at all altitudes. But as altitude increases r decreases, so ta maintain a constant value of IAS, the value of V, which is TAS, must increase. This means that the TAS equating to any given IAS increases with increasing altitude. Climbing at constant mach number above the tropopause therefore causes IAS to decrease and TAS to remain constant.

CLlMB 144. b. The troposphere is that part of the atmosphere belaw the tropopause at 36000 feet. As altitude increases in the troposphere, temperature falls at a rate of approximately 2°C / 1000 ft. The local speed of sound is proportional to temperature so as altitude increases in the troposphere! the iocal speed of sound decreases. The mach number of an aircraft represents its speed as a fraction of the local speed of sound. The TAS equating to any given mach number therefore decreases with increasing altitude.

CLlMB 142. b. The troposphere is that part of the atmosphere below the tropopause at 36000 feet. As altitude increases in the troposphere, temperature falls at a rate of approximately 2°C / 1000 ft. The local speed of sound is proportional to temperature so as altitude increases in the troposphere, the local speed of sound decreases. Above the tropopause temperature and the local speed of sound remain constant. The mach number of an aircraft represents its speed as a fraction of the local speed of sound. So a constant mach number CLlMB means climbing at a constant fraction of the local speed of sound. Climbing at constant mach number below the tropopause causes TAS to decrease to match the decreasing speed of sound, whilst above the tropopause it causes TAS to remain constant to match the constant speed of cound. The overall effect of a climb is that TAS decreases up to 360(10 feet then remains constant. Decreasing Constant TAS

CLlMB 143. b. IAS is proportional to %pV2,where r is Decreasing Constant air density and V is TAS. This means that for any given value of %pV2,the IAS will be constant at all altitudes. But air density decreases with increasing altitude. So for any given IAS, the TAS must increase to maintain a constant value of KpV2. This means that as altitude increases the IAS equating to any given TAS decreases. in a cGnstant TAS climb, IAS will

ALT


287

therefore decrease. This effect is illustrated in the diagram at the right. It should be noted that this effect continues above the troposphere.

CLIMB 144. a. IAS is proportional to %pV2, where r is air density and V is TAS. This means that for any given value of %pV2, the IAS will be constant at all altitudes. But air density decreases with increasing altitude so for any given IAS, the TAS must increase to maintain a constant value of %pV2. This means that as altitude increases the IAS equating to any given TAS decreases. In a constant IAS climb, TAS will therefore decrease. This effect continues even above the tropopause because the decrease in air density is caused by decreasing pressure. This effect is illustrated in the diagram at the right. It should be noted that this effect continues above the troposphere.

ALT

Increasing speed

r

CLIMB 145. d. IAS is proportional to MpV2, where r is air density and V is TAS. This means that for any given value of %pV2,the IAS will be constant at all altitudes. But air density decreases with increasing altitude so for any given IAS, the TAS must increase to maintain a constant value of MpV2. This means that as altitude increases the IAS equating to any given TAS decreases. In a constant TAS climb, IAS will therefore decrease. Below the tropopause the reduction in air density caused by falling air pressure is partly offset by the reduction in air temperature. Above the tropopause however temperature is constant so the density decreases at a greater rate. This increased rate of density reduction increases the rate of IAS decrease. The overall effect of a constant TAS CLIMB is therefore reducing IAS up to the tropopause, above which IAS reduces at a greater rate. Constant IAS Increas~ng TAS

CLIMB 146. b. In addition to the changing ratio between IAS and TAS, increasing altitude in the troposphere also causes the local speed Mach of sound to decrease. This means that when climbing at constant IAS, both TAS and mach number increase This effect is illustrated in the diagram at the right.

ALT

Increasing values -b

CLIMB 147. c. The maximum operating speed of a modern sub-sonic jet aircraft is limited by the effects of air loads due to dynamic pressures acting on the structure, and the buffeting and high speed stall that occur at , and .M ,, V, is the maximum allowable transonic speeds. These limiting conditions are termed,V CAS. ,M , is the limiting mach number. At any given altitude the limiting condition is the lower of VM0 and MMo. ,M , is the limiting factor at high altitude where the speed of sound is lowest. ,V , is the limiting factor at low altitude where air density and hence dynamic pressure are highest. If a descent is , there is a danger that ,V , will be exceeded as conducted at a constant mach number close to ,M altitude decreases.

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CLlMB 148. a. The CLlMB gradient is the ratio of height gained to distance travelled over the ground. It is usua!ly expressed as a percentage. In calculating climb gradient it is convenient to use the rate of c!imb (ROC) as a measure of height gained , and TAS as a measure of distance travelled. This method is reasonably accurate in still air for gradients up to 15%. To provided an accurate calculation of gradient, both ROC and TAS must be in the same units, such as Wmin. An approximate value of gradient is often used based on the assumption that 1 Kts equals 100 Wmin, and the following equation: Gradient = (ROC 1 TAS) x 100%

CLlMB 149. d. Both IAS and drag are proportional to % rV2 SO drag at any given IAS is constant at all altitudes. V, is the speed at which excess thrust is maximum and excess thrust is thrust available minus drag. Although thrust available decreases with increasing altitude, the relationship between thrust and IAS does not vary. The IAS value of V, is therefore constant at all altitudes. Vy is the speed at which excess power is maximum. Excess power is power available minus power required and power required is drag multiplied by TAS. But the TAS equating to any given IAS increases with altitude, causing the power required at any given IAS to increase. This causes the IAS value Vy to decrease with increasing altitude, until they are equal at the absolute ceiling.

CLlMB 150. c. At the absolute ceiling climbing is not possible and there is only one speed at which an aircraft can maintain straight and level flight. For both jet and propeller aircraft this speed is Vx. Absolute Also for both aircraft types, the CAS Ceiling value of V, and VMD remain constant at all altitudes. For propeller aircraft V, is less than VMDand because the Aliitude two remain constant with changes in atiitude, they do not converge at the absolute ceiling. But because V, is the only practicable flight speed at the absolute , VMa, and Vy all converge on V, at this ceiling,,,V altitude whilst VMDdoes not. This is illustrated in the diagram at the right.

I

CLlMB 151. c. At the absolute ceiling the only speed at which a prope!ler aircraft can maintain straight and level flight is V., At lower altitudes V, is the speed at which excess thrust is at a maximum value, thereby facilitating the greatest CLlMB gradient. As altitude increases, reducing air density decreases the power produced by the engine and the thrust produced by the prope!ler. This causes excess thrust and maximum climb gradient to decrease with increasing altitude. At the absolute ceiling excess thrust is zero, so no climb gradient is possible. But increasing altitude does not affect the CAS at which maximum excess thrust occurs, so V, remains constant at all altitudes. VMDis the speed at which the drag force is at its minimum value. Because both drag and CAS are proportional to dynamic pressure (WpV2), the CAS value of V,4, remains constant at ali altitudes. So for a propeller aircraft both V, and VMDremain constant as it climbs to its absolute ceiling.


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CLlMB 152. d. CLlMB gradient is the ratio of height gained divided by distance flown over the ground. It is normally expressed as a % gradient, where 100% represents a vertical climb and 0% represents level flight. For small gradient (up to 15%) in still air, a reasonably accurate estimate can be made using the: %Gradient = ROCflAS x 100%. But at any given TAS, the groundspeed is decreased in a headwind and decreased in a tailwind. A more accurate equation is therefore: % gradient = ROCIgroundspeed x 100%

CLIMB 153. a. The forces acting on an aircraft in a steady CLIME are illustrated in the upper diagram Ate the right. From this it can be seen that forces acting up the slope are (T-D) and the force acting vertically downwards is the weight of the aircraft. So the sine of the CLlMB angle (a) is (T-D)M. The velocities of an aircraft in a steady CLlMB are illustrated in the lower diagram at the right. From this it can be seen that the CLlMB TAS gradient is equal to the ROCIGroundspeed. But this is also the sine of the angle of climb.

TAS

So gradient = ROCIGroundspeed = (T-D)/W.

/

Groundspeed

I

GI-IMB 154. b. The maximum operating speed of an aircraft is specified as VMo in terms of CAS and MMo in terms of mach number. As altitude increases, the TAS equating to VMo increases whilst that equating to MMo decreases. At low altitudes the TAS equating to VMo is the limiting factor and at high altitude MMo is the limiting factor. In order to avoid exceeding these limits aircraft are climbed at constant CAS at low altitude and at constant Mach number at high altitude. .......................................................

The crossover altitude is the altitude at which the selected CAS is equal to the selected mach number. If a higher CAS is selected for the CLlMB schedule, the crossover altitude will be reduced as illustrated in Crossover at 27510.81 is the diagram at the right. than at 30010.81

.............................

TAS at 275 Kts CAS

TAS at 300 Kts CAS

CLlMB 155. c. The maximum operating speed of an aircraft is specified as VMo in terms of CAS and MMo in terms of mach number. As altitude increases, the TAS equating to VMo increases whilst that equating to MMo decreases. At low altitudes VMois the limiting factor and at high altitude MM0is the limiting factor. In order

290


to avoid exceeding these limits aircraft are climbed at constant CAS at low altitude and at constant mach , is equal number at high altitudes. The crossover altitude is the altitude at which the TAS equating to ,V to that at MMo. So CLIMB below the crossover is at constant CAS and above crossover it is at constant rnach. For the vast majority of aircraft the crossover altitude is below the tropopause at 36000 ft. So when climbing above the tropopause aircraft fly at constant mach number. Mach number is equal to TASILSS. The local speed of sound (LSS) is proportional to temperature, such that decreasing temperature decreases the LSS. Above the tropopause in the ISA, temperature and LSS remain constant up to 58000 h. So climbing at constant mach in this area means climbing at constant TAS. Few if any aircraft operate above 58000 ft so this area can be ignored in answering this question.

CLIMB 156. a. The maximum operating speed of an aircraft is specified as ,V , in terms of CAS and ,M , in terms of mach number. As altitude increases, the TAS equating to ,V , increases whilst that equating to ,M , decreases. At low altitudes,V , is the limiting factor and at high altitude ,M , is the limiting factor. In order to avoid exceeding these limits aircraft are climbed at constant CAS at low altitude and at constant mach , is equal number at high altitudes. The crossover altitude is the altitude at which the TAS equating to ,V to that at .M ,, Mach number is equal to TASILSS. The local speed of sound (LSS) is proportional to temperature such that increasing temperature at any altitude increases the LSS. So increasing temperature at any altitude decreases the mach number at any given TAS. So in higher temperatures an aircraft must CLIMB to a greater altitude in order to reach appoint where ,V , is equal to .M ,, This means that increasing temperature increases the crossover altitude. This effect is illustrated in the diagram overleaf. TAS at mach 0.81 at high temp

TAS at mach 0.81 at low temperature Increased crossover altitude at high temperature Crossover altitude at low temperature

Altitude

1 / TAS at 280 Kts CAS

CLIMB 157. a. To achieve maximum glide endurance it is necessary to fly at the speed producing the minimum rate of descent. When the engines of an aircraft fail, the aircraft possesses a store of kinetic energy due to its velocity, and potential energy due to it height. Throughout the subsequent glide this energy is consumed in doing work, pushing the aircraft forward against the drag force. Because this energy store cannot be replenished, the maximum glide endurance will be achieved by flying at the speed requiring the lowest energy consumption rate. But energy consumption rate is power required, so maximum glide endurance is achieved by flying at the minimum power speed. Power required is equal to drag multiplied by TAS and although VIM, produces the

Climbing and Descendi~g

291

lowest drag force, it does not produce the lowest power required. This occurs at VIM, which is lower than ,,, for all aircraft. For propeller aircraft Vx is lower than VIM, and V, is higher than V., For the minimum V ,,, and V, is higher than V .,, So Vx and V, do not provide minimum rate of decent in a jet aircraft Vx is V a glide for any aircraft type.

CLIMB 158. a. Indicated airspeed (IAS) are proportional to dynamic pressure (%pV2) where: p is air density, and V is TAS. So climbing at constant IAS means climbing at constant %pV2. But as altitude increases, density decreases so the TAS must increase to balance the falling density, such that MpV2 remains constant. So as altitude increases, the TAS at any given IAS increases. But lift is also proportional to dynamic pressure such that low speed stall occurs at the same dynamic pressure and the same IAS at all altitudes. So stalling TAS increases as altitude increases. Mach number is the TAS divided by the local speed of sound (LSS). But LSS is proportional to temperature and temperature decreases as altitude increases. So as altitude increases the LSS decreases, causing the mach number at any given TAS to increase. The overall effect of climbing is therefore an increase in low speed stalling TAS and an increase in mach number at any given TAS. This means that the low speed stall mach number increases in a climb.

CLIMB 159. a. The crossover altitude is the altitude at which a climbing aircraft changes from constant CAS to constant mach number. The crossover is typically below the tropopause. so as altitude increases, temperature and LSS decrease. But mach number is TASILSS, so as LSS decreases, TAS must also decrease in order to maintain constant mach number. Decreasing TAS in turn means that CAS decreases. But lift and drag are proportional to CAS, so as CAS decreases, decreasing drag would cause the aircraft to accelerate to a higher mach number. In order to maintain constant mach number the angle of attack must be increased. But this increased angle of attack also increases lift, which causes the CLIMB gradient to increase. So climbing above the crossover altitude causes climb gradient to increase.

CLIMB 160. c. The maximum operating speed of an aircraft is specified as ,V , in terms of CAS and ,M , in terms of , increases whilst that equating to ,M , mach number. As altitude increases, the TAS equating to ,V decreases. At low altitudes,V , is the limiting factor and at high altitude ,M , is the limiting factor. In order to avoid exceeding these limits aircraft are climbed at constant CAS at low altitude and at constant mach number at high altitudes. The crossover altitude is the altitude at which the TAS equating to V, is equal ,, So CLIMB below the crossover is at constant CAS and above crossover it is at constant to that at .M mach.

CLIMB 161. c. The maximum operating speed of an aircraft is specified as ,V , in terms of CAS and ,M , in terms of mach number. As altitude increases, the TAS equating to ,V , increases whilst that equating to ,M , decreases. At low altitudes ,V , is the limiting factor and at high altitude ,M , is the limiting factor. The , is equal to that at .M ,, In order to avoid crossover altitude is the altitude at which the TAS equaiing to,V , at low altitudes aircraft are climbed at constant CAS. In order to avoid exceeding ,M , at exceeding ,V high altitude they are dimbed at constant mach number.

POWER REQUIRED AND POWER AVAILABLE

POWER 1. c. Power required = Drag x TAS and Drag = CD%pV2S,where V = TAS So power required = C,%%pV2S x V This means that power required is proportional to V x V2 which is V3 So power required is proportional to V3 In accelerating from 300 Kts to 400 Kts TAS is increased to 133% of its previous value. So power required is increased to (1.33)3 times its previous value. So power required is 2.35 times its previous value which is an increase of 135%

POWER 2. c. At VMDthe value of total drag is at a minimum, so as the aircraft accelerates drag will increase. Power required = Drag x TAS and Drag = C,%pV2S where V = TAS So power required = CD%p(TAS)2Sx TAS This means that power required is proportional to TAS x (TAS)2which is (TAS)3 So power required increases in proportion to (TAS)3

POWER 3. b. Maximum endurance is achieved by flying at the speed that requires the minimum fuel consumption rate. Fuel flow in a jet is proportional to thrust, which in straight and level flight is equal to drag. So maximum endurance in a jet occurs at the minimum drag speed (V,). Although,V , varies with aircraft weight, load factor, configuration and altitude, it is typically in the order of 132% of the stalling speed corresponding to the existing state of these variables. So maximum endurance in a jet is achieved by flying at approximately 132% of stalling speed.

POWER 4. a. Lift = CcApV2S. C, is proportional to angle of attack, so if angle of attack remains constant, so will C,. As altitude increases, air density (p) decreases so to maintain the same lift at the same angle of attack V must increase. The aircraft must therefore fly at a higher TAS to create the same lift at the same angle of attack in the less dense atmosphere. Power required is TAS x Drag. Drag = CD%pV2S,so increasing TAS will increase both power required and drag. The aircraft must therefore use more power and fly at a higher TAS .

POWER 5. b. Lift = C
.Power required and Power available

293

increase. Power required is TAS x Drag. Drag = CD%pV2S,so increasing TAS will increase both drag and power required. The aircraft must therefore use more power and fly at a higher TAS.

POWER 6. a. For any given airspeed, a headwind will reduce ground speed and hence range. Maximum range will therefore be achieved by reducing the time during which the wind is affecting the aircraft. This is done by increasing airspeed.

POWER 7. b. Flap deployment increases both lift and drag, but in most cases causes a reduction in the lift : drag ratio. This means that for any given value of lift, deploying flaps incurs a greater drag force. Drag is therefore lowest with flaps retracted. Power required = thrust x TAS. In straight and level flight drag = thrust, so power = drag x TAS. At high altitude the TAS equating to any given IAS increases, so for a given drag force, the power required is lowest at low altitude. The lowest value of power required at any given IAS is therefore achieved at zero flap and low altitude.

POWER 8. a. Maximum range in 2ny aircraft type is achieved when the ratio between distance covered and fuel flow is maximum. That is to say, maximum range requires minimum fuel flow, for a given distance covered. Fuel flow in a propeller aircraft is proportional to power, so maximum range will be achieved when the power required for a given distance covered is at a minimum. Power required in straight and level flight = drag x TAS, and increasing TAS equates to increasing rate of distance covered. For maximum range, it is therefore necessary to fly at the speed at which the ratio of TAS to , and can be found drag is at a maximum. This speed is,V on a power required curve by drawing a tangent from the origin to the power required curve as indicated in the diagram.

Power required

POWER 9. a. Power required in level flight = Drag x TAS.

..

Drag = CD%pV2S,

where

C, = coefficient of total drag p = air density V = TAS.

So power required = Drag x TAS is proportional to (TAS)2x TAS Which means that power required is proportional to (TAS)3 Indicated air speed (IAS) is also proportional MpV2such that any given IAS will occur at the same value of %pV2at all altitudes, and the drag force at any given IAS will also be the same at altitudes. But as altitude increases, r decreases, so the TAS equating to any given IAS must increase with altitude to maintain a constant value of %pV2at any given IAS. At 40000 feet pressure altitude in a standard atmosphere, p is 5/4 of its mean sea level value, so V2 must be 4 times its sea level value. This means that at 40000 feet, the TAS equating to any given IAS is twice its sea level value.

294


But power required at sea level = C, %p (TAS)% x TAS Which is C,%pS

x TAS3 ............................................. .Equation 1

p is % of its sea level value , while TAS is twice its sea level value. So power required at 40000 feet = C, % (114 p)S (2TAS)3 Which is C, 118pS x 8 (TAS)3 This simplifies to give C, pS TAS3 ................................Equation2. Comparing equations 1 and 2 reveals that the power required at 40000 feet ISA at any given indicated airspeed is twice that at sea level. Put more simply, if power required = Drag x TAS and between sea level and 40000 feet drag remains constant while TAS doubles, then power required also doubles.

POWER 10. d. Power required in level flight Drag = C, %pV2S, where

= Drag x TAS. C, = coefficient of total drag p = air density V = TAS.

So power required = Drag x TAS = C, %PS(TAS)~ Which means that power required is proportional to TAS3.

POWER 11. d. Work is done when a force moves its pcint of application in the direction of the force. Power is a measure of the rate at which work is done. In the case of an aircraft in level flight, the work done is the thrust multiplied by the distance flown. Thrust required is equal to the drag force so the work done is equal to the drag multiplied by the distance flown. The true airspeed of an aircraft is the rate at which it covers the distance flown through the air. Power required is therefore equal to drag multiplied by TAS. Others forms of airspeed including IAS, CAS and EAS cannot be employed in this equation because they do not provided a true indication of the rate at which distance is flown through the air.

POWER 12. c. The power of all air breathing aircraft engines is proportional to the mass flow of air passirrg through them. As altitude increases air density decreases causing mass flow and power output to decrease.

POWER 13. a. The power required in any flight condition is proportional to AS3 and as altitude increases, the ratio of TAS to IAS increases. This means that when climbing at any given IAS, the TAS and power required both increase with increasing altitude. The overall effect of these changes is that the power required : TAS curve moves upwards and to the right with increasing altitude. Also, the power available for all air breathing engines is proportional to mass airflow. This decreases with increasing altitude due to reducing air density. The power available curve therefore moves down with increasing altitude. These effects at sea level at high altitude are illustrated in the diagram at the right.

I

Power available at sea

'Owe'

/

Power required at sea level

/

.

Power avarlable at high altitude


I

POWER 14. d. The power required by any aircraft type in any given flight condition increases with increasing altitude. The power output of aircraft engines is proportional to the mass flow of air passing through them, As air density decreases with increasing altitude, power output also decreases. Excess power is the proportion of the power available that is not required to maintain straight and level flight. The combined effect of these factors is that as altitude increases, power required increases whilst power available decreases. The overall effect therefore is that excess power also decreases with increasing altitude. These effects are illustrated in the diagram at the right. Because the increase in power required and Rate of increase in Rate of reduction decrease in power available are caused by power required in power available decreasing air density, the rate of change of decreases these variables is proportional to the rate of altitude increases. altitude increases. change of air density. As altitude increases the weight of air above any given point decreases, causing static pressure and hence density to decrease. But because air in the Altitude lower atmosphere is more dense than that in the higher atmosphere, the rate of reduction in static pressure per thousand feet of altitude increase is greater at low altitudes that at high altitudes. This means that the rate of change of air density, power available, power required and excess power all decrease as altitude increases. The effects on power available and power required with changing altitude is illustrated in the diagram at the right. The overall effect of the above factors is that as altitude increases, excess power decreases at a rate that decreases with increasin-g altitude.

1

Power availably at sea level

Power required at sea eve1

\/i

1

Power available at igh altitude Rate of increase in er required

1

I

Power required at high altitude Rate of reductior! inpower available decreases as altitude increases

I

te of reduction in excess power descreases as altitude increases

Power available Power required

POWER 15. c. Power required in level flight is equal to drag multiplied by TAS.

-295

Excess power

But drag = CD'%pV2S,where V is TAS. This means that power required is proportional to TAS3 so if TAS is doubled and all other factors remain unchanged, then power required is multiplied by z3 which is 8.

POWER 16. c. VMpis the speed at which an aircraft requires the lowest power. As indicated in the diagram at the right, VMpis higher than V, but lower than VMD.V, is the CAS for best climb gradient and for a propeller aircraft is close to the lowest safe flight speed and therefore just above V.,

. .

V, Propeller

% ,,

296


POWER 17. b. For maximum range it is necessary to achieve the best ratio of fuel consumption to distance flown. For a jet aircraft fuel consumption is proportional to thrust, which in straight and level flight is equal to drag. Best jet aircraft range is therefore achieved at the speed Drag providing the best ratio of drag to airspeed. This occurs at the speed where a tangent drawn from the origin just touches the drag : EAS curve. As indicated in the diagram, this speed is higher than .V ,, EAS

i

"MD

POWER 18. b. Excess power is equal to power available minus Best excess IAS at sea level power required. The power output of both and 40000 feet piston and jet engines is proportional to the mass flow of air passing through them. As altitude increases, both air density and power output decrease. The overall effect of increasing altitude is that the power available : IAS curve moves downwards with increasing altitude. Power required in constant speed straight and level flight is equal to drag multiplied by TAS. Although the drag at any given IAS remains constant, the TAS equating 0 100 150 200 250 to any given IAS increases. This means that the power required : IAS curve moves upwards 0 50 75 100 125 and to the left with increasing attitude. The IAS at 40000 feet overall effect of these changes is that the maximum excess power IAS (V,) decreases with V, decreases from 150 Kts increasing altitude as illustrated in the diagram. at msl, to 100 Kts at 40000 ft

I

Jet best range

Best excess IAS at sea level and 40000 feet

... .. ..

0

100

150;

.... ...... . j200

250

IAS at MSL

250 IAS st 40000 feet V, decreases from 150 Kts at msl, to I00 Kts at 40000 ft

POWER 19. a. Excess power is equal to power available minus Best excess IAS at sea level power required. The power output of both and 40000 feet piston and jet engines is proportional to the mass flow of air passing through them. As Power Available at altitude increases both air density and power output decrease. The overall effect of these changes is that the power available : TAS curve moves downwards and to the right with increasing altitude. Power required in constant speed level flight is equal to drag multiplied by TAS. Although the drag at any given IAS remains constant, the TAS equating 0 100 150 to any given IAS increases. This means that the power required : TAS curve moves upwards 0 50 75 and to the right, with increasing altitude. The IAS at 40000 feet overall effect of these changes is that the maximum excess power IAS (V,), decreases with increasing altitude. These effects are at msl to 75 Kts at 40000 feet.illustrated for a jet engine in the diagram.

Best excess IAS at sea level and 40000 feet Available at

0

\\ A

100

IAS at MSL

0

50

IAS at 40000 feet V, decregses from 130 Kts at msl, to 75 Kts at 40000 ft


297

But at 40000 feet TAS is twice IAS, so the decrease from 130 Kts to 75 Kts IAS represents an increase from 130 to 150 kts TAS. The figures used in this explanation are merely examples and do not apply to all aircraft. It is however true that for all aircraft, increasing altitude decreases the IAS value of Vy and increases the TAS value. This process continues until the absolute ceiling where for a jet aircraft, V, = I

Power required

I

Altitude

POWER 20. a. As altitude increases, power available decreases and power required for any given Altitude flight profile increases. The overall effect of increasing altitude is that excess power and Absoluterate of climb decrease. At the absolute ceilingceiling, power available is equal to power required, so both excess power and rate of climb are zero. This condition is illustrated for a piston engine aircraft in the diagram at the right.

/ '

Power available Airspeed

Power available and power required curves are tangential and power available is equal to power required at the absolute ceiling

POWER 21. a. At the absolute ceiling power available equals power required, so excess power is zero. Flight at this altitude, is possible at only one speed. Flight at any other speed will cause power required to increase and power available to decrease. Flight at other speeds at this altitude requires more power than is available and will therefore be impossible. For a jet aircraft at this altitude, the single practicable airspeed therefore constitutes VMp,VMD,V, V, best endurance speed and best range speed.

POWER 22. c. As altitude increases, power available decreases and power required for any given Altitude flight profile increases. The overall effect increasing altitude is that excess power and Absoluterate of climb decrease. At the absolute ceilingceiling power available is equal to power required. Both excess power and rate of climb are zero and the two curves are parallel as indicated in the diagram.

POWER 23. a.

Power available

7---

Airspeed

Power available and power required curvks are parallel at the absolute ceiling

At all altitudes below the absolute ceiling, the power available and power required curves cross at two points, as indicated in Power the diagram. These points represent the requiredmaximumand minimum flight speeds, for which sufficient power is available. Minimum flight Speed

Maximum fliiht speed

POWER 24. b. When an aircraft's engines are shut down in flight the aircraft possesses a limited store of potential and kinetic energy, which cannot be replenished. Throughout the subsequent glide this energy is expended in

298


overcoming drag, until the energy reduces to zero at the end of the landing run. The rate at which energy is expended by an aircraft in flight is the power required, so for maximum glide endurance the aircraft must be flown at .V ,, The power required curve is therefore highly relevant in gliding flight. The actual power consurr~ptionrate and hence endurance, depend on the speed and angle of attack chosen. The power available curve is however irrelevant in gliding flight became the engine can produce no power when in the shut-down condition.

POWER 25. b. The speed providing the best ratio of power required : TAS is used when flying for maximum still air range in a propeller driven aircraft. This speed occurs where a tangent drawn from the origin just touches the power required curve. As indicated in the diagram below, the speed is also.,V,

POWER 26. a. Maximum endurance in any aircraft type is achieved when the rate of fuel flow is minimum. Fuel flow in a jet aircraft is proportional to Thrust, so maximum jet endurance is achieved when the thrust required for a given TAS is at a minimum.

,V ,

and speed for best Power required: TAS ratio

In straight and level flight thrust = drag, so best straight and level endurance will be achieved at minimum drag speed, which is .,V , This is the lowest point on the total drag : EAS curve as indicated in the diagram.

POWER 27. b. Power required = drag x TAS, and in straight and level flight lift = weight. A decrease in weight therefore results in a decrease in the lift required. Induced drag is proportional to lift, so decreasing left results in a decrease in induced drag. The total drag force at any given TAS therefore reduces with reducing weight, rn in a reduction in power which in t ~ ~ results required at that TAS. Induced drag is proportional to 1/V2, so any reduction in induced drag is most prominent at the low end of the speed range. This means that reductions in weight, also tend to move the drag curve, and hence power required curve to the left The overall effect therefore is that reducing weight moves the power required curve down and to the left. TAS

required

TAS


299

POWER 28. c. The diagram in this question is whole aircraft C,:C, polar, which indicates how the L:D ratio varies throughout the speed range. If a line is drawn horizontally left to right from the origin, this indicates the condition of zero lift and infinite drag. This constitutes a L:D ratio of zero. If a line is drawn vertically upwards from the origin, this indicates the condition of infinite lift and zero drag. This constitutes an infinitely large L:D ratio. The value of the L:D ratio at any point on the diagram is directly proportional to the gradient of any straight line from the origin to that point. The steeper the gradient, the higher the L:D ratio. The curve indicates those L:D ratios that are attainable and the best L:D ratio is achieved where a straight line drawn from the origin, just touches the curve. This is indicated by point C on the diagram. An aircraft achieves its best L:D ratio when flying at,,V, so point C represents this speed.

POWER 29. c. Work is done when a force moves its point of application in the direction of the force. In the case of an aircraft in flight, the work done equals the thrust multiplied by the distance flown through the air. Power is the rate at which work is done and velocity is the rate at which distance is travelled. In the case of an aircraft, TAS equals the rate at which distance is flown through the air. The power required by an aircraft in flight is therefore equal to the thrust force acting on it multiplied by its TAS. Power required = Thrust x TAS. But in straight and level constant speed flight, thrust equals drag, so in this condition the work done may also be described as the drag force multiplied by the distance flown. So in the case of level constant speed flight power required = Drag x TAS. It should be noted that IAS can be used in place of TAS only when flying at ISA MSL where TAS = IAS.

POWER 30. d. The process of gaining height involves app!ying an upward force equal to the weight of the aircraft and using this force to move the aircraft upwards. Work is done when a force moves its point of application in the direction of the force, so moving the aircraft upwards in this manner involves doing work. Power is the rate at which work is done, so the rate of climb depends upon the amount of power available to move the aircraft upwards. The power available to move the aircraft upwards is the excess power available. This is the total power available minus the power required to overcome drag at the chosen airspeed. This situation is described in the following equation: Power expended = (aircraft weight x height gained) / time taken But height gained / time taken = rate of climb and for maximum rate of climb, power expended is the excess power, so maximum rate of climb = excess power / weight

POWER 31. c. If the C of G is forward of the C of P, the aircraft will be subject to a nose down moment, which must be balanced by a downward tailplane force. The generation of this downward tallplane force will cause additional trim drag. Because extra fuel must be burned to overcome this trim drag, the range of the aircraft will be reduced. If the C of G is moved stdl further forward of the C of P, the nose down moment will increase, necessitating a greater tailplane down force, greater trim drag and hence still less range. If however the C of G is initially aft of the C of P, the aircraft will be subject to a nose up moment, which must be balanced by an upward tailplane force. The generation of this force will of course produce trim drag and hence reduce range. If the C of G is then moved forward, the nose up pitching moment will reduce, necessitating

300

Power required and Power avai/ab/e

a reduction in tailplane up force, and trim drag. This process will therefore increase range. Forward movement of the C of G will therefore decrease range only if the C of G is ahead of the C of P. It should be noted that a C of G aft of the C of P would make the aircraft longitudinally unstable and would therefore not be permitted. The condition is considered here, only to demonstrate the full effect of C of G position on range.

POWER 32. b. When the C of G is at its aft limit, aft of the C of P, the aircraft will be subject to a nose up moment, which must be balanced by an upward tailplane force. The generation of this force will produce trim drag and hence reduce range. Forward movement of the C of G to the C of P will reduce the nose up moment to zero, permitting the tailplane up force and trim drag to be reduced to zero, thereby increasing range. If the C of G is then moved further forward to its forward limit, the original nose up moment will be replaced by an increasing nose down moment. This will necessitate a balancing downward tailplane force. The generation of this downward tailplane force will cause additional trim drag and hence decrease range. The overall effect of moving the C of G from aft of the C of P to forward of the C of P will be to increase then decrease range. It should be noted that a C of G aft of the C of P would make the aircraft longitudinally unstable and would therefore not be permitted. The condition is considered here, only to demonstrate the full effect of C of G position on range.

POWER 33. c. If the C of G is forward or aft of the C of P the resulting nose down or nose up moments will necessitate longitudinal trimming. This will increase trim drag and hence decrease range. When the C of G and C of P coincide, the aircraft will not be subject to any nose up or down moments, so no longitudinal trimming will be necessary. This will minimise trim drag thereby maximising range.

POWER 34. a. For very small climb angles, the climb gradient is approximately equal to the height gained divided by the distance flown as indicated in the diagram at the right. But height gained Idistance flown is also the Distance flown sine of the climb angle. A gradient of 2.5% therefore represents a sine of 0.025 and a gradient of 3.0% represents a sine of 0.03. In steady flight the sine of the climb angle is equal to the excess thrust divided by the aircraft weight.

Distance flown

,

Climb angle

So in a 2.5% climb, the sine of the climb angle = Excess thrust I aircraft weight Rearranging this gives Excess thrust = Sine of climb angle x aircraft weight And inserting data provided in the question gives: For a 2.5% climb gradient Excess thrust = 0.025 x .I20000 bf So Excess thrust = 3000 Ibf. The weight at which this excess thrust will give a 3% climb gradient may be found using the same equation as follows: Sine of climb angle = Excess thrust I weight Rearranging this equation gives weight = Excess thrust I sine of climb angle.


301

Inserting .03 for the sine of the climb angle and 3000 lbf excess thrust gives: Weight = 3000 lbf 10.03 which is 100000 Ibf This aircraft will therefore achieve a 3% climb gradient at a weight of 100000 Ibf.

POWER 35. a. In the initial condition 2.5% is proportional to excess power 1 120000 Ibf. Rearranging this equation gives excess power = 2.5% x 120000 Ibf. .... ..... Equation 1 In the second condition X% is proportional to excess power 1 150000 Ibf. Rearranging this equation gives excess power = X% x 150000 Ibf ....... .. Equation 2 Increasingweight will decrease excess power, but if this is ignored as stated in the question, then equations 1 and 2 are equal. This means that 2.5% x 120000 = X% x 150000 Rearranging this equation gives X% = (2.5% x 120000) 1 150000 Which is X% = 2% so the maximum gradient available at a weight of 150000 is 2%.

POWER 36. b. Power required = total drag x TAS. Assuming the aircraft maintains constant TAS, flight profile and configuration, its C, must increase in direct proportion to its weight. A 25% increase in weight will therefore require a 25% increase in C,. But C, is proportional to C : SO if CL is increased by 25% (to 1.25 of its original value), then C, and hence induced drag, will increase by a factor of 1.252,to become 1.5625 of its original value. This is an increase of 56.25% in induced drag. The increase in weight will not however alter C ,, significantly, so profile drag will remain approximately constant at 100% of its initial value. This constitutes a 0% increase.

, induced drag = profile drag, so each represents 50% of total drag. The overall effect on total drag At ,V is therefore equal to the average increase of induced drag and profile drag. That is (56% increase in induced drag +0% increase in profile drag ) 1 2 which is a 28.125% increase in total drag. Power = total drag x TAS, so assuming TAS remains unchanged, power required will also increase by 28.125%

POWER 37. b. If the vector sum of all forces acting on a body in any given direction is not zero, then the body will accelerate in the direction of the resultant of those forces. In climbing or descending flight the forces of lift and drag are added to by the component of weight acting along the flight path. In such circumstances if thrust equals drag then this component of weight will cause the aircraft to accelerate if descending or decelerate if climbing. Options a and c are therefore incorrect. In straight and level flight the only horizontal forces acting on an aircraft are thrust and drag so if these are equal, the vector sum will be zero and speed (IAS) will remain constant. Option b is therefore correct which means that option d is incorrect.

302


POWER 38. a. If the mass of an aircraft is increased, then lift must increase by an equal amount to maintain level flight. Lift = CLMpV2S.This question specifies that angle of attack remains unchanged. But CL is proportional to angle of attack so CL will also remain unchanged. This means that speed (V) or wing area (S) must increase. Drag is proportional to speed and wing area so if either of these are increased then extra power will be required to overcome the higher drag force. Of the options offered in this question, only option a satisfies these requirements.

POWER 39. b. For jet aircraft fuel flow is proportional to thrust. In straight and level flight thrust equals drag. To solve this problem it is therefore necessary to calculate the increase in drag. As aircraft weight increases the additional drag is principally lift induced drag. At any given airspeed C, is proportional to weight and C, is proportional to C,2. TO calculate the increase in fuel consumption it is therefore necessary to calculate the new C, C, and induced drag. The ratio of weights of aircraft B to aircraft A = 2500001200000 = 1.25 So the C, of aircraft B = 1.25 times that of aircraft A and the C : aircraft A. : But C, is proportional to C aircraft A.

SO

the C,

of aircraftB is 1.5625 times that of

and induced drag of aircraft B are also 1.5625 of those of

Aircraft B therefore requires 1.5625 times as much thrust and fuel consumption as aircraft A. This means that the fuel consumption of aircraft b is 5000 Kglhr x 1.5625. The fuel consumption of aircraft B is therefore 7812.5 Kglhr. The above answer is something of a simplification in that it assumes that the relationship between fuel consumption and thrust is linear. It is however the best answer that can be produced on the basis of the information provided in the question.

POWER 40. b. Vx is the IAS at which an aircraft will achieve its best angle of climb and is the speed at which maximum excess thrust is available. Because the thrust of a propeller decreases rapidly with increasing airspeed, Vx for a propeller aircraft is very close to the minimum safe flying speed. V,, is the IAS corresponding to the minimum value of power required. Because fuel consumption in a propeller aircraft is proportional to power output, VIM, is also the lowest fuel consumption airspeed. VIMpfora propeller aircraft is greater than V2 and V ,, but less than V .,

POWER 41. a. Vx is the IAS at which an aircraft will achieve its best angle of climb and is the speed at which maximum excess thrust is available. Because the thrust o i a propeller decreases rapidly with increasing airspeed, Vx for a propeller aircraft is very close to the minimum safe flying speed. As ifidicated in the diagram V, for a propeller aircraft is lower than VIM,. V ,, is the IAS corresponding to the minimum value of power required. Because fuel consumption in a prope!ler aircraft is proportional to power output, V ,, is also the lowest fuel consumption airspeed. V, is the IAS for best rate of climb and is greater than VIM, V2 is the take-off safety speed. It is lower than V.,

I

Power required


303

POWER 42. c. Rate of climb is determined by excess power. As altitude increases power available decreases and power required increases, causing both excess power and rate of climb to decrease. The service ceiling is the pressure altitude at which the maximum rate of climb is 500 feet per minute for a jet and 100 feet per minute for a pistonlpropeller aircraft. But power required is equal to drag multiplied by TAS and drag increases with aircraft weight, so power required at any Altitude given altitude also increase with weight. This means that the reduction of excess power to the level that is just sufficient to achieve the service ceiling rates of climb, occurs at a lower altitude as weight increases.

Altitude

The effect of increasing aircraft weight is therefore to lower the service ceiling. So the service ceiling at 75000 Kg will be lower than that at 50000 Kg. The effect of increasing or 500 fpm (jet) weight is illustrated in the diagram at the right.

O I00 fpA (piston) or 500 fpm Uet)

POWER 43. c.

Low altitude

Power required is equal to drag multiplied by TAS. At any given IAS drag remains constant with increasing altitude, but the TAS corresponding to that IAS increases. This means that as an aircraft climbs at any given IAS, the multiplication of constant drag by increasing TAS causes power required to increase. The overall effect of these changes is that the power required : TAS curve moves up and to the right with increasing altitude as illustrated in the diagram.

POWER 44. a.

ROC

Power -equired

TAS

The term "the back of the drag curve" refers to all speeds I Power At such speeds the gradient of the drag : lower than .,V, EAS curve is negative so decreasing speed causes drag to increase. But to maintain any given speed, thrust must equal drag, so reducing speed at the back of the drag curve requires an increase in thrust. Power required equais drag (cr thrust) multiplied by TAS. Although reducing IAS also reduces TAS, the rate of increase in drag tends to exceed the rate of decrease in speed. The overall effect is that power required increases as speed decreases. However, as indicated in the diagram, the minimum power speed is lower than the minimum drag speed, so when decelerating below,V , drag increases whilst power required decreases until VMp,then increases as speed reduces below .V ,, This option is not offered in this question however, so the selection of the most appropriate answer must be , is much closer to ,V , than to V, so power required based on the predominant or overall effect. ,V increases with decreasing speed over the majority of the back of the drag curve. Option a, is therefore the most appropriate.

304

Power rewired and Power available

POWER 45. b. When operating at the back of the drag curve, all aircraft tend to be speed unstable. This is because increasing speed, decreases drag, causing the aircraft to accelerate further. Conversely, decreasing speed increases drag, causing the aircraft to decelerate further. Although both jet and propeller aircraft exhibit this speed instability, propeller aircraft are less affected than jets. This is because the thrust produced by a jet engine remains approximately constant with increasing TAS, whereas that of a propeller decreases rapidly. This means that as a jet accelerates below VMD,its thrust remains constant, so decreasing drag causes it to continue to accelerate. In the case of a propeller aircraft increasing speed decreases drag, but also decreases thrust, so the excess thrust and hence subsequent acceleration is less. Similarly, when a propeller aircraft is decelerating below VMD,the increasing drag is partly offset by increasing thrust, thereby reduce the effect of speed instability. A propeller aircraft is therefore inherently more speed stable than a jet because its thrust decreases with increasing airspeed.

POWER 46. c. The tangent drawn on the diagram touches the curve at the point at which the ratio of EAS : whatever is represented by the curve is at a maximum. Although the point is marked as VMD,this occurs at the bottom of the drag : EAS curve, so the diagram does not represent a drag : EAS curve. But VMDis also the speed at which the ratio of EAS : power required is maximum, so the curve can be identified as the power required : EAS curve.

POWER 47. c. The tangent touches the curve at the point where the ratio of EAS : whatever is represented by the curve is maximum. This point is however iden.tified as U D Max which occurs at VMD.But VMDoccurs at the bottom of the Drag : EAS curve, so this must be some other curve. VMDis also the point at which the ratio of EAS : power required is maximum, so the curve represents Power required : EAS

POWER 48. a. VMp is the speed at which power required is at its minimum value. The diagram in this question illustrates a Drag : EAS curve and this can be converted into a power required curve by multiplying all drag values by TAS.The Power required : EAS curve produced by this process is illustrated at the right. VMp is represented by the lowest point on the Power required : EAS curve. This is point A on the original diagram. It should be

Drag


305

noted this is also the point at which a straight line drawn from the origin intersects the Drag: EAS curve at right angles (90 degrees).

POWER 49. a. The best angle of climb speed, V, for a propeller aircraft at the lowest safe flying speed. This is higher than V, but lower than VMp.The diagram in this question illustrates a Drag : EAS curve and VMp occurs where a line drawn from the origin intersects such a curve at right angles. This occurs at point B on the diagram. The only point on the diagram that might represent V, for a propeller aircraft is therefore point A.

POWER 50. c. The best rate of climb speed, Vy is the speed at which excess power is maximum. For a propeller aircraft this occurs between VMpand VMD.VMplies at the point where a straight line drawn from the origin intersects the Drag : EAS curve at right angles. This ispoint B on the diagram. VMDis the speed at which drag is minimum. This is point D on the diagram. The point most representative of Vy for a propeller aircraft is therefore point C.

Drag

POWER 51. b. VMp is the speed at which power required is minimum so as speed decreases below VMp, the power required increases.,V , is the speed at which drag is lowest so when decreasing speed below VMDthe drag force increases. But VMp is lower than VMD,so decreasing speed below VMpcauses both drag and power required to increase. Power required

POWER 52. b. VMp is the speed at which power required is minimum, so as speed increases above VMp, the power required increases. VMD is the speed at which drag is minimum, so when increasing speed, drag decreases below VMD but increases above VMD. But VMp is lower than VMD, so increasing speed above VMp means first approaching, then exceeding VMD. The effect of such acceleration is that power required increases whilst drag decreases up to VMDthen increases at higher speeds. This is illustrated in the diagram at the right.

VMP above VM P

Drag decreases up to VMD

I

VMD

\

Drag increases above VMD

306


POWER 53. b. ,,C , occurs at the low speed stali, which is the lowest speed at which an aircraft is capable of ge~eratingsufficient lift for straight and level flight. VMo is the maximum speed at which an aircraft can be flown in normal circumstances. This question therefore concerns the shapes of the Drag : EAS and power required : EAS curves, between V, and .,V , As illustrated in the diagram at the right, both drag and power required decrease then increase.

Drag

Drag decrease up to, V ,

then increases

POWER 54. b. Power required = Drag x TAS. IAS is proportional to %pV2 such that climbing at constant IAS means climbing at constant %pV2. But as altitude increases density decreases, so when climbing at constant IAS, TAS must increase such that the increase in V2 balances the decrease in r. If density at 40000 feet is 1/4 of that at sea level, then in a constant IAS climb, TAS must double its sea level value, so that V2 increases by a factor of 4. This means that at 40000 feet, although drag remains unchanged, TAS is double its sea level value. Because power required is equal to drag multiplied by TAS, this doubling of TAS will cause power required to be doubled. Power required will therefore increase by a factor of 2. It should be noted that although power required at constant density is proportional to TAS3 and would therefore increase by a factor of 8 if TAS were doubled, the decreasing density in this question offsets some of the increase in power required.

POWER 55. d. IAS is proportional to %pV2(where r is air density and V is TAS), such that climbing at constant IAS means climbing at constant MpV2. But as altitude increases density decreases, so when climbing at constant IAS, TAS must increase such that the increase in V2 balances the decrease in r. If density at 40000 feet is %of that at sea level, then in a constant IAS climb, TAS must double IAS so that V2 increases by a factor of 4. This means that at 40000 feet, IAS is half of TAS. So in a constant TAS climb up to 40000 feet IAS decreases to half of its sea level value. But power required is proportional to IAS such that it decreases then increases between ,V , and V., This means that in a constant TAS climb, between VMo and ,,V, If however airspeed becomes power required will decrease provided the airspeed remains above .,V, then power required will increase with reducing IAS. lower than ,,V,

POWER 56. b. Vx is the airspeed at which an aircraft will achieve its best climb gradient. The climb gradient is approximately equal to excess thrust divided by aircraft weight. But power output at any given power lever setting is proportional to the mass flow of air through the engine. As altitude increases, air density, air mass flow and hence power output decrease. So if an aircraft climbs at a constant power lever setting, the power output, excess power and climb gradient will all decrease with increasing altitude. The rate of climb is proportional to TAS and as power output decreases TAS and rate of climb both decrease. The absolute ceiling of an aircraft is the altitude at which excess thrust, excess power, climb gradient and rate of climb have all reduced to zero.


%

307

POWER 57. b. V, is the airspeed at which an aircraft will achieve its best rate of clir~b.The rate of climb is equal to excess power divided by aircraft weight. But power output at any given power lever setting is proportional to the mass flow of air through the engine. As altitude increases, air density, air mass flow and hence power output decrease. So if an aircraft climbs at a constant power lever setting, the power output, excess power and rate of climb will all decrease with increasing altitude. The absolute ceiling of an aircraft is the altitude at which excess thrust, excess power, climb gradient and rate of climb have all reduced to zero. POWER 58. b. The thrust power output of any aircraft propulsion system LOW alt~tude equal to the product of its thrust multiplied by itsTAS. When the aircraft is stationary on the Ground, TAS and thrust horsepower are zero. As airspeed increases, thrust Power horsepower varies in proportion to the changes in thrust and available TAS. In the case of jet aircraft thrust is approximately constant at all speeds, so power available increases linearly with TAS. In the case of a propeller aircraft, thrust decreases with EAS increasing TAS. The combined effect of increasing TAS and decreasing thrust, as a propeller aircraft accelerates from rest is that the power available is initially zero at zero speed, then increases with increasing TAS. before reaching a peak value at about 300 Kts, above which it gradually decreases. Power output is also proportional to the air mass flow passing through an engine. As altitude increases, air density, air mass flow and power output all decrease. The diagram in this question represents the variation of power available from a propeller aircraft with increasing altitude.

POWER 59. d. The power required by an aircraft in constant speed level flight is equal to drag multiplied by TAS. At any given IAS the drag remains constant with increasing altitude, but the TAS increases. This means that power required at Powerany given speed, and the speed at which minimum power is required, both increase with increasing altitude. The effect of these changes is that the power required : EAS curve moves upwards and to the right as altitude increases. The diagram in this question illustrates these effects.

Power required

EAS

'

POWER 60. c. The power required by an aircraft in constant speed level flight is equal to drag multiplied by TAS. At any given IAS the drag remains constant with increasing altitude but the TAS increases. This means that power required at Power any given speed and the speed at which minimum power is required both increase with increasing altitude. But increasing aircraft weight has a similar effect to increasing altitude in that it moves the power required curve upwards and to the right. The difference is that increasingweight increases power required at all speeds, whereas increasing altitude does not. Increasing altitude causes the new and old power required curves to cross as in the diagram in question POW59, but,

H ~ g hweight

power requlred

& LOW

weicht

EAS

308


as indicated above, increasing weight does not cause the curves to cross. The diagram is therefore most representative of option c, an increase in weight.

POWER 61. c. The power available from any aircraft engine is equal to Thrust multiplied by TAS. In the case of a jet engine thrust is approximately constant with increasing TAS so power available increases linearly with increasing TAS. Options A and C in the diagram might therefore be representative of jet power available. As altitude increases air density decreases causing power available to reduce. The line representingjet power available at high altitude would therefore be of a shallower gradient than that representing sea level performance. This is most accurately represented by line C. Lines B and D are representative of power available from piston engines at high and low altitudes respectively.

A

EAS

POWER 62. d. Power available from any aircraft engine is equal to Thrust multiplied by TAS. In the case of a propeller aircraft, thrust decreases with increasing altitude such that power available increases from zero at zero TAS, reaching a maximum value at about 300 Kts, then reducing as speed increases further. This effect might be represented by curves B and Don the diagram. As altitude increases, air density decreases causing power available to reduce. The gradient of the power available curve is therefore shallower at high altitude than at low altitude. Curve D is therefore most representative of the power available from a propeller at high altitude.

pAv

khz?z E -AS -

A

POWER 63. a. Jet power available increases linearly with increasing speed and so might be represented by lines A and C on the diagram. But as altitude increases, decreasing air density causes power available to decrease, reducing the gradient of the power available : EAS line. Line A therefore best representsjet power available at low altitude.And line C best representsjet power available at high altitude.

P~~

EAS

POWER 64. b. Propeller power available is zero at Zero airspeed, increases to a maximum value at about 300 Kts then decreases as speed increases further. This might be represented by curves B and D on the diagram. But as altitude increases air ,P density reduces causing power available to decrease. This has the effect of reducing the gradient of the power available curve at high altitude.

pAv

c

EAS


309

Curve B therefore best represents propeller power available at low altitude. Curve D best represents propeller power available at high altitude. I

POWER 65. d. Propeller thrust decreases with increzsing €AS in a manner similar to curves C and D in the diagram. But as altitude increases thrust decreases, such that the high altitude thrust curve is lower than the low altitude T curve. Curve 0 is therefore the most representative of propeller thrust at high altitude.

T

D EAS

POWER 66. c. Propeller thrust decreases with increasing €AS in a manner similar to curves C and D in the diagram. But as altitude increases thrust decreases, such that the high altitude thrust curve is lower than the low altitude T curve. Curve C is therefore the most representative of propeller thrust at low altitude.

T

D EAS

POWER 67. b. Jet thrust available is maximum at zero EAS, then decreases to a minimum value at about 250 Kts before again increasing at higher speeds as indicated in curves A and B. As altitude increases, air density T and thrust decrease so the curve is lower

POWER 68. a.

T

EAS

Jet thrust available is maximum at zero €AS, then decreases to a minimum value at about 250 Kts before again increasing at higher speeds as indicated in curves A and B. As altitude increases, air density T and thrust decrease so the curve is lower

POWER 69. d.

D

EAS

In straight and level flight lift equals weight so if weight increzses then lift must increase by the same amount. But Lift = CL%pV2S where CL is the coefficient of lift p is the air density V is the TAS S is the wing area p is beyond the control of the pilot and S is normally varied only by extending flaps for take-off and landing, so if weight and lift are to be increased while continuing to fly at the same speed (V) then C, must be increased by increasing angle of attack. At ,V , the total drag is made up of equal proportions of induced drag and profile drag. Increasing CL will cause induced drag to increase in proportion to C?, but will have no significant effect on profile drag. So to increase lift by 50% whilst maintaining constant speed, C, must also increase by 50% to 1.5 of its

310


previous value. The resulting increase in induced drag will be proportional to CL2SO if CL becomes 1.5 times its previous value, then induced drag will become 1.52 times its previous value. This means that induced drag will increase to 2.25 times its previous value. This represents a 125% increase. The overall effect on total drag will be a 125% increase in induced drag but no increase in profile drag. And because each constitutes half of the total drag the average rise will be (125/2)%, which is 62.5% Power required is equal to total drag multiplied by TAS so increasing total drag by 62.5% whilst maintaining constant speed will increase power required by 62.5%.

POWER 70. c. In straight and level flight lift equals weight so if weight increases then lift must increase by the same amount. where C, is the coefficient of lift But Lift = C,1/2pV2S p is the air density V is the TAS S is the wing area p is beyond the control of the pilot and S is normally varied only by extending flaps for take-off and landing, so if weight and lift are to be increased while continuing to fly at the same speed (V), then C, must be increased by increasing angle of attack. Increasing C, will cause induced drag to increase in proportion to CL2,but will have no significant effect on profile drag. So to increase lift by 50% whilst maintaining constant speed, C, must also increase by 50%-to 1.5 of its previous value. The resulting increase in induced drag will be proportional to CL2SO if CL becomes 1.5 times its previous value, then induced drag will become 1.52 times its previous value. This means that induced drag will increase to 2.25 times its previous value. This represents a 125% increase. Profile drag will not however increase significantly. At ,V , the total drag is made up of equal proportions of induced drag and profile drag. The overall effect on total drag will be a 125% increase in induced drag but no effect on profile drag. And because each constitutes half of the total drag the average rise will be (125/2)%, which is 62.5%. But at speeds above ,,V, induced drag represents less than half of total drag so the 125% increase in induced drag will increase total drag by less than 62.5%. The figure of 42.5% offered in option a is correct , whereas option c is true for all speeds above,V , and is therefore the most for only one speed above,,V appropriate.

POWER 71. a. In straight and level flight lift equals weight, so if weight increases then lift must increase by the same amount. where But Lift = C,%pV2S C, is the coefficient of lift p is the air density V is the TAS S is the wing area p is beyond the control of the pilot and S is normally varied only by extending flaps for take-off and landing, so if weight and lift are to be increased while continuing to fly at the same speed (V) then C, must be increased by increasing angle of attack.


311

increasing C, will cause induced drag to increase in proportion to C z , but will have no significant effect on profile drag.

So to increase lift by 50% whilst maintaining constant speed, CL must also increase by 50% to 1.5 of its : SO if CL becomes 1.5 previous value. The resulting increase in induced drag will be proportional to C times its previous value, then induced drag will become 1.5* times its previous value. This means that induced drag will increase to 2.25 times its previous value. This represents a 125% increase. Prafile drag will not however increase significantly. , the total drag is made up of equal proportions of induced drag and profile drag. The overall effect At ,V on total drag will be a 125% increase in induced drag but no effect on profile drag. And because each constitutes half of the total drag the average rise will be (125/2)%, which is 62.5%. But at speeds below,,V , induced drag represents more than half of total drag so the 125% increase in induced drag will increase total drag by more than 62.5%. The relative position of the speed chosen and ,V , has no significance in determining the qualitative effects of weight increases so options c and d are both incorrect.

POWER 72. c. At the absolute ceiling the maximum power available is just sufficient to permit straight and level flight at .,V, At all other flight speeds more power will be required than is available so the aircraft will loose height. If an engine fails then total power available will decrease causing, the absolute ceiling to decrease and so the aircraft will descend. In attempting to maintain height the options of increasing or decreasing speed will be ineffective as both will result in a power deficit. Because the absolute ceiling by definition involves the use of maximum power available, option d, "increase power setting" is not possible. Had the engines not been at maximum power setting prior to the engine failure, the aircraft would not have been at its absolute ceiling. The only practicable option to maintain height following single engine failure is therefore to decrease weight in order to decrease power required.

POWER 73. b.

-

At the absolute ceiling the maximum power availab!e is just sufficient to permit straight and level flight at .,V, At all other flight speeds more power will be required than is available so the aircraft will loose height. If an engine fails then total power available will decrease making continued flight at that altitude and speed impossible. In attempting to maintain speed the options of climbing, decelerating or accelerating will be ineffective as all will result in a power deficit. The only practicable option to maintain speed following single engine failure, is therefore to descend in order to decrease power required and increase power available.

POWER 74. d. At the absolute ceiling the maximum power available is just sufficient to permit straight and level flight at .,V, At all other flight speeds more power will be required than is available so the aircraft will loose height. If reheat is ihen selected then all of the additional thrust will be excess to that required for that given combination of altitude and speed. This excess thrust can be employed to climb. The sine of the maximum angle of climb of an aircraft in steady flight is equal to the excess thrust divided by the weight of the aircraft. In this case Sine of climb angle = 5000 Ibf I 50000 Ibf which is 0.1 giving an angle of approximately 6 degrees. But the sine of an angle is also approximately equal to the gradient of that angle so the maximum achievable climb gradient is 0.1 or 10%.

312


POWER 75. c. The power required by an aircraft in constant speed level flight is equal to drag multiplied by TAS. At any given IAS the drag remains constant with increasing altitude but the TAS increases. This means that power required at any given speed increases with increasing altitude.

POWER 76. b. The drag force acting on an aircraft is the sum of two components parts, the induced drag and the profile drag. Induced drag is proportional to IN2whilst profile drag is proportional to V2. The relationship between speed and drag is therefore a complex one. The total drag curve is typically a bucket shape as illustrated in the diagram below. The is the speed at which total drag minimum drag speed (V), is lowest.

Mi,

/!

Min drag

The power required to maintain any given flight condition is equal to the drag force multiplied by the TAS. This means that as speed increases, the factor by which the drag is "IMP V~~~ SPEED -) multiplied also increases. The power curve is therefore a different shape Min power and tilted upwards at the high speed end of the range. The minimum drag occurs at the bottom of the power required curve. As illustrated in the diagram, VIM, is also speed (V), the speed at which a tangent from the origin just touches the power required curve. VIMPis therefore always lower than VIM,.

POWER 77. b. The power required to maintain any given flight condition is equal to the drag force multiplied by the TAS. This means that as speed increases, the factor by which the drag is multiplied also increases. The power curve is therefore a different shape Min power and tilted upwards at the high speed end of the range. The minimum drag speed (VIM,) occurs at the bottom of the power required curve. As illustrated in the diagram, VIM, is the speed at which a tangent from the origin just touches the power required curve.

POWER 78. a.

Min

....................,.,,. Min drag

VIMP

V,MD SPEED

-+

The optimum altitude for operation of a jet aircraft is that at which the airframe achieves the best TAS:Drag ratio and the engines achieve their lowest specific fuel consumption (SFC). Drag is proportional to mass such that as mass decreases due to fuel usage during flight, the altitude for best TAS:Drag ratio gradually increases. This means that the optimum operating altitude increases throughout any given flight. Although minimising fuel costs requires cruise at the optimum altitude, the gradual increase in altitude that this would involve is usually unacceptable for air traffic control purposes. Commercial aircraft will therefore operate at their optimum altitude whenever ATC considerations permit them to do so. The step climb procedure is a method by which efficiency and range are maximised whilst avoiding a gradual upward drift.

POWER 79. a. Jet engine specific fuel consumption (SFC) is the amount of fuel that is consumed per hour to produce each pound (or Newton) of thrust. SFC is lowest when the engines are operating within the 85% to 95%


313

RPM range. As RPM increases above or decreases below this range SFC increases. SFC is not therefore directly related to aircraft C of G position. Movement of the C of G will however affect SFC if it necessitates a change in RPM. Although not specifically stated, it should be assumed that the aircraft is initially cruising at its optimum SFC. If the C of G is forward of the C of P, it will produce a nose down pitching moment. The trimming of the horizontal stabiliser to balance out this moment will generate a certain amount of trim drag, which will affect the thrust and RPM required. If the C of G moves further forward the nose down moment will increase, thereby necessitating a greater degree of trimming. This in turn will increase drag so increase the RPM required. Assuming the aircraft is already within its optimum RPM band, this increase will take RPM outside of that band and hence increase SFC. If however the C of G is initially behind the C of P the situation would be reversed. The aircraft would require nose up trim which would increase drag and RPM required. If the C of G then moved forward, the need for trimming would reduce. This would reduce drag and required RPM. It should be noted that this RPM reduction would increase SFC if it caused RPM to move outside of the optimum 85% to 95% RPM range. This question is therefore defective in that a number of options might be correct, depending upon the precise circumstance. Its has however been included in this book in order to illustrate the problems sometimes posed by poorly worded question. In this particular case the question is actually probing the effect of C of g position on total drag. The most appropriate answer is therefore option a, forward movement will increase SFC only when C of G is already ahead of C of P.

POWER 80. c. Jet engine specific fuel consumption (SFC) is the amount of fuel that is consumed per hour to produce each pound (or Newton) of thrust. SFC is lowest when the engines are operating within the 85% to 95% RPM range. As RPM increases above or decreases below this range SFC increases. SFC is not therefore directly related to aircraft C of G position. Movement of the C of G will however affect SFC if it necessitates a change in RPM. Although not specifically stated, it should be assumed that the aircraft is initially cruising at its optimum SFC. The entire authorised C of G for all aircraft types is forward of the C of P. If this condition the 1ift:weight couple produces a nose down pitching moment. The trimming of the horizontal stabiliser to balance out this moment will generate a certain amount of trim drag, which will affect the thrust and RPM required. If the C of G moves further forward the nose down moment will increase, thereby necessitating a greater degree of trimming. This in turn will increase drag so increase the RPM required. Assuming the aircraft is already within its optimum RPlVl band, this increase will take RPM outside of that band and hence increase SFC.

POWER 81. b. Maximum range in still air is achieved when flying such that the TAS:drag ratio is maximised and the SFC is minimised. In still air the ground speed will be equal to the TAS. But a headwind will cause the mass of air through which the aircraft is flying, to be blown in the opposite direction to flight. This will reduce ground speed to equal the TAS minus the headwind velocity. This will reduce the distance flown over the ground in any given time. Although the range through the air will be maximum, range over the ground will be reduced. A headwind will therefore decrease range.

POWER 82. b. Maximum range in still air is achieved when flying such that the TAS:drag ratio is maximised and the SFC is minimised. In still air the ground speed will be equal to the TAS. But a headwind will cause the mass of air through which the aircraft is flying, to be blown in the opposite direction to flight. This will reduce ground

-

314


speed to equal the TAS minus the headwind velocity. This will reduce the distance flown over the ground in any given time. So although the aircraft will still be achieving its maximum range through the air, its range over the ground will be reduced. A headwind will therefore decrease range. In still air, the optimum range speed increases with increasing altitude. In the case of jet aircraft, the maximum range is achieved by flying slightly faster than VIM, at an altitude where VIM, is equal to .M ,,,, Under these circumstances any increase in TAS will cause a disproportionate increase in drag, thereby reducing range. If however the aircraft is flying at any lower altitude its optimum range speed will be lower ,,, Under these circumstances it is possible to increase range in a headwind, by increasing TAS than .,M slightly. It should however be noted that the required TAS increase is less than the headwind velocity. This technique cannot therefore fully restore the still air range when flying into a headwind.

POWER 83. a. A cost index represents the ratio of fuel costs to total flight costs. The closer the cost index gets to 1, the greater the overall efficiency of the aircraft will be. The chosen cost index is put into the flight management computer which then adjusts the speed of the aircraft to achieve the desired efficiency. A cost index greater than 1 will give a mach number greater than that at which distance/Kg of fuel is greatest.

POWER 84. b.

Power required

To achieve maximum endurance it is necessary to minimise the rate at which fuel is consumed. Fuel consumed by a piston is proportional to the power output, so fuel consumption is minimised by minimising power output.This means that maximum piston engine endurance is achieved by flying at the speed requiring the minimum power V, V, VIM, (V,). The speed for piston engine best angle of climb (V,) is slightly less than V ,, whereas the speed for best rate of climb (V,) . .. .. .. .. . . . is slightly greater than VIM,. V, is however closer to VIM, \ix ! v, VIM, than is V ., The minimum drag speed VIM, is considerably VIM, greater than V, V ,, or V., The relationship between these speeds is illustrated in the diagram at the right. The most appropriate answer to this question is option b, approximately the speed for best rate of ., climb V

POWER 85. c. This question is of dubious quality in that the term "output" is used without any clear definition of what this means. The question has however been reported by a number of students and is included in this book to illustrated the type of problems that might be encountered in the examination. If it is assumed that the term "output" is intended to mean fuel consumption in a given condition, then the question can be solved using the following equation: SFC = Fuel consumption /Thrust, which can be rearranged to give: Thrust = Fuel consumption / SFC Comparing the two engines it can be said that: For engine A

Thrust A = Fuel consumption A / SFC A

And for engine B

Thrust B = Fuel consumption B / SFC B

And when both engines are producing the same thrust:


315

Fuel consumption A I SFC A = Fuel consumption B I SFC B This car1 be rearranged to give : Fuel consumption B = (Fuel consumption A I SFC A) x SFC B Inserting the data provided in the question gives: Fuel consumption B = (14 I.0.6) x 0.035 So fuel consumption of engine B = 8.167 Kglh

POWER 86. b. The entire authorised C of G for all aircraft types is forward of the C of P. If this condition the liftweight couple produces a nose down pitching moment. The trimming of the horizontal stabiliser to balance out this moment will generate a certain amount of trim drag, which will affect the thrust required and hence fuel consumption. If the C of G movesfurther aft, the nose down moment will decrease, thereby necessitating a lesser degree of trimming. This in turn will decrease drag, so decreasing the thrust required and fuel consumption. But reducing fuel consumption whilst maintaining constant speed will increase the range achieved for any given amount of fuel. Moving the C of G aft will therefore increase range. It should be noted that aft movement of the C of G also decreases both stalling speed and longitudinal stability so options c and d are untrue.

POWER 87. b. -

Specific air rarlge (SAR) is a measure of the efficiency of an aircraft. More specifically it is equal to the distance flown through the air divided by the fuel consumed. But distance flown through air = TAS x time

/

r

And fuel consumption = fuel flow (FF) x time. So SAR = (TAS x time) I (FF x time) Which simplifies to give SAR = TAS IFF

POWER 88. a. Jet engine specific fuel consumption (SFC) is the amount of fuel that is consumed per hour to produce each pound of thrust. SFC is lowest when the engines are operating within the 85% to 95% RPM range. As RPM increases above or decreases below this range, SFC increases. Maximum jet aircraft range in still air is achieved when flying such that the TAS:drag ratio is maximised This and the SFC is minimised. The best TAS:Drag ratio occurs at 1.32 of the minimum drag speed (V,). is the maximum range cruise speed (MRC). So maximum range cruise in a jet aircraft is achieved by flying at MRC at the altitude at which the drag at this speed is equal to the thrust produced at 85% to 95% RPM. Because of the shape of the TAS:drag curve it is possible to increase speed within a narrow band, without incurring significant drag increases. At 1. I 5 MRC, for example, the range is still close to -99% of its maximum value. Similarly. Within the 85% to 95% RPM band, changes in RPM cause very little change in SFC. So there is a narrow band of speeds around the MRC, in which the range is very close to maximum. /

But the total cost of operating an aircraft is not just the fuel costs. It also includes other factors such as equipment running costs and crew costs. Getting passengers to their destinations more quickly than competitors can, also has a cash value. 1.32 VIM, is a comparatively low speed, so non-fuel costs can be reduced by flying faster and reducing flight duration. For greatest overall efficiency (lowest total

316


costs) it is necessary to fly faster in order to trade off increasing fuel costs against decreasing non-fuel costs So aircraft usually fly at a speed slightly higher than the MRC to reduce flight time whilst retaining a good level of efficiency. This is achieved by flying at the long range cruise speed (LRC). The actual value of the best speed for a given Right depends upon the distance to be flown and the relationship between fuel costs and non-fuel costs.

CURVES

CURVES 1. a. The diagram is a standard representation of the drag forces affecting an aircraft. It illustrates the total drag curve, together with the two main components of which it is comprised. Curve A represents lift induced drag. Below the stalling angle, lift induced drag is proportional to angle of attack. This is greatest at low speeds where very high angles of attack are required to generate the necessary lift force. As airspeed increases, the required angle of attack, and hence lift induced drag decreases.

CURVES 2. b. The diagram is a standard representation of the drag forces affecting an aircraft. It illustrates the total drag curve, together with the two main components, induced drag and profile drag, of which it is comprised. Curve B is the total drag. At any airspeed total drag is equal to the sum of the lift induced drag and profile drag. The total drag curve is therefore bucket shaped, with the lowest value at the airspeed at which the induced drag and profile drag curves cross.

CURVES 3. c. The diagram is a standard representation of the drag forces affecting an aircraft. It illustrates the total drag curve, together with the two main components, induced drag and profile drag, of which it is comprised. Curve C represents profile drag. Profile drag is not related to lift, but is caused by the motion of the aircraft through the air. It is proportional to the square of the EAS. Profile drag is therefore greatest at high speeds and reduces to almost zero at very low speeds.

The diagram is a standard representation of the drag forces affecting an aircraft. It illustrates the total drag curve, together with the two main components, induced drag and profile drag, of which it is comprised. Curve 6 is the total drag. At any airspeed total drag is equal to the sum of the lift induced drag and profile drag. The total drag curve is therefore bucket shaped, with its miniml-~rn value at point D. This corresponds to the airspeed at which the induced drag and profile drag curves cross. This speed is ,V, and the total drag at this speed is minimum total drag.

CURVES 5. b. The diagram is a standard representation of the drag forces affecting an aircraft. It illustrates the total drag curve, together with the two main components, induced drag and profile drag, of which it is comprised. Curve B is the total drag. At any airspeed total drag is equal to the sum of the lift induced drag and profile drag. The total drag curve is therefore bucket shaped, with its minimum value at point D. -TI- is corresponds to the airspeed at which the induced drag and profile drag curves cross. This speed is V ,, and the total drag at this speed is minimum total drag. In JAR ATPL aircraft performance examinations this speed is termed V .,, Options a (V,) and c V (),, in this question are not commonly used terms.

CURVES 15. d. V, is the speed at which an aircraft is able to achieve its best angle of climb. Angle of climb is proportional to excess thrust, which is equal to thrust available minus drag. In the case of jet aircraft, thrust is approximately constant throughout the operating speed range. As illustrated in the diagram referred to in this question, drag is high at low speeds, decreases up to a minimum value at VMD,before increasing at higher speeds. The maximum excess thrust therefore occurs at VMDwhere drag is at a minimum. This is at point D in the diagram.

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CURVES 6. b. In this diagram something has caused the lift induced drag to increase over the low speed range, whllst profile drag has remained unchanged. This increase in induced drag has caused the total drag curve to move upwards and to the right, increasing both VMDand total drag at all speeds. Induced drag is proportional to lift, and in any given flight condition lift is proportional to weight. The effect illustrated in this diagram is typical of that of a weight increase. Option a (flap deployment) is incorrect because this would increase both induced and profile drag. Option c (weight decrease) is incorrect because this would cause induced drag to decrease. Option d (gear deployment) is incorrect because this would not alter induced drag but would increase profile drag.

CURVES 7. d. In this diagram something has caused the profile drag to increase over the high speed range, whilst lift induced drag has remained unchanged. This increase in profile drag has caused the total drag curve to move upwards and to the left, decreasing VMDand increasing total drag at all speeds. Profile drag is not related to lift but is proportional to the surface area of the aircraft exposed to the airflow. The effect illustrated in this diagram is typical of that of landing gear deployment. Option a (flap deployment) is incorrect because this would increase both induced and profile drag. Options b (weight increase) and c (weight decrease) are incorrect because these would not affect profile drag significantly, but would cause induced drag to change. Option d (gear deployment) is correct because this would not alter induced drag but would increase profile drag.

CURVES 8. a. VMp is the speed at which the power required is minimum. Power required in level flight is equal to drag multiplied by TAS. On a total drag diagram the power required at any speed is proportional to the area under the curve between origin and the speed under consideration. The minimum power speed can be found by drawing a straight line from the origin such that it intersects the total drag at right angles. This speed equates to point A in the diagram in this question.

CURVES 9. d. The diagram in this question illustrates induced drag, profile drag and total drag curves. In level flight the thrust required is equal to the total drag. The ratio of TAS : total thrust required at any point on the total drag curve, can be found by drawing a straight line from the origin to the point under consideration. The ratio of TAS : thrust required is then proportional to the slope of this line. If for exarr~plethe line were vertical, it would equate to infinite thrust required at zero TAS. This would indicate zero TAS : thrust required ratio. A horizontal line would equate to zero thrust required at infinite TAS. This would indicate an infinitely high TAS : thrust required ratio. Neither a vertical, nor horizontal line will touch the total drag curve however, so these infinitely high and inrinitely low ratios are not achievable. The greatest ratio of TAS : thrust required occurs where the line of shallowest slope just touches the total drag curve. This is point D in the diagram.

CURVES 10. a. For maximum endurance an aircraft must fly at the speed requiring minimum fuel consumption rate. Fuel consumption in a propeller aircraft is proportional to power output, so maximum endurance occurs at the speed requiring minimum power. This is VMp.Power required in level flight is equal to drag multiplied by TAS. On a total drag diagram as illustrated in this question, the power required at any speed is proportional to the area under the curve between origin and the speed under consideration. The minimum power

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speed can be found by drawing a straight line from the origin such that it intersects the total drag at right angles. This speed equates to point A in the diagram in this question.

CURVES 11. c. For maximum range an aircraft must fly at the speed providing the best ration between TAS and fuel consumption. Fuel consumption is a propeller aircraft is proportional to power required, so best range requires the best ratio of TAS : power required. The diagram provided in this question illustrates a total drag curve. The best ratio of TAS : power required occurs at the minimum drag speed .,V, This is indicated at point c in the diagram.

CURVES 12. a. Vx is the speed at which an aircraft is able to achieve its best angle of climb. Angle of climb is proportional to excess thrust, which is equal to thrust available minus drag. In the case of propeller aircraft, thrust reduces rapidlywith increasing airspeed. The maximum excess thrust therefore occurs close to the minimum which is located at point B in the diagram. The most probable safe flying speed. This is lower than ,,V, location of V, is therefore point A.

CURVES 13. c. V, is the airspeed at which an aircraft is able to achieve its best rate of climb. Rate of climb is proportional to excess power, which is power available minus power required for level Right. In the case of propeller aircraft, power available increases from zero at zero airspeed, reaches a maximum at about 200 to 250 Kts, before decreasing at higher speeds. Power required is high at very low speeds, decreases to a minimum at ,,V, before increasing rapidly at higher speeds. The greatest excess power available and but lower than .,V, In hence best rate of climb occurs at a speed (V,), which is slightly higher than ,,V, the diagram provided in this question ,V , is at point B and ,V , is at point D. the most probable location of V, is therefore point C.

CURVES 14. e. V, is the airspeed at which an aircraft is able to achieve its best rate of climb. Rate of climb is proportional to excess power, which is power available minus power required for level flight. In the case of a jet aircraft, power available increase approximately linearly with increasing airspeed. Power required is high at very low speeds, decreases to a minimum at ,V ,, before increasing rapidly at higher speeds. The greatest excess power available and hence best rate of climb speed (V,), occurs at a speed slightly greater than ,,V, ,V , is point D on the diagram in this question. The most probable location of Jet aircraft V, is therefore point E.

CURVES 15. d. V, is the speed at which an aircraft is able to achieve its best angle of climb. Angle of climb is proportional to excess thrust, which is equal to thrust available minus drag. In the case of jet aircraft, thrust is approximately constant throughout the operating speed range. As illustrated in the diagram referred to in this question, drag is high at low speeds, decreases up to a minimum value at ,V ,, before increasing at , where drag is at a minimum. higher speeds. The maximum excess thrust therefore occurs at ,V This is at point D in the diagram.

CURVES 16. d. For maximum endurance an aircraft must fly at the speed requiring minimum fuel consumption rate. Fuel consumption in q jet aircraft is proportional to thrust. Minimum fuel consumption rate and hence maximum endurance are therefore achieved by flying at the speed requiring minimum thrust. But in level flight thrust

required is equal to drag. Maximum jet endurance is therefore achieved when flying at the minimum drag speed.,V, This is point D in the diagram.

CURVES 17. d. For maximum range an aircraft must fly at the speed providing the best ratio between TAS and fuel consumption. In a jet aircraft, fuel consumption is proportional to thrust, which in level flight is equal to drag. Maximum jet range is therefore achieved by flying at the speed providing the best ratio of TAS to drag. The diagram referred to in this question includes a total drag curve. The ratio of TAS : drag at any point on a total drag curve, can be found by drawing a straight line from the origin to the point under consideration. The ratio of TAS : drag is proportional to the slope of the line. If for example the line were vertical, it would equate to infinite thrust required at zero TAS. This would indicate zero TAS : drag. A horizontal line would equate to zero thrust required at infinite TAS. This would indicate an infinitely high TAS : drag. Neither a vertical, nor horizontal line will touch the total drag curve however so these infinitely high and infinitely low ratios are not achievable. The greatest ratio of TAS : drag occurs where the line of shallowest slope just touches the total drag curve. This is point E in the diagram, which corresponds to option d. Jet aircraft best range therefore occurs as indicated in option d.

CURVES 18. c. The diagram is a standard representation of the drag forces affecting an aircraft. It illustrates the total drag curve, together with the two main components of which it is comprised. Curve A represents lift induced drag. Below the stalling angle, lift induced drag is proportional to angle of attack. This is greatest at low speeds where very high angles of attack are required to generate the necessary lift force. As airspeed increases, the required angle of attack, and hence lift induced drag decreases such that induced drag is proportional to 1/EAS2.

CURVES 19. a. Curve B represents total drag, which is the sum of induced drag plus profile drag. Induced drag is proportional to I/EAS~, whilst profile drag is proportional to EAS2. Total drag is therefore proportional to EAS2 + I / EAS2, as described in option a.

CURVE 20. a. Curve C represents profile drag. This is proportional to E A S ~Curve . C is therefore proportional to EAS2, as described in option a.

CURVES 21. a. The diagram represents a series of total drag curves. In each case the value of minimum drag and the shape of the CURVES are identical. Each successive curve occurs at a higher TAS however. This means that whatever caused the changes from curve 1 to 2 to 3, caused the same values of drag to occur but at higher values of TAS. Eoth total drag and indicated airspeed (IAS) are proportional to dynamic pressure (%pV2). This means that any given IAS will occur at the same value of (%pV2) at all altitudes. This in turn means that when climbing at constant IAS, the drag force will remain constant. But climbing at constant IAS causes TAS to increase, so as altitude increases, the same value of drag occurs at the same value of IAS but at increasing values of TAS. The diagram therefore represents the effects of increasing altitude, on the relationship between drag and TAS.

CURVES 22. b. Although the curves in this diagram might represent either drag or power required, only power required is considered in the various options. It can therefore be assumed that the curves refer to power required.

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Whatever has caused the changes from curve 1 to 2 to 3, has caused the minimum value of power required to increase, and to occur at a higher EAS. It should also be noted that the range between minimum and maximum speeds (at the ends of the curves) becomes narrower with each successive curve. The cause of these changes must therefore be something that increases power required at any speed, increases minimum power, increases,V ,, and reduces the available speed range. Of the options offered in this question, only option b, increasing weight has these effects. Option a, decreasing weight will have the opposite effects. Options c and d refer to changes in altitude. Although these would have broadly similar effects, each successive curve would cross over its neighbours. That is to say, if weight increases it will increase the power required at all speeds. If altitude increases it will increase the power required at low speeds, but decrease the power required at some high speeds.

CURVES 23. d. Although the curves in this diagram might represent either drag or power required, only power required is considered in the various options. It can therefore be assumed that the curves refer to power required. Whatever has caused the changes from curve 1 to 2 to 3, has caused the minimum value of power required to increase, and to occur at a higher EAS. It should also be noted that the range between minimum and maximum speeds (at the ends of the curves) becomes narrower with each successive curve, and that the curves cross at speeds above .V ,, The cause of these changes must therefore be something that increases power required at low speeds, increases minimum power required and,V ,, but decreases the power required at some high speeds, and reduces the ava~lablespeed range. Of the options offered in this question, only option d, increasing altitude, has these effects. Option c, decreasing altitude will have the opposite effects. Options a and b refer to changes in weight. Although these would have broadly similar effects, each successive curve would not cross over its neighbours. That is to say, if weight increases it will increase the power required at all speeds. If altitude increases it will increase the power required at low speeds, but decrease the power required at some high speeds. The curvescurves in this diagram therefore represent the effects of increasing altitude.

CURVES 24. b. The curves in this diagram might represent either drag or power required. Because both power and drag are offered in the options, both must be considered. Option a refers to the effects of changing altitude on the drag force at any given EAS. But Both total drag and equivalent airspeed (EAS) are proportional to dynamic pressure (%pV2).This means that any given EAS will occur at the same value of (%pV2) at all altitudes. This in turn means that when climbing at constant EAS, the drag force will remain constant. The CURVES in this diagram cannot therefore represent the relationship between drag and EAS with changing altitudes. But as altitude increases at any given EAS, the TAS increases. This means that as altitude increases, the same drag curves occur, but at greater values of TAS. This effect is accurately illustrated by the curves in the diagram. Options c and d refer to the relationship between power required and changes in altitude. Power required is equal to drag multiplied by TAS. As altitude increases or decreases at any given EAS, the equivalent TAS increases or decreases, thereby causing power required to increase or decrease. But the CURVES in the diagram show no such changes in power required so these options must be incorrect.

CURVES 25. c. Curve B could represent either total drag or power required. Curve A starts at a high point then decrease with increasing airspeed. This is typical of the relationship between propeller thrust and airspeed. In the

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case of a jet aircraft however, the thrust is highest at zero speed, decreases slightly up to about 250 Kts, before again increasing with increasing airspeed. In the case of both propeller and jet aircraft, power available is zero at zero airspeed. This diagram therefore most probably represents thrust available and drag for a propeller aircraft.

CURVES 26. a. Curve A is zero at zero airspeed, increases up to a maximum, then decreases at higher speeds. This is typical of the power available from a propeller aircraft. Had the curve been representative of propeller thrust it would have been at a maximum at zero speed. Curve B is generally bucket shaped, with values at the right side or high speed range being higher than those at the left side or low speed range. This is typical of a power required curve. The most appropriate option is therefore a, power available and power required for a propeller aircraft.

CURVES 27. d. Curve A is approximately a straight horizontal line. Closer inspection however reveals a gradual reduction in values up to the mid speed range, followed by a gradual increase at higher airspeeds. This is typical of the thrust available from a jet engine. Jet thrust is maximum at zero airspeed, and decreases with increasing speed up to about 250 Kts. It then increases at higher airspeeds as ram effect increases the mass flow of air through the engine. Curve B is generally bucket shaped, with values at the right side or high speed range being higher than those at the left side or low speed rarrge. This is typical of a power required curve. These curves most probably represent thrust available and power required for a jet aircraft. This option is not available however, so option d, thrust available and drag for a jet aircraft, is the most appropriate option available.

CURVES 28. b. Curve A is an approximately straight line angled upwards from left to right. Closer inspection however reveals a slight dip in the area of the middle of this curve. This is typical of the power available from a jet engine. The propulsive power available from any propulsion system is equal to the thrust produced multiplied by the TAS. The power available from both jets and propellers is therefore zero at zero airspeed, where thrust is maximum but TAS is zero. Curve A, with values increasing throughout the airspeed range is typical of the power available from a jet aircraft. Curve B is generally bucket shaped, with similar values at both ends of the speed range. This is typical of a drag curve. These curves most probably represent power available and drag for a jet aircraft. This option is not available however, so option b, power ava~lable and power required for a jet aircraft, is the most appropriate option available.

CURVES 29. c. The diagram shows a single drag curve, so whatever has caused the changes has not affected drag. This rules out changes in weight as these would affect drag at all airspeeds. The horizontal axis represents EAS, and changes in altitude do not alter the drag at any given EAS. The changes might therefore be caused by an increase in altitude. Curves 1 and 2 are approximately straight horizontal lines. Closer inspection however reveals a gradual reduction in values up to the mid speed range, followed by a gradual increase at higher airspeeds. This is typical of the thrust available from a jet engine. Jet thrust is maximum at zero airspeed, and decreases with increasing speed up to aboc;t 250 Kts. It then increases at higher airspeeds as ram effect increases the mass flow of air through the engine. These curves therefore probably represent thrust available from a jet aircraft.

But curve 2 is lower than curve 1, indicating lower values of thrust at all airspeeds. This is typical of the effect of increasing altitude which, by reducing air density, reduces the thrust available at all airspeeds. The diagram therefore most probably represents the effects of increasing altitude, which decreases jet thrust but does not affect the drag : EAS relationship.

CURVES 30. d. The diagram shows two bucket shaped curves with values approximately equal at both ends. This is typical of the shape of a drag curve. There are however two of these curves, both at the same vertical position, but with nurnber two displaced towards the low speed end of the airspeed range. If these curves represent values of drag, then whatever caused the change from curve 1 to curve 2, has not changed the values of drag but has caused these values to occur at lower values of TAS. This is typical of the effect of a decrease in altitude. As altitude decreases the drag at any given EAS does not change but the TAS at any EAS decreases. This means that any given drag value occurs at a lower TAS as altitude decreases. The other two curves are approximately straight horizontal lines. Closer inspection however reveals a gradual reduction in values up to the mid speed range, followed by a gradual increase at higher airspeeds. This is typical of the thrust available from a jet engine. Jet thrust is maximum at zero airspeed, and decreases with increasing speed up to about 250 Kts. It then increases at higher airspeeds as ram effect increases the mass flow of air through the engine. These curves therefore probably represent thrust available from a jet aircraft. But curve 2 is higher than curve 1, indicating that whatever caused the change has increased the thrust available at all airspeeds. This is typical of the effect of an altitude decrease which, by increasing air density, increases the mass flow of air through the engines. The most probable cause of these effect is therefore a decrease in altitude.

CURVES 31. c. The diagram shows a single drag curve, so whatever has caused the changes has not affected drag. This rules out changes in weight as these would affect drag at all airspeeds. The horizontal axis represents EAS, and changes in altitude do not alter the drag at any given EAS. The changes might therefore be caused by an increase or decrease in altitude. The other two curves are approximately straight lines, angled upwards from left to right. Closer inspection however reveals a slight dip in the area of the middle of this curve. This is typical of the power available from a jet engine. The propulsive power available from any propulsion system is equal to the thrust produced multiplied by the TAS. The power available from both jets and propellers is therefore zero at zero airspeed, where thrust is maximum but TAS is zero. Curves 1 and 2, with values increasing throughout the airspeed .range are typical of the power ava~lablefrom a jet aircraft. But the gradient of curve 2 is shallower than that of curve 1, indicating lower values of power available at all non zero airspeeds. 'This is typical of the effect of increasing altitude which, by decreasing air density, decrease the mass flow of air through the engines. The effects illustrated in this diqgram were therefore most probably caused by an increase in altitude.

CURVES 32. c. The vertical axis is marked P, indicating that the curves represent power. The upper curve is generally bucket shaped with the minimum value part way between the two ends. This is typical of a power required curve. The lower curve is zero at zero airspeed, reaches a maximum at a medium airspeed, then decreases with further increases in airspeed. This is typical of the power available from a propeller aircraft. It can be distinguished from that of a jet aircraft which is approximately straight and angled upwards from left to right. So the upper curve is most probably a power required curve, whilst the lower one is power available for a propeller aircraft.

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The two curves do not cross at any point , but are tangential at a single point. This means that power available is less than power required at all speeds, except that at which the two are equal. This occurs at the absolute ceiling (option c), where power available is equal to power required and level flight is possible at only one airspeed.

CURVES 33. b. The vertical axis is marked P, indicating that the curves represent power. The upper curve is generally bucket shaped with the minimum value part way between the two ends. This is typical of a power required curve. The lower curve is zero at zero airspeed, reaches a maximum at a medium airspeed, then decreases with further increases in airspeed. This is typical of the power available from a propeller aircraft. It can be distinguished from that of a jet aircraft which is approximately straight and angled upwards from left to right. So the upper curve is most probably a power required curve, whilst the lower one is power available for a propeller aircraft. The two curves do not cross at any point , but are tangential at a single point. This means that power available is less than power required at all speeds, except that at which the two are equal. This occurs at the absolute ceiling, where power available is equal to power required and level flight is possible at only , and V,, and one airspeed. For a propeller aircraft this speed is Vx, which is slightly lower than ,V Option b, (Vx and )V ,, is therefore the most appropriate. considerably lower than .,V,

CURVES 34. b. The vertical axis is marked P, indicating that the curves represent power. The upper curve is generally bucket shaped with the minimum value part way between the two ends. This is typical of a power required curve. Point A is where a straight line drawn from the origin just touches the power required curve. This represents the speed providing the best ratio of TAS : power required. This speed is also the minimum which in turn is the best angle of climb speed (Vx) for jet aircraft. The minimum power drag speed (V), (), is much lower, at the bottom of the power required curve (point B). The most appropriate speed V option is therefore b, (Vx Jet and V .),

CURVES 35. c. The vertical axis is marked P, indicating that the curves represent power. The upper curve is generally bucket shaped with the minimum value part way between the two ends. This is typical of a power required curve. The lower curve is a straight line, startirlg at a value of zero at zero airspeed, with values increasing with increasing airspeed. This is a reasonable approximation of the power available from a jet aircraft. It can be distinguished from that of a propeller aircraft which is more curved, with a maximum value around the mid speed range. So the upper curve is most probably a power required curve, whilst the lower one is power available for a jet aircraft. The two curves do not cross at any point , but are tangential at a single point. This means that power available is less than power required at all speeds, except at that one point, where the two are equal. This occurs at the absolute ceiling (option c), where power available is equal to power required and level flight is possible at only one airspeed.

CURVES 36. b. The diagram shows a single drag curve, so whatever has caused the changes has not affected drag. But the horizontal axis represents TAS, so this rules out charlges in altitude, as these would affect drag at all true airspeeds. If the altitude had increased, it would have caused the same drag curve to have been repeated at higher values of TAS, causing a second drag curve to appear to the right of the first.

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325

The other two curves are approximately straight lines, angled upwards from left to right. This is a reasonable approximation of the power available from a jet engine. The propulsive power available from any propulsion system is equal to the thrust produced multiplied by the TAS. The power available from both jets and propellers is therefore zero at zero airspeed, where thrust is maximum but TAS is zero. Curves 1 and 2, with values increasing throughout the airspeed range are typical of the power available from a jet aircraft. But the gradient of curve 2 is shallower than that of curve 1, indicating lower power available at all airspeeds. Although this might have been caused by an increase in altitude, this is ruled out by the absence of a second drag curve. Of the options offered in this question, the most probable cause of the changes is therefore option b, the failure of a single jet engine. This would reduce power available, without significantly affecting drag. It should however be noted that in reality, the subsequent trimming of the aircraft to overcome asymmetric thrust caused by the engine failure would increase drqg.

CURVES 37. a. The diagram shows two bucket shaped curves with values approximately equal at both ends. This is typical of the shape of a drag curve. There are however two of these curves, one of which is higher, narrower and further to the right, than the other. If these are drag curves, then points 1 and 2 represent the for two different flight conditions. Whatever caused minimum drag values and minimum drag speeds (V), , to the change from point 1 to point 2, has decreased the values of drag at all speeds, and caused ,V occur at a lower EAS. This effect would not be achieved by a change in altitude, as such a change would not alter the drag at any given EAS. A decrease in weight would however cause the effects shown. The other curve starts at a high point then decreases with increasing airspeed. This is typical of the relationship between propeller thrust and airspeed. It is totally unlike the jet thrust curve or the power curves for jets or propellers. Any change in altitude would change the thrust available, causing two thrust curves to be shown in the diagram. The lack of a second thrust curve is further evidence that the effects were not caused by changes in altitude. Changes in weight would not however alter the thrust available, so only a single thrust curve, together with two drag curves, would be required to illustrate the effect of weight changes. The diagram therefore most probably represents the effects of a decrease in weight.

CURVES 38. b. The diagram shows a two bucket shaped curves with values approximately equal at both ends. This is typical of the shape of a drag curve. There are however two of these curves, one of which is higher, narrower and further to the right, than the other. Whatever caused the change from point 1 to point 2, has increased the values of drag at all speeds, and caused,V , to occur at a higher EAS. This effect would not be achieved by a change in altitude, as such a change would not alter the drag at any given EAS. An increase in weight would however cause the effects shown. The other curve starts at a high point then decreases with increasing airspeed. This is typical of the relationship between propeller thrust and airspeed. It is totally unlike the jet thrust curve or the power curves for jets or propellers. If the first two curves are drag curves, then points 1 and 2 represent the minimum speeds at which sufficient thrust is available to balance drag. Any change in altitude would change the thrust available, causing two thrust curves to be shown in the diagram. The lack of a second thrust curve is therefore further evidence that the effects were not caused by changes in altitude. Changes in weight would not however alter the thrust available, so only a single thrust curve, together with two drag curves, would be required to illustrate the effect of weight changes. In the case of an increase in weight, the drag curve would move upwards and to the right as indicated in the diagram. The diagram therefore most probably represents the effects of a decrease in weight. The change from point 1 to point 2 therefore probably represents the increase in minimum flying speed caused by an increase in weight.

CURVES 39. a. The vertical axis in this diagram is marked D, so the bucket shaped curve is most probably a total drag curve. The second curve starts at a high point then decreases with increasing airspeed. This is typical of the relationship between propeller thrust and airspeed. It is totally unlike the jet thrust curve or the power curves for jets or propellers. The second curve therefore probably represent thrust available for a propeller aircraft.

Vx is the airspeed at which an aircraft is able to achieve its best angle of climb. In the case of a propeller aircraft Vx is close to the lowest safe fly speed and is lower than the minimum power speed,.,V, ,V , can be found on a total drag curve by drawing a straight line from the origin to intersect the drag curve at right angles. This is at point B in the diagram. The most probable location of the best angle of climb speed on this diagram is therefore point A.

CURVES 40. c. The vertical axis in this diagram is marked D, so the bucket shaped curve is most probably a total drag curve. The second curve starts at a high point and is approximately straight and parallel with the horizontal axis. This is a reasonable approximation of the relationship between jet thrust and airspeed. It is totally unlike the propeller thrust curve or the power curves for jets or propellers. The second curve therefore probably represent thrust available for a jet aircraft.

Vx is the airspeed at which an aircraft is able to achieve its best angle of climb. In the case jet aircraft where thrust is constant, Vx occurs at the minimum drag speed,.,V, This is at point C in the diagram.

CURVES 41. d. The vertical axis in this diagram is marked P, so it may be assumed that the curves represent power required and power available. The convex nature of the lower curve suggests that it is a power available curve for a propeller aircraft. The best rate of clirr~bspeed (V,) is the speed at which the excess power is at its greatest. In the case of a propeller aircraft this occurs at a speed slight greater than the minimum power speed .V ,, ,V , occurs at the lowest point on the power required curve, which is point C in the diagram. The most probable location of the best climb speed in the diagram is therefore point D.

CURVES 42. c. The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. A vertical line for example, would indicate lift with no drag, giving an infinitely great lift : drag ratio. A horizontal line would represent drag with no lift, giving zero lift : drag ratio. A vertical line will not however touch any part of the curve, so there is no flight condition at which an infinitely high lift : drag ratio can be achieved. The best lift : drag ratio occurs at the point where the line of greatest gradient just touches the curve. This is point C in the diagram.

CURVES 43. b. When the engines of an aircraft fail in flight, the aircraft possesses a store of potential energy due to its height, and kinetic energy due to its TAS. This energy cannot be replenished following total engine failure. From the moment of engine failure this store of energy is expended in driving the aircraft forward against the drag force. For maximum glide endurance it is necessary to minimise the rate at which the aircraft sinks. This is achieved by flying at the speed at which the rate of energy dissipation is lowest. Power required is a measure of the energy dissipation rate, so for minimum sink rate an aircraft must glide at the ,, minimum power speed,.V

The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. A vertical line for example, would indicate lift with no drag, giving an infinitely great lift : drag ratio. A horizontal line would represent drag with no lift, giving zero lift : drag ratio. A vertical line will not however touch any part of the curve, so there is no flight condition at which an infinitely high lift : drag ratio can be achieved. The best lift : drag ratio occurs at the point where the line of greatest gradient just which occurs at point C in the diagram. touches the curve. This is the minimum drag speed,,V, The point at which the value of C, is maximum is point A. This represents the low speed stalling condition. The speed in this condition is lower than ,,V, so it can be seen that the airspeed increases from right to , is higher left across the diagram. That is to say point C represents a higher airspeed than point A. ,V , for all aircraft types. The only point on the diagram than the stalling speed (V,) and lower than ,V corresponding to this condition is point 6.Minimum sink rate will therefore be achieved at point 6.

CURVES 44. b. For maximum endurance an aircraft must fly at the speed requiring minimum fuel consumption rate. Fuel consumption in a propeller aircraft is proportional to power output, so minimum fuel consumption rate, and ,, hence maximum endurance occur at the speed requiring minimum power. This is .V The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just which occurs at point C in the diagram. touches the curve. This is the minimum drag speed ,,V, The point at which the value of C, is maximum is at point A. This represents the low speed stalling condition. so it can be seen that the airspeed increases from right to left The speed in this condition is lower than,,V, , is greater than the across the diagram. That is to say point C represents a higher airspeed than point A. ,V , for all aircraft types. The only point on the diagram corresponding to stalling speed (V,) and less than ,V this condition is point 6.lblaximum propeller aircraft endurance therefore occurs at point B.

CURVES 45. c. For maximum range an aircraft must fly at the speed providing the best ratio between TAS and fuel consumption. Fuel consumption is a propeller aircraft is proportional to power required, so best range ,, This is also requires the best ratio of TAS : power required. This occurs at the minimum drag speed, .V the speed at which the lift : drag ratio is greatest. The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just which occurs at point C in the diagram. Best touches the curve. This is the minimum drag speed ,,V, propeller aircraft range therefore occurs at point C in the diagram.

CURVES 46. b. When the engines of an aircraft fail in flight, the aircraft possesses a store of potential energy due to its height, and kinetic energy due to its TAS. This energy cannot be replenished following total engine failure. From the moment of engine failure this store of energy is expended in driving the aircraft forward against the drag force. For maximum glide endurance it is necessary to minimise the rate at which this store of energy is expended. The rate of energy expenditure is equal to the power required, so maximum glide ,, endurance is achieved by flying at the minimum power speed, .V

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Curves

The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just which occurs at point C in the diagram. touches the curve. This is the minimum drag speed,,V, The point at which the value of C, is maximum is at point A. This represents the low speed stalling so it can be seen that the airspeed increases from condition. The speed in this condition is lower than,,V, , is right to left across the diagram. That is to say point C represents a higher airspeed than point A. ,V , for all aircraft types. The only point on the diagram greater than the stalling speed (V,) and less than,V corresponding to this condition is point B. Maximum glide endurance therefore occurs at point B.

CURVES 47. c. When the engines of an aircraft fail in flight, the aircraft possesses a store of potential energy due to its height, and kinetic energy due to its TAS. This energy cannot be replenished following total engine failure. From the moment of engine failure this store of energy is expended in driving the aircraft forward against the drag force. For maximum glide range it is necessary to fly at the speed providing the best ratio of TAS: , ,V , is also the speed at which energy consumption rate. This occurs at the minimum drag speed,.,V the best lift : drag ratio is achieved. The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just touches the curve. This is the minimum drag speed,,V , which occurs at point C in the diagram.

CURVES 48. c. The term minimum glide angle means the gliding flight path closest to level flight. In this condition an aircraft will achieve its best glide range. This question therefore requires the identification of the best glide range speed. Glide range is equal to the lift : drag ratio multiplied by the height at the start of the glide. For maximum range from any given height it is necessary to fly at the speed providing the best lift : drag ratio. This is the minimum drag speed,.,V, The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just touches the curve. This is the minimum drag speed,,V, which occurs at point C in the diagram. The minimum glide angle is therefore achieved at point C.

CURVES 49. a. ,

Assuming that sufficient power is available, the lowest speed at which an aircraft can fly is its low speed stalling speed, V., This is the speed at which the value of C, is at its maximum. The diagram in this question is a whole aircraft lift : drag polar, indicating the C, and C, throughout the speed range. The point at which the value of C, is maximum is at point A. This represents the low speed stalling condition.

CURVES 50. d. The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just which occurs at point C in the diagram. touches the curve. This is the minimum drag speed,,V,

The point at which the value of C, is maximum is at point A. This represents the low speed stalling condition. The speed in this condition is lower than,,V, so it can be seen that the airspeed increases from right to left across the diagram. That is to say point C represents a higher airspeed than point A. From the options offered in this question, the point on the diagram indicating the highest speed is therefore point D.

CURVES 51. c. The maximum angle of climb that an aircraft can achieve is proportional to its excess thrust divided by its weight. Excess thrust is equal to thrust available minus drag. The thrust available from a jet aircraft is approximately constant throughout the speed range, so maximum excess thrust occurs at the minimum drag speed. Best jet angle of climb therefore occurs when flying at the minimum drag speed,.,V, The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just touches the curve. This is the minimum drag speed ,,V, which occurs at point C in the diagram.

CURVES 52. d. The maximum rate of climb that an aircraft can achieve is equal to its excess power divided by its weight. Excess power is power available minus power required. Jet aircraft power increases approximately linearly with airspeed, whilst power required increases up to a maximum close to the middle of the speed range, then decrease with increasing speed. Maximum excess power occurs at a speed greater than.,V, The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just touches the curve. This is the minimum drag speed,,V, which occurs at point C in the diagram. The only point representing a speed greater than ,V , in the diagram is point D. The most probable location of the best jet rate of climb speed is therefore point D.

CURVES 53. d. Jet aircraft invariably cruise at speeds considerable greater than their minimum drag speed, .,V, The diagram in this question is a whole aircraft lift : drag polar, indicating the variation in the lift : drag ratio. The lift : drag ratio at any point on the curve is proportional to the gradient of a straight line joining that point with the origin. The best lift : drag ratio occurs at the point where the line of greatest gradient just touches the curve. This is the minimum drag speed,,V, which occurs at point C in the diagram. The only point , in the diagram is point D. The most probable location of the best jet representing a speed greater than ,V cruising speed is therefore point D.

CURVES 54. a. The diagram is a whole aircraft L:D polar. Two points can be accurately located. The First ,is, ,C (point , which occurs where a straight line, A), which occurs at C, the low speed stall (V,). The second is,,V drawn from the origin, just touches the curve. This is point D.

CURVES 55. d. The diagram is a whole aircraft L:D polar. Two points can be accurately located. The first is ,,C , (point which occurs where a straight line, drawn A), which occurs at the low speed stall (V,). The second is,,V, ,From the origin, just touches the curve. This is point D. The possible identities of all other points can then To identify other pointsfitshould first be be identified on the basis of their positions relative to V, and.,V,

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noted that V, (point A) is lower than ,V , (point D), so the speed scale must increase from ieft to right. Point E is therefore at a speed greater than.,V, Of the options provided in this question, only d, VyJe,, is greater than,,V , so point E must be VyJe,.

CURVES 56. d. The diagram is a whole aircraft L:D polar.Two points can be accurately located. The first is ,,C , (point A), which occurs which occurs at the low speed stall (V,). The second is,,V, where a straight line, drawn from the origin, just touches the curve. This is point D. The possible identities of all other points can then be identified on the basis of their positions relative to V, and.,V ,

I CD

To identify other points it'should first be noted that V, (point A) is lower than ,V , (point D), so the ssked scale must increase from left to right. Point C is therefore at a speed greater than V, but lower,than.,V, Of the options provided in this question, b (V,) c (VxProD), and d (VyProD), all satisfy this condition. But Vxpropis less than,V ,, which is less than Vyprop,and only two points, 5 and Care indicated in the diagram, in this part of the speed range. Point 5 is marked ,V ,, so point C can only be Vyprop,

CURVES 57. a. The diagram is a whole aircraft L:D polar. Two points can be accurately located. The greatest value of C, occurs at point A. In all aircraft the greatest value of C,, ,is, ,C which occurs at the stalling speed V., Point A is therefore V.,

CD

CURVES 58. b. The diagram is a whole aircraft L:D polar. Two points can be ,,C , (point A), which occurs accurately located. The first is which occurs at the low speed stall (V,). The second is,,V, where a straight line, drawn from the origin, just touches the curve. This is point D. The possible identities of all other points can then be identified on the basis of their positions relative to V, and.,V , To identify other points it should first be noted , (point D), so the speed that V, (point A) is lower than ,V scale must increase From right to left. Point B is therefore at Of the options a speed greater than V, but lower than .,V, satisfies thiscondition. provided in this question, only b (V),

CD

CURVES 59. b. The diagram is a whole aircraft L:D polar. Two points can be ,,C , (point A), which occurs accurately located. The first is , which at C, the low speed stall (V,). The second is,,V occurs where a straight line, drawn from the origin, just touches the curve. This is point D. The possible identities of all other points can then be identified on the basis of their positions relative to V, and.,V,

CD

To identify other points it should first be noted that V, (point A) is lower than,V , (point D), so the speed so point B is therefore at scale must increase from right to left. The question specifies that point c is,,V, Of the options provided in this question, only b ,),V ,(,, a speed greater than V, but lower than .,V, satisfies this condition.

CURVES 60. b. The diagram is a whole aircraft L:D polar. Two points can be ,,C , (point A), which occurs accurately located. The first is at C, the low speed stall (V,). The second is,,V, which occurs where a straight line, drawn from the origin, just touches the , is also prop best range speed. curve. This is point D. But,V

CD

CURVES 61. c.

CL

The power available from any aircraft engine is equal to thrust multiplied by TAS. In the case of a jet engine, thrust is approximately constant with increasing TAS, so power available increases linearly with increasing TAS. Options A and C ill the diagram might therefore be representativeof jet power available. As altitude increases air density decreases causing power available to reduce. The line representing jet power available at high altitude would therefore be of a shallower gradient than that representing sea level performance. This is most accurately represented by line C. Lines B and D are representative of power available from piston engines at high and low altitudes respectively.

A

pAv

c

EAS

A

CURVES 62. d. Power available from any aircraft engine is equal to thrust multiplied by TAS. In the case of a propeller aircraft, thrust decreases with increasing speed such that power available increases from zero at zero TAS, reaching a maximum value at about 300 Kts, then reducing as speed increases further. This effect rr~ightbe represented by curves B and D on the diagram. As altitude increases, air density decreases, causiqg power available to reduce. The gradient of the power available curve is therefore shallower at high altitude than at low altitude. Curve D is therefore most representative of the power available from a propeller at high altitude.

CURVES 63. a. Jet power available increases linearly with increasing speed and so might be represented by lines Aand C on the diagram. But as altitude increases, decreasing air density causes power available to decrease, reducing the gradient of the power available : EAS line. Line A therefore best representsjet power available at low altitude. And line C best represents jet power available at high altitude.

'AV

C

EAS

I

EAS

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Curves

CURVES 64. b. Propeller power available is zero at zero airspeed, increases to a maximum value at about 300 Kts then decreases as speed increases further. This might be represented by curves B and D on the diagram. But as altitude increases, air PA, density reduces, causing power available to decrease. This has the effect of reducing the gradient of the power available curve at high altitude.

PA"

Curve B therefore best represents propeller power available at low altitude. Curve D best represents propeller power available at high altitude.

CURVES 65. d. Propeller thrust decreases with increasing EAS in a manner similar to curves C and D in the diagram. But as altitude increases, thrust decreases, such that the high altitude thrust curve is lower than the low altitude curve. Curve D is therefore the most representative of propeller thrust at high altitude.

EAS

CURVES 66. c. Propeller thrust decreases with increasing EAS in a manner similar to curves C and D in the diagram. But as altitude increases, thrust decreases, such that the high altitude thrust curve is lower than the low altitude curve. Curve C is therefore the most representative of propeller thrust at low altitude.

D EAS

CURVES 67. b. Jet thrust available is maximum atzero EAS, then decreases to a minimum value at about 250 Kts before again increasing at higher speeds as indicated in curves A and B. As altitude increases, air density T and thrust decrease so the curve is lower at high altitude. Option B is therefore the high altitude thrust available curve.

CURVES 68. a. Jet thrust available is maximum at zero EAS, then decreases to a minimum value at about 250 Kts before again increasing at higher speeds as indicated in curves A and B. As altitude increases, air density T and thrust decrease so the curve is lower at high altitude. Option A is therefore the low altitude thrust available curve.

CURVES 69. c.

T

D I

T

D EAS

Whatever has brought about the change from curve 1 to curve 2, has caused the aircraft to require a longer take-off run and to achieve a lower climb gradient. This question can best be solved by considering

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each of the options in turn. Option a, increased power setting would decrease the take-off run, but would increase the maximum climb gradient. Option b, decreased flap angle from the normal take-off setting would increase the take speed and hence increase the take-off run. The reduced drag at the lower flap setting would however increase the climb gradient. Option c, increased weight would increase the take-off run and decrease the climb gradient as indicated in the diagram. Option d, decreased weight would have the opposite effect. Option c, increased weight is therefore the most appropriate.

CURVES 70. a. Whatever has caused the change from take-off path 1 to path 2 has decreased both the take-off run and climb gradient. Option a, increased flap deployment beyond the optimum angle would increase both the lift and drag forces at any given airspeed. This would decrease both the lift off speed and acceleration rate. Provided the increased flap angle were not too great however, the overall effect would be to decrease both the take-off run and the climb gradient. The effect of decreased flap angle would be a reduction in the lift and drag at any given airspeed, but an increase in lift : drag ratio. The overall effect would be an increase in both the take-off run and climb gradient. Option c, increased weight would decrease acceleration rates and increase the lift off speed. This would increase the take-off run and decrease the climb gradient. Option d, decrease in weight would have the opposite effect, decreasing take-of run and increasing climb gradient. Option a, increased flap angle is therefore the most probable cause of the effects illustrated. It should however be noted that the use of large flap angles would cause a disproportionately large increase in drag, thereby increasing the take-off run and decreasing climb gradient.

CURVES 71. b. Whatever has caused the change from take-off path 1 to path 2 has increased both the take-off run and climb gradient. Option a, increased flap deployment beyond the optimum angle would increase both the lift and drag forces at any given airspeed. This would decrease both the lift off speed and acceleration rate. Provided the increased flap angle were not too great however, the overall effect would be to decrease both the take-off run and the climb gradient. The effect of decreased flap angle would be a reduction in the lift and drag at any given airspeed, but an increase in lift : drag ratio. The overall effect would be an increase in both the take-off run and climb gradient. Option c, increased weight would decrease acceleration rates and increase the lift off speed. This would increase the take-off run and decrease the climb gradient. Option d, decrease in weight would have the opposite effect, decreasing take-of run and increasing climb gradient. Option b, decreased flap angle is therefore the most probable cause of the effects illustrated.

CURVES 72. a. Whatever has brought about the change from climb path 1 to climb path 2 has increased the take-off run and decreased the climb gradient. Option a, increased temperature, would decrease air density, thereby reducing both the thrust provided by the engines and t'.e lift at any given combination of angle of attack and airspeed. The reduced thrust would reduce the rate of acceleration and hence increase the take-off run, whilst the reduced lift would reduce the climb gradient. Increased temperature would therefore have the effects shown in the diagram. Option b, decreased temperature, would have the opposite effect, increasing both thrust and lift, and hence decreasing the take-off run and increasing the climb gradient. Option c, decreased flap angle, would increase stalling speed, thereby increasing lift off speed. Although the reduced drag would increase acceleration rate thereby offsetting the increased lift off speed, it would also increase the climb gradient.

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Curves

The overall effect on take-off run would however depend upon the relative magnitude of the decrease in drag and the decrease in lift. Option d, decreased wejg ht, would decrease the take-off run and increase the climb gradient. Option a, increased temperature is.therefore the most probable cause of the effect shown in the diagram.

CURVES 73. b. Whatever has brought about the change from climb path 1 to climb path 2 has decreased the take-off run and increased the climb gradient. Option a, increased temperature, would decrease air density, thereby reducing both the thrust provided by the engines and the lift at any given combination of angle of attack and airspeed. The reduced thrust would reduce the rate of acceleration and hence increase the take-off run, whilst the reduced lift would reduce the climb gradient, Increased temperature would therefore have the opposite effect to that illustrated in the diagram. Option b, decreased temperature, would have the opposite effect to that described above, increasing both thrust and lift, and hence decreasing the take-off run and increasing the climb gradient. Option b would therefore have the effect illustrated. Option c, increased flap angle, would decrease stalling speed, thereby decreasing lift off speed. Although the increased drag would decrease acceleration rate thereby offsetting the decreased lift off speed, it would also decrease the climb gradient. Option d, increased weight, would increase the take-off run and decrease the climb gradient. Option b, decreased temperature is therefore the most probable cause of the effect shown in the diagram.

CURVES 74. a. Whatever has brought about the change from climb path 1 to climb path 2 has increased the take-off run and decreased the climb gradient. Option a, increased altitude, would decrease airdensity, thereby reducing both the thrust provided by the engines and the lift at any given combination of angle of attack and airspeed. The reduced thrust would reduce the rate of acceleration and hence increase the take-off run, whilst the reduced lift would reduce the climb gradient. Increased altitude would therefore have the effects shown in the diagram. Option b, decreased temperature, would have the opposite effect, increasing both thrust and lift, and hence decreasing the take-off run and increasing the climb gradient. Option c, decreased flap angle, would increase stalling speed, thereby increasing lift off speed. Although the redticed drag would increase acceleration rate, thereby offsetting the increased lift off speed, it would also increase the climb gradient. The overall effect on take-off run would however depend upon the relative magnitude of the decrease in drag and the decrease in lift. Option d, decreased weight, would decrease the take-off run and increase the climb gradient. Option a, increased altitude is therefore the most probable cause of the effect shown in the diagram.

CURVES 75. a. Whatever has brought about the change from climb path 1 to climb path 2 has increased the take-off run but has not affected the climb gradient, Option a, early rotation would increase the drag force, thereby decreasing the post rotation acceleration rate. This would increase the time taken to reach lift off speed, and hence increase the take-off run. Early rotation would not however affect the climb gradient. Option a, would therefore have the effects illustrated. Option b, decreased temperature, would increase air density, thereby increasing both thrust and lift. This would decrease the take-off run and increase the climb gradient. Option c, decreased flap angle, would increase stalling speed, thereby increasing lift off speed. Although the reduced drag would increase

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acceleration rate thereby offsetting the increased lift off speed, it would also increase the climb gradient. The overall effect on take-off run would however depend upon the relative magnitude of the decrease in drag and the decrease in lift. Option d, decreased weight, would decrease the take-off run and increase the climb gradient. Option a, early rotation is therefore the most probable cause of the effect shown in the diagram.

CURVES 76. a. Whatever has brought about the change from take-off path 1 to path 2 has increased the take-off run but increased the climb gradient. Option a, late rotation would increase the distance covered before lift-off, thereby increasing the take-off run. The increased period of acceleration prior to rotation would however increase the speed at which lift-off occurs. This increased speed would increase the climb gradient. Option a, would therefore have the effects illustrated. Option b, decreased temperature, would increase air density, thereby increasing both thrust and lift. This would decrease the take-off run and increase the climb gradient. Option c, decreased flap angle, would increase stalling speed, thereby increasing lift off speed. Although the reduced drag would increase acceleration rate thereby offsetting the increased lift off speed, it would also increase the climb gradient. The overall effect on take-off run would however depend upon the relative magnitude of the decrease in drag and the decrease in lift. Option d, decreased weight, would decrease the take-off run and increase the climb gradient. Option a, delayed rotation is therefore the most probable cause of the effect shown in the diagram.

CURVES 77. b. Whatever has brought about the change from take-off path 1 to take-off path 2 has decreased the take-off run and increased the climb gradient. Option a, increased density altitude would decrease the air density, thereby decreasing the thrust produced by the engines. This would increase the take-off run and decrease the climb gradient. Option b, decreased density altitude would however have the opposite effect. By increasing air density and thrust, it would decrease the take-off run and increase the climb gradient, as illustrated in the diagram. Option c, decreased pitch attitude would decrease the lift generated at any given airspeed. This would increase the take-off run and decrease the climb gradient. Although option d, increased pitch attitude would increase the lift generated at any given airspeed, it would produce a proportionally greater increase in drag. This would increase the take-off run and decrease the climb gradient. Option b, decreased density altitude is therefore the most probable cause of the effects illustrated.

CURVES 78. a. Whatever has brought about the change from take-off path 1 to take-off path 2 has decreased the take-off run, but not affected the climb gradient. This effect would result from anything which increased the acceleration rate without altering the lift : drag ratio of the aircraft, or thrust available. Option a, increased down slope would increase the acceleration rate during the take-off run, thereby enabling the aircraft to reach lift-off speed over a shorter distance. Down slope does not however affect the subsequent climb gradient. This option would therefore have the effects illustrated. Option b, decreased down slope would have the opposite effect. By decreasing acceleration rate the takeoff run would be increased, but the climb gradient would not be affected. Options c, and d, increasing and decreasing pitch angle, would both increase the take-off run and decrease the climb gradient. The most probable cause of the effects illustrated is therefore option a, increased down slope.

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Curves

CURVES 79. b. Whatever has caused the change from take-off climb path 1 to take-off climb path 2 has increased the take-off run without affecting the subsequent climb gradient. This effect might be caused by any factor that reduced the rate of acceleration during the take-off run, but did not affect the lift : drag ratio, or thrust available. Option a, increased down slope would increase the acceleration rate during the take-off run, thereby enabling the aircraft to reach lift-off speed over a shorter distance. Down slope does not however affect the subsequent climb gradient, Option b, increased up slope would have the opposite effect. By decreasing acceleration rate the take-off run would be increased, but the climb gradient would not be affected. These are the effects illustrated in the diagram. Option c3decreased density altitude'would increase air density thereby increasing the thrust produced by the engines. This would decrease the take-off run and increase the climb gradient. Option d, increased density altitude would have the opposite effect. By decreasing the air density and hence thrust, it would increase the take-off run and decrease the climb gradient. Option b, increased up slope is therefore the most probable cause of the effects illustrated.

CURVES 80. b. Whatever has brought about the change from take-off path 1 to path 2 has decreased the take-off run and increased the climb gradient. Option a, an increased tailwind would increase the ground speed equating to the lift-off airspeed. Although this would decrease aerodynamic drag during the take-off run, such drag is relatively small compared to the rolling friction caused by the undercarriage. The overall effect would therefore be an increase in the takeoff run. Also, because the mass of air through which the aircraft climbed would be moving for~vardwith it, the climb gradient would be decreased. Option b, decreased tailwind would have the opposite effects, decreasing take-off run and increasing climb gradient as illustrated in the diagram. Option c, decreased pitch attitude would decrease the lift generated at any given airspeed. This would increase the take-off run and decrease the climb gradient. Although option d, increased pitch attitude would increase the lift generated at any given airspeed, it would produce a proportionally greater increase in drag. This would increase the take-off run and decrease the climb gradient. Option b, decreased tailwind is therefore the most probable cause of the effects illustrated.

CURVES 81. a. Whatever has brought about the change from take-off path 1 to path 2 has decreased the take-off run and increased the climb gradient. Option a, an increased headwind would decrease the ground speed equating to the lift-off airspeed. This would decrease in the takeoff run. Also, because the mass of air through which the aircraft climbed would be moving in the opposite direction, the climb gradient would be increased. Option b, decreased headwind would have the opposite effects, increasing take-off run and decreasing climb gradient. Option c, decreased pitch attitude would decrease the lift generatec! at any given airspeed. This would increase the take-off run and decrease the climb gradient. Although option d, increased pitch attitude would increase the lift generated at any given airspeed, it would produce a proportionally greater increase in drag. This would increase the take-off run and decrease the climb gradient. Option a, decreased tailwind is therefore the most probable cause of the effects illustrated.

CURVES 82. a. Whatever has brought about the change from take-off path 1 to 2 has increased the take-off run but not affected the climb gradient. Option a, contaminated runway might increase the take-off run if the contaminant increases rolling resistance. This would occur with contaminants such as soft snow or standing water. A

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contaminated runway would not however affect the subsequent climb gradient, provided the degree of contamination was not of such magnitude as to require a reduction in take-off weight. If this were the case then the take-off distance might remain unchanged whilst the climb gradient increased, compared with a clean runway. Option b, a decreased tailwind would decrease the ground speed equating to the lift-off airspeed. Although this would increase aerodynarr~icdrag during the take-off run, such drag is relatively small compared to the rolling friction caused by the undercarriage. The overall effect would therefore be a decrease in the takeoff run. Also, because the mass of air through which the aircraft climbed would be moving forward with it at a lower speed, the climb gradient would be increased. Option c, increased density altitude would decrease air density and hence thrust available. This would increase the take-off run and decrease climb gradient. Option d, increased weight would necessitate a greater lift-off speed. This would increase the take-of run and decrease climb gradient. Overall option a, contaminated runway is therefore the one most likely to have caused the effects illustrated.

CURVES 83. c. Whatever has brought about the change from take-off path 1 to path 2 has increased the take-off run and decreased the climb gradient. Option a, decreased tailwind would decrease the ground speed equating to the lift-off airspeed. Although this would increase aerodynamic drag during the take-off run, such drag is relatively small compared to the rolling friction caused by the undercarriage. The overall effect would therefore be a decrease in the takeoff run. Also, because the mass of air through which the aircraft climbed would be moving forward with it at a lower speed, the climb gradient would be increased. Option b. increased headwind would have similar effects to a decreased tailwind. Option c, increased field elevation would decrease air density at any given ambient temperature. The would decrease the thrust available, thereby increasing take-off run and decreasing climb gradient. Option d, decreased weight would decrease lift-off speed and lift and drag at any given speed. This would decrease take-off run and increase climb gradient. Option c, increased field elevation is therefore the most probable cause of the effects illustrated.

CURVES 84. a. Whatever has brought about the change from take-off path 1 to path 2 has decreased the take-off run and increased the climb gradient. Option a, contaminated runway might cause this effect if the contaminant were such that a weight reduction was required. In this case the take-off run might be reduced and the climb gradient would be increased. Option b, decreased headwind would increase the ground speed equating to lift-off airspeed. This would increase the take-off run. Also because the mass of air in which the aircraft was climbing would be moving in the opposite direction at a reduced speed, the climb gradient would be decreased. Option c,- lncreased density altitude would decrease air density and thrust. This would increase the takeoff run and decrease climb gradient. lncreased weight would have a similar effect. Option a, contaminated runway is therefore the most probable cause of the effects illustrated.

CURVES 85. a. Whatever has brought about the change from take-off path 1 to path 2 has decreased the take-off run and increased the climb gradient. Option a, anti-skid system failure would necessitate a weight reduction to reduce the take-off run if the ASDA was insufficient to allow for the greater stopping distance. This weight reduction would decrease take-off run but increase the subsequent climb gradient. Option b, increased

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tailwind would increase the ground speed equating to lift-off airspeed. This would increase tlie take-off run. Also because the air mass in which the aircraft was climbing would move in the opposite direction at a greater speed, the climb gradient would be reduced. Option c, increased field elevation would decrease air density and hence thrust. This would increase the take-off run and decrease climb gradient. Option d, increased weight would have similar effects. Option a, anti-skid system failure is therefore the one most likely to cause the effects illustrated.

SPEED

SPEED 1. c. V, is the design manoeuvre speed. It is the speed at which an aircraft will stall when subjected to the limiting load factor. This question specifies a change in weight, but no change in limiting load factor. The problem can therefore be solved using the following equation: V, at new weight = V, at old weight d (New weight /Old weight) So V, at 70000 Kg = 250 Kts d(70000 1 60000) Which is V, at 70000 Kg = 270 Kts. It should be noted that this method that this method can be employed to calculate the effect of weight changes on all speeds that are fixed multiples of the stalliqg speed VS.

SPEED 2. b. V, is the design manoeuvre speed. It is the speed at which an aircraft will stall when subjected to the limiting load factor. This question involves both a change in weight and a change in flap setting. For JAR certificated passenger aircraft the limiting load factor is 2.5 with flaps up and 2 with flaps down. It must therefore be addressed in two stages. Firstly flaps up V, at the new weight can be calculated using the following equation: V, at new weight = Flaps up V, at old weight d(New weight /Old weight) So V, at 70000 Kg = 250 Kts d(70000 / 60000) Which is V, at 70000 Kg = 270 Kts. The effect of the reduction in limiting load factor must then be calculated using the following equation: Flaps down V, = Flaps up V, d ( ~ l a p sup load factor I Flaps down load factor) So flaps up V, = 270 Kts d(2 12.5) Which is 241.5 Kts The closest option to this figure is option b, 240 Kts. It should be noted that although flap deployment would alter the shape of the C,:a curve, this has been ignored as specified in the question.

SPEED 3. b. V, is the design manoeuvre speed. It is the speed at which an aircraft will stall when subjected to the limiting load factor. This question specifies a change in weight, but no change in configuration or limiting load factor. The problem can therefore be solved using the following equation: V, at new weight = V, at old weight x d(New weight /Old weight) Although no weight are specified in this question, an increase of 25% constitutes a change from an old weight of 1 to a new weight of 1.25. Putting these figures into the equation gives:

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V, after 25% increase in weight = Original V, x v'(1.25 11) Which means that V, after 25% increase in weight = 1.I18 x Original V, This means that a 25% increase in weight will cause an 11.8% (or approximately 12%) increase in V, This question illustrates a more general fact that for small % changes in weight, the % change in stalling speed (or any speed that is a multiple of stalling speed), will be approximately half of the % change in weight.

SPEED 4. b. VMo is the maximum operating speed of an aircraftexpressed as CAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at low altitudes where air density is high, VMo is the limiting parameter. MMo is the maximum operating speed of an aircraft expressed as a mach number. If the aircraft is operated at higher mach numbers the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. MMo is some specific fraction of the local speed of sound. But the local speed of sound is proportional to air temperature and hence increases with decreasing altitude. This means that the CAS equating to MMo increases as altitude decreases. The altitude at which the CAS equating to VMo is the same as that equating to MMo is called the crossover altitude. The overall effect of these factors is that VMo is the limiting speed below the crossover altitude and MMois the limiting speed above the crossover altitude. When descending at constant mach number there is a danger that VMo will be exceeded below the crossover altitude.

SPEED 5. a. VMo is the maximum operating speed of an aircraft, expressed as IAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at low altitudes where air density is high, VMo is the limiting parameter. MMo is the maximum operating speed of an aircraft, expressed as a mach number. If the aircraft is operated at higher mach numbers, the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. ,M , is some specific fraction of the local speed of sound. But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the IAS equating to MMo deceases as altitude increases. The overall effect of these factors is that VMo is the limiting speed at low altitude and MMois the limiting speed at high altitude. The altitude at which the TAS equating to VMo is equal to that equating to MMo is called the crossover altitude. When climbing at constant IAS, both TAS and Mach number increase. There is therefore a danger that MMowill be exceeded in a constant IAS climb above the crossover altitude.

SPEED 6. c. Vc is the design cruise speed and V, is the design dive speed. These speeds are both used by aircraft designers when constructing flight envelopes. Although no fixed values of either V, or V, are specified by regulating authorities, it is importact that cruising aircraft do not inadvertently exceed V., The margic betweec the two must therefore be sufficient to avoid unintentional over-speeds.

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SPEED 7. b. V, is the take-off safety speed. It is related to aircraft weight, atmospheric conditions and flap setting. Increasing flap setting decreases the stalling speed and this decreases the value of V,. But in order to is set at 1.1 ensure that control is maintained in the event of engine failure, a minimum value of V, (V),, So increasing flap setting from 10 degrees to 30 degrees will decrease of the minimum control speed (V)., V, unless it has reached it lower limit of 1. I.,V,

SPEED 8. a. The buffet boundaries mark the upper and lower limits of the speed range in which an aircraft can operate without incurring buffet. The speed at which low speed buffet commences is related to lift force being generated. This in turn is proportional to the aircraft weight, load factor and wing lift. Any factor that increases weight, load factor or wing lift required will increase the low speed buffet boundary speed, thereby causing the boundaries to converge. The position of the C of G determines the longitudinal pitching moment. If the C of G is forward of the C of P it will produce a nose down pitching moment. This pitching moment must be balanced by a tailplane down force in order to maintain level flight. But to maintain altitude the lift generated by the wings must equal the weight plus the tailplane down force. Forward movement of the C of G increases the nose down pitching moment and hence increases the tail down force and wing lift required. This increases the stalling speed and hence increases the low speed buffet speed. The overall effect is that forward movement of the C of G causes the buffet boundaries to converge.

SPEED 9. d. The buffet boundaries mark the upper and iower limits of the speed range in which an aircraft can operate without incurring buffet. The speed at which low speed buffet commences is related to the magnitude of the lift force being generated by the wings. Any factor that increases the required wing lift will increase the low speed buffet boundary speed, thereby causing the boundaries to converge. As an aircraft banks its lift force is angled in the direction of the bank. This directs some of the lift force horizontally, thereby reducing its vertical component. But only the vertical component of lift supports the weight. So as bank angle increases, the total wing lift must be increased to maintain an upward lift component equal to weight. This increases the stalling speed and hence increases the low speed buffet speed. The overall effect is that increasing bank angle causes the buffet boundaries to converge.

SPEED 16. a. The maximum operating speed of an aircraft is specified as VMo in terms of CAS and ,M , in terms of , increases whilst that equating to ,M , mach number. As altitude increases, the TAS equating to ,V decreases. At low altitudes the TAS equating to ,V , is the TAS at mach 0.81 , is the limiting factor. limiting factor and at high altitude ,M In order to avoid exceeding these limits aircraft are climbed at constant CAS at low altitude and at constant Mach number at high altitude. The crossover altitude is the altitude at which the selected CAS is equal to the selected mach number. If a higher CAS is selected for the climb schedule, the crossover altitude will be reduced as illustrated in the diagram at the right. higher than at 30010.81 TAS at 275 Kts

CAS TAS at 300 Kts

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SPEED 11. a. VMcLis the minimum speed at which control can be maintained in the event of a critical engine failure in the landing configuration, with landing power selected. The control authority available is proportional to dynamic pressure, whilst the yawing moment that must be resisted is proportional to thrust. Dynamic pressure and thrust are proportional to air density, which is proportional to altitude and humidity. Thrust is also proportional to power setting. So altitude, power setting and humidity will all affect VMc,. Option a, aircraft weight is the only one that will not do so.

SPEED 12. d The buffet margins are the speed ranges between the speed at which an aircraft is operating and the boundaries at which low and high speed buffet commences. The closer together the high and low speed boundaries, the smaller will be the margins. For any given combination of weight, load factor and configuration, the low speed buffet commences at a specific CAS. The high speed buffet commences at a specific fraction of the local speed of sound. Increasing altitude increases the TAS:CAS ratio and decreases the speed of sound. The increasing TAS:CAS ratio increase the TAS equating to the low speed boundary. The decrease in the local speed of sound decrease the TAS equating to the high speed boundary. This causes the boundariesto converge asaltitude increases. 'The converging of the boundaries decrease both the low speed and high speed margins. This effect is illustrated in the diagram at the right. at low altitude at high altitude

Alt

Large margins at low altitude

Smaller margins at high altitude

SPEED 13. a. Mach number is the speed of an aircraft as a fraction of the local speed of sound (LSS). Mach 1 for example means a speed equal to the LSS, whereas mach 0.5 means a speed only half of the LSS. The speed of sound in the atmosphere is not constant but is proportional to the square root of absolute temperature (LSS = 38.94 d~bsolutetemperature ). This means that if air temperature remains constant then the TAS at any given mach number remains constant. Above the tropopause at 36000 ft ISA, the temperature and LSS remain constant. So when climbing at constant mach number above the tropopause, TAS remains constant. But the ratio of IAS:TAS is related to air density, which is primarily related to air pressure. So as altitude increases, density decreases and the IAS at any given TAS decreases. Because this effect is related to air pressure rather than to temperature, it continues above the tropopause. So when climbing at constant mach number above the tropopause, IAS decreases and TAS remains constant. It should be noted that above 58000 ft the temperature increases, causing the TAS at any given mach number to increase. But very few aircraft operate above 58000 ft so this factor can be ignored in JAR examinations unless specifically stated in the questions.

SPEED 14. d. V, , V, and V, are some of the speeds used by aircraft designers in developing and specifying flight envelopes. V, is the design maximum gust intensity speed. It is the speed at which a JAR certificated passenger aircraft must be capable of surviving a vertical gust of 65 ft per second without structural

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damage. Although JAA regulations do not specify any fixed relationship between V, and Vc, V, is usually lower than Vc. Vc is the design cruise speed and V, is the design dive speed. These speeds are both used by aircraft designers when constructing flight envelopes. Although no fixed values of either Vc or V, are specified by regulating authorities, it is important that cruising aircraft do not inadvertently exceed V., The margin between the two must therefore be sufficient to avoid unintentional over-speeds. So the normal relationship between these speeds is that Vc is more than V, and V, is less than V? ,

SPEED 15. b. V, is the speed at which an aircraft is capable of achieving its greatest climb gradient. V, is the speed at ,, which an aircraft is capable of achieving its greatest rate of climb. For all aircraft types V, is less than V except at the absolute ceiling where the two converge. VMcGis the minimum speed at which it is possible to maintain control on, or close to the ground, following a critical engine failure in the take-off configuration. VMcGis lower than both V, and V, for all aircraft. The correct relationship is therefore V, is less than V, and more than VMcG?

SPEED 16. b. VMcGis the minimum speed at which it is possible to maintain control on, or close to the ground, following a critical engine failure in the take-off configuration. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing the take-off or aborting it, in the event of a critical engine failure. In order for such a decision to be made, the aircraft must be capable of being controlled. V, is therefore greater than VMcG. VMcAis the minimum speed at which it is possible to maintain control following a critical engine failure in the air in the climb configuration. V, is the speed at which an aircraft must be rotated into the lift-off attitude during a take-off run. In order to do this safely the aircraft must already be above VMcA,to ensure that it remains controllable in the initial climb out. Taking the above factors into account it can be seen that VMcAis less than V,, and V, is more than VMcG.

SPEED 17. b. V, is the stalling speed, or if no stall occurs V, is the minimum speed at which it is possible to control an aircraft in flight with all engines operating. VMcAis the minimum speed at which it is possible to maintain control following a critical engine failure in the air in the climb configuration. The effectiveness of aerodynamic flying controls is proportional to airspeed, such that control becomes more effective as speed increases. But the asymmetric thrust following a single engine failure makes control more difficult. V, (with all engines operating) is therefore lower than VMcA(with the critical engine failed). V, is the speed at which an aircraft must be rotated into the lift-off attitude during a normal take-off run. VMu,the minimum unstuck speed is the minimum speed at which an aircraft can lift-off the ground following ., ,,V ,, is the minimum acceptable speed that an aircraft rotation at V., VMu is therefore greater than V must attain when reaching screen height with all engines operating. But rotation and lift-off must both ,,, So V, < VMcA VZmin. occur before screen height, so V, is less than VMu, which is less than.,V

SPEED 18. b. Mach number is the speed of an aircraft as a fraction of the local speed of sound (LSS). Mach 1 for example means a speed equal to the LSS, whereas mach 0.5 means a speed only half of the LSS. The speed of sound in the atmosphere is not constant but is proportional to the square root of absolute temperature (LSS = 38.94 d~bsolutetemperature). This means that if air temperature decreases, the

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LSS also decreases. But if the LSS decreases, then the TAS equating to any given mach number also decreases. Decreasing temperature therefore causes the TAS at any given mach number to decrease. As pressure altitude increase in the troposphere, the air temperature decreases at a rate of approximately 1.98 degrees per 1000 ft. This means that as pressure altitude increases in the troposphere, the local speed of sound decreases. SoTAS decreases when climbing at constant mach number in the troposphere.

SPEED 19. b. Mach number is the speed of an aircraft as a fraction of the local speed of sound (LSS). Mach 1 for example means a speed equal to the local speed of sound, whereas mach 0.5 means a speed only half of the local speed of sound. The speed of sound in the atmosphere is not constant but is proportional to the temperature). This means that if air temperature square root of absolute temperature (LSS = 38.94 d~bsolute decreases, the local speed of s o u ~ dalso decreases. But if the local speed of sound decreases, then the TAS equating to any given mach number also decreases. Decreasing temperature therefore causes the TAS at any given mach number to decrease. As pressure altitude increases up to the tropopause at 36000 ft, the air temperature decreases at a rate of approximately 1.98 degrees per 1000 f t . Above 36000 ft the temperature remains constant. This means that as pressure altitude increases up to 36000 f t , the local speed of sound decreases, then remains constant at higher altitudes. So when climbing at constant mach number TAS will decrease then remain constant.

SPEED 20. a. V, is the take-off safety speed. It is the minimum acceptable speed that an aircraft must attain when reaching screen height with all engines operating.

SPEED 21. a. V, is the stalling speed, or if no stall occurs V, is the minimum speed at which it is possible to maintain control of an aircraft in flight with all engines operating. VMcGis the minimum speed at which it is possible to maintain control following a critical engine failure on the ground, using only aerodynamic controls, in the take-off configuration. The effectiveness of aerodynamic flying controls is proportional to airspeed, such that control effectiveness increases with increasing airspeed. But the asymmetric thrust following a single engine failure makes control more difficult. V, (with all engines operating) is therefore lower than VMcG (with the critical engine failed). V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing the take-off or of aborting it, in the event of a critical engine failure. In order for such a decision to be made, the aircraft must be capable of being controlled. V, must therefore greater than V, and at least equal to VMcG. V, is the speed at which an aircraft must be rotated into the lift-off attitude during a normal take-off run. But the decision to continue or abort the take-off must be made before rotation, so V, cannot be greater than V., V, is the minimum acceptable speed at screen height and V, is the initial climb speed, both of which are greater than V., So the minimum value for V, is VMcGand the maximum value is V.,

SPEED 22. b. An airspeed indicator gives an indication proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so climbing at constant IAS means climbing at constant dynamic pressure. But dynamic pressure is equal to X p v , where p is air density and V is TAS. But as altitude increases, the air density decreases. So when climbing at constant IAS, the value of TAS must increase to compensate for decreasing density, in order to maintain a constant dynamic pressure

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and airspeed indication. The TAS equating to any given IAS therefore increases as altitude increases. This in turn means that in a constant TAS climb the IAS decreases with increasing altitude. Because this change is caused by decreasing density, which is caused by decreasing pressure, the effect continues throughout the entire atmosphere.

SPEED 23. a. An airspeed indicator gives an indication proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so climbing at constant IAS means climbing at constant dynamic pressure. But dynamic pressure is equal to XpV2, where p is air density and V is TAS. As altitude increases, the air density decreases. So when climbing at constant IAS, the value of TAS must increase to compensate for decreasing density, in order to maintain a constant dynamic pressure and airspeed indication. The TAS equating to any given IAS therefore increases as altitude increases. So in a constant IAS climb the TAS increases with increasing altitude. Because this change is caused by decreasing density, which is caused by decreasing pressure, the effect continues throughout the entire atmosphere.

SPEED 24. d. An airspeed indicator gives an indication proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so climbing at constant IAS means clirr~bingat constant dynamic pressure. But dynamic pressure is equal to 1%pV2,where p is air density and V is TAS. As altitude increases, the air density decreases. So when climbing at constant IAS, the value of TAS must increase to compensate for decreasing density, in order to maintain a constant dynamic pressure and airspeed indication. The TAS equating to any given IAS therefore increases as altitude increases. This in turn means that in a constant TAS climb the IAS decreases with increasing altitude. Because this change is caused by decreasing density, which is in turn caused by decreasing pressure, the effect continues throughout the entire atmosphere. But below 36000 ft, the temperature also decreases as altitude increases. This tends to contract the air, thereby partly offsetting the effect of decreasing pressure. Above 36000 ft the temperature is constant up to 58000 ft, so the offsetting effect is lost. The overall effect is that the rate of decrease in density is greatest above 36000 ft. The overall effect of climbing at constant TAS is therefore that IAS decreases up to 36000 ft then decreases more quickly. Few commercial aircraft operate above 58000 ft so this part of the atmosphere can be ignored.

SPEED 25. b. An airspeed indicator gives an indication proportional to dynamic pressure. That is to say the indication will always be the same for any given dynamic pressure, so climbing at constant IAS means climbing at constant dynamic pressure. But dynamic pressure is equal to WpV2,where p is air density and V is TAS. As altitude increases, the air density decreases. So when climbing at constant IAS, the value of TAS must increase to compensate for decreasing density, in order to maintain a constant dynamic pressure and airspeed indication. The TAS equating to any given IAS therefore increases as altitude increases. So in a constant IAS climb the TAS increases with increasing altitude. Mach number is the speed of an aircraft as a fraction of the local speed of sound (LSS). Mach 1 for example means a speed equal to the LSS, whereas mach 0.5 means a speed only half of the LSS. The LSS in the atmosphere is not constant but is proportional to the square root of absolute temperature (LSS = 38.94 d~bsolutetemperature). This means that if air temperature decreases, the LSS also decreases. But if the LSS decreases, then the TAS equating to any given mach number also decreases. Decreasing temperature therefore causes the TAS at any given mach number to decrease.

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As pressure altitude increase in the troposphere, up to the tropopause at 36000 ft, the air temperature decreases at a rate of approximately 1.98 degrees per 1000 ft. This means that as pressure altitude increases in the troposphere, the local speed of sound decreases. So when climbing at constant IAS the increasing TAS and decreasing LSS cause mach number to increase.

SPEED 26. a. VMo is the maximum operating speed of an aircraft, expressed as IAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic , is the limiting pressure is proportional to air density, so at low altitudes where air density is high, ,V parameter. MMo is the maximum operating speed of an aircraft, expressed as a mach number. If the aircraft is operated at higher mach numbers, the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. ,M , is some specific fraction of the local speed of sound (LSS). But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the IAS equating to M,, deceases as altitude increases. The overall effect of these factors is that,V , is the limiting speed at low altitude and MMois the limiting speed at high altitude. The altitude at which the TAS equating to VMo is equal to that equating to MMo is called the crossover altitude. ,V, (the maximum demonstrated flight velocity) is the maximum speed at which controlled flight has been demonstrated. McDR(the critical drag rise mach number) is the lowest mach number at which a significant increase in wave drag occurs. Both of these are higher than VMo/MMo.So VMois the limiting factor at low altitude.

SPEED 27. b. VMo is the maximum operating speed of an aircraft, expressed as IAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at low altitudes where air density is high, VMo is the limiting parameter. MMo is the maximum operating speed of an aircraft, expressed as a mach number. If the aircraft is operated at higher mach numbers, the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. ,M , is some specific fraction of the local speed of sound (LSS). But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the IAS equating to ,M , deceases as altitude increases. The overall effect of , is the limiting speed at high altitude. these factors is that VMo is the limiting speed at low altitude and ,M The altitude at which the TAS equating to VMo is equal to that equating to MMo is called the crossover altitude. ,V, (the maximum demonstrated flight velocity) is the maximum speed at which controlled flight has been demonstrated. McDR(the critical drag rise mach number) is the lowest mach number at which a significant increase in wave drag occurs. But both of these speeds are prohibited for normal operations. So MMois the limiting factor at high altitude.

SPEED 28. c. VMo is the maximum operating speed of an aircraft, expressed as IAS. It is determined by the ability of the structure to withstand the aerodynamic forces produced by the dynamic pressure acting upon it. Dynamic pressure is proportional to air density, so at low altitudes where air density is high,,V , is the limiting parameter. MMo is the maximum operating speed of an aircraft, expressed as a mach number. If the

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aircraft is operated at higher mach numbers, the generation of supersonic airflows and shock waves will cause high-speed buffet and eventually high-speed stall. MMois some specific fraction of the local speed of sound (LSS). But the local speed of sound is proportional to air temperature and hence decreases with increasing altitude. This means that the IAS equating to MMo deceases as altitude increases. The overall effect of these factors is that VMo is the limiting speed at low altitude and MMois the limiting speed at high altitude. The altitude at which the TAS equating to VMo is equal to that equating to MMo is called the crossover altitude. When descending at constant mach number the CAS increases, so below the crossover altitude there is therefore a danger that VMo will be exceeded.

SPEED 29. b. V, is the speed at which an aircraft must be rotated into the lift-off attitude during a normal take-off. For safety reasons it is necessary to ensure that V, is more than the minimum control speed VMc (also called .),V ,, In order to ensure this, JAR 25.107 (e) specifies a minimum value of 1.05 VMc. This means that V, must be at least VMc plus 5%. VMcLis the minimum control speed in the landing configuration. This has no relevance is setting limits for V.,

SPEED 30. a. Increasing aircraft mass increases the lift force required. This in turn increases the induced drag but does not significantly affect the parasite drag. So increasing mass by 10% will increase induced drag but not change parasite drag. In order to increase lift at VIM, to match the 10% mass increase, it is necessary to increase C, by 10% to 1. I of its initial value. lnduced drag is proportional to the coefficient of induced drag (C,,), which is in turn proportional to .C : SO if CL increases to 1. I of its initial value, C, increases to 1. I 2 or 1.21 of its initial value. This means that a 10% increase in mass causes a 21% increase in induced drag at the same speed. So options c and d are incorrect. The new VIM, can be calculated using the following equation: VIM, at new mass = VIM, at old mass x ./(new mass I old mass) So VIM, after 10% mass increase = initial VIM, d(1. II1) Which is 1.05 x initial VIM,. So a 10% increase in mass increase VIM, by 5%. This illustrates a more general point that changing mass by any given small % changes VIM, by half that %. Taking into account all of the above factors, option a is the most appropriate.

SPEED 31. a. Decreasing aircraft mass decreases the lift force required. This in turn decreases the induced drag, but does not significantly affect the profile drag. So decreasing mass by 20% will decrease induced drag but not change profile drag. In order to decrease lift at VIM, to match the 20% mass decrease, it is necessary to decrease C, by 20% to 0.8 of its initial value. lnduced drag is proportional to the coefficient of induced drag (C,,), which is in turn proportional to C,2. SO if CL decreases to 0.8 of its initial value, C, decreases to 0.82 or 0.64 of its initial value. This means that a 20% decrease in mass causes a 36% decrease in induced drag at the same speed. So options b and c are incorrect. The new VIM, can be calculated using the following equation:

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V,,,

Speeds

at new mass = VIM, at old mass x d(new mass Iold mass)

So VIM, after 20% mass decrease = initial VIM, 40.8 11) Which is 0.895 x initial VIM,. So a 20% decrease in mass decreases VIM, by approximately 10%. This illustrates a more general point that changing mass by any given small % changes VIM, by half that %. Taking into account all of the above factors, option a is the most appropriate.

SPEED 32. a. VMcGis the minimum speed at which it is possible to maintain control following a critical engine failure on the ground, using only aerodynamic controls, in the take-off configuration. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing the take-off or aborting it, in the event of a critical engine failure. In order for such a decision to be made, the aircraft must be capable of being controlled. V, must therefore be equal to or greater than VMcG. V, is the speed at which an aircraft must be rotated into the lift-off attitude during a normal take-off run. But the decision to continue or abort the take-off must be made before rotation, so V, cannot be greater than V., V, is the minimum acceptable speed at screen height and V, is the initial climb speed, both of ., So the minimum value for V, is VMcGand the maximum value is V ., And V, which are greater than V must be equal to or greater than V ., The correct sequence is therefore VMCG< V, < V, < V., It should be noted that VMcAand VMc are in fact the same speed and are greater than V, but less than V.,

SPEED 33. c. Vx is the speed at which an aircraft is able to achieve its best angle of climb. For a propeller aircraft it is slightly lower than VIM, whilst for a jet aircraft it is VIM,. V, is the speed at which an aircraft is able to ,,, whilst for a jet it is slightly achieve its best rate of climb. For a propeller aircraft it is slightly higher than ,V higher than VIM,. As altitude increases the IAS equating to Vx remains constant, whilst that equating to V, slowly decreases, such that at the absolute ceiling Vx is equal to V ., So for all aircraft types, Vx is always ., less than or equal to V It should be noted however that the TAS equating to both Vx and V, increase with altitude, but are equal ., The changing at the absolute ceiling. So the TAS value of Vx is also always less than or equal to V relationship between propeller aircraft Vx, V, VIM, and VIM, with increasing altitude is illustrated in the diagram below. ,,V ,,

VIM, Vy and , ,V

Altitude

I

converge on V, at the absolute ceiling

Absolute Ceiling

- IAS Minimum speed for which power is available

V,

VIM,

Vy

VIM,

Maximum speed for which power is available

Speeds

349

SPEED 34. b. Vx is the speed at which an aircraft is able to achieve its best angle of climb. For a propeller aircraft it is slightiy lower than VIM, whilst for a jet aircraft it is VIM,. V, is the speed at which an aircraft is able to achieve its best rate of climb. For a propeller aircraft it is slightly higher than VIM,, whilst for a jet it is slightly higher than VIM,. As altitude increases the IAS equating to Vx remains constant, whilst that equating to V, slowly ., So for all aircraft types, Vx is always less than decreases, such that at the absolute ceiling Vx is equal to V ., It should be noted however that the TAS equating to both Vx and V, increase with altitude, but or equal to V l., are equal at the absolute ceiling. So the TAS value of Vx is also always less than or e q ~ ~toa V V, is the stalling speed, which is considerably lower than both Vx and V ., So the correct sequence in order ., of increasing magnitude is V, Vx V

SPEED 35. d. Vx is the speed at which an aircraft is able to achieve its best angle of climb. For a propeller aircraft it is slightly lower than VIM, whilst for a jet aircraft it is VIM,. V, is the speed at which an aircraft is able to ,, whilst for a jet it is slightly achieve its best rate of climb. For a propeller aircraft it is slightly higher than V higher than VIM,. As altitude increases the IAS equating to Vx remains constant, whilst that equating to V, slowly decreases, such that at the absolute ceiling Vx is equal to V ., So for all aircraft types, Vx is always less than or equal to V., It should be noted however that the TAS equating to both Vx and V, increase with altitude, but are equal at the absolute ceiling. But Vx and V, are conventionally quoted in terms of IAS so option d is the most appropriate answer to this question.

SPEED 36. a. ,,V , is the minimum speed at which it is possible to maintain control following a critical engine failure on the ground, using only aerodynamic controls, in the take-off configuration.The effectiveness of aerodynamic flying controls is proportionalto airspeed, such that control effectiveness increases with increasing airspeed. But the asymmetric thrust following a single engine failure makes control more difficult.

The degree of control power required and hence the value of the minimum control speed is proportional to . , ,V , But thrust is proportional to the thrust being generated. So any factor that reduces thrust will reduce . , ,V , Such factors include high air density, so any factor that reduces air density will reduce thrust and ambient temperature humidity and pressure altitude, so a combination of all three will produce a minimum . , ,V , value of

SPEED 37. a. Vx is the speed at which an aircraft is able to achieve its best angle of climb. For a jet aircraft it is VIM,. V, is the speed at which an aircraft is able to achieve its best rate of climb. For a jet aircraft it is slightly higher than VIM,. As altitude increases the IAS equating to Vx remains constant, whilst that equating to V, slowly .,,,V , and ,V ,, are all equal. So at the absolute ceiling decreases, such that at the absolute ceiling VxJET, a jet aircraft must fly at VIM, in order to maintain altitude. Vx for a propeller aircraft it is slightly lower than VIM,. V, is the speed at which an aircraft is able to achieve its best rate of climb. For a propeller aircraft it is slightly higher than VIM,. As altitude increases ,, and V, slowly decreases, such the IAS equating to Vx remains constant, whilst that equating to ,V that they converge at the only possible flight speed at the absolute ceiling. So at the absolute ceiling a in order to maintain altitude. So a jet aircraft must I'ly at VIM, and a propeller aircraft must fly at Vxp,, , , ,V , to maintain altitude in straight and level flight at the absolute ceiling. propeller aircraft must fly at , , ,V ,V ,,, VIM, Vx V, and , ,V for a jet aircraft is illustrated in the The changing relationship between diagram below.

350

Speeds

V ,,

Altitude

Absolute Ceiling

V, and V,

4"

converge on Vx and Vp ,

at the absolute ceiling

and V ,,

which power is available

remains

Maximum speed for which power is available

SPEED 38. c. Vx is the speed at which an aircraft is able to achieve its best angle of climb. For a propeller aircraft it is slightly lower than VIM,. V, is the speed at which an aircraft is able to achieve its best rate of climb. For a propeller aircraft it is slightly higher than VIM,. As altitude increases the IAS equating to V, remains constant, and V, slowly decreases, such that they converge at the only possible flight whilst that equating to ,V,! speed at the absolute ceiling. But the IAS equating to VIM, remains constant and does not converge with the other speeds. But only one flight speed is possible at the absolute ceiling so the minimum flight speed V ,,, and the maximum llight speed, ,V also converge on ,V, V, and V ., V, the stalling speed also remains constant at all altitudes. So option c is the only one that is true in this question.

SPEED 39. c. Vx is the speed at which an aircraft is able to achieve its best angle of climb. For a propeller aircraft it is slightly lower than VIM,. It is also the speed at which excess thrust is maximum. Excess thrust is equal to thrust available minus drag. The drag at any given IAS does not change with altitude, but the thrust decreases due to decreasing air density. This causes excess thrust to decrease, but V, to remain constant.

VIM, is the IAS at which total drag is minimum. The drag at any given IAS does not vary with altitude, so both minimum drag and VIMDremain constant as altitude increases. So as altitude increases, the IAS equating to V, and VIMDboth remain constant.

SPEED 40. b. V, is the speed at which an aircraft is able to achieve its best climb gradient. Climb gradient is proportional to excess thrust, so V, is the speed at which excess thrust is greatest. V, is the speed at which an aircraft is able to achieve its best rate of climb. Rate of climb is proportional to excess power, so V, is the speed at which excess power is greatest. Increasing aircraft weight increases both the minimum drag speed (VIM,) and minimum power speed (VIM,), and the drag and power required at any speed. This increases the speeds at which both excess thrust and excess power are maximum. So increasing weight increases both V, and V., This effect is illustrated for a jet aircraft in the diagrams belcw.

Speeds

351

Drag at high

Excess

V, at IoWf weight

:

: . :.:.

: :

zyVx

weight at mgh weight

Vx at low weight

v,\,t

high weight

SPEED 41. c. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. But the decision to complete or abandon the take-off must be made before rotation. So the maximum possible value of V, is V., VLo,, the lift-off speed and V, the take-off safety speed are both greater than V., So the maximum value of V, is limited by the value of V.,

SPEED 42. a. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available.

SPEED 43. b. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude duriqg the take-off run. But the decision to complete or abandon the take-off must be made before rotation. VLo, is the speed at which an aircraft will lift off the ground after being rotated into the lift-off attitude at V., lncreasing air density increases the thrust available from the engines. The increased thrust increases the rate of acceleration, thereby reducing the time and distance necessary to accelerate from V, to V., This decreases the minimum acceptable value of V,. But increasing density also increases the dynamic pressure and hence lift available at any given TAS. This decreases the TAS necessary to take-off. So increasing air density decreases V,, V, and VLo,.

SPEED 44. a. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. But the decision to complete or abandon the take-off must be made before rotation. VLo, is the speed at which an aircraft will lift off the ground after being rotated into the lift-off attitude at V., lncreasing altitude decreases air density. This decreases the thrust available from the engines. The decreased thrust decreases the rate of acceleration, thereby increasing the time and distance necessary to accelerate from V, to V,. This increases the minimum acceptable value of V,. But decreasing density

352

Speeds

also decreases the dynamic pressure and hence lift available at any given TAS. This increases the TAS necessary to take-off. So increasing altitude increases V,, V, and VLoF.

SPEED 45. a. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft' must be rotated into the lift-off attitude during the take-off run. But the decision to complete or abandon the take-off must be made before rotation. VLoFis the speed at which an aircraft will lift off the ground after being rotated into the lift-off attitude at V., Increasing aircraft weight decreases the rate of acceleration, thereby increasing the time and distance necessary to accelerate from V, to V., This increases the minimum acceptable value of V,. But increasing weight also increase the speed necessary to generate sufficient lift to lift-off. So increasing weight increases V,, V, and VLoF.It should however be noted that increasing weight also increases the stopping distance required in the event of an abandoned take-off. This decreases the maximum value of V, for any given ASDA.

SPEED 46. c. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. But the decision to complete or abandon the take-off must be made before rotation. VLoFis the speed at which an aircraft will lift off the ground after being rotated into the lift-off attitude at V., A headwind reduces the ground speed at any given IAS. This decreases the distance required to stop following a decision to abort a take-off. This in turn means that the decision to abort can be taken at a higher speed. The maximum acceptable value of V, is therefore increased by a headwind. By V, and VLoF are related to the generation of lift, which is related to dynamic pressure, and hence to IAS. Dynamic pressure at any given IAS is not affected by a headwind. So a headwind will not affect the value of V, or VLOF.

SPEED 47. b. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. But the decision to complete or abandon the take-off must be made before rotation. VLoFis the speed at which an aircraft will lift off the ground after being rotated into the lift-off attitude at V., lncreasing flap angle increases the lift force at any given IAS. This reduces stalling speed and hence reduces the speed at which an aircraft can lift-off safely. This means that the decision to complete the take-off can be made at a lower speed, the aircraft can be rotated into the lift-off attitude at a lower speed, and will lift-off at a lower speed. So increasing flap angle decrease V,, V, and VLo,.

SPEED 48. a. V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completing or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. But the decision to complete or abandon the take-off must be made before rotation. VLoFis the speed at which an aircraft will lift off the ground after beirrg rotated into the lift-off attitude at V .,

Speeds

353

lncreasing air temperature decreases air density. This decreases the thrust available from the engines. The decreased thrl~stdecreases the rate of acceleration, thereby increasing the time and distance necessary to accelerate from V, to V., This increases the minimum acceptable value of V,. But decreasing density also decreases the dynamic pressure and hence lift available at any given TAS. This increases the TAS necessary to take-off. So increasing air temperature increases V,, V, and VLoF.

SPEED 49. c V, is the take-off decision speed. It is the maximum speed at which an aircraft is equally capable of completirlg or abandoning a take-off within the distances available. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. But the decision to cornplete or abandon the take-off must be made before rotation. VLoFis the speed at which an aircraft will lift off the ground after ., being rotated into the lift-off attitude at V lncreasing runway upward slope will decrease the rate of acceleration. This will increase the distance needed to accelerate from V, to flying speed. So the decision to complete a take-off must be made at a greater speed. This means that the minimum value of V, is increased. Increased upward slope also increases the deceleration rate, so the decision to abandon a take-off can also be made at a higher speed. So the maximum value of V, is also increased. But V, and VLOFare related to dynamic pressure which is independent of acceleration rates. So increasing upward runway slope does not affect V, or VLoF.

SPEED 50. b. V, is the take-off safety speed. It is the minimum speed that must be achieved by screen height will all engines operating. Vx is the speed at which an aircraft is able to achieve its best climb gradient. V, is the speed at which an aircraft is able to achieve its best rate of climb. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration. For safety reasons V, must be not less than 1.1 VMcA.So any factor that decreases VMcAwill decrease the minimum value of V., Following a critical engine failure the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing tendency is proportional to the thrust being produced. So any factor that reduces the take-off thrust will reduce VMcAand hence V., As altitude increases, decreasing air density decreases the thrust produced, and hence decreases VMCAand V., So as altitude increases V, decreases. Vx is also the IAS at which excess thrust is maximum. As altitude increases the drag at any giver1 IAS is constant but the thrust decreases. The overall effect of this is that as altitude increases, excess thrust decreases but Vx remains constant. V, is the speed at which excess power is maximum. As altitude increases the power required at any given IAS increases whilst the power available decreases. The overall effect is that as altitude increases, excess power decreases and V, decreases. It should however be noted that the rate of decrease in V, is extremely small. JAR performance examinations therefore generally assume that V, is constant at all altitudes

SPEED 51. b. V, is the take-off safety speed. It is the minimum speed that must be achieved by screen height will all engines operating. Vx is the speed at which an aircraft is able to achieve its best climb gradient. V, is the speed at which an aircraft is able to achieve its best rate of climb. VMcA is the minimum speed at which control of an aircraft can be maintained folluuing a critical engine failure in the air in the climb configuration. For safety reasons V, must be not less than 1.1 VMcA.So any factor that decreases VMcAwill decrease the minimum value of V,. Following a critical engine failure, the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll.

.

354

Speeds

The magnitude of this yawing tendency is proportional to the thrust being produced. So any factor that reduces the take-off thrust will reduce VMcAand hence V,. As temperature increases, decreasing air density decreases the thrust produced, and hence decreases VMcAand V,. So as temperature increases, V, decreases. Vx is also the IAS at which excess thrust is maximum. IAS is proportional to density. As temperature increases, the drag at any given IAS is constant but the thrust decreases. The overall effect of this is that as temperature increases, excess thrust decreases but Vx remains constant. V, is the speed at which excess power is maximum. As temperature increases, the power required at any given IAS increases whilst the power available decreases. The overall effect is that as altitude increases, excess power and V, both decrease. So increasing air temperature decrease V, and Vy, but does not affect Vx.

SPEED 52. a. V, is the take-off safety speed. It is the minimum speed that must be achieved by screen height will all engines operating. Vx is the speed at which an aircraft is able to achieve its best climb gradient. Vy is the speed at which an aircraft is able to achieve its best rate of climb. lncreasing the weight of an aircraft increases its stalling speed V., For safety reasons the minimum value for V, is 1.2 V, so as V, increases, so does V., Vx is the speed at which excess thrust is maximum and V, is the speed at which excess power is maximum. These speeds are related to the minimum drag speed VIM, and the minimum power speed VIM, respectively. But as weight increases, so does VIM, and VIM,. So increasing aircraft weight increases V, Vx and V.,

SPEED 53. b. Although runway slope affects the rate of acceleration during the take-off run, it has no effect on the aircraft after it has lifted-off. V, Vx, and Vy are all speeds attained in the air and so they are not affected by runway slope.

SPEED 54. b. Although a headwind affects the rate of acceleration during the take-off run, it has no effect on the IAS of an aircraft after it has lifted-off. V, Vx, and V, are all speeds attained in the air, so they are not affected by headwinds.

SPEED 55. b. V, is the take-off safety speed. It is the minimum speed that must be achieved by screen height will all engines operating. Vx is the speed at which an aircraft is able to achieve its best climb gradient. Vy is the speed at which an aircraft is able to achieve its best rate of climb. lncreasing the flap angle of an aircraft decreases its stalling speed V., For safety reasons the minimum value for V, is 1.2 V, so as V, decreases, so does V., Vx is the speed at which excess thrust is maximum and V, is the speed at which excess power is maximum. These speeds are related to the minimum drag speed VIM, and the minimum power speed VIM, respectively. But as flap angle increases, both VIM, and VIM, decrease. So increasing flap angle decreases V, V, and V.,

SPEED 56. d. VMc, is the minimum speed at which it is possible to maintain control following a critical engine failure on the ground in the take-off configuration. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration.

Speeds

355

Following a critical engine failure the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportional to the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. So any factor that reduces the take-off thrust will reduce the minimum control speeds, VMcGand VMcA.As temperature increases, the decreasing air density reduces the thrust produced, and hence decreases VMcGand VMcA. V, is the maximum speed at which it is equally possible to continue or abort a take-off following a critical within engine failure. In order to continue a take-off the aircraft must accelerate to the lift-off speed,,,V, the remaining take-off run available. The rate at which an aircraft can accelerate is proportional to its thrust, so any factor that decreases thrust will increase the minimum acceptable value of V,. So by reducing air density and hence thrust, increasing temperature increases V,. It should however be noted that decreasing density also increases the ground speed at any given IAS. This increases the stopping distance required in the event of an aborted take-off. So increasing temperature also decreases the maximum allowable value of V, for any given combination of aircraft mass and ASDA. But this effect is not included in any of the options in this question, so option d is the most appropriate.

SPEED 57. d. VMcGis the minimum speed at which it is possible to maintain control following a critical engine failure on the ground in the take-off configuration. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration. Following a critical engine failure, the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportional to the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. So any factor that reduces the take-off thrust will reduce the minimum control speeds, VMcGand VMcA.As altitude increases, the decreasing air density reduces the thrust produced, and hence decreases VMcGand VMcA. V, is the maximum speed at which it is equally possible to continue or abort a take-off following a critical within engine failure. In order to continue a take-off the aircraft must accelerate to the lift-off speed,,,V, the remaining take-off run available. The rate at which an aircraft can accelerate is proportional to its thrust, so any factor that decreases thrust will increase the minimum acceptable value of V,. So by reducing air density and hence thrust, increasing altitude increases V,. It should however be noted that decreasing density also increases the ground speed at any given IAS. This increases the stopping distance required in the event of an aborted take-off. So increasing altitude also decreases the maximum allowable value of V, for any given combination of aircraft mass and ASDA. But this effect is not included in any of the options in this question, so option d is the most appropriate.

SPEED 58'. c. VMc, is the minimum speed at which it is possible to maintain control following a critical engine failure on the ground in the take-off configuration. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration. Following a critical engine failure the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportional to the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. But none of these factor is dependent upon aircraft weight, so increasing weight will have no effect on VMcGand VMcA. V, is the maximum speed at which it is equally possible to continue or abort a take-off following a critical ,, within engine failure. In order to continue a take-off the aircraft must accelerate to the lift-off speed,,V

356

Speeds

the remaining take-off run available. The rate at which an aircraft can accelerate decreases with increasing weight. So increasing weight will increase the minimum acceptable value of V,. It should however be noted that increasing weight also increases the stopping distance required in the event of an aborted take-off. So increasing weight also decreases the maximum allowable value of V, for any given combination of aircraft mass and ASDA. But this effect is not included in any of the options in this question, so option c is the most appropriate.

SPEED 59. c. VMcGis the minimum speed at which it is possible to maintain control following a critical engine failure on the ground in the take-off configuration. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration. Following a critical engine failure the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportionalto the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. But none of these factor is dependent upon runway slope, so increasing runaway up slope will have no effect on VMcGand VMCA. V, is the maximum speed at which it is equally possible to continue or abort a take-off following a critical engine failure. In order to continue a take-off the aircraft must accelerate to the lift-off speed,,V ,, within the remaining take-off run available. The rate at which an aircraft can accelerate decreases with increasing runway upward slope. So increasing runway up slope will increase the minimum acceptable value of V,. It should however be noted that increasing runway up slope also decreases the stopping distance required in the event of an aborted take-off. So increasing up slope also increases the maximum allowable value of V, for any given combination of aircraft mass and ASDA. Option c is therefore the most appropriate.

SPEED 60. c. VMcGis the minimum speed at which it is possible to maintain control following a critical engine failure on the ground in the take-off configuration. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration. Following a critical engine failure the asyrr~metricthrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportional to the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. But none of these factor is dependent upon headwind velocity, so increasing headwind will have no effect on VMCGand VMCA. V, is the maximum speed at which it is equally possible to continue or abort a take-off following a critical engine failure. In order to abort a take-off the aircraft must stop within the remaining ASDA. The distance required to do this is proportional to its ground speed. But as headwind velocity increases, the ground speed at any given IAS decreases. So a headwind will decrease the stopping distance required at any given V,. This in turn will increase the maximum allowable value of V, for any given combination of aircraft mass and ASDA. Headwind has no effect on the minimum value of V,.

SPEED 61. b. ,,V , is the minimum speed at which it is possible to maintain control following a critical engine failure on the ground in the take-off configuration. VMcAis the minimum speed at which control of an aircraft can be maintained following a critical engine failure in the air in the climb configuration.

Speeds

357

Following a critical engine failure the asyrr~metricthrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportional to the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. But none of these factor is dependent upon flap angle, so increasing flap angle will have no effect on VMCGand VMCA* V, is the maximum speed at which it is equally possible to continue or abort a take-off following a critical engine failure. In order to continue a take-off the aircraft must accelerate to the lift-off speed VLoF,within the remaining take-off run available. But increasing flap angle decreases VLoF,enabling the take-off to be completed from a lower V,. So increasing flap angle decreases the minimum value of V,. In order to abort a take-off the aircraft must stop within the remaining ASDA. The distance required to do this is proportional to the deceleration rate. But as flap angle increases, the additional drag increases the rate at which an aircraft can decelerate. So increasing flap angle also increases the maximum value of V, for any given combination of aircraft mass and ASDA.

SPEED 62. c. VATis the threshold speed. JAR 25.123 requires that it be not less than 1.23 times the reference stalling speed VsR.So option c is correct and option d incorrect. But VsR is proportional to flap position and weight, so options a and b are also incorrect.

SPEED 63. a. The purpose of the landing climb speed is to ensure that an aircraft can achieve a safe climb gradient in the event of a rejected landing. JAR 25.11 Sspecifies that the landing climb speed be equal to 1.08VsRfor 4 engine class A aircraft, and 1. I 3 VsR for all other class A aircraft. It must also be VREFfor all class B aircraft. Its must not however be less than VMcLnor more than VREF,which is 1.23 VsR for class A aircraft. So option a, is the only one that is true.

SPEED 64. b. The purpose of the landing climb speed is to ensure that an aircraft can achieve a safe climb gradient in the event of a rejected landing. JAR 25.119 specifies that the landing climb speed be equal to 1.08VsRfor 4 engine class A aircraft, and 1.I 3 VsR for all other class A aircraft. It must also be VREFfor all class B aircraft. Its must not however be less than VMcLnor more than VREF,which is 1.23 VsR for class A aircraft. So option b is the only one that is true.

SPEED 65. d. The purpose of the landing clirnb speed is to ensure that an aircraft can achieve a safe climb gradient in the event of a rejected landing. JAR 25.119specifies that the landing climb speed be equal to 1.08VsRfor 4 engine class A aircraft, and 1. I 3 VsR for all other class A aircraft. It must also be VREFfor all class B aircraft. Its must not however be less than VMcLnor more than VREF,which is 1.23 VsR for class A aircraft. So option d is the only one that is true.

SPEED 66. b. The purpose of the landing climb speed is to ensure that an aircraft can achieve a safe climb gradient in the event of a rejected landing. JAR 25.1 19 specifies that the landing climb speed be equal to 1.08VsRfor 4 engine class A aircraft, and I.I 3 VsR for all other class A aircraft. It must also be VREFfor all class B aircraft. Its must not however be less than VMcLnor more than,,,V , which is 1.23 VsR for class A aircraft. So option b is the only one that is true.

358

Speeds

SPEED 67. b. JAA 25.121 states that the one engine inoperative approach climb speed must be not greater than 1.4 VsR for class A aircraft.

SPEED 68. c. JAA 23.67 states that the discontinued climb speed must be not greater than I.5 V,

for class B aircraft.

SPEED 69. b. JAA 25.1 07 states that ,,V ,, must be not less than 1.I 32VvsRfor aircraft with 2 and 3 turbo-propeller engines, and 1.08 VsR for all 4 engine turbo-props and all jet aircraft, and 1.IVMc.

SPEED 70. c. JAA 25.107 states that ,,V ,, must be not less than 1.132VvsR for aircraft with 2 and 3 turbo-propeller engines, and 1.08 VsR for all 4 engine turbo-props and all jet aircraft, and 1.IVMc.

SPEED 71. d. JAA 25.107 states that ,,V ,, must be not less than 1.132VvsRfor aircraft with 2 and 3 turbo-propeller .,V , engines, and 1.08 VsRfor all 4 engine turbo-props and all jet aircraft, and 1.I

SPEED 72. a. JAR 25.107states that V, must be not less than: 1. v,. 2. 1.05VMc, 3. The speed required to reach V, at 35 ft ,, being 1.IV ,, with a!l engines operating or 1.05 ,V, 4. That which results in ,V inoperative.

with one engine

SPEED 73. b. JAR 25.1 07 states that V, must be not less than: 1. v,. 2. 1.05VM, 3. The speed required to reach V, at 35 ft 4. That which results in V ,,, being 1. I V ,, with all engines operating or 1.05 V ,, inoperative.

with one engine

SPEED 74. c. JAR 25.1 07states that V, must be not less than: 1. v,. 2. 1.05VM, 3. The speed required to reach V,at 35 ft 4. That which results in ,V ,, being 1.1 V ,, with all engines operating or 1.05 V ,, inoperative.

with one engine

SPEED 75. d. JAR 25.1 07states that V, must be not less than: 1. v,. 2. I.05vMC 3. The speed required to reach V, at 35 ft ,,, being 1. I V ,, with all engines operating or 1.05 V ,, 4. That which results in V inoperative.

with one engine

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359

SPEED 76. d. JAR 25.1 07states that V, must be not less than: 1. v,. 2. 1.05VMc 3. The speed required to reach V, at 35 ft 4. That which results in VLoFbeing 1. I VMuwith all engines operating or 1.05 VMuwith one engine inoperative.

SPEED 77. a. Vl/Dmax?is the condition in which the ratio of IAS to drag is greatest. It is the speed at which the SAR is greatest. Its actual speed depends upon the shape of the drag curve so it has a different value for each ,, type of aircraft. But for jet aircraft it is typically in the region of 1.97 .V

SPEED 78. c. V,/Dmax?is the condition in which the ratio of IAS to drag is greatest. It is the speed at which the SAR is greatest. Its actual speed depends upon the shape of the drag curve so it has a different value for each type of aircraft. But is typically in the region of 1.32 VIM,.

SPEED 79. d. V,/Dm,,? is the condition in which the ratio of IAS to drag is greatest. It is the speed at which the SAR is greatest. Its actual speed depends upon the shape of the drag curve so it has a different value for each type of aircraft. But for propeller aircraft it is typically in the region of 1.7 V.,

SPEED 80. a. JAR 25.1 03 defines ,V, as being the greater of: 1. The Ig stall speed. 2. Not less than 2 kts or 2% above stick pusher activation speed

SPEED 81. d. JAR 25.103 defines ,V, as being the greater of: 3. The l g stall speed. 4. Not less than 2 kts or 2% above stick pusher activation speed

SPEED 82. d. JAR 25.107 specifies that the minimum value of V, must be not less than: 1. 1.23 V ,, for 2 and 3 engine turbo-props. ,, for 4 engine turbo-props and all jets. 2. 1.08 V 3. 1.1 VMC.

SPEED 83. c. JAA regulations state that V, must be not less than: 1. v,. 2. 1.05VMc 3. The speed required to reach V, at 35 ft 4. That which results in VLoFbeing 1.I,V , with all engines operating or 1.05 VMuwith one engine inoperative.

360

Speeds

SPEED 84. d. JAA regulations state that V, must be not less than: 1. v,. 2. 1.O5VM, 3. The speed required to reach V, at 35 ft 4. That which results in,,V, being 1.1 V ,, with all engines operating or 1.05 ,V, inoperative.

with one engine

SPEED 85. a. In effect this question is asking what decrease in VIM, will be produced by a 10% decrease in weight. This can be calculated using the following equation: VIM, at new weight = VIM, at old weight ;(new weight Iold weight) Although the old and new weights are not known, if the old weight is represented by 1, then following a 10% decrease the new weight should be represented by 0.9. Putting these figures into the equation gives the following: ,V ,,

at new weight = ,V ,,

at old weight d(0.911)

Which is VIM, at new weight = 0.949 ,V ,,

at old weight.

This represents a decrease of approximately 5% in VIM, for a 10% decrease in weight. This answer illustrates a more general point that the change in for small % changes in weight, the resulting % change in V ,,, or V, is approximately half the % change in weight.

SPEED 86. b. In effect this question is asking what decrease in VIM, will be produced by a 20% decrease in weight. This can be calculated using the following equation: VIM, at new weight = VIM, at old weight d(new weightiold weight) Although the old and new weights are not known, if the old weight is represented by 1, then following a 20% decrease the new weight should be represented by 0.8. Putting these figures into the equation gives the following: VIM, at new weight = VIM, at old weight d(0.811) Which is VIM, at new weight = 0.895 VIM, at old weight. ,,, This represents a decrease of approximately 10% in V

for a 20% decrease in weight.

This answer illustrates a more general point that the change in for small % changes in weight, the resulting % change in VIM, or V, is approximately half the % change in weight.

SPEED 87. c. ,,V , is the minimum speed at which it is possible to maintain control following a critical engine failure on the grol~ndin the take-off configuration. Following a critical engine failure, the asymmetric thrust from the remaining engines causes the aircraft to yaw and roll. The magnitude of this yawing and rolling tendency is proportional to the thrust being produced. The ability of the control surfaces to maintain control reduces with reducing speed. But none of these factors is dependent upon aircraft weight, so increasing weight will have no effect on . , ,V , But increasing thrust, temperature, or altitude will all do so.

Speeds

361

SPEED 88. a. This question relates to the threshold speed.,V ,, , at a 35 ft screen height. 1.23, ,V

JAR 25.125 specifies that this must not be less than

SPEED 89. a. V, is the speed at which it is equally possible to complete or abort a take-off within the remaining distance available following a critical engine failure. In some circumstance there may be a range of speeds within which this is possible. In such cases the minimum decision speed from which it is possible to complete the take-off within the remaining TODA is called VGo and the maximum decision speed from which it possible to stop within the remaining ASDA is called VsTop.

SPEED 90. c. V, is the speed at which it is equally possible to complete or abort a take-off within the remaining distance available following a critical engine failure. In some circumstance there may be a range of speeds within which this is possible. In such cases the minimum decision speed from which it is possible to complete the take-off within the remaining TODA is called VGo and the maximum decision speed from which it possible to stop within the remaining ASDA is called VsTop.

SPEED 91. b. JAA regulations define V, as the steady initial climb speed, which must be achieved by screen height, when taking-off with all engines operating. It must be not less than V2 + 10 Kts.

SPEED 92. d. UFTois the final take-off speed. JAR 25.107 states that it must be not less than: 1. 1.18VSR 2. High enough to provide the manoeuvring capability defined in JAR 25.143.

SPEED 93. b. JAA regulations define V, as the steady initial climb speed, which must be achieved by screen height, when taking-off with all engines operating. It must be not less than V, + 10 Kts.

SPEED 94. c. VFTois the final take-off speed. JAR 25.107 states that it must be not less than: 1. 1.18VSR 3. High enough to provide the manoeuvring capability defined in JAR 25.143.

SPEED 95. d. JAA regulations define V, as the steady climb takeoff speed for noise abatement procedures with all engines operating, using the scheduled techniques. It must be achieved by minimum flap retraction height. ,, and is such that the gross flight path does not fall below the It must be not less than 1.2 VMCAnor 1.3,,V gross flight path from which the net flight path has been derived.

SPEED 96. b. VATis the target threshold speed. With all engines operating it is called VAT,, and must be not less than 1.23 Us, following a steady descent to screen height at a gradient not greater than 5%. With one engine inoperative it is called VAT,.

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Speeds

SPEED 97. d. VATis the target threshold speed. With all engines operating it is called VAT,, and must be not less than 1.23,,V , following a steady descent to screen height at a gradient not greater than 5%. With one engine inoperative it is called VAT,.

SPEED 98. a. JAR 25.125 defines VREFas the reference threshold speed. It must be not less than 1.23 VsR, and not less than VMcL.

SPEED 99. a. ,V , is the minimum unstuck speed. It is the minimum speed at which an aircraft can lift-off and continue a take-off safely under any conditions. It is therefore the minimum possible value of VLoF under any circumstances.

SPEED 100. b. ,,V , is the maximum brake energy speed. It is the maximum CAS at which it is possible to abandon a take-off and stop the aircraft without exceeding the energy absorption capacity of the brakes. The energy to be absorbed increases with increasing aircraft weight and increasing TAS. But as altitude increases the TAS equating to any given CAS increases, thereby increasing the energy that must be absorbed to stop ,(,, from which it is possible to stop. the aircraft. So increasing altitude decreases the maximum CAS )V

SPEED 101. a. JAR 25.107 states that ,V ,,, must be not less than 1.13VsR for aircraft with 2 and 3 turbo-propeller engines, 1.08 VsR for all 4 engine turbo-props and all jet aircraft, and 1. I.,V,

SPEED 102. b. JAR 25.107 states that ,,V ,, must be not less than 1.13VsR for aircraft with 2 and 3 turbo-propeller engines, 1.08 VsR for all 4 engine turbo-props and all jet aircraft, and 1.IVMc.

SPEED 103. c. JAR 25.107 states that ,,V ,, must be not less than 1.13VsR for aircraft with 2 and 3 turbo-propeller engines, 1.08 VsR for all 4 engine turbo-props and all jet aircraft, and 1. I VMc.

SPEED 104. d. JAR 25.107 states that ,,V ,, must be not less than 1.13VsR for aircraft with 2 and 3 turbo-propeller engines, 1.08 VsR for all 4 engine turbo-props and all jet aircraft, and 1.IVMc.

SPEED 105. b. JAR 25.125 defines VREFas the reference speed to be attained by screen height in a specified configuration. VAT,VAT, and VAT, are examples of VREF. It must be at least 1.23 VsRo. But VsRo is proportional to aircraft mass so if mass increases, then,,V , will also increase.

SPEED 106. a. JAA regulations define VREFas the reference speed to be attained by screen height in a specified configuration. VAT,VAT, and VAT, are examples of VREF. It must be at least 1.3 VS. But Vs is proportional to aircraft mass so if mass increase then VREFwill also increase.

Speeds

363

SPEED 107. a. The minimum acceptable value of,,V , is 1.3 Vs, But Vs is proportional to the square root of the mass, so ,, In effect this question is asking what increase in,,V , will be increasing landing mass will increase.,V produced by a 10% increase in weight. This can be calculated using the following equation: ,,V ,

at new weight = VREFat old weight d(new weight Iold weight)

Although the old and new weights are not known, if the old weight is represented by 1, then following a 10% increase the new weight should be represented by 1.1. Putting these figures into the equation gives the following: ,,V , at new weight =,,V , at old weight d(1 .I 11) Which is,,V ,

at new weight = 1.049,,V ,

at old weight.

, This represents an increase of approximately 5% in,,V

for a 10% increase in weight.

This answer illustrates a more general point that the change in for small % changes in weight, the resulting % change in,,V , or V, is approximately half the % change in weight.

SPEED 108. d. The minimum acceptable value of,,V , is 1.3 V, But increasing flap angle decreases V, and hence The % decrease cannot in,,V , cannot however be calculated based on the information decreases.,,V, given in this question.

SPEED 109. c. The rr~inimumacceptable value of,,V , is 1.3 V, Both V, and,,V , in altitude do not affect V, and hence do not affect.,,V,

are specified in terms of CAS. Changes

SPEED 110. c. V, is the CAS at which an aircraft must be rotated into the flight attitude during the take-off run. V, is the initial climb speed that must be achieved by screen height. One of the factors limiting the minimum value of V, is that it must be at least 1.Iof the minimum control speed VMc. VMc is proportional to engine thrust, which decreases with increasing altitude. So by decreasing the value of VMc, increasing altitude also decreases the minimum value of V., But by decreasing thrust, increasing altitude also reduces the rate at which the aircraft accelerates from V, to V., So in order to achieve V, by screen height, V, must be increased to bring it closer to V., So as altitude increases the decreasing V, moves closer to the increasing V.,

SPEED 111. a. As pressure altitude increases, decreasing air density reduces the thrust available. This reduces the rate This in turn increases the distance required. So at which an aircraft can accelerate from V, and V, to.,,V, in order to ensure that the take-off can be completed within the remaining TORA, the minimum acceptable values of V, and V, must be increased. This effect can be seen in the CAP 698 MRJTI take-off speed charts.

SPEED 112. a. JAA regulations state that V, must be not less than: 1. v,. 2. 1.05VM, 3. The speed required to reach V, at 35 ft 4. That which results in ,V ,, being 1. I VMuwith all engines operating or 1.05 VMuwith one engine inoperative.

364

Speeds

SPEED 113. d. The length of surface capable of bearing the weight of an aircraft is the ASDA. This is made up of the runway plus the stopway. In an abandoned take-off the aircraft must therefore be brought to a halt before the end of the stopway. The latest point at which a decision can be made to abandon a take-off following an engine failure, is V, plus a two second interval during which the engine failure at V, is identified. So bringing all of the above factors together, the ASDA must be long enough to enable an aircraft to accelerate to V, plus 2 seconds, then decelerate to a stop without going off the end of the stopway.

SPEED 114. a. JAA regulations require that V, not be less than 1.I5 V., lncreasing flap angle decreases V, thereby permitting the minimum value of V, to be reduced. But the regulation also require V, to not less than 1.I VMcA.So as flap angle is increased, V, decreases until limited by VMcA. High mass, and low ambient pressure both increase V,. Low ambient temperature increases both VMcAand V., Option a is therefore the most appropriate.

SPEED 115. c. JAA regulations state that V, must be not less than VMcA,nor greater than either V, or.,,V , The same regulations also state that V, must be greater than V., VMcGis less than VMcA.So VMcG< VMcA< V, < V, < V., This means that as V, increases, V, will be the first limiting factor.

SPEED 116. b. JAA regulations state that V, must be not less than VMcA,nor greater than either V, or.,,V , The same ., VMc, is less than VMcA.So VMcG< VMcA< V, < V, regulations also state that V, must be greater than V < V,. This means that as V, increases, V, will be the first limiting factor. V, is the minimum speed from which a take-off can be safely completed within the remaining TODA following an engine failure. Decreasing mass increases acceleration rates and allows the take-off to be completed from a lower V,. So decreasing mass decreases V,. In extreme cases this reducing V, might equal its lowest acceptable value equal to VMcA.Small flap angles and low pressure increase V,. Although low temperatures increase VMcA and decrease V,, the effect is less marked than that of decreasing aircraft mass. So the most appropriate option is b but d is also true.

SPEED 117. c. JAR regulations define V ,, (Velocity Landing Gear Operating), as the maximum speed at which the landing gear may be lowered or raised.

SPEED 118. c. JAR regulations state that,,V ,, must be not less than 1.IVMc. lncreasing ambient pressure increases air density which increases engine thrust. But the minimum control speeds are proportional to thrust, so increasing pressure increases VMc. This in turn increases the minimum value of V., High mass and small ,, taking it further away from VMcAso options a and b are incorrect. Low flap angles both increase,,V temperature increases both VMcAand V, so option c is the most appropriate.

SPEED 119. d. JAR regulations define VLE(Velocity Landing Gear Extended), as the maximum speed for flight with the landing gear deployed (extended).

Speeds

365

SPEED 120. d. JAR regulations state that ,,V ,, must be not less than 1.1 VMc. Decreasing density altitude increases air density which increases engine thrust. But the minimum control speeds are proportional to thrust, so decreasing density altitude increases VMc. This in turn limits the minimum value of V., High mass and small flap angles both increase V, taking it further away from ,,,V , so options a and b are incorrect. Low ambient pressures have the opposite effect to that of low density altitudes. So option d is the most appropriate.

SPEED 121. b. This question specifies only a change in weight, so the problem can be solved using the following equation: VA at new weight = VA at old weight x 4 ( new weight 1 old weight) Although neither the new weight nor the old weight are known the question specifies that the new weight is 25% greater than the old weight. This fact can be used to solve the problem by assigning a value of 1 to the old weight, and a value of 1.25 to the new weight. Inserting these values into the equation gives the following:

This resolves to give V, at new weight = VA at old weight x 1.118 V, has therefore increased by approximately 12%. This answer demonstrates the more general point that for small weight changes, the % change in speed (V, V, ,V , etc.), is approximately %the % change in weight. Using this shortcut, a 50% weight increase gives a 25% speed increase, whereas the true figure is 22.5%. The accuracy of this shortcut method increases, as the % weight change decreases. For a 10% change in weight for example, the shortcut gives 5% change in speed, whereas the true figure is 4.9%.

SPEED 122. b. The lowest possible value for V, is.,V ,,, This is whichever is the higher of 1.15V, and 1.1 VMc. Extending flaps from the 1O0 to the 30° setting reduces Vs and so reduces V ,,, provided this would not cause it to be lower than 1.1 VMc.

SPEED 123. b. Vx is the CAS at which an aircraft will achieve the best angle of climb. It is the speed at which excess thrust is maximum. For a propeller aircraft, it is just above the minimum safe flying speed, and for a jet aircraft it is.,V , V, is the CAS at which an aircraft will achieve the best rate of climb. It is the speed at which excess power is maximum. For a propeller aircraft it is slightly higher than,V ,, and for a jet it is somewhat greater than .V ,, Vx is always less than V, at low altitudes, but the two converge until meeting at the absolute ., VMcGis the minimum control speed on the ground ceiling. Vx is therefore always less than or equal to V in the take-off configuration. This is lower than both Vx and V, for all aircraft types. The best option provided in this question is therefore b.

SPEED 124. b. , ,V , is the minimum CAS at which control can be maintained following a critical engine failure in the air in the take-off configuration. VMcGis the minimum CAS at which control can be maintained following a critical engine failure on the ground in the take-off configuration. V, is the CAS at which an aircraft is

366

Speeds

rotated into the flight position during take-off. It must be at least 110% of . , ,V , V, which is more than .,,V ,

,,V ,

is therefore less than

SPEED 125. b. ,V , is the minimum CAS at and above which an aircraft can safely lift-off the ground and continue take,,V , is the minimum CAS at which control can be maintained following a critical engine failure in the off. air in the take-off configuration. The term, , V , is being superseded by the term.,V , V, is the take-off ,, which must be at least 1. I 5 V, or 1. I safety speed. The minimum acceptable value of V, is termed,,V ,,V, whichever is the greater. Placed in order of increasing magnitude these speeds are, V, ,,,V , ,,,V , V,, V, ,,V, ,,,V, .,V ,,, The correct equation is therefore b, V, < ,,V , <.,V ,,,

SPEED 126. a. JAR 25 defines V, as "the take-off safety speed".

SPEED 127. a. JAR 25 states that V, must not be less than, , V , and that V, must not be less than V,. This implies that the minimum value of V, is, , V , and the maximum value is V.,

SPEED 128. b. The various parameters that limit the value of V, are detailed in JAR 25.1 07. Of the options offered in this , is included in the JAR. question only option b, 105% of ,V

SPEED 129. a. ,V , is the minimum drag IAS and for any given aircraft and configuration it is a fixed multiple of V., The effect of a mass increase on,V , can therefore be calculated by calculating the effect on V., This problem can be solved using the following equation: New V, = Previous V, d(new mass Iprevious mass) Although neither the previous mass nor the new mass are known, it is known that the new mass is 1.I times the previous mass. This ratio can be put into the equation to give the following: New V, = Previous V, d ( l . I 1 I ) which is 1.05 Previous V., The ,V , after the mass increase is therefore 5% greater than that before the mass increase so ,V , also increased by 5%.

has

The increase in mass will necessitate an increase in lift in any given flight condition and this in turn will cause an increase in lift induced drag. , then the C, will be unchanged. The new induced drag will If the speed is increased to the new,,V therefore be proportional to IIEAS,, which is 1/1.052. That means induced drag will decrease by 10%. If the speed remains unchanged then C, must increases by 10%. But C, is proportional to C?, SO a 10% increase in C, from 1 to 1. I will give a 1.12 increase in C,. This will give a 21 % increase in induced drag. Neither of thee effects are predicted in the options offered, A 10% increase in lift at constant speed would not have any significant effect on parasite drag, whereas a 5% increase in speed would increase profile drag by a factor of 1.05,, which is a 10% increase. So if , will increase by 5% but parasite drag will not speed does not change, then option a is correct, ,V change.

Speeds

367

SPEED 130. a. VMDis the minimum drag IAS and for any given aircraft and configuration it is a fixed multiple of V., The effect of a mass increase on VMDcan therefore be calculated by calculating the effect on V., This problem can be solved using the following equation: New V, = Previous V, d(new mass I previous mass) Although neither the previous mass nor the new mass are known it is known that the new mass is 0.8 times the previous mass. This ratio can be put into the equation to give the following: New V, = Previous V, d(0.8 1 I) which is 0.9 Previous V., The VMDafter the mass decrease is therefore 10% less than that before the mass decrease, so VMDhas decreased by 10%. The increase in mass will necessitate an increase in lift in any given flight condition and this will in turn cause an increase in lift induced drag. Such increases in lift do not however have any significant effect on profile drag. The correct answer is therefore option a, VMDwill decrease by 10% but parasite drag will not change.

SPEED 131. a. VMcGis the minimum speed at which control can be maintained on the ground in the take-off configuration, following a critical engine failure. V, is the speed at which the pilot must decide to continue or abort the take-off. In order for him to be able to make this decision the aircraft must have exceeded its VMcG.V, must therefore be greater than VMc.. V, is the speed at which the aircraft must be rotated into the takeoff attitude, after the decision has been made to continue the take-off. V, must therefore be greater than V,. V, is the take-off safety speed. This must be at least V, plus the speed increment achieved before reaching a height of 35 feet above the take-off surface. The correct sequence is therefore VMcG,V,,

v, v,. SPEED 132. c. V, is the CAS at which an aircraft is capable of achieving its best angle of climb. It is the speed at which excess thrust is maximum. V, is the CAS at which an aircraft is capable of achieving its best rate of climb. It is the speed at which excess power is maximum. At low altitudes V, is less than V, for all aircraft types. As altitude increases however, V, remains constant whilst V, decreases until the two coincide at the ., absolute ceiling. V, is therefore always less than or equal to V It should be noted that although the above description is correct, in JAR ATPL Performance examinations, V, and V, are both assumed to be unaffected by altitude. This is because the rate of change df V, with increasing altitude is very small, and V, and V, are of significance only during climb out at low altitudes. ,, it was more appropriate to use the Because this question concerned the relative magnitudes of V, and V true effects of altitude.

SPEED 133. b. V, is the stalling speed and is the lowest speed at which an aircraft can produce lift equal to its weight. V, is the CAS at which an aircraft is capable of achieving its best angle of climb. It is always greater than V., V, is the CAS at which an aircraft is capable of achieving its best rate of climb. It is always more than or ., equal to Vx. The correct sequence is therefore V, < V, < V

SPEED 134. a. VMcGis the lowest CAS at which it is possible to maintain control in the take-off configuration, on or close to the ground following failure of a critical engine.

368

Speeds

It is the minimum CAS at which control authority equals the yawing and rolling moments caused by asymmetric thrust. The control authority is constant at any given CAS, but the asymmetric thrust moments are proportional to engine thrust. Any factor that reduces thrust will therefore reduce VMcG.Such factors include high ambient temperature, humidity and pressure altitude.

SPEED 135. c. JAR 25.107 defines a number of take-off speeds. These include the take-off safety speed is V, the threshold speed is, ,V the take-off decision speed V,, and the minimum control speed in the take-off configuration is VMcG.V, is the speed at which an aircraft is equally able to complete or abort a take-off safely following a critical engine failure.

SPEED 136. a. ,V, is the stalling speed. VMcAis the minimum speed at which control can be maintained in the air following the failure of the critical engine in the take-off configuration. V, is the maximum speed at which a decision to abandon a take-off can be made. V, is the take-off safety speed that must be achieved by screen height. The aircraft must exceed Vs before leaving 'rhe ground so V, is less than V., But it must also be possible to maintain control before committing to the take-off, so VMcA is lower than V,. The decision to take-off must also be made before leaving the ground so V, must be less than V., So option a, V, < VMcA<,,V ,, is correct.

SPEED 137. a. VMcGis the minimum speed at which it is possible to maintain control of an aircraft on the ground, in the take-off configuration, following the failure of the critical engine. V ,, is the speed at which the critical engine is assumed to fail during take-off, for the purposes of calculating take-off performance. For safety ,, must not be less than VMcG.V, is the maximum speed at which a reasons JAR 25.1 07 specifies that V decision can be made to continue or abort a take-off. In order to make such a decision safely the aircraft must have exceeded its minimum control speeds. So option a, VMcG< V ,, < V, is correct.

SPEED 138. d. Fuel consumption is a piston engine is proportional to power output. So endurance is maximum when .flyingat the minimum power required speed VIM,. When ordered to maintain a holding pattern it is necessary .,, to maxirrlise endurance, so the holding speed is typically the minimum power speed V The minimum drag speed VIM, is always greater than VIM,for all aircraft types. V, IS the speed at which an aircraft is able to achieve its best rate of climb. For piston aircraft V, is slightly greater than VIM, but lower than VIM,. For the aircraft referred to in this question, VIM, is 100 Kts and VIM, is 130 Kts. So option d, 100 Kts < V, < 130 Kts is correct.

SPEED 139. a. V, is the take-of safety speed and is must be achieved by screen height. The take-off decision speed is V,. The speed at which a critical engine is assumed to fail is V,,. The speed at which an aircraft first leaves ,, So option a, is the most appropriate for this question. the grol~ndis.,V

SPEED 140. c. JAR 25 defines the following speeds.,,V , is the reference speed that should be attained at screen height , therefore has no relevance in determining any of the take-off speeds. V ,, is the during a landing.,,V speed at which the critical engine is assumed to fail during take-off performance planning calculations. V, is the speed at which an aircraft must be rotated to (he lift-off attitude during the take-off run. V, is the take-off safety speed that must be achieved by screen height. The decision to complete or abort a take-off

Speeds

369

must be made before the aircraft is rotated into the lift-off attitude. JAR 25.1 07 therefore specifies that V, must not exceed V., So V M I, is V .,

SPEED 141. a. ,,V , is the reference speed that should be attained at screen height during a landing. JAR 25.125 specifies that,,V , must be not less than 1.23 VsRo.So the minimum acceptable margin between,,V , and, ,V , is 23% of VsRo.

SPEED 142. d. V, is the take-off safety speed that must be achieved by screen height. VMcAis the minimum speed at which control can be maintained in the air in the take-off configuration following a critical engine failure. In order to ensure that control can be maintained during a take-off JAR 25.107 specifies that V, must be at least 1. I VMcA.So the minimum acceptable margin between the two is 10% of VMcA.

SPEED 143. c. VMcAis the minimum speed at which control can be maintained in the air in the take-off configuration following a critical engine failure. VMc, is the minimum speed at which control can be maintained on the ground in the take-off configuration following a critical engine failure. V, is the maximum speed at which a decision must be made to continue or abort a take-off following a critical engine failure. V, is the stalling speed. V, is the speed at which an aircraft must be rotated into its lift-off attitude during the take-off run. For safety reasons it is essential that an aircraft is not rotated into its take-off attitude before VMc, VMcA, V, or V,. JAR 25.1 07 therefore specifies that V, must be at least I.0 V, and 1.05 VMcA.So option c is the most appropriate in this question.

SPEED 144. a. JAR 25.107 defines,,V, the Minimum Unstick Speed, as the minimum calibrated airspeed at which an aircraft can safely lift off the ground and continue the take-off under any circumstances. ,

SPEED 145. d. JAR 25.1 07 defines the initial steady climb speed V, as the speed that must be achieved by screen height with all engines operating. It must be not less than V, + 10 Kts.

TAKE-OFF

TAKE-OFF 1. d 'The field limited take-off mass is the maximum mass at which an aircraft can take-off safely within the available field length. The climb limited take-off mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient. This question relates to the effects of increasing or decreasing flap angle. The fact that the aircraft mass is greater than that originally estimated, has no bearing on these effects. Increasing flap angle has two principal effects. Firstly, by increasing the coefficient cf lift, the lift generated at any given airspeed is increased. This decreases the speed necessary for take-off and hence decreases the take-off distance required for any given mass, thereby increasing the field length limited mass. But increasing flap angle also increases drag and worsens liftdrag ratio. This in turn decreases the maximum attainable climb gradient at any given mass. So increasing flap angle to 15 degrees will decrease the climb limited take-off mass and increase the field limited take-off mass. Decreasing flap angle to 5 degrees will have the opposite effects, increasing climb limited mass and decreasing field limited mass.

TAKE-OFF 2. C. The first segment commences at screen height at the end of the TODR at which point the undercarriage up button is pressed. The speed at this point must be at last V2 and the power setting is that for maximum take-off power. This segment ends as soon as the undercarriage is fully retracted.

TAKE-OFF 3. a. Take-off performance is proportional to thrust, such that increasing thrust improves take-off performance. Thrust output at any given RPM is proportional to air density, so increasing air density increases both thrust and take-off performance. Low altitude, low temperature and low humidity all increase air density. So best take-off performance is attained under this combination of conditions.

TAKE-OFF 4. C. The JAA regulations governing minimum acceptable climb gradients, refer to still air conditions. This means that the ability of an aircraft to achieve these minimum standards is not affected by headwinds or tailwinds. But such winds do affect the actual gradients achieved, so this is taken into account when calculating obstacle clearance.

TAKE-OFF 5. C. The energy that must be absorbed to stop an aircraft is proportional to ground speed.,,V , is the maximum velocity at which it is possible to within the remaining TORA, in an aborted take-off. The maximum speed , which must not exceed.,,V , at which a decision to abort is permitted is V,, so, ,V

TAKE-OFF 6. a. Maximum tyre speed is the ground speed above which tyres will structurally fail. It is therefore necessary to ensure that this speed is not exceeded during the take-off run. The greatest ground speed in the takeoff run is achieved at V,- so the maximum tyre speed is referenced to this.

.,,

TAKE-OFF 7. C. Contamination increases the rolling resistance as the wheels are pushed during the take-off run. This increases the TORR. The TODR is the TORR plus the horizontal distance between lift-off and screen height, so increasing the TORR also increases the TODR. But part of the TODR is in the air, and contamination does not affect an aircraft after lift-off, so the TODR is affected less than the TORR.

TAKE-OFF 8. d. The length of the take-off run and the climb gradient are proportional to the thrust available. lncreasing thrust increases the acceleration rate during the take-off run, thereby decreasing the take-off run required to achieve VLo,. Climb gradient is proportional to excess thrust, so increasing thrust increases climb gradient. Thrust is proportional to the mass flow of air passing through the engines, which is in turn proportional to air density. lncreasing ambient pressure compresses the air, increasing both density and thrust. So increasing ambient pressure decreases the take-off run and increases climb gradient.

TAKE-OFF 9. b. ,The

length of the take-off run and the climb gradient are proportional to the thrust available. Decreasing thrust decreases the acceleration rate during the take-off run, thereby increasing the take-off run required to achieve VLo,. Climb gradient is proportional to excess thrust, so decreasing thrust decreases climb gradient. Thrust is proportional to the mass flow of air passing through the engines, which is in turn proportional to air density. lncreasing pressure altitude decreases ambient pressure. This allows the air to expand, thereby decreasing both density and thrust. So increasing pressure altitude increases the takeoff run and decreases climb gradient.

TAKE-OFF 10. d. The field limited take-off mass is the maximum mass at which an aircraft can take-off within the distances available in a given set of ambient conditions. So for a field limited take-off, the actual mass of the aircraft can be equal to, but not greater than the field limited mass. This is not directly related to the climb limited mass. The calculation of the field limited mass takes,,V , into account, so,,V , cannot be too high in a filed limited take-off.

TAKE-OFF 11. d. V, is the maximum speed at which an aircraft is equally capable of completing or aborting a take-off safely following a critical engine failure. Engine failure at speeds below V, necessitates aborting, whilst failures above V, require the take-off to be continued. So if an engine failure is detected at V, the pilot can decide to do either. But it takes time to detect a failure, so if a failure actually occurs at V, it will be detected at some higher speed. In which case the take-off must be continued.

TAKE-OFF 12. b V, is the maximum speed at which an aircraft is equally capable of completing or aborting a take-off safely following a critical engine failure. The ability to decelerate is dependent upon a number of factors including aerodynamic drag. If the flaps were inadvertently set at 25" instead of 35" degrees, the deceleration rate would be decreased. This would reduce stopping ability and hence reduce the true value for V,. The value of V, actually used would therefore be too high, causing the aircraft to overrun the stopway if the take-off was aborted at the calculated V,.

TAKE-OFF 13. a. The ability of an aircraft to take-off at a given mass and from a given field, depends upon two factors. These are the aircraft's ability to reach screen height in the distance available and its ability to stop in the

event of an aborted take-off. The distance required to achieve screen height (TODR) must not be greater than the take-off distance available (TODA). The accelerate stopping distance required (ASDR) must not exceed the emergency distance available (EMDA). Under some circumstances it is possible for one of these factors to be more restrictive than the other. A balanced field is one in which the EMDA is equal to the TODA. Under these circumstances it is sometimes possible to simplify the take-off performance calculation process.

TAKE-OFF 14. a. The take-off distance is the horizontal distance from the point at which the brakes are released (BRP) at the start of the take-off run, to the point at which the aircraft attains screen height. The take-off distance available (TODA) is the horizontal distance from the BRP to the end of the clearway. The take-off run available (TORA) is the horizontal distance from BRP to the end of the runway. The distance from the BRP to the end of the stopway is the accelerate stop distance available (ASDA), or the emergency distance available (EMDA). The 400 ft point has no relevance to the TD, TODA, TORA, ASDA or EMDA.

TAKE-OFF 15. a. lncreasing runway upward slope will decreases the rate at which an aircraft can accelerate during the take-off run. This in turn will mean that a longer take-off run is required t o reach lift-off speed and screen height. So increasing upward slope will increase both the take-off run required (TORR) and the take-off distance required to achieve screen height (TODR).

TAKE-OFF 16. a. lncreasing flap angle from 1O0 to 20° will increase the coefficient of lift, thereby increasing the lift generated at any given airspeed. This in turn will reduce the stalling speed of the aircraft, and all other speeds related to the stalling speed. Both V, and VLoFare fixed multiples of the stalling speed, so increasing flap angle will decreases V, and VLoF.

TAKE-OFF 17. b. Decreasing flap angle will decrease the coefficient of lift at any given angle of attack, thereby decreasing the lift generated at any given airspeed. This in turn will increase the stalling speed of the aircraft, and all other speeds related to the stalling speed. Both V, and VLoFare fixed multiples of the stalling speed, so decreasing flap angle from 15O to 5O will increase V, and VLoF.

TAKE-OFF 18. C. lncreasing flap angle will increase the coefficient of lift at any given angle of attack, thereby increasing the lift generated at any given airspeed. This in turn will reduce the stalling speed of the aircraft, and all other speeds related to the stalling speed. VLoFand V, are fixed multiples of the stalling speed, so increasing flap angle will decrease VLoFand V., This decrease in VLoFtends to decrease the take-off run required (TORR) to reach VLoF. The effect on the take-off distance required (TODR) is more complex, due to the fact that increasing flap angle also increases drag. The increase in drag reduces the rate at which an aircraft can accelerate during the take-off run. This in turn tends to increase the take-off run required (TORR). The increased drag also decreases climb perfgrmance, thereby increasing the distance between lift-off and achieving screen height. The overall effect depends upon the balance between increasing TODR due to reduced acceleration and climb rates, and decreasing TODR due to reduced VLoFand V., But the recommended flap angle is intended to provided best take-off performance, so increasing flap angle beyond optimum will

generally increase TODR. So increasing flap angle beyond the normal setting decreases VLoFbut increases TODR.

TAKE-OFF 19. C. The take-off run required (TORR) is the distance from BRP to attaining VLoF. The Take-off distance required (TODR) is the distance from BRP to attaining at ieast V, at screen height. Both VLoFand V, are airspeeds, so anything that increases the airspeed will reduce the distances required to attain them. In a headwind the aircraft already possesses an airspeed equal to the headwind, before the brakes are released. This decreases the acceleration required and hence decreases both the TORR and the TODR.

TAKE-OFF 20. d. V, is the take-off safety speed and VLoF is the lift-off speed. Both are related to dynamic pressure and hence indicated airspeed (IAS). The airspeed indicator operates by sensing dynamic pressure and producing a reading proportional to it. So at any given IAS the dynamic pressure is constant. Although a headwind decreases the ground speed equating to any give IAS, it does not change the dynamic pressure. So a headwind does not affect V, or VLoF.

TAKE-OFF 21. a. The take-off run required (TORR) is the distance from BRP to attaining VLOF.The Take-off distance required (TODR) is the distance from BRP to attaining at least V, at screen height. Both VLoFand V, are airspeeds, so anything that increases the airspeed will reduce the distances required to attain them. In a tailwind the aircraft possesses a negative airspeed equal to the tailwind, before the brakes are released. This increases the acceleration required, and hence increases both the TORR and the TODR.

TAKE-OFF 22. d. V, is the take-off safety speed and,,V , is the lift-off speed. Both are related to dynamic pressure and hence indicated airspeed (IAS). The airspeed indicator operates by sensing dynamic pressure and producing a reading proportional to it. So the dynamic pressure is constant at any given IAS. Although a tailwind increases the ground speed equating to any given IAS, it does not change the dynamic pressure. So a tailwind does not affect V, or VLoF.

TAKE-OFF 23. C. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. V, is the take-off safety speed that must be achieved by screen height. Both V, and V, are related to the ability of the wings to generate sufficient lift. This in turn is related to dynamic pressure and hence indicated airspeed (IAS). When taking-off from a contaminated runway the acceleration rate is reduced. This is due to higher rolling resistance as the wheels must move through the contaminant, and also extra drag as contaminant is thrown up and strikes the landing gear, flaps and fuselage. Although this lower acceleration rate increases the time and distance required to attain V, or V, it does not affect the magnitude of these speeds. So the standard values of V, and V, are used when taking-off from contaminated runways.

TAKE-OFF 24. a. V, is the speed at which an aircraft must be rotated into the, lift-off attitude during the take-off run. It is related to both the lift force that needs to be generated for take-off, and the dynamic pressure. lncreasing aircraft mass increases the lift force that must be generated, and hence increases V., lncreasing flap angle increases the coefficient of lift and hence decreases the speed necessary to generate the required lift force. This in turn decreases V.,

But V, must not be less than the speed required to attain V, by screen height. lncreasing air density increases the thrust produced by the engines, thereby increasing the rate of acceleration from V, to V,. So for any given V, increasing density will decrease the minimum value of V., lncreasing temperature decreases density, and hence increases V., Although runway slope will affect the time and distance required to attain V, it will not affect the value of V., Options b, c and d are therefore incorrect.

TAKE-OFF 25. b. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. It is related to both the lift force that needs to be generated for take-off, and the dynamic pressure. lncreasing aircraft mass increases the lift force that must be generated, and hence increases V., lncreasing flap angle increases the coefficient of lift, thereby decreasing the speed necessary to generate the required lift force. This in turn decreases V., But V, must not be less than the speed required to attain V, by screen height. lncreasing air temperature, decreases air density. Decreasing air density decreases the thrust produced by the engines, thereby decreasing the rate of accelerationfrom V, to V., So for any given V, increasingtemperature will increase the minimum value of V., Although runway slope and condition will affect the time and distance required to attain V, it will not affect the value of V., Options a, c and d are therefore incorrect

TAKE-OFF 26. C. The climb limited take-off mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient. So any factor that increases climb performance will increase the climb limited take-off mass. The field length limited take-off mass is the maximum mass at which an aircraft can safely take-off within the field length (TORA, TODA and ASDA) available. This is primarily affected by its ability to accelerate and decelerate and the value of its V,, V, and VLoF.Anything that increases V,, V, and VLoF, or which decreases the attainable acceleration rates, will decreases the field length limited mass. Decreasing flap angle will decrease both lift and drag at any given speed. The decreased lift will require a greater V, V, and VLoF, and hence will tend to increase the distances required to accelerate to these speeds. This will tend to decrease the field limited mass. But the decreased drag will increase both the acceleration rates and the climb performance. This will tend to increase both the field limited mass and the climb limited mass. The overall effect will depend upon the balance between the above effects, but will generally be an increase in climb limited mass and a decrease in field length limited mass. lncreasing flap angle will have the opposite effects.

TAKE-OFF 27. d. The climb limited take-off mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient. So any factor that decreases climb performance will decrease the climb limited take-off mass. The field length limited take-off mass is the maximum mass at which an aircraft can safely take-off within the field length (TORA, TODA and ASDA) available. This is primarily affected by its ability to accelerate and decelerate and the value of its V,, V, and VLoF.Anything that decreases V,, V, and ,,,V, or which increases the attainable acceleration rates, will increase the field length limited mass. lncreasing flap angle will increase both lift and drag at any given speed. The increased lift will require a lower V, V, and VLoF, and hence will tend to decrease the distances required to accelerate to these speeds. This will tend to increase the field limited mass. But the increased drag will decrease both the acceleration rates and the climb performance. This will tend to decrease both the field limited mass and the climb limited mass. The overall effect will depend upon the balance between the above effects, but will generally be a decrease in climb limited mass but an increase in field length limited mass. Decreasing flap angle will have the opposite effects.

TAKE-OFF 28. a. The take-off performance of an aircraft can be considered to comprise of two parts. The climb limited take-off mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient. Any factor that decreases climb performance will decrease the climb limited take-off mass. The field length limited take-off mass is the maximum mass at which an aircraft can safely take-off within the field length (TORA, TODA and ASDA) available. This is primarily affected by its ability to accelerate to V,, V, and.,,V, Both climb performance and acceleration rates are proportional to the thrust available. Any factor that decreases thrust will degrade take-off performance. The thrust available at any given RPM is determined by the mass flow of air passing through the engines. This in turn is dependent upon air density. Increasing ambient temperature decreases air density, thereby decreasing mass flow, and ultimately take-off performance. So take-off performance will be degraded if OAT increases

TAKE-OFF 29. b. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. It is related to dynamic pressure and is hence not affected by runway condition. V, is the maximum speed at which it is equally possible to complete or abort a take-off, within the remaining TODA and ASDA. lf takeoff is aborted then the aircraft must be brought to a halt before reaching the end of the ASDA. Slush will reduce friction between the tyres and the runway and hence increase the distance required to bring the aircraft to a halt. This in turn means that a lower V, must be used when operating from a contaminated runway. If however the same V, is to be used, then the mass of the aircraft must be decreased to reduce stopping distance required. Reducing mass will reduce V., So a slush contaminated runway will decrease V,, maximum take-off weight, and V., In considering this result it should be noted that contamination has no direct effect upon V, but can affect it indirectly.

TAKE-OFF 30. b. Although this questions includes a significant amount of data, which might suggest the use of CAP 698 charts, the options relate only to general effects. It can therefore be answered without using these charts. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. It is related to dynamic pressure and hence is not affected by runway condition. V, is the maximum speed which it is equally possible to complete or abort a take-off, within the remaining TODA and ASDA. If takeoff is aborted, the aircraft must be brought to a halt before reaching the end of the ASDA. Standing water will reduce friction between the tyres and the runway and hence increase the distance required to bring the aircraft to a halt. This in turn means that a lower V, must be used when operating from a contaminated runway. If however the same V, is to be used, then the mass of the aircraft must be decreased to reduce stopping distance required. Any reduction in mass will reduce V., So a runway contaminated with standing water might decrease V,, maximum take-off weight, and V., This effect is illustrated in CAP 698 chart 4.14, which indicates mass reductions of between 6200 and 8400 Kg and V, reductions of between 13 and 8 Kts depending on field pressure altitude. It should be noted that questions of this form can often be answered by examining the general trends in the CAP 698 charts.

TAKE-OFF 31. C . The anti-skid system improves the ability of an aircraft to decelerate to a halt. In the case of an aborted take-off, anti-skid will reduce the stopping distance, and hence decrease the ASDR. This will increase the field limited take-off mass. During landing, an anti-skid system will decrease the landing run, thereby

increasing the maximum landing mass. So a defective anti-skid system will reduce both take-off and landing performance.

TAKE-OFF 32. b. Decreasing flap angle decreases both lift and drag at any given airspeed. The decreased lift requires a greater V,, V, VLoFand V, thereby tending to increase the take-off run required. But the reduced drag increases acceleration rates, thereby tending to partly offset the increase in take-off run required. The reduced drag also increases the maximum gradient of climb, enabling the aircraft to clear close-in obstacles more easily. This can be particularly important at high field pressure altitudes, where low air density reduces thrust and climb gradient. The overall effect of reducing flap angle from 10 to 5 degrees would be an increased take-off run and an improved climb gradient. The longer take-off run would require a longer runway, whilst the improved climb gradient would improve clearance of close-in obstacles. So provided sufficient field length is available, reducing flap angle might irr~provethe ability of an aircraft to clear close-in obstacles when taking-off at high pressure altitudes. It should however be noted that the examiners intentions in drafting this question are not entirely clear. The combination of long runway, and low pressure altitude would also increase the ability to clear close-in obstacles. This technique would however be less necessary at low pressure altitudes where increased density would increase climb performance without the need to alter flap settings. Options c and d are incorrect because short runways are unsuitable for low flap angle take-offs.

TAKE-OFF 33. a. lncreasing flap angle Increases both lift and drag at any given airspeed. The increased lift requires a lower V,, V, VLoFand V, thereby tending to decrease the take-off run required. But the increased drag decreases acceleration rates, thereby tending to increase the take-off run required. The increased drag also decreases the maximum gradient of climb, degrading the ability of the aircraft to clear obstacles. The overall effect of increasing flap angle from 10 to 15 degrees would be a decreased take-off run and a degraded obstacle clearance. The effect of TODR to achieve screen height is a balance between reduced TORR and reduced climb gradient. The result of which can be a slight increase or slight decrease in TODR. Option a, decrease take-off run, TODR and obstacle clearance, provides the best overall assessment of the effects of using the 15 degree flap setting.

TAKE-OFF 34. d. Take-off distance and climb performance are dependent upon a number of variables, including thrust and power available. lncreasing thrust increases the acceleration rate during the take-off run, thereby reducing the take-off distance. Maximum attainable climb gradient is proportional to thrust, whilst maximum attainable rate of climb is proportional to power available. Both the power and thrust available from aircraft engines are proportional to the mass flow of air passing through them. This in turn is proportional to air density. But increasing atmospheric pressure compresses the air, causing its density to increase. This increases thrust and power available at take-off RPM, and hence decreases take-off distance and increases climb performance.

TAKE-OFF 35. b. ,,V , is the maximum calibrated airspeed (CAS) at which an a take-off can be aborted and the aircraft brought to a halt within the remaining ASDA without exceeding the energy absorption capabilities of the wheel brakes. It is related to mass and ground speed, so any factor that increases ground speed will

reduce the maximum acceptable take-off mass. Decreasing ambient ternperature contracts the air causing its density to increase. This reduces the TAS at any given CAS, thereby reducing ground speed at V,. This allows a greater mass to take-of from any given ASDA. So decreasing temperature increases,,V , limited mass. The climb limited mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient after take-off. Any factor that increases climb gradient will increase climb limited mass. 'The field length limited take-off mass is the maximum mass at which an aircraft can take-off within the field length available. Any factor that increases acceleration rates will increase field limited take-off mass. Take-off distance and climb performance are dependent upon a number of variables, including thrust and power available. Increasing thrust increases the acceleration rate during the take-off run, thereby reducing the take-off distance. Maximum attainable climb gradient is proportional to thrust, whilst maximum attainable rate of climb is proportional to power available. Both the power available and thrust produced by aircraft engines are proportional to the mass flow of air passing through them. This in turn is proportional to air density. But decreasing ambient temperature contracts the air, causing its density to increase. This increases thrust and power available at take-off RPM, and hence decreases take-off distance and increases climb performance. So decreasing ambient temperature increases both field length limited take-off mass and climb limited take-of mass.

TAKE-OFF 36. d. Increasing flap angle from 5 degrees to 15 degrees will increase the lift at any given airspeed. This will decrease the stalling speed (V,) and hence decrease all other speeds that are based on the stalling speed. V, must be at least 1.2 V, so by decreasing V, increasing flap angle will decrease the minimum , So increasing flap angle will decrease V, unless value of V,. But V, must also be not less than 1..I,,V limited by .,,V ,

TAKE-OFF 37. a. The maximum ground speed that may be safely used during take-off or landing is limited by the ability of the tyres to maintain their structural integrity. If these limiting tyre speeds are exceeded there is a danger that the tyres will fail. Ground speed is related to TAS whereas lift-off speed is related to CAS. Increasing ambient temperature or pressure altitude causes the air to expand, thereby decreasing its density. This in turn increases the ratio of TAS to CAS. ,, This means that as temperature or altitude increase, the TAS and hence ground speed equating to ,,V both increase. Tyre speeds therefore become more limiting as temperature and pressure altitude increase.

TAKE-OFF 38. d. V, is the maximum speed at which an aircraft is equally capable of completing or aborting a take-off safely, followirrg a critical engine failure. When an engine fails during the take-off run, a finite amount of time is required before it can be detected by the pilot. When making take-off performance calculations, there is therefore an interval of 2 seconds between the speed at which an engine is assumed to fail and the maximum value of V,. This period is allowed regardless of engine type.

TAKE-OFF 39. d. The measured take-off distance (TOD) is established during certification test flying. It is the horizontal distance from the brake release point (BRP) to the point at which an aircraft attains screen height when taking-off from a level hard surface. This figure is then increased by a factor to establish the gross takeoff distance required (Gross TODR) for an average aircraft of that type. This in turn is increased by various factors reflecting such things as contamination, slope and temperature, in order to calculate the

TODR for a given aircraft using a given runway on a given day. The TOR , or take-off run is the distance . , ,V from BRP to

TAKE-OFF 40. b. The take-off run required is the distance that an aircraft is calculated to require to lift off the ground. It is the calculated horizontal distance between the BRP and the point at which the aircraft attains its lift-off The distance actually taken to achieve lift-off is the take-off run, or TOR. The distance speed,.,,V, actually taken from BRP to screen height is the take-off distance (TD or TOD), The screen weight referred to in option d, is not a recognised term.

TAKE-OFF 41. d The take-off distance required is the distance that an aircraft is calculated to require to clear the screen. It is the calculated horizontal distance between the BRP and the point at which the aircraft is expected to attain screen height. The distance actually taken to achieve this is the take-off distance, or TOD. The , is the take-off run, or TOR. actual distance from BRP to attaining,,V

TAKE-OFF 42. a. JAR 25.107 specifies the criteria limiting the minimum acceptable value for V., These are that it must be , It should be noted that VMCmeans the minimum air control speed, . , ,V , not less than, 1. I.,V

TAKE-OFF 43. b. JAR 25.1 07 specifies the criteria limiting the minimum acceptable value for V., These are that it must be not less than, 1.2 Vs for 4 engine turbo-props and jets with provision for significant reduction in oneengine-inoperative stalling speed.

TAKE-OFF 44. C. JAR 25.107 specifies the criteria limiting the minimum acceptable value for V,. These are that it must be not less than, 1. I 5 Vs for 2 and 3 engine turbo-props and jets without provision for significant reduction in one-engine-inoperative stalling speeds.

TAKE-OFF 45. a. JAR 25.1 07 specifies the criteria limiting the minimum acceptable value for V., not less than, V,.

These are that it must be

TAKE-OFF 46. b. JAR 25.107 specifies the criteria limiting the minimum acceptable value for V., not less than, 105%.,V ,


TAKE-OFF 47. C. JAR 25.107 specifies the criteria limiting the minimum acceptable value for V., not less than, The speed required to attain V, by screen height.


TAKE-OFF 48. d. JAR OPS 1.495 specifies the conditions to be met in planning and executing turns during the take-off and initial climb. This specifies that turns involving bank angles between 15' and 25' must not be commenced below 400 ft, and that obstacles within the accountability area must be cleared by a vertical separation of at least 50 ft during such turns.

TAKE-OFF 49. a. V, is the speed at which an aircraft must be rotated into its lift-off attitude during the take-off run. It is related to the generation of lift and hence to dynamic pressure. A contaminated runway affects the ability of an aircraft to accelerate for take-off and to decelerate in the event of an aborted take-off. Because of these effects it is necessary to adjust the take-off distances (TORR and TODR) to ensure safety. But a contaminated runway does not affect the ability of the wings to generate lift, so V, is not affected. So when taking-off from a contaminated runway the aircraft is rotated at the normal V., V, will be affected by runway contamination, only if conditions are such that the aircraft take-off mass needs to be reduced.

TAKE-OFF 50. a. V, maximum is the speed at which an aircraft is equally able to complete or abandon a take-off safely, following a critical engine failure during the take-off run. At speeds below V, such a failure would prevent the aircraft from attaining screen height within the remaining TORA, but the take-off could be aborted safely. At speeds above V, such a failure would cause the aircraft to overrun the ASDA in the event of a decision to abort the take-off, but the take-off could be completed safely. In this question the calculated V, is 160 Kts and the actual V, is 150 Kts. So at any speed above 150 Kts, a decision to abandon the takeoff would cause the aircraft to overrun the ASDA. At all speeds below 150 Kts, the take-off could be abandoned safely. So only option a is correct.

TAKE-OFF 51. d. V, is the maximum speed at which an aircraft is equally able to complete or abandon a take-off safely, following a critical engine failure during the take-off run. At speeds below V, such a failure would prevent the aircraft from attaining screen height within the remaining TORA, but the take-off could be aborted safely. At speeds above V, such a failure would cause the aircraft to overrun the ASDA in the event of a decision to abort the take-off, but the take-off could be completed safely. In this question, the calculated V, is 140 Kts and the actual V, is 150 Kts. So at arly speed below 150 Kts, a takeoff can be safely abandoned. The calculated V, is less than 150 kts so engine failures at any speed up to the calculated V, can be dealt with safely by abandoning the take-off.

TAKE-OFF 52. d. V, is the maximum speed at which an aircraft is equally able to complete or abandon a take-off safely, following a critical engine failure during the take-off run. At speeds below V, such a failure would prevent the aircraft from attaining screen height within the remaining TORA, but the take-off could be aborted safely. At speeds above V, such a failure would cause the aircraft to overrun the ASDA in the event of a decision to abort the take-off, but the tak'e-off could be completed safely. In this question, the calculated V, is 160 Kts and the actual V, is 150 Kts. So at any speed above 150 Kts, the take-off can be conti~iued safely. So engine failures at any speed above the calculated V, can be dealt with safely by continuing the ta ke-off.

TAKE-OFF 53. b. JAR regulations require that when calculating take-off performance not more than 50% of any headwind component and not less than 150% of any tailwind component should be taken into account.

TAKE-OFF 54. d. JAR regulations require that when calculating take-off performance not more than 50% of any headwind component and not less than 150% of any tailwind component should be taken into account.

TAKE-OFF 55. b. The clearway commences at the end of the runway and ends at the first obstacle, or at a distance equal to 50% of the runway length, whichever is the shorter. In this case the runway is 4000 m long so the maximum clearway is 2000 m. But the first obstacle is at 1500 m, so the clearway is 1500 m. The maximum TODA is equal to the runway plus the clearway, which in this case is 4000 m plus 1500 m. So the maximum TODA is 5500 m.

TAKE-OFF 56. C. The anti-skid system improves the ability of an aircraft to decelerate to a halt. In the case of an aborted take-off, anti-skid will reduce the stopping distance, and hence decrease the ASDR. This will increase the field limited take-off mass. During landing, the anti-skid will decrease the landing run, thereby increasing the maximum landing mass. So a defective anti-skid system will reduce both take-off and landing performance. This will decrease the maximum field length limited TOW and the maximum landing mass.

TAKE-OFF 57. C. The anti-skid system improves the ability of an aircraft to decelerate to a halt. In the case of an aborted take-off, anti-skid will reduce the stopping distance, and hence decrease the accelerate stop distance required (ASDR). So a defective anti-skid system will increase the ASDR. The take-off run required (TORR) is the calculated distance required to accelerate from BRP to VLoF.This is not affected by anti-skid system serviceability. Defective anti-skid will also decrease deceleration after touch down, thereby increasing the landing distance required (LDR). The take-off run available (TORA) is equal to the combined lengths of the runway plus the stopway. These are a functions of the airfield design and as such are not affected by anti-skid.

TAKE-OFF 58. C. The calculated horizontal distance between the brake release point (BRP) and attaining lift-off speed (VLoF)is the take-off run required (TORR). The distance actually taken is the take-off run (TOR). The minimum unstuck speed V (), is the minimum speed at which an aircraft can safely lift-off and complete , has no specific title or relevance. a take-off under any circumstance. The distance between BRP and,V The calculated distance between BRP and attaining screen height is the take-off distance required (TODR).

TAKE-OFF 59. d. The calculated horizontal distance between the brake release point (BRP) and attaining lift-off speed (VLoF)is the take-off run required (TORR). The distance actually taken is the take-off run (TOR). The minimum unstuck speed V (), is the minimum speed at which an aircraft can safely lift-off and complete a take-off under any circumstance. The distance between BRP and,V , has no specific title or relevance. The calculated distance between BRP and attaining screen height is the take-off distance required (TODR).

TAKE-OFF 60. C. JAR OPS 1.495 specifies that that class a aircraft must clear all obstacles within the obstacle domain by a minimum vertical separation of 35 feet, or 50 ft when conducting turns using between 15O and 25O of bank above 400 feet.

-TAKE-OFF 61. a. JAR OPS 1.525 specifies the minimum acceptable climb gradients. For a twirl piston (class B) aircraft this minimum is 4% with both engine operating.

TAKE-OFF 62. d The situation is illustrated at the right. The clearance is equal to the height of the aircraft as it passes over the obstacle, minus the height of the obstacle. The height of the aircraft as it passes the obstacles is its initial height of 15m plus its height gain prior to reaching the obstacle. Climb gradient = 10% = height gained 1 horizontal distance ,flown x 100% So height gained = 10% x horizontal distance to obstacle Which gives a height gain of 1O0/0 x 3280 m which is 328 m So aircraft height at obstacle = 15 m + 328 m or 343 m. Clearance is equal to aircraft height minus obstacle height So clearance = 343 m - 900 m which is - 557 m So the aircraft will pass 557 m below the top of the obstacle.

TAKE-OFF 63. d. JAR 25.1.13 states that the TORR for a class A aircraft must be the greater of the following: ,, 1. 1. I 5 times the sum of the gross distance from the start of the take-of run to the point at which ,V is reached, plus W of the distance taken to attain a screen height of 35 ft, with all engines operating. 2. The gross distance from the start of the take-off run to a point equidistant between the points at ,, and a 35 ft screen height are attained, with one engine inoperative on a dry hard which ,V runway. ,, is reached with one 3. The gross distance from the start of the take-of run to the point at which ,V engine inoperative on a wet hard runway. It should be noted that the question does not specify runway condition, nor does it define what exactly the quoted distances refer to. The question must however be answered on the basis of the limited information provided. The question indicates a gross distance of 1547 m for the all engines operating condition. Multiplying this by 1. I 5 gives a TORR of 1779 m. A gross figure of 1720 m is quoted for the one engine inoperative condition. Although the question does not state the runway conditions at the time of the test, this is not important because the all engines figure of 1779 m is the greater of the two. JAR specifies that the TORR must be the greater of the three possible conditions, so it can be seen that the answer must be at last 1779 m, which is option d. The only other available figure that is greater than this and hence viable, is option c 1978 m. But there is no specific data to indicate that the 1978 is correct. The most appropriate option is therefore d in that it corresponds to the quoted safety factor of 1.15.

TAKE-OFF 64. b. It should be noted that this question refers to the use of the incorrect V,, but does not imply that the runway is actually wet. The question therefore relates only to the effects of the use of the incorrect V, , rather than to the effects of a wet runway. V, is the maximum speed at which an aircraft is equally capable of continuing a take-off or aborting it safely within the remaining ASDA in the event of a critical engine failure. The ability to take-off is related to

dynamic pressure, whilst the ability to stop is related to deceleration rate. A wet runway will reduce the friction between tyres and surface, thereby reducing braking efficiency. This in turn increases the distance required to abort at any given combination of V, and mass. So a wet runway V, is lower than a dry runway V, for the same aircraft. This question therefore relates to using too low a value of V,. Although this will not directly affect TODR or climb performance, this result is not offered as an option. It must therefore be assumed that the error is also applied to V, and initial climb speed. The rotation speed and initial climb speeds are fixed multiples of V, so it is reasonable to assume that the aircraft in this question will also use the incorrect V, and V., Using too low a value for these speeds will degrade climb performance. TODR is the distance from BRP to screen height, so the reduced climb performance will also increase TODR.

TAKE-OFF 65. d. V, is the maximum speed at which a decision to abort a take-off can be made. Above V, the aircraft is committed to take-off. In order to take-off safely an aircraft must remain under control, so control must be guaranteed at all speeds above V,. VMcGis the minimum speed at which it is possible to maintain control in the event of a critical engine failure on the ground in the take-off configuration. So VMcGmust not be greater than V,. The processes for calculating V, specified in the CAP 698 state that if the calculated V, is less than VMcG,then a V, equal to VMcGmust be used. This is requirement is also reflected in JAR 25.107.

TAKE-OFF 66. a. JAA regulations define V2 as the take-off safety speed that must be attained by screen height. The other speeds referred to in this question are option b, the final climb speed, option c, VMcGor VMc,, and option dJVLOF.

TAKE-OFF 67. C. If the take-off distance required at any given mass is considerably shorter than the take-off distan~e available, then the excess clearway may be used to increase the maximum TOW. To do this the aircraft is held on the ground longer so that V, and V, are increased. This brings the speed of the aircraft closer to the optimum climb gradient speed Vx, and the optimum rate of climb speed V,, thereby improving climb performance. So the overall effect of using the increased V, procedure is an increase in TODR, maximum TOW and climb performance. This effect is indicated in option c. Although option a might also appear to be true, this is not the case if the TOW has been limited by the field length. Options b and d are both untrue.

TAKE-OFF 68. d. A headwind or tailwind does not affect the gradient at which an aircraft can climb through the air. It does however cause the mass of air through which the aircraft is climbing, to move relative to the ground. This causes the effective climb gradient relative to the ground to increase in a headwind and decrease in a tailwind. So a tailwind will decrease the effective climb gradient and a headwind will increase it.

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The obstacle limited TOW is the maximum weight at which an aircraft can clear obstacles. By reducing climb gradient, a tailwind will reduce the obstacle limited mass, whilst a headwind will increase it. The field limited TOW is the maximum weight at which an aircraft can safely take-off within the distances available. A tailwind decreases the airspeed at any given ground speed and so increases the distances required. So a tailwind reduces the field limited TOW, whilst a headwind increases it. Rate of climb is equal to excess power divided by aircraft weight. Neither of these factors are affected by a headwind or a tailwind, so option d is correct.

TAKE-OFF 69. C. The take-off distance available (TODA) is the distance from the start of the take-off run, to the first obstacle. The clearway runs from the end of the runway to the first obstacle. So the TODA is equal to the runway plus the clearway. The take-off run available is the distance from the start of the take-off run to the end of the runway. The accelerate stop distance available (ASDA) is the TORA plus the length of the stopway. There is no specific title for the stopway plus the clearway indeed the two overlap. These distances are illustrated overleaf.

First obstacle

RUNWAY I

L:l 7:-

f

STOPWAY

4

H

f

TORA

TAKE-OFF 70. a. The first segment commences at the end of the take-off distance when the landing gear up button is pressed. It ends as soon as the landing gear is fully retracted.

TAKE-OFF 71. b. The take-off run required is the distance from the first point at which the take-off run can start and the end of the runway. Although the stopway is able to support the weight of an aircraft, it is to be used only in the event of an abandoned take-off. It cannot therefore be included in the TORR calculations.

TAKE-OFF 72. d. When the TOW is less than the field limited TOW, not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). JAA regulations prohibit the use of this technique when the runway is contaminated, icy or very slippery, or when the anti-skid system, reverse thrust, or power management computers are unserviceable.

TAKE-OFF 73. a. Anti-skid systems reduce stopping distance by ensuring that maximum braking effort can be applied without allowing the wheels to skid. The system can be used both when landing'and when abandoning a ta ke-off.

TAKE-OFF 74. C. The field limited take-off weight (TOW) is the maximum weight at which an aircraft can take-off safely within the existing TORA , ASDA and TODA. Anything that increases the stalling speed or decreases the rate of acceleration, increases the distances required. This in turn decreases the field limited TOW for any given field length. The climb limited TOW is the maximum weight at which an aircraft can achieve the minimum acceptable climb gradient after take-off. Anything that reduces climb performance will therefore decrease the climb limited TOW. Increasing outside air temperature (OAT) reduces air density. This reduces the lift at any given TAS and reduces engine thrust. The reduced lift increases stalling TAS, whilst the reduced thrust reduces both

acceleration rate and climb performance. So increasing OAT decreases both the field limited TOW and climb limited TOW.

TAKE-OFF 75. a. The accelerate stop distance available (ASDA) is the distance from the earliest point at which the take-off run can start, to the end of the stopway. It is a function of the physical structure of the airfield and hence is not affected by the serviceability of the anti-skid or any other aircraft system. If the anti-skid system becomes unserviceable the ASDA might become inadequate but this is simply because the ASDR has increased. The ASDA itself will not have been affected in any way.

TAKE-OFF 76. a. Aircraft in flight produce turbulent wakes due primarily to the vortices trailing from their wingtips. When aircraft follow each other during take-off and landing, it is possible that the wake turbulence from the leading aircraft will cause loss of control of the trailing aircraft. Wake turbulence is greatest when aircraft are heavy and when flying slowly. Although landing speeds are slower than take-off speeds, the much greater weight at take-off is the overriding factor. The spacing required to avoid wake turbulence problems is therefore greatest when taking-off at high weights. It should however be noted that there are some exceptions to this rule. The Boeing 777, for example produces greater wake turbulence when landing than when taking-off.

TAKE-OFF 77. d. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). JAA regulations prohibit the use of this technique when the runway is contaminated, icy or very slippery, or when the anti-skid system, reverse thrust, or power management computers are unserviceable.

TAKE-OFF 78. b. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This reduces the rateof acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). JAA regulations prohibit the use of this technique when the runway is contaminated, icy or very slippery, or when the anti-skid system, reverse thrust, or power management computers are unserviceable.

TAKE-OFF 79. b. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). JAA regulations prohibit the use of this technique when conducting increased V2 take-offs. This is because the increased V2 procedure is intended to improve climb performance. This objective is incompatible with reducing take-off thrust.

TAKE-OFF 80. C. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This is sometimes termed the assumed temperature take-off procedure. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA).

JAA regulations prohibit the use of this technique when the runway is contaminated, icy or very slippery, or when the anti-skid system, reverse thrust, or power management computers (PMC) are unserviceable or switched off.

TAKE-OFF 81. C. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This is sometimes termed the assumed temperature take-off procedure. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). JAA regulations prohibit the use of this technique when the runway is contaminated, icy or very slippery, or when the anti-skid system, reverse thrust, or power management computers are unserviceable.

TAKE-OFF 82. C. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This is sometimes termed the assumed temperature take-off procedure. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). JAA regulations specify a maximum thrust reduction of 25% but tighter lirr~itsmay apply to individual aircraft types.

TAKE-OFF 83. b. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This is sometimes termed the assumed temperature take-off procedure. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). This technique may be used only when the distances available greatly exceed the distances required.

TAKE-OFF 84. a. When the TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This is sometimes termed the assumed temperature take-off procedure. This reduces the rate of acceleration and hence increases both the take-off run required (TORR) and the take-off distance required (TODA). The primary purpose of this technique is to preserve engine life.

TAKE-OFF 85. a. The maximum brake energy speed)V ,(,, is the maximum speed at which it is possible to abort a take-off and bring the aircraft to a halt within the available ASDA, without exceeding the brake energy absorption capabilities. Attempting to abort at greater speeds will cause the brakes to fail due to overheating. Anything that increases the ratio of ground speed to CAS, or reduces aerodynamic drag during take-off, will take the aircraft closer to its.,,V , In a tailwind the aircraft is initially subjected to a negative airspeed and hence must attain a greater ground speed during the take-off run. This increase in ground speed causes,,V , to become more limiting. In the most extreme case the IAS at,,V , might be less than the stalling speed. By reducing the IAS at any given ground speed, a tailwind will also reduce the drag during the take-off run. This will also cause,,V , to become more limiting. In strong tailwinds aircraft mass might need to be reduced to facilitate take-off without exceeding.,,V , A headwind, low pressure altitude and low temperatures will have all the opposite , to be less restrictive. effect, causing,,V

TAKE-OFF 86. C. The maximum brake energy speed )V ,(,, is the maximum speed at which It is possible to abort a take-off and bring the aircraft to a halt within the available distances without exceeding the brake energy absorption capabilities. Attempting to abort at greater speeds will cause the brakes to fail due to overheating. Anything that increases the ratio of ground speed to CAS, or reduces aerodynamic drag during take-off, will take , the aircraft closer to its.,,V Increasing pressure altitude increases the ratio of TAS to CAS. This in turn increase the ground speed at any given CAS. But the CAS equating to stalling speed does not vary with increasing altitude. So to attain , So the same stalling speed an aircraft requires a greater ground speed, thereby moving closer to its.,,V ,,V , is most restrictive at high pressure altitudes. When operating from very high altitude airfields it may be necessary to reduce weight in order to take-off without exceeding.,,V,

TAKE-OFF 87. d. The maximum brake energy speed )V ,(,, is the maximum speed at which it is possible to abort a take-off and bring the aircraft to a halt within the available distances without exceeding the brake energy absorption capabilities. Attempting to abort at greater speeds will cause the brakes to fail due to overheating. Anything that increases the ratio of ground speed to CAS, or reduces aerodynamic drag during take-off will take the aircraft closer to its.,,V , Increasing air temperature decreases air density, thereby causing the ratio of TAS to CAS to increase. This in turn increases the ground speed at any given CAS. But the CAS equating to stalling speed does not vary with increasing altitude. So to attain the same stalling speed, an aircraft requires a greater ground speed, thereby moving closer to its.,,V , So,,V , is most restrictive at high ambient temperatures. When operating , from hot airfields it may be necessary to reduce weight in order to take-off without exceeding.,,V

TAKE-OFF 88. a. The tyre speed during a take-off is the ground speed of the aircraft. Ground speed is determined by the ,, constitutes increasing pressure altitude. This causes value of VLoFand the TAS:CAS ratio. Decreasing V both VLoFand the TAS:CAS ratio to increase, thereby increasing tyre speed. High QFE, high QlVH and low OAT each have the opposite effect.

TAKE-OFF 89. d. The aircraft referred to in this question is class A. JAR 25.11 3 and the CAP 698 state that the TORR for a class A aircraft must be the greater of the following: 5 times the sum of the gross distance from the start of the take-off run to the point at which ,V ,, 1. 1.I is reached, plus W of the distance taken to attain a screen height of 35 ft, with all engines operating. 2. The gross distance from the start of the take-off run to a point equidistant between the points at which ,V ,, and a 35 ft screen height are attained, with one engine inoperative on a dry hard runway. 3. The gross distance from the start of the take-of run to the point at which ,V ,, is reached with one engine inoperative on a wet hard runway. It should be noted that the question does not specify runway condition. It must however be answered on the basis of the limited information provided. The question indicates a gross distance of 1500 m to reach VLoF in the all engines operating condition. Multiplying this by 1. I 5 as stated in paragraph 1 above gives 1725 m. It should be noted that this is an underestimate for TORR because it does not include W the distance between VLoFand screen height. We can however conclude that the true value of TORR is greater than 1725 m.

A gross figure of 2000 m is quoted for BRP to VLoFin the one engine inoperative condition. Although the question does not state the runway conditions at the time of the test. If this figure related to a dry runway then the TORR would be the distance from BRP to the midway point between VLoFand screen height. So the figure of 2000 m would be too low. If the runway was wet, then the required distance would be from BRP to screen height. The quoted figure of 2000 m would again be too low. We can therefore conclude that the required TORR is greater than 2000 m. Only option d satisfies this condition.

TAKE-OFF 90. b. The specification in this question indicates a class A aircraft. JAR 25.113 and the CAP 698 state that the TORR for a class A aircraft must be the greater of the following: 1. 1.15 times the sum of the gross distance from the start of the take-of run to the point at which VLoF is reached, plus %of the distance taken to attain a screen height of 35 ft, with all engines operating. 2. The gross distance from the start of the take-off run to a point equidistant between the points at ,, and a 35 ft screen height are attained, with one engine inoperative on a dry hard which ,V runway. 3. The gross distance from the start of the take-of run to the point at which VLoFis reached with one engine inoperative on a wet hard runway. The question also indicates that the runway is wet, so only conditions 1 and 3 need to be considered. The , in the all engines operating condition. Multiplying question indicates a gross distance of 1500 m to reach,,V this by 1.15 as stated in paragraph 1 above gives 1725 m. It should be noted that this is an underestimate for TORR because it does not include %the distance between VLoFand screen height. We can however conclude that the true value of TORR is greater than 1725 m. A gross figure of 2000 m is quoted from BRP to VLoFin the one engine inoperative condition. The runway was wet during the tests, so the required distance is that from BRP to screen height. So the quoted figure of 2000 m is correct. The regulations specify the greater of the two figures so 2000 m is the more appropriate of the two derived above.

TAKE-OFF 91. C. The specification in this question indicates a class A aircraft. JAR 25.113 and the CAP 698 state that the TODR for a class A aircraft must be the greater of the following: 1. 1.15 times the horizontal distance travelled, with all engines operating, to reach screen height if 35 ft . 2. The horizontal distance from BRP to the point at which the aeroplane attains screen height, when taking-off from a dry hard surface with one engine inoperative after V,. 3. The horizontal distance from BRP to the point at which the aeroplane attains a screen height of 15 ft, when taking-off from a wet or contaminated hard runway with one engine inoperative after V1. This questions refers to a dry hard runway, so only conditions 1 and 2 need to be considered. The question indicates a gross distance of 1500 m to reach screen height, in the all engines operating condition. Multiplying this by 1.15 as stated in paragraph 1 above gives a TODR of 1725 m. A gross figure of 2000 m is quoted for BRP to screen height in the one engine inoperative condition. The surface is dry so this can be used as a second estimate for the TODR. The regulations specify that the TODR must be the greater of the two, so option c, 2000 m is the correct answer.

TAKE-OFF 92. d. The specification in this question indicates a class A aircraft. JAR 25.113 and the CAP 698 state that the TODR for a class A aircraft must be the greater of the following: 1. 1. I 5 times the horizontal distance travelled, with all engines operating, to reach screen height if 35 ft. 2. The horizontal distance from BRP to the point at which the aeroplane attains screen height, when taking-off from a dry hard surface with one engine inoperative after V,. 3. The horizontal distance from BRP to the point at which the aeroplane attains a screen height of 15 ft, when taking-off from a wet or contaminated hard runway with one engine inoperative after V1. This questions refers to a contaminated hard runway, so only conditions 1 and 3 need to be considered. The question indicates a gross distance of 1500 m to reach screen height, in the all engines operating condition. Multiplying this by 1. I 5 as stated in paragraph 1 above gives a TODR of 1725 m. A gross figure of 2000 m is quoted for BRP to screen height in the one engine inoperative condition. The surface is contarrtinated so this can be used as a second estimate for the TODR. It should be noted however that condition 3 actually specifies a screen height of 15 ft for the one engine inoperative configuration. The figure of 2000 m might therefore be an overestimate, but cannot be an underestimate. So the TODR for this aircraft is more than 1725 m and not more than 2000 m. Option d, (1850 m) is within these limits, whereas option c (2000 m) is on the upper limit. Option d is therefore the most appropriate.

TAKE-OFF 93. a. In addressing this problem it should be noted that the questions asks which speed w o ~ ~be l d FOUND to be greater during the take-off. The values of V,, V, and V, actually used during a take-off are controlled entirely by the pilot. They will therefore be whatever the pilot chooses to make them. But VLo, is the lowest speed at which the lift is sufficient to support the mass of the aircraft. This is beyond the control of the pilot. So,,V , will be found to be greater than expected if mass is greater than calculated. The distances required will also be found to be greater.

TAKE-OFF 94. C . Decreasing flap angle decreases both lift and drag at any given airspeed. The decreased lift requires a greater V, , V, ,,V , and V, thereby tending to increase the take-off run required. But the reduced drag increases acceleration rates, thereby tending to partly offset the increase in take-off run required. The reduced drag also increases the maximum gradient of climb, enabling the aircraft to clear close-in obstacles more easily. This can be particularly important at high field pressure altitudes or high ambient temperatures, where low air density reduces thrust and climb gradient. The overall effect of reducing flap angle would be an increased take-off run and an improved climb gradient. The longer take-off run would require a longer runway, whilst the improved climb gradient would improve clearance of close-in obstacles. So provided sufficient field length is available, reducing flap angle might improve the ability of an aircraft to clear close-in obstacles when taking-off at high pressure altitudes or high ambient temperatures. It should however be noted that the examiners intentions in drafting this question are not entirely clear. The use of low flap angles in conjunct~onwith a combination of long runway, and low pressure altitude might be necessary in order to clear close-in obstacles in conditions of high ambient temperatures. This technique would however be less necessary for distant obstacles, and totally unnecessary if no obstacles existed. Overall option c is therefore the most appropriate, but option a might also be true.

TAKE-OFF 95. a. 5 mm of standing water on a runway qualifies as a contaminated surface. Standing water affects take-off performance in two ways. Firstly, the additional rolling resistance as the wheels are pushed through the contaminant reduces the rate of acceleration, thereby increasing the take-off run required to attain VLOF. Secondly the reduced friction between tyres and the runway degrades the ability of the aircraft to stop in the event of an abandoned take-off. This increases the accelerate stop distance required. Both of these effects will decrease the mass that can take-off safely from a given field length. So 5 mm of standing water w ~ ldecrease l the field lirrlited take-off mass.

TAKE-OFF 96. d. The specification in this question indicates a class A aircraft. JAR 25.113 and the CAP 698 specify a number of criteria for determining the TORR for class A aircraft. These include the gross distance from the start of the take-off run to a point equidistant between the points at which VLoFand a 35 ft screen height are attained, with one engine inoperative on a dry hard runway.

TAKE-OFF 97. d. This question offers a number of options that might appear to be equally correct. It should however be noted that both the Gross TODR and Net TODR are calculated distances, rather than actual distances used. The Gross TOD is the measured TOD factorised to ensure that at any given aircraft has a 50% probability of achieving it. The question makes no reference to calculations or safety factors, so option d, the measured TOD is the most appropriate.

TAKE-OFF 98. a. The measured TOD is the horizontal distance actually taken between BRP and attaining screen height. The distances from the mid point between VLoFand screen height (option b), and from VLoFto screen height (option d) have no specific titles. The distance from BRP to the first obstacle is the take-off distance available (TODA).

TAKE-OFF 99. d. If runway 09 has a 1% downward slope, then runway 27 has a 1% upward slope. ,,V , is the maximum speed at which it is possible to stop an aircraft within the remaining ASDA, in the event of an aborted take-off, without exceeding the energy absorption capabilities of the wheel brakes. A decision to abort at a greater speed will cause the brakes to fail due to overheating, thereby causing the aircraft to overrun the stopway. A down slope makes stopping more difficult because part of the weight of the aircraft acts down the slope, tending to add to its speed. A down slope therefore decreases, ,V , for any given combination of aircraft mass and field length. When changing to runway 27 the slope changes from 1% down to 1% up. This makes it more easy to stop the aircraft and hence increases.,,V, This in turn might allow a greater take-off mass to be used because the aircraft will be able to attain a greater speed before lift-off.

TAKE-OFF 1Q0.a. Reference point zero is located at the point at which an aircraft attains screen height. All subsequent points in the take-off flight path are specified relative to this point.

TAKE-OFF 101. b. JAR 25.107 specifies that V, must not be less than 1.. ,I ,V , This regulation also specifies a minimum value for V, in terms of the stalling speed V, depending on aircraft type. The lowest of these is 1. I5 V.,

So V, must be the greater of 1.1 VMcAor 1.15 V., Increasing flap angle will increase the lift generated at any given speed and hence decrease the value of V, But this will not allow V, to be reduced because it is already less than the 1.IVMcAminimum. Of the options offered in this question, only option b is correct.

TAKE-OFF 102. C. 5 mm of standing water on a runway qualifies as a contaminated surface. Standing water affects take-off performance in two ways. Firstly, the additional rolling resistance as the wheels are pushed through the water, reduces the rate of acceleration, thereby increasing the take-off run required to attain VLoF.Secondly the reduced friction between tyres and the runway degrades the ability of the aircraft to stop in the event of an abandoned take-off. This increases the accelerate stop distance required. Both of these effects will decrease the mass that can take-off safely from a given field length. So 5 mm of standing water will decrease the field limited take-off mass. V, is the maximum speed at which it is possible to stop an aircraft within the remaining TORA following a critical engine failure. So by degrading the deceleration rate available, 5 mm of standing water will reduce the maximum allowable V,, for any given combination of aircraft mass and field length. The overall effect is therefore that when taking-off from a runway with 5 mm of standing water, field limited mass will decrease and V, will decrease compared to a damp runway?

TAKE-OFF 103. a. V, is the maximum speed at which it is possible to stop an aircraft within the remaining ASDA following a critical engine failure. The purpose of an anti-skid system is to improve stopping ability by ensuing maximum braking effort is applied without causing the tyres to skid. So by degrading the deceleration rate available, a defective anti-skid system will reduce the maximum allowable V,, for any given combination of aircraft mass and field length. The purpose of a reduced thrust take-off (option b) is to preserve engine life by using a lower value of takeof power. But this does not affect the stopping ability of an aircraft so V, is not affected. Noise abatement procedures (option c) and obstacle clearance (option d) both apply after lift-off and hence do not affect V,.

TAKE-OFF 104. b. If runway 09 has a 1% upward slope, then runway 27 has a 1% downward slope. VMBEis the maximum speed at which it is possible to stop an aircraft within the remaining ASDA, in the event of an aborted take-off, without exceeding the energy absorption capabilities of the wheel brakes. A decision to abort at a greater speed will cause the brakes to fail due to overheating, thereby causing the aircraft to overrun the stopway. An upward slope makes stopping more easy because part of the weight of the aircraft acts down the slope, tending to reduce its speed. An upward slope therefore increases VMBE for any given combination of aircraft mass and field length. When changing to runway 27 the slope changes from 1% up to 1% down. This makes it more difficult to stop the aircraft and hence decreases VMBE.The reduced VMBEwill also reduce the maximum take-off mass that can be used because of the reduced maximum value of VLoFthat can be achieved.

TAKE-OFF 105. C. The effect of the additional passenger will depend upon the relative masses of the passenger and the aircraft. For very small aircraft one additional passenger will make a significant difference to take-off performance. For a very large aircraft the effect will be less significant. It can however be said that regardless of the size of the aircraft, provided an equivalent mass of fuel is burned before take-off, and provided the C of G of the aircraft is not affected, then the take-off performance with the additional passenger will be unchanged.

TAKE-OFF 106. a Maximum take-off mass is limited by the ability of an aircraft to reach VLoFwithin the TORA, to stop within the remaining ASDA in the event of a decision to abort, and to achieve the minimum acceptable climb gradient after lift-off. All of the options offered in this question relate to the take-off run, so climb gradients can be ignored. Increasing take-off mass increases stalling speed V., The decision speed V, and lift-off speed VLo, are both fixed multiples of V, so increasing mass increases V, and.,V ,, But increasing these speeds increases the ground speed during the take-off run. The maximum allowable ground speed is limited by the ability of the tyres to retain their structural integrity as wheel rotation speed increases. So maximum take-off mass is limited by the wheel rotation speed.

TAKE-OFF 107. b. The gross flight path is the profile of the path of an aircraft after it has reached the end of the TODR. It is plotted using gross performance data. Gross performance is the performance that will be achieved by the average aircraft of that type when flown by an average pilot. So the gross flight path can be said to be that of an average aircraft flown by an average pilot. The net flight path is the gross flight path reduced by various factors to reflect performance reductions due to factors not included in the gross performance. The performance of a new aircraft is of no particular significance in that, as they age the majority of aircraft of that type will not achieve it.

TAKE-OFF 108. a. The gross flight path is the profile of the path of an aircraft after it has reached the end of the TODR. It is plotted using gross performance data. Gross performance is the performance that will be achieved by the average aircraft of that type when flown by an average pilot. So the gross flight path can be said to be that of an average aircraft flown by an average pilot. The net flight path is the gross flight path reduced by various factors to reflect performance reductions due to factors not included in the gross performance. The net flight path is the one that is used in planning to avoid obstacles. By reducing the flight path in this way, it can be ensured that the vast majority of aircraft will clear all obstacles.

TAKE-OFF 109. a. An Airbus A320 is a twin turbojet class A aircraft. JAR 25.115 specifies the degree by which gross flight paths must be reduced to determine net flight paths. For twin jet class A aircraft this reduction is to be 0.8%.

TAKE-OFF 11). d. An Boeing 747 is a four engine turbojet class A aircraft. JAR 25.115 specifies the degree by which gross flight paths must be reduced to determine net flight paths. For four engine turbojet class A aircraft this reduction is to be 1.0%.

TAKE-OFF 111. b. JAR 25.115 specifies the degree by which gross flight paths must be reduced to determine net flight paths. For three engine turbojet class A aircraft this reduction is to be 0.9%.

TAKE-OFF 112. a. The gross flight path is the profile of the path of an aircraft after it has reached the end of the TODR. It is plotted using gross performance data. Gross performance is the performance that will be achieved by the average aircraft of that type when flown by an average pilot. So the gross flight path can be said to be that of an average aircraft flown by an average pilot. The net flight path is the gross flight path reduced by

various factors to reflect performance reductions due to factors not included in the gross performance. The net flight path is therefore always less than the gross flight path.

TAKE-OFF 113. a. JAR 25.115 specifies the degree by which gross flight paths must be reduced to determine net flight paths. For twin jet class A aircraft this reduction is to be 0.8%. JAR 25.121 specifies the minimum acceptable gradients for each segment. The specified minima for the first segment are 0% gross, which equates to 0.8% net.

TAKE-OFF 114. a. VMCGis the minimum speed at which it is possible to maintain control of an aircraft, on the ground in the take-off configuration, following a critical engine failure. In the event of such an engine failure the pilot will use all available means to maintain control. These include the primary flying controls, the wheel brakes and the nose-wheel steering system. But when using contaminated runways the effectiveness of the wheel brakes and nose-wheel steering system is reduced. The effectiveness of these systems is also reduced as speed increases, even on a dry runway. In order to avoid accidents, it is necessary that the specified VMcGerrs on the side of safety. The definition of VMcGtherefore specifies that it be the minimum speed at which control can be maintained using only the primary flying controls. This ensures that control can actually be maintained at slightly lower speeds.

TAKE-OFF 115. a. The minimum acceptable climb gradients for class B aircraft are detailed in Appendix 1 to JAR OPS 1.525. These include a minimum of 4% with all engines operating and measurably positive at 400 ft and 0.75% at 1500 ft, with one engine inoperative. Option a is therefore the most appropriate of the options available in this question.

TAKE-OFF 116.6. The data provided in this question indicate that the mass of the aircraft is in excess of its maximum allowable climb limited mass and equal to its field length limited mass. There is therefore a need to increase the climb limited mass, without decreasing the field limited mass. Decreasing flap angle will decrease the lift and drag at any airspeed. The decreased drag will improve climb performance and hence increase climb limited mass. But the decrease in lift will increase VLo, and hence increase the take-off run required. This will decrease the field limited mass. But the aircraft mass is already on its field limited mass limit, so any reduction will be unacceptable. So decreased flap angle cannot be used to permit take-off at this mass. The effect of increasing flap angle will be a decrease in climb limited mass and an increase in filed limited mass. But the aircraft is already above its climb limited mass, so this option is also unacceptable. So the take-off cannot be carried out as planned and no change of flap setting will overcome this problem

TAKE-OFF 117. C . In a headwind the aircraft is subjected to an airflow over its wings before the brakes are released. This reduces the amount of acceleration required to reach VLo, and hence decreases the take-off run required (TORR) and the take-off distance required (TODR). A headwind will not affect the rate of climb or climb gradient through the air. But it will cause the mass of air through which the aircraft climbs, to move backwards relative to the path of the aircraft. This will increase the effective climb gradient relative to the ground. So a headwind will decrease the TODR and increase the climb gradient. This in turn will increase both the field limited mass and the climb limited mass.

TAKE-OFF 118. d. The data provided in this question indicate that the mass of the aircraft is in excess of its maximum allowable clirr~blimited mass and equal to its field length limited mass. There is therefore a need to increase the climb limited mass, without decreasing the field limited mass. Decreasing flap angle will decrease the lift and drag at any airspeed. The decreased drag will improve climb performance and hence increase climb limited mass. But the decrease in lift will increase VLoFand hence increase the take-off run required. This will decrease the field limited mass. But the aircraft mass is already on its field limited mass limit, so any reduction will be unacceptable. So decreased flap angle cannot be used to permit take-off at this mass. The effect of increasing flap angle will be a decrease in climb limited mass and an increase in field limited mass. But the aircraft is already above its climb limited mass, so this option is also unacceptable. So the take-off cannot be carried out as planned and no change of flap setting will overcome this problem. But a headwind will increase climb gradient and decrease the TORA. This will increase both the climb limited mass and the field limited mass. So a headwind might make take-off possible at this mass.

TAKE-OFF 119. d. Increasing flap angle will increase the lift and drag at any airspeed. The increased drag will degrade climb performance and hence decrease climb limited mass. But the increase in lift will decrease VLoFand hence decrease the take-off run required (TORR) and the take-off distance required (TODR).

TAKE-OFF 120. a. The 60 hPa reduction in ambient QFE and 3.96O C decrease in ambient temperature both equate to an increase of 2000 ft in pressure altitude in the ISA. The overall effect of this will be an increase in the TAS:IAS ratio and a reduction in thrust. The initial calculations indicated that the aircraft was at its maximum take-off mass. But the field was balanced so this means that the aircraft was at both its field limited TOM and its climb limited TOM. The fact that it was .,,V , But one of the effects of the increased at its field limited TOM suggests that V, might be close tc~ pressure altitude would be an increase in the TAS:IAS ratio.,,V , is related to ground speed and hence to , But V, was already close to,,V , so there is TAS, so the increased TAS:IAS ratio would reduce.,,V danger that,,V , will be exceeded if the take-off is carried out without first reducing mass.

TAKE-OFF 121. d. A downhill slope increases the rate at which an aircraft accelerates during the take-off run. This decreases the distances required to achieve V,, V,and VLo,. A downhill slope also decreases the rate at which an aircraft can decelerate to a halt in the event of a decision to abort a take-off. This means that a greater distance is required to bring the aircraft to a halt. V, is the maximum speed at which it is possible to bring an aircraft to a halt without overrunning the stopway, in the event of an aborted take-off. So by increasing the distance required to stop an aircraft, a downhill slope decreases the maximum value of V, for any given combination of aircraft mass and field length. V, is the take-off safety speed that must be achieved by screen height. Although a downhill slope will reduce the distance required to attain V, it will not affect the value of V.,

TAKE-OFF 122. C. When the actual TOW is less than the field limited TOW not all of the TORA or TODA will be used. It is therefore possible to preserve engine life by using a lower level of take-off thrust. This is sometimes termed the assumed temperature take-off procedure. This reduces the rate of acceleration and hence

increases both the take-off run required (TORR) and the take-off distance required (TODR). JAA regulations prohibit the use of this technique when the runway is contaminated, icy or very slippery, or when the anti-skid system, reverse thrust, or power management computers are unserviceable.

TAKE-OFF 123. d. The specification in this question indicates a class A aircraft. JAR 25.113 and the CAP 698 state that the TODR for a class A aircraft must be the greater of the following: 1. 1.15 times the horizontal distance travelled, with all engines operating, to reach screen height of 35 ft. 2. The horizontal distance from BRP to the point at which the aeroplane attains screen height, when taking-off from a dry hard surface with one engine inoperative after V,. Applying condition 1 gives a minimum TODR of 1.15 x 2555 m = 2938.25 m. Applying condition 2 gives a minimum TODR of 3050 m. But the TODR must be the greater of the two figures, so TODR is 3050 m.

TAKE-OFF 124. d. The climb performance of an aircraft is determined by the thrust and power available and the 1ift:drag ratio. Thrust and power available are proportional to the mass flow of air passing through the engines. Mass flow is proportionalto air density, so any factor that decreases air density, decreases both thrust and power available, and hence climb performance. Increasing altitude and ambient temperature each decrease density, so climb performance deteriorates as altitude or temperature increase. Although increasing flap setting increases lift, it also increases drag, such that the 1ift:drag ratio decreases. So increasing flap setting degrades climb performance. Option d, would produce the worst climb performance because it includes high altitude, high temperature and high flap setting.

TAKE-OFF 125. a. Decreasing flap angle decrease both lift and drag. The decreased drag would decrease the rate at which an aircraft could decelerate to a halt following a decision to abort a take-off. So decreasing flap angle would increase the accelerate stop distance required (ASDR). But the decreased lift would also increase the values of V,, V, and VLoF, and hence increase the distance required to attain these speeds. So decreasing flap angle would also increase the take-off distance required (TODR).

TAKE-OFF 126. C. Decreasing flap angle will decrease both lift and drag and increase the 1ift:drag ratio at any given airspeed. The increased 1ift:drag ratio will improve climb performance and hence increase the climb limited take-off mass (option b). The decreased lift will increase the values of V, V, and VLoF, and hence increase the length of the take-off run. But after lift-off the improved 1ift:drag ratio will increase the attainable climb gradient. So provided there is sufficient distance between the BRP and the obstacle, reducing flap setting might improve the mass at which the aircraft can clear the obstacles (option a). So both options a and b are correct, making c, the most appropriate. But the increased take-off run will worsen the field limited take-off mass (option d).

TAKE-OFF 127. C. Slush will increase the rolling resistance as the wheels of an aircraft are pushed through it duriqg the takeoff run. This will reduce the acceleration rates and hence increase the take-off run required at any given mass. This in turn will reduce the maximum mass that can take-off from any given field length.

TAKE-OFF 128. C. A downhill slope increases the rate at which an aircraft accelerates during the take-off run. This decreases the distances required to achieve V,, V, and.,V ,, A downhill slope also decreases the rate at which an aircraft can decelerate to a halt in the event of a decision to abort a take-off. This means that a greater distance is required to bring the aircraft to a halt. V, is the maximum speed at which it is possible to bring an aircraft to a halt without overrunning the stopway, in the event of an aborted take-off. So by increasing the distance required to stop an aircraft, a downhill slope decreases the maximum value of V, for any given aircraft mass and field length. V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. It is a function of dynamic pressure and coefficient of lift, and hence not affected by runway slope.

TAKE-OFF 129. a. V, is the maximum speed at which it is possible to bring an aircraft to a halt without overrunning the stopway, in the event of an aborted take-off. If V, is increased, the distance required to stop is also increased. And increasing V, will also increases the distance required to achieve V,. So increasing V, increases the accelerate stop distance required (ASDR). V, is the speed at which an aircraft must be rotated into the lift-off attitude during the take-off run. And V, must not be les than V,, so any increase in V, might require an increase in V., This in turn will require that the aircraft be held on the ground longer, thereby increasing the take-off run required (TORR) and take-off distance required (TODR). So increasiqg V, might increase both the ASDR and the TODR.

EN-ROUTE

EN-ROUTE 1. d. The optimum altitude for operation of a jet aircraft is that at which the airframe achieves the best TAS:Drag ratio and the engines achieve their lowest specific fuel consumption (SFC). Drag is proportional to mass, so as mass decreases due to fuel usage during flight, the altitude for best TAS:Drag ratio gradually increases. This means that the optimum operating altitude increases throughout any given flight. Although minimising fuel costs requires cruise at the optimum altitude, the gradual increase in altitude that ihis would involve is usually unacceptable for air traffic control purposes. But corr~mercialaircraft will operate at their optimum altitude whenever ATC considerations permit them to do so. The step climb procedure is a method by which efficiency is maximised whilst avoiding a gradual upward drift.

EN-ROUTE 2. a. Fuel consumption in a jet engine is approximately proportional to thrust such that decreasing thrust decreases fuel consumption. Thrust in the cruise is equal to drag, so cruising fuel consumption is propor.!ional to drag. Any factor that decreases drag at any given TAS will decrease fuel consumption. If mass is reduced by 5% then lift must also be reduced by 5% in order to maintain altitude. This is done by reducing angle of attack, which reduces both the coefficient of lift (C,) and the coefficient of induced drag (C,,). Changes in angle of attack do not however significantly affect the coefficient of profile drag (C,). Induced drag is proportional to C, and profile drag is proportional to C .,, So reducing angle of attack to compensate for reducing mass, reduces induced drag but does not affect profile drag. At any given TAS and altitude, induced drag is proportional to C, whilst lift is proportional to C,. C, is proportional to C?. SO if mass is reduced by 5%, then C, must also be reduced by 5%, to 0.95 of its previous value. This will cause C, to be reduced to 0.952 or 0.9025 of its previous value. So reducing mass by 5% causes C, and hence induced drag to reduce to 0.9025 of its previous value. This constitutes ,, is not affected by angle of attack changes, profile a 9.75% decrease in induced drag. But because C drag will remain unchanged. The total drag force at any speed is the sum of induced drag and profile drag. At the minimum drag speed (V,,,), induced drag is equal to profile drag, so each makes up half of the total drag. But as speed ,, induced drag decreases whilst profile drag increases. Jet aircraft cruising speed is increases above V greater than VIM, so induced drag constitutes considerably less than half of the total drag. So although a 5% decrease in drag would cause a 9.75% decrease in induced drag, the overall change in total drag would be less than half of this figure. Of the options offered in this question, only a (4.5%) satisfies this condition.

EN-ROUTE 3. b. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. Immediately following the engine failure the thrust available will be less than the drag. This will cause the aircraft to lose airspeed and this loss of speed will cause a reduction in lift force. This reduction in lift force will then cause the aircraft to lose altitude, gradually drifting downwards. But the loss of sufficient airspeed to commence the drift down

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following engine failure, requires a finite amount of time, so the drift down commences several minutes after engine failure.

EN-ROUTE 4. a. Fuel consumption in a jet engine is approximately proportional to thrust such that decreasing thrust decreases fuel consumption. Thrust in the cruise is equal to drag, so cruising fuel consumption is proportional to drag. Any factor that decreases drag at any given TAS will decrease fuel consumption. If mass is reduced by 10% then lift must also be reduced by 10% in order to maintain altitude. This is done by reducing angle of attack, which reduces both the coefficient of lift (C,) and the coefficient of induced drag (C,,). Changes in angle of attack do not however significantly affect the coefficient of profile drag (C,). lnduced drag is proportional to C, and profile drag is proportional to C .,, So reducing angle of attack to compensate for reducing mass, reduces induced drag but does not affect profile drag. At any given TAS and altitude, induced drag is proportional to C, whilst lift is proportional to C,. C, is proportional to .C : SO if mass is reduced by lo%, then C, must also be reduced by lo%, to 0.9 of its previous value. This will cause C, to be reduced to 0.g2 or 0.81 of its previous value. So reducing mass by 10% causes C, and hence induced drag to reduce to 0.81 of its previous value. This constitutes a 19% decrease in induced drag. But because C ,, is not affected by angle of attack changes, profile drag will remain unchanged. The total drag force at any speed is the sum of induced drag and profile drag. At the minimum drag speed (VIM,), induced drag is equal to profile drag, so each makes up half of the total drag. So in this case a 10% decrease in mass at ,V ,, will decrease half of the drag (induced drag) by 19% but will not affect the other half (profile drag). The overall effect will therefore be the average of these two effects, which is a reduction of 9.5% in total drag. But fuel consumption is approximately proportional to drag, so a 10% reduction in mass, causing a 9.5% decrease in total drag, will produce approximately 9.5% decrease in fuel consumption. In reality the thermal efficiency of the engines will decrease at reduced thrust settings so the reduction will be less than 9.5% so option a (9%) is the most accurate.

EN-ROUTE 5. a. Fuel consumption in a jet engine is approximately proportional to thrust such that decreasing thrust decreases fuel consumption. Thrust in the cruise is equal to drag, so cruising fuel consumption is proportional to drag. Any factor that decreases drag at any given TAS will decrease fuel consumption. If mass is reduced by 1094 then lift must also be reduced by 10% in order to maintain altitude.. This is done by reducing angle of attack, which reduces both the coefficient of lift (C,) and the coefficient of induced drag (C,,). Changes in angle of attack do not however significantly affect the coefficient of profile drag (C,). lnduced drag is proportional to C, and profile drag is proportional to .C ,, So reducing angle of attack to compensate for reducing mass, reduces induced drag but does not affect profile drag. At any given TAS and altitude, induced drag is proportional to C, whilst lift is proportional to C,. C, is : So if mass is reduced by lo%, then C, must also be reduced by lo%, to 0.9 of its proportional to .C previous value. This will cause CDlto be reduced to 0.g2 or 0.81 of its previous value. So reducing mass by 10% causes C, and hence induced drag to reduce to 0.81 of its previous value. This constitutes a 19% ,, is not affected by angle of attack changes, profile drag will decrease in induced drag. But because C remain unchanged. The total drag force at any speed is the sum of induced drag and profile drag. At the minimum drag speed (VIM,), induced drag is equal to profile drag, so each makes up half of the total drag. So in this case a 10% decrease in mass at VIM, will decrease half of the drag (induced drag) by 19% but will not affect the other half (profile drag). The overall effect will therefore be the average of these two effects, which is a reduction

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of 9.5% in total drag. But in this case the aircraft is flying at 1.5VIM, where induced drag is considerably less than M of the total drag. So the reduction in total drag will be even less than 9.5% Fuel consumption at any given TAS is approximately proportional to drag, so a 10% reduction in mass, causing a less than 9.5% decrease in total drag, will produce a less than 9.5% decrease in fuel consumption. In reality the thermal efficiency of the engines will decrease at reduced thrust settings so the reduction will be even less. So option a (less than a 19% decrease) is the most accurate.

EN-ROUTE 6. C. Fuel consumption in a jet engine is approximately proportional to thrust such that decreasing thrust decreases fuel consumption. Thrust in the cruise is equal to drag, so cruising fuel consumption is proportional to drag. Any factor that decreases drag at any given TAS will decrease fuel consumption. If mass is reduced by 10% then lift must also be reduced by 10% in order to maintain altitude. This is done by reducing angle of attack, which reduces both the coefficient of lift (C,) and the coefficient of induced drag (C,,). Changes in angle of attack do not however significantly affect the coefficient of profile drag (C,). Induced drag is proportional to C, and profile drag is proportional to .C ,, So reducing angle of attack to compensate for reducing mass, reduces induced drag but does not affect profile drag. At any given TAS and altitude, induced drag is proportional to C, whilst lift is proportional to C,. C, is : SO if mass is reduced by lo%, then C, must also be reduced by lo%, to 0.9 of its proportional to .C previous value. 'This will cause C, to be reduced to 0.g2or 0.81of its previous value. So reducing mass by 10% causes C, and hence induced drag to reduce to 0.81 of its previous value. This constitutes a 19% decrease in induced drag. But because ,C , is not affected by angle of attack changes, profile drag will remain unchanged. The total drag force at any speed is the sum of induced drag and profile drag. At speeds lower than the minimum drag speed (VIM,), induced drag makes up more than half of the total drag. So a 10% decrease in mass at 0.9VIM, will decrease more than half of the drag (the induced drag) by 19% but will not affect the remaining profile drag (less than of the total). The overall effect will therefore be the average of these two effects. This will be a reduction of more than 9.5% in total drag. But fuel consumption is approximately proportional to drag, so a 10% reduction mass, causing a reduction of more than 9.5% in total drag, will produce more than 9.5% decrease in fuel consumption. In reality the thermal efficiency of the engines will decrease at reduced thrust settings so the reduction will be slightly less than 9.5%. Option c (more than 9% decrease) is therefore the most accurate.

EN-ROUTE 7. b. Turbojet Specific Fuel Consumption (SFC) is equal to Thrust Ifuel consurr~ptionrate. And in cruising flight thrust is equal to total drag, so SFC = total drag Ifuel consumption. Total drag is the sum of induced drag and profile drag. At any given speed, induced drag is proportional to the square of lift, whilst profile drag is not affected by lift. In cruise flight lift is equal to mass, so increasing mass by 33% from 150000 kg to 200000 kg will increase induced drag to I. 332of its previous value, but will not affect profile drag. This would increase induced drag to 1.7689 of its previous value. So the induced drag on aircraft B would be 1.7689times that of aircraft A. But jet aircraft cruise at speeds greater than VIM,. At such speeds the induced drag makes up less than half of the total drag. So the overall effect of increasing mass from 150000 kg to 200000 kg would be to increase less than half of the drag by 76.89% whilst not affecting the remaining larger profile drag. The overall effect on total drag would therefore be an increase of less than 76.89% 1 2. So the total drag affecting aircraft B would be less than 1.385that of aircraft A.

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But for aircraft A SFC A = drag A Ifuel consumption A And for aircraft B SFC B = drag B Ifuel consumption B And drag B is less than 1.385 drag A so : So SFC B = (less than 1.385 drag A) Ifuel consumption B. But the SFC of the two aircraft are equal, so combining the two equations gives: Drag A Ifuel consumption A = (less than 1.385 drag A) Ifuel consumption B. Rearranging this gives the following: Fuel consumption B = (Fuel consumption A 1 Drag A) x (less than 1.385 Drag A) Which is Fuel consumption B = 5000 Kglhr x less than 1.385. So fuel consumption B is less than 6925 kglh. Of the options offered in this question, only a (5000 kglh) and b (6667 kglh) satisfy this condition. But option a is the same as the fuel consumption for aircraft A and so cannot be the correct figure for the heavier aircraft B. The most appropriate option is therefore b.

EN-ROU'TE 8. b. Turbojet Specific Fuel Consumption (SFC) is equal to Thrust Ifuel consumption rate. And in cruising flight thrust is equal to total drag, so SFC = total drag Ifuel consumption. Total drag is the sum of induced drag and profile drag. At any given speed, induced drag is proportional to the square of lift, whilst profile drag is not affected by lift. In cruise flight lift is equal to mass, so decreasing mass by 14.3% from 175000 kg to 150000 kg will decrease induced drag to 0.857*of its previous value, but will not affect profile drag. This will increase induced drag to 0.735 of its previous value. So the induced drag on aircraft B will be 0.735 times that of aircraft A. At V,, induced drag makes up half of .the total drag. So the overall effect of decreasing mass from 175000 kg to 150000 kg would be to decrease half of the drag by a factor of 0.735, whilst not affecting the other half. The overall effect on total drag would therefore be to decrease it by a factor of (1 + 0.735 12. So the total drag affecting aircraft B would be 0.865 that of aircraft A. But for aircraft A SFC A = drag A Ifuel consumption A.. . .......equation 1. And for aircraft B SFC B = drag B Ifuel consumption B And drag B is 0.865 drag A So SFC B = (0.865 drag A) Ifuel consumption B .........................equation 2. But the SFC of the two aircraft are equal, so combining equations 1 and 2 gives: Drag A /.fuel consumption A = (0.865 drag A) Ifuel consumption B. Rearranging this gives the followiug: Fuel consumption B = (Fuel consumption A IDrag A) x (0.365 Drag A) Fuel consumption A = 2000 kg.h, so fuel consumption B = 2000 kglh x 0.865 Which is fuel consumption B = 1730 kglh.

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EN-ROUTE 9.

C.

As altitude increases, the power required ro sustain any given flight condition increases, whilst the power available decreases. At the absolute ceiling the power available is just equal to the power available and the excess power is zero. But rate of climb is equal to excess power divided by aircraft weight. So at the absolute ceiling where excess power is zero, the maximum rate of climb is also zero (option c). It should be noted that option a (100 fpm) and b (500 fpm) are the gross service ceilings for propeller and jet aircraft respectively.

EN-ROUTE 10. C. Fuel consumption in a turbojet aircraft is proportional to thrust, and in level flight, thrust is equalto drag. Maximum endurance requires minimum fuel Drag consumption, so maximum turbojet endurance occurs at the minimum drag speed V .,, But to achieve maximum range it is necessary to achieve the best ratio of TAS : fuel consumption, which in turn means the best ratio of TAS:drag. This speed is found by drawing a tangent to the drag curve as illustrated in the diagram at the right. For a typical turbojet aircraft the best endurance speed is approximately 1.32 VIM,.

Drag

.

Best range speed approximately 1.32 .V ,,,

EN-ROUTE 11. b. As altitude increases, the power required to sustain any given flight condition increases, whilst the power available decreases. Excess power is equal to power available minus power required, so as altitude increases, excess power decreases. But rate of climb is equal to excess power divided by aircraft mass. So as altitude increases the decrease in excess power causes maximum rate of climb to decrease. The gross service ceiling is the altitude at which the excess power available is just sufficient to achieve a rate of climb of 100 fprn in a piston aircraft or 500 fprn in a jet aircraft. The rate at which power available diminishes is a function of engine type and air density lapse rate, and is therefore independent of aircraft mass. But the power required at any given altitude is proportional to mass. The greater the mass of the aircraft, the greater will be the power required. So increasing mass from 3000 kg to 3500 kg will increase the power required at any altitude, but will not affect power available. This will reduce the excess power available at any altitude and hence reduce the altitude at which there is just sufficient power to achieve the 100 fprn or 500 fprn climb rates. The heavier aircraft will therefore have a lower service ceiling.

EN-ROUTE 12. a. The power required to sustain any given flight condition increases with increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to prodilce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to'reduce speed, altitude, or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. Immediately following the engine failure the thrust available will be less than the drag. This will cause the aircraft to lose airspeed and this loss of speed will cause a reduction in lift force. This reduction in lift force will then cause the aircraft to lose altitude, gradually drifting downwards until it reaches its stabilising altitude. For a twin jet this will be the single engine service ceiling.

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EN-ROUTE 13. C. Specific Air Range (SAR) is a measure of the efficiency of an aircraft in flight. It is the distance that an aircraft can fly through the air, for each kg of fuel burned. SAR depends upon two factors, one relating to engine efficiency and the other relating to aerodynamic efficiency. The first of these is the Specific Fuel Consumption (SFC). This is a measure of the amount of fuel burned per hour, for each Newton of thrust generated. It is measured in KglNlh. The lower the SFC, the less fuel will be consumed for each Newton of thrust. For jet engines the lowest SFC is achieved when operating at 85% to 95% RPM. The second factor is the ratio of TAS:drag. The greater the TAS:drag ratio, the less will be the thrust required to sustain any given TAS. This ratio tends to increase with increasing altitude. As altitude increases the thrust produced at any given RPM decreases. The optimum altitude is that at which the thrust produced when operating the engines in the 85% to 95% RPM range, is equal to the drag at the TAS producing , maximum TAS:drag ratio. Under these circumstance the SAR will be maximum.

EN-ROUTE 14. d. Specific Air Range (SAR) is a measure of the efficiency of an aircraft in flight. It is the distance flown through the air, for each kg of fuel burned. Maximum efficiency is achieved by operating at the altitude providing optimum SAR. But as a flight progresses, the burning of fuel reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift to match the reducing mass. But this also reduces drag, and so it is also necessary to reduce engine RPM to reduce thrust, so that TAS remains constant. But reducing angle of-attack reduces TAS:drag ratio, whilst reducing RPM increases speciric fuel consumption (SFC). Both of these effects reduce SAR. In order to regain optimum SAR it is necessary to increase altitude. This means that the optimum altitude gradually increases throughout cruising flight. Maximum efficiency would be achieved by cruise climbing such that the aircraft gradually drifts upwards, always remaining at its optimum altitude. But this gradual change in altitude is unacceptable from the ATC viewpoint, which requires strict control of altitude in order to avoid traffic conflicts. Aircraft therefore employ step climb techniques. This involves cruising for a short period at one altitude before climbing to a slightly higher one. By repeating this clirr~b- cruise - climb sequence, the time at or close to optimum altitude is maximised.

EN-ROUTE 15. C. The power required to sustain any given flight condition increases with increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude, or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. When an engine fails the thrust available immediately decreases, whilst the drag remains constant. This difference between thrust and drag causes the aircraft to decelerate. This loss of speed reduces the lift that can be generated and so the aircraft drifts downwards. But the induced drag at any speed is proportional to aircraft mass. So a heavy aircraft will generate more drag than a light one. This means that a heavy aircraft will decelerate to drift down speed lift more quickly and hence will commence its drift down sooner

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than a light one. This is reflected in the CAP 698 drift down graphs which indicate immediate drift down at 70000 kg mass but delayed drift down at 35000 kg.

EN-ROUTE 16. b. The power required to sustain any given flight condition increases with increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. But power required also increases with increasing mass, so the stabilising altitude at which power available equals power required, decreases with increasing mass. When an engine fails the power available immediately decreases, whilst the power required remains constant. This difference between power available and power required causes the aircraft to drift downwards. The rate of descent is equal to the power deficit (power available minus power required) divided by the mass of the aircraft. So a heavy aircraft will drift down more quickly than a light one. So increasing mass increases drift down rate and decreases stabilising altitude.

EN-ROUTE 17. b. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. Power available is proportional to the mass flow of air passing through the engines. tn icing conditions some of the engine compressor air must be bled away for de-icing and anti-icing purposes. This reduces the power available and hence increases drift down rate and deaeases stabilising altitude. This effect is illustrated in the CAP 698 drift down charts.

EN-ROUTE 18. C. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remainirlg engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. Power available is proportional to the mass flow of air passing through the engines. Air conditioning packs consume a large volume of engine compressor air and so their use reduces the power available. If these are switched off below 17000 ft the extra power available will reduce drift down rate and increase stabilising altitude.

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EN-ROUTE 19. C. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high aititude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and attitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. The rate of descent during the drift down will be equal to the power deficit (power available minus power required) divided by the mass of the aircraft. This will not be affected by a headwind or tailwind. But the distance over the ground covered during the drift down will be affected. This is because although the aircraft will follow a set descent path through the air, a tailwind will cause the air mass to be blown in the same direction as the aircraft. The overall effect of a tailwind is that drift down rate and stabilising altitude are not affected but drift down distance is increased.

EN-ROUTE 20. b. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available e is just sufficient to maintain cruise speed. When an engine fails with all engines operating at c r ~ ~ i spower, during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. The rate of descent during the drift down will be equal to the power deficit (power available minus power required) divided by the mass of the aircraft. This will not be affected by a headwind or tailwind. But the distance over the ground covered during the drift down will be affected. This is because although the aircraft will follow a set descent path through the air, a headwind will cause the air mass to be blown in the opposite direction to that of the aircraft. The overall effect of a headwind will therefore be that drift down rate and stabilising altitude are not affected but drift down distance is decreased.

EN-ROUTE 21. C. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. The rate of descent during the drift down will be equal to the power deficit (power available minus power required) divided by the mass of the aircraft. This will not be affected by a headwind or tailwind. But the gradient is the height loss divided by the distance flown over the ground covered during the drift down. This will be affected because although the aircraft will follow a set descent path through the air, a headwind will cause the air mass to be blown in the opposite direction to that of the aircraft. The overall effect of a headwind will therefore be that drift down rate is not affected but drift down gradient is increased.

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EN-ROUTE 22. a. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. Both power available and power required are proportional to air density such that increasing density increases power available and decreases power required. So as altitude decreases, the increasing air density reduces the power deficit (power available minus power required) following engine failure. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which increased air density enables sufficient power to be produced from the remaining engines to maintain level flight. Because of this relationship between air density, power available and power required, anything that increases air density will decrease drift down rate and increase stabilising altitude. But increasing density altitude constitutes a decrease in density, which causes drift down rate to increase. So increasing density altitude will increase drift down rate.

EN-ROUTE 23. C. The power required to sustain any given flight condition increases with increasing mass and increasing altitude. Commercial aircraft are designed such that when cruising at high altitude, the power available with all engines operating at cruise power, is just sufficient to maintain cruise speed. When an engine fails during high altitude cruising flight, the remaining engines are unable to produce'sufficient power to maintain flight at that speed and altitude. It is therefore necessary to reduce speed, altitude or both. In order to achieve maximum efficiency the process of drift down is used, whereby the aircraft drifts slowly down to the maximum altitude at which sufficient power is available from the remaining engines. Power available is proportional to the mass of air flowing through the engines. Air conditioning packs consume a large volume of engine compressor air and so their use reduces the power available. If these are switched off below 17000 ft the extra power available will reduce drift down rate and increase the stabilising altitude. But packs must always be switched on at or above 17000 ft, thereby increasing drift down rate and decreasing stabilising altitude. The drift down performance therefore jumps from a steep curve to a shallower one at 17000 ft. This results in a break in the CAP 698 drift down curves.

EN-ROUTE 24. a. The power output of a jet engine is directly proportional to N, and inversely proportional to pressure altitude and ambient temperature. Maximum continuous N, is limited by maximum allowable power output, and pressures and temperatures within the engine. As pressure altitude increases, air density decreases thereby reducing power output. The engine can therefore be operated at higher values of N, without exceeding the power output limits. Increasing ambient temperature has a similar effect. The overall effect is that increasing pressure altitude and increasing air temperature both increase the maximum continuous N,.

EN-ROUTE 25. b. The regulations regarding positive load factor limits for JAR certificated passenger aircraft are specified in JAR 25. Although JAR 25 specifies a method of calculating these load factors based on aircraft weight, questions in the JAR ATPL examinations are usually based on certain values. These are the minimum allowable values, which are 2 with flaps down and 2.5 with flaps up.

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EN-ROUTE 26. b. V, is the CAS at which an aircraft will stall whilst just attaining the limiting load factor. This means that at lower speeds the aircraft will stall before attaining the limiting load factor. At higher speeds it can be damaged due to overloading by exceeding the limiting load factor before stalling. An aircraft cannot therefore be damaged by overlcading due to the use of the flying controls at or below V.,

EN-ROUTE 27. b. 'The maximum operating altitude of a JAR certificated passenger aircraft is that altitude at which it can just attain a positive load factor of 1.3 before the onset of buffet. Load factor greater than 1.3 when flying at this altitude will cause buffeting.

EN-ROUTE 28. b. V, is the CAS at which an aircraft will stall whilst just attaining the limiting load factor. This means that at lower speeds the aircraft will stall before attaining the limiting load factor. At higher speeds it can be damaged due to overloading by exceeding the limiting load factor before stalling. V, is therefore the maximum speed at which full nose up control inputs are permitted.

EN-ROU'FE 29. d. V, is the CAS at which an aircraft will stall whilst just attaining the limiting load factor. For the purposes of JAR ATPL examinations, the limiting load factor is normally taken to be 2.59 with flaps retracted. Load factor = lift 1 weight, so in order to increase load factor at any given weight, the magnitude of the lift force must be increased. Lift is proportional to C, x CAS so at any given speed an increase in lift (and hence load factor) requires an increase in C,. Any change which reduces the gradient of the C, : angle of attack curve and decreases C, ,,wil increase V ., Of the options offered in this question only option d, increasing the wing sweep back angle will have and this effect. Of the other options, increasing aspect ratio and increasing camber both increase C, ., Increasing wing area increases the lift force attained at any given C, and therefore also hence reduce V ., decreases V

,,,

EN-ROUTE 30. d. Turbulence results in the wings being subjected to random variations in upward and downward gusts of air. These cause random changes in the angle of attack and hence C,. The varying lift force produced by these changes in C, cause the aircraft to climb and descend in an erratic manner. The degree to which flight path and hence ride quality is affected depends upon the magnitude of the variations in C,. The magnitude of theses variations in C, depends upon the gradient of the C,:a curve. The steeper the C,:a curve, the greater will be the change in C, for any given change in a. The smoothest ride will be provided by an aircraft exhibiting a shallow C,:a curve. Of the options provided in this question only option d, increased sweep back angle, will have these effects.

EN-ROUTE 31. a. V, is the design manoeuvre speed of an aircraft. This is the CAS at which the aircraft stalls just as it attains the limiting load factor. At lower speeds the aircraft will stall at lower load factors, whilst at higher speeds it will exceed the limiting load factor before stalling. The recommended rough air speed is normally set approximately midway between the low speed and high speed stall speeds, in order to provided the widest possible safety margin when flying in rough air. The maximum operating speeds of JAR certificated passenger aircraft are VMo 1 MMo.

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EN-ROUTE 32. d. Turbulence results in the wings of an aircraft being subjected to random upward and downward gusts of air. These cause random changes in the angle of attack and hence C,. If up-gusts are sufficiently severe, the angle of attack can exceed the stalling angle, thereby causing low speed stall. Also if the gusts are of sufficient intensity the C, can be increased to such an extent that the limiting load factor of the aircraft is exceeded. Deployment of trailing edge flaps reduces the stalling angle, and so makes the wing more prone to low speed stall in turbulence. Trailing edge flap deployment also increases C, and hence increases the danger of exceeding limiting load factor. In order to compensate for this effect the limiting toad factor of JAR certificated passenger aircraft is reduced to 2.09 when flaps are deployed.

.,,

Although the above effects are mitigated to some extent by the deployment of leading edge devices, the most probable overall effect is still an increase in stalling hazard and overload hazard.

EN-ROUTE 33. C. At any given altitude an aircraft can be flown only at those speeds between the low speed buffet and high speed buffet boundaries. Because the speed of sound reduces with decreasing temperature, the value of the high speed buffet boundary decreases, causing the available speed range to decrease with increasing altitude. The aerodynamic ceiling is that altitude at which the low speed and high speed buffet boundaries meet, leaving only a single flight speed available. Banking an aircraft increases load factor and hence the magnitude of the lift force required. This increase in required lift force increases the low speed stalling speed, causing the buffet boundaries to converge further. The magnitude of this additional convergence is proportional to the load factor and hence bank angle. At high altitudes where the speed margin is very narrow, the ability of an aircraft to bank is therefore very limited. It should however be noted that increasing altitude also decreases stability, thereby increasing the roll, pitch and yaw rates-that an aircraft is capable of achieving. The use of these higher rates will however cause stalling, so the overall practical effect is that manoeuvrability of an aircraft decreases with increasing altitude.

EN-ROUTE 34. C. Turbulence results in the wings being subjected to random variations in upward and downward gusts of air. These result in random changes in the angle of attack and hence C,. The variations in lift force produced by these-changes in C, cause the aircraft to rise and descend in an erratic manner. The degree to which flight path and hence ride quality is affected, depends upon the magnitude of the variations in C,. The magnitude of theses variations in C, depends upon the gradient of the C, : angle of attack curve. A steeper curve gives larger variations in C, for any given change in angle of attack. Because swept wings exhibit shallow C, : angle of attack curves, they are less affected by turbulence than straight wings. Options a and b in this question are incorrect, because dihedral and anhedral have no effect on the C, : angle of attack curve or response to turbulence.

EN-ROUTE 35. b. 'The V-n envelope illustrates the range of speeds and load factors at which an aircraft is capable of operating. The gust envelope illustrates the range of speeds and vertical gusts within which an aircraft is capable of operating. In both envelopes the airspeed increases from left to right along the horizontal axis. The left boundary of these envelopes is the low speed stall line

EN-ROUTE 36. b. 'The V-n envelope illustrates the range of speeds and load factors at which an aircraft is capable of operating. The gust envelope illustrates the range of speeds and vertical gusts within which an aircraft is

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capable of operating. In both envelopes the airspeed increases from left to right along the horizontal axis. The right boundary indicates V, which is the design dive speed.

EN-ROUTE 37. b. The range of speeds at which an aircraft can be operated is limited by the low and high speed buffet boundaries. As altitude increases these boundaries converge, such that the available speed range becomes more limited. Manoeuvringthe aircraft increases ioad factor and so causes the low speed buffet boundary to move closer to the high speed boundary. The maximum operating altitude for JAR certificated passenger aircraft is that altitude at which a 1.39 manoeuvre is just possible without incurring buffet. Exceeding the maximum operating altitude is therefore likely to cause buffet at lower load factors. If this altitude is exceeded by a sufficient margin the aerodynamic ceiling will be reached-at which high speed buffet will occur in straight and level flight.

EN-ROUTE 38. d. JAR 25 states that for certification purposes, buffeting in passenger aircraft must not be evident at speeds ,, and VMo/MMo. between 1.2 V

EN-ROUTE 39. d. The aerodynamic ceiling is that altitude at which the low speed and high speed buffet boundaries cross -leaving only a single flight speed available. Any manoeuvre that increases load factor above one or increases or decreases TAS will cause buffet.

EN-ROUTE 40. d. The power required to maintain level flight at any given airspeed increases with altitude, whilst the power available from an engine decreases with increasing altitude. The absolute ceiling is the altitude at which the entire power output of an aircraft's engines is just sufficient to permit level flight. At this altitude further climbing is not possible and even level flight is possible at only one speed. This speed is VMDfor a jet and Vx for a propeller aircraft. All manoeuvres that increase the load factor will increase the power required. Any manoeuvres that do not increase load factor but cause a change in airspeed will also require more power. All such manoeuvres will therefore cause loss of altitude.

EN-ROUTE 41. a. The ability of an aircraft to conduct a steady climb is determined by the amount of excess power available. Excess power is that proportion of available power that is not required to push the aircraft forward against the drag force. Excess power is therefore equal to the power available minus the power required for straight and level constant speed flight. At all altitudes below the absolute ceiling, the power available and power required curves cross at two points. These points represent the minimum and maximum level flight speeds, unless speed is limited by other factors such as low and high speed buffet or stall. At the speeds at which the power available and power required curves coincide there is no excess power available. At the low speed crossing point, acceleration will reduce the power required thereby enabling the aircraft to climb. At the high speed crossing point however, speed increases will increase the power required so acceleration is not possible. The performance an aircraft can achieve at the high speed crossover is therefore unaccelerated straight and level flight. The only method of climbing possible at this speed is zoom climbing in which altitude is increased at the expense of decreasing airspeed. The aircraft can of course conduct an accelerating descent as this requires less power than straight and level flight. Of the four options offered in this question only a, "Constant speed level flight is possible", is true at both points, so this represents the best answer.

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EN-ROUTE 42. C. In this question fuel consumption has been defined as Kglnm, which is in fact the Specific Air Range (SAR). This means that any action that increases the fuel flow required to maintain speed, constitutes an increase in fuel consumption. If the C of G is ahead of the C of P, then the resulting nose-down moment must be balanced by a down force provided by the tailplane. For level flight the sum of all down forces (weight plus tailplane down force) must be balanced by lift, so any increase in tailplane down force will increase the magnitude of the wing lift required. The production of this additional lift will cause an increase in lift induced drag. In addition to this, the generation of the tailplane down force will increase trim drag. This increased drag will necessitate the consumption of more fuel to create additional thrust to balance drag. Forward movement of the C of G when it is already ahead of the C of P will therefore increase fuel consumption. If however the C of G is aft of the C of P (which is not normally permitted), it will produce a nose-up moment that must be balanced by a tailplane up force. Forward movement of the C of G towards the C of P will therefore reduce trim drag and fuel consumption. Of the options offered in this question only c is fully consistent with the effects described above.

EN-ROUTE 43. C. To increase airspeed above,V , in straight and level flight, it is necessary to increase the power output of the engines. The maximum speed that any aircraft can possibly achieve in straight and level flight is that at which the power required is equal to the power available. In this condition no excess power is available, so option d is incorrect. Option b describes the situation existing at the minimum power speed,V ,, which is much lower than maximum speed. The power available from a propeller aircraft increases with airspeed before decreasing at higher speeds, so option a is also untrue.

EN-ROUTE 44. d.

DRAG

The angle of attack required to maintain level flight decreases as speed increases.V,, is the speed at which total drag is at a minimum value and the L:D ratio is at a maximum value. Maximum range in a jet aircraft is achieved when flying at the speed providing the best TAS:Drag DRAG ratio. This occurs at the speed at which a tangent drawn from the origin touches the drag curve as indicated in the diagram at the right. This is higher than ,V , and so flying at this speed requires a lower angle of attack than that providing best L:D ratio at.,V,

Best range speed

I

Power required

EN-ROUTE 45. a.

Drag

,V , is the speed at which the power required is at its minimum , any increase or decrease in speed value. When flying at,V will cause an increase in power required.,V , is the speed at which the drag force is at a minimum value. Any speed , will cause drag to change that takes an aircraft closer to,V decrease. therefore decrease drag Changing speed from,V , to V,wli and increase power required. This effect is illustrated in the diagram at the right. Drag

............................................ --.......................

-

-

.:

.

I

Increase in power required

.

.

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EN-ROUTE 46. a. The relationships between CAS, TAS and Mach number vary with altitude. Also the stalling speed and buffet onset speeds vary with aircraft weight and C of G position. The buffet boundary chart indicates the relationship between altitude, mass, C of G position and the low and high speed stall mach numbers.

EN-ROUTE 47. b. At low altitudes the margin between low speed buffet and high speed buffet is wide, so the ability of an aircraft to manoeuvre is limited primarily by its structural strength. At higher altitudes the low speed buffet boundary increases and the high speed buffet boundary decreases. The convergence of these two boundaries causes the manoeuvrabilityof an aircraft to be degraded as altitude increases. At the maximum operating altitude a JAR certificated passenger aircraft can achieve only 1.3g without incurring buffet.

EN-ROUTE 48. b. The stability of an aircraft depends to a considerable extent upon the magnitude of the aerodynamic damping forces caused by changes of attitude. As altitude increases, these damping effects decrease causing stability to be degraded.

EN-ROUTE 49. a. ,V , is the speed at which total drag is at a minimum value and the L:D ratio is at a maximum value. Maximum range in a propeller aircraft is achieved at the Drag speed providing best EAS:Power Required ratio. This occurs at the speed at which a tangent drawn from the origin touches the power required curve, as indicated in the diagram at the right. This is at VMD,so flying at this speed requires the angle of attack that gives best L:D ratio. Best range speed=V,,

EN-ROUTE 50. a. At the absolute ceiling the only speed at which a propeller aircraft can maintain straight and level flight is V., At lower altitudes V, is the speed at which excess thrust is at a maximum value, thereby facilitating the greatest climb gradient. V, is the speed at which excess power is maximum, facilitating maximum rate of climb. As altitude increases, reducing air density decreases the power produced by the engine and the thrust produced by. the propeller, thereby causing excess power, excess thrust, maximum climb gradient, and maximum rate of climb, to reduce. At the absolute ceiling both excess thrust and excess power are zero, so further climbing is impossible. So at the absolute ceiling thrust available is equal to drag and power available is equal to power required. This situation is illustrated in the diagrams overleaf.

Thrust available at

Drag is constant at all altitudes

Reduced thrust available at The

Thrust is equal to drag at vx at the

,ragisgreaterthan thrust at all speeds ~ other than V x . "x

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Drag is constant at all altitudes Thrust available at sea level Reduced thrust available at The absolute ceiling Thrust is equal to drag at Vx at the absolute ceiling Drag is greater than thrust at all speeds other than Vx.

vx

EN-ROUTE 51. b. ~ a x i ~ uendurance m occurs when the fuel consumption rate is minimum. Fuel consumption in a propeller aircraft is proportionalto power. So maximum endurance speed in a propeller driven aircraft is the minimum , But this is lower than the minim drag speed,,V , and reducing speed requires increasing power speed,.,V C, So flight at,V , requires a greater C, than that at.,V, C, at,V , is that giving best L:D ratio, so best endurance speed in a propeller aircraft requires a C, greater than that for best L:D ratio.

EN-ROLITE 52. a. Maximum endurance requires minimum fuel consumption rate. In a jet aircraft fuel consumption is proportional to thrust and in straight and level flight thrust is equal to drag. So maximum endurance speed in a jet aircraft is the minimum drag speed.,V , But,V , is the speed at which the L:D ratio is maximum. Maximum jet endurance therefore requires the C, giving best L:D ratio.

EN-ROLITE 53. c In order to achieve maximum efficiency jet aircraft must operate at the altitude producing lowest specific fuel consumption and best TAS:Drag ratio. For jet aircraft this is at 1.32 VIM, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. But as flight progresses, reducing fuel load reduces aircraft mass. So to maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift. But this also reduces drag, and so it is also necessary to reduce engine RPM in order to reduce thrust. But reducing angle of attack reduces TAS:drag ratio, whilst reducing RPM increases specific fuel consumption (SFC). Both of these effects reduce efficiency. In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 VIM, once again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight. Maximum possible efficiency would be achieved by allowing the aircraft to drift steadily upwards at constant mach number, angle of attack and RPM as mass decreases, such that it was always at the optimum altitude. This is however unacceptable from the ATC standpoint, which requires altitude to be strictly controlled. In order to achieve the best compromise between efficiency and ATC considerations, operators err~ploystep climb techniques. These comprise of short periods at a fixed altitude followed by a climb to a

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slightly higher one. After a further short period of cruise the aircraft again climbs and the process is repeated. In this way the time spent at or close to their steadily increasing optimum altitude is maximised.

EN-ROUTE 54. a. According to the JAR classification system all jet powered aircraft are assigned to Class A, regardless of passenger numbers or aircraft mass. For Class A aircraft, JAR OPS 1.500( c ) specifies that the minimum allowable en-route obstacle clearance following an engine failure is 2000 ft.

EN-ROUTE 55. d. For all normal operations aircraft must be flown such that the onset of low and high speed buffet is avoided. The speed boundaries at which buffet commences vary with aircraft mass, altitude, C of G position and load factor. Buffet is avoided by ensuring that aircraft are operated within, and do not cross these boundaries. A buffet boundary chart indicates the location of these boundaries for various masses, altitudes, C of G positions, and load factors. It should be noted that upon crossing either boundary, the onset of buffet occurs before reaching either the stall or VMo/MMo.Options a, b, and c are therefore incorrect.

EN-ROUTE 56. C. The maximum attainable rate of climb at any altitude is equal to the excess power divided by the aircraft weight. Excess power is equal to power available minus power required. As altitude increases, the power required to maintain any given flight condition increases, whilst the power available decreases. This reduces the excess power and hence reduces the maximum rate of climb. The absolute ceiling for any type of aircraft is the altitude at which power available is equal to power required. Under these circumstances both excess power and rate of climb are zero. At the absolute ceiling, sustained level flight is possible only at V., In the case of a jet aircraft V, is.,V,

EN-ROUTE 57. b. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 V ,,, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. But as flight progresses, reducing fuel load reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift. But this also reduces drag, so it is also necessary to reduce engine RPM in order to reduce t h r ~ s tReducing . angle of attack reduces TAS:drag ratio, whilst reducing RPM increases SFC, Both of these effects reduce efficiency. In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 V ,,, once again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight. Maximum possible efficiency would be achieved by allowing the aircraft to drift steadily upwards at constant mach number, angle of attack RPM, as mass decreases, such that it was always at the optimum altitude. This procedure is termed a cruise climb and employs constant N,.

EN-ROUTE 58. d. In order to achieve maximum efficiency aircraft must operate at the altitude producing lowest specific fuel ,,, at the altitude where the consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 V thrust at 85% to 95% RPM is equal to the drag force. Under these circumstance the engines are operating close to their maximum continuous power output. The failure of one or more engine will therefore reduce power available to less than power required.

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It will therefore be necessary to reduce altitude or speed or both. The best efficiency is achieved by increasing power on the remaining engines to the maximum continuous setting and allowing the aircraft to drift down to the altitude at which power available again equals power required. The drift down rate is proportional to weight, such that decreasing weight will decrease drift down rate. If it is believed that an aircraft drift down path will not clear an obstacle, the ditching of fuel might overcome this problem. It is however important to ensure that sufficient fuel remains to reach an appropriate landing site with sufficient reserves remainirlg. The correct procedure is therefore to assess fuel requirements before ditching sufficient fuel to clear the obstacles. It should however be noted that a typical drift down path is initially very steep, becoming more shallow as altitude decreases. This means that the greatest clearance of close obstacles is gained by dumping fuel as soon as possible.

EN-ROUTE 59. C. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel consumption )SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 ,V ,, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. But as flight progrssses, reducing fuel load reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift. But this also reduces drag, and so it is also necessary to reduce engine RPM in order to reduce thrust. Reducing angle of attack reduces TAS:drag ratio, whilst reducing RPM increases SFC. Both of these effects reduce efficiency. In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 VIM, once again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight. Increasing altitude decreases outside air temperature, causing the thermal efficiency of the engines to increase. This in turn will decrease SFC. So provided the aircraft remains at its optimum altitude the SFC will gradually decrease.

EN-ROUTE 60. C. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 VIM, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. Under these circumstance the engines are operating close to their maximum continuous power output. The failure of one or more engine will therefore reduce power available to less than power required. It will therefore be necessary for the ai~craftto continue the flight at an altitude lower than optimum. Also the trimming control inputs required to overcome the asymmetric thrust effect, will increase trim drag. Both the reduced altitude and increased drag will reduce efficiency. Cruise range will therefore be reduced following a single engine failure.

EN-ROUTE 61. a. Optimum efficiency and hence range are achieved by flying such that the TAS:drag ratio is maximised and the SFC minimised. This involves flight at 1.32 ,V ,,, at an altitude at which ,M ,,, is slightly less than.,V ,, LRC therefore involves flight at a fixed mach number. But as flight progresses reducing weight causes optimum altitude to increase, so the aircraft must climb. Mach number is the ratio of TAS:LSS, and as altitude increases in the troposphere the local speed of sound (LSS) decreases. So climbing at a fixed mach number means TAS decreases to match the decrease in LSS. But as altitude increases, the IAS at any given TAS decreases. So increasing altitude below the

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tropopause at fixed mach number, decreases both TAS and IAS. Above the tropopause temperature and LSS remain constant up to approximately 58000 ft. So increasing altitude at constant mach number in this part of the atmosphere causes IAS to decrease, while TAS remains constant. Option a is therefore the most accurate.

EN-ROUTE 62. C. In order to achieve maximum efficiency aircraft must operate at the altitude producing lowest specific fuel at the altitude where the consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at I.32V,, thrust at 85% to 95% RPM is equal to the drag force. Under these circumstance the engines are operating close to their (typically 100% RPM) maximum continuous power output. The failure of one or more engine will therefore reduce power available to less than power required. It would therefore be necessary to reduce altitude or speed or both. The best efficiency is achieved by increasing power on the remaining engines to the maximum continuous setting and allowing the aircraft to drift down to the altitude at which power available again equals power required. The drift down rate is proportional to weight such that decreasing weight wiil decrease drift down rate and increase stabilising altitude. If it is believed that an aircraft drift down path will not clear an obstacle, the ditching of fuel might overcome this problem. It is however important to ensure that sufficient fuel remains to reach an appropriate landing site with sufficient reserves remaining. The correct procedure is therefore to assess fuel requirements before ditching sufficient surplus fuel as soon as possible, to clear the obstacles.

EN-ROUTE 63. C. Glide range in still air is equal to 1ift:drag ratio multiplied by the height at which the glide is commenced. Lift:drag ratio is a function of aerofoil section, surface condition and planform, none of which are dependent of aircraft mass. Increasing mass will therefore have no effect on glide range. But increasing mass does increase the glide speed. So increasing mass causes a gliding aircraft to fly down the same slope but at a greater speed. This means that glide range is unaffected but glide endurance decreases as mass increases.

EN-ROUTE 64. b. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel ,, at the altitude where the consumption SFC) and best TAS:Drag ratio. For jet aircraft this is at I.32 ,V thrust at 85% to 95% RPM is equal to the drag force. Under these circumstance the engines are operating close to their maximum continuous power output. The failure of one or more engine will therefore reduce power available to less than power required. It will therefore be necessary to reduce altitude or speed or both. The best efficiency is achieved by increasing power on the remaining engines to the maximum continuous setting and allowing the aircraft to drift down to the stabilising altitude at which power available again equals power required. But rate of descent in a drift down is equal to the power deficit (power available minus power required) divided by the mass of the aircraft. At the absolute stabilising altitude this deficit is zero so the rate of descent is also zero. The aircrafi will therefore level out at its stabilising altitude.

EN-ROUTE 65. C. For all normal operations aircraft must be flown such that the onset of low and high speed buffet is avoided. The speed boundaries at which buffet commences vary with aircraft mass, altitude, C of G position and load factor. Buffet is avoided by ensuring that aircraft are operated within, and do not cross these boundaries. A buffet boundary chart indicates the location of these boundaries for various masses,

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altitudes,-C of G positions, and load factors. It should be noted that when crossing either boundary, the onset of buffet occurs before reaching either the stall or VMo/MMo.Options a, b, and d are therefore incorrect.

EN-ROUTE 66. d. A number of factors limit the maximum altitude at which an aircraft can be operated. The aerodynamic ceiling is the altitude at which the low speed and high speed buffet boundaries cross, Flight at this altitude is possible only at one speed and flight at any speed or at any higher altitude will induce buffet. The absolute ceiling is that at which the power available is equal to the power required and maximum possible rate of climb is zero. It is therefore the maximum altitude that can be achieved using all of the available power. Steady flight above this altitude is not possible because insufficient power is available to sustain it. A further potential limitation is the maximum altitude at which the cabin structure can sustain the loads exerted upon it by internal pressurisation. The highest altitude at which an aircraft could theoretically be operated is the lowest of the three altitudes described above. But if aircraft were to operate at any of these altitudes there would be a constant danger of buffet, structural failure or loss of altitude. In order to ensure that flights can be conducted safely, a lower maximum operating altitude is specified. At this altitude an aircraft can attain a load factor of 1.3 without buffet, has sufficient structural strength to prevent failure and has sufficient power to maintain a range of speeds.

EN-ROUTE 67. a. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 VIM, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. Under tliese circumstance the engines are operating close to their maximum continuous power output. The failure of one or more engine will therefore reduce power available to less than power required. It will therefore be necessary to reduce altitude or speed or both. The best efficiency is achieved by increasing power on the remaining engines to the maximum continuous setting and allowing the aircraft to drift down to the stabilising altitude at which power available again equals power required. The drift down procedure is therefore employed following engine failure at any altitude above that at which reduced power can maintain level flight.

EN-ROUTE 68. a. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 ,V ,, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. But as flight progresses, reducing fuel load reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift. But this also reduces drag, and so it is also necessary to reduce engine RPM in order to reduce thrust. Reducing angle of attack reduces TAS:drag ratio, whilst reducing RPM increases SFC. Both of these effects reduce efficiency. ,,, once In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 V again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight.

EN-ROUTE 69. d. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel ,, at the altitude where the consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 ,V thrust at 85% to 95% RPM is equal to the drag force. But as flight progresses, reducing fuel load reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift.

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But this also reduces drag, and so it is also necessary to reduce engine RPM in order to reduce thrust. But reducing angle of attack reduces TAS:drag ratio, whilst reducing RPM increases SFC. Both of these effects reduce efficiency. In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 V!,, once again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight. Maximum possible efficiency would be achieved by allowing the aircraft to drift steadily upwards at constant mach number, angle of attack and RPM as mass decreases, such that it was always at the optimum altitude. This is however unacceptable from the ATC standpoint, which requires altitude to be strictly controlled. So in order to achieve the best compromise between efficiency and ATC considerations, operators employ the step climb technique. This involves repeated cycles of short periods of constant altitude cruise followed by short climbs to higher altitudes, to maximise their time at or close to their steadily increasing optimum altitude.

EN-ROUTE 70. a. In order to achieve maximum efficiency, aircraft must operate at the altitude producing lowest specific fuel consumption (SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 VIM, at the altitude where the thrust at 85% to 95% RPM is equal to the drag force. But as flight progresses, reducing fuel load reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attack in order to reduce lift. But this also reduces drag, and so it is also necessary to reduce engine RPM in order to reduce thrust. But reducing angle of attack reduces TAS:drag ratio, whilst reducing RPM increases SFC. Both of these effects reduce efficiency. In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 VIM, once again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight. Maximum possible efficiency would be achieved by employing a cruise climb technique by allowing the aircraft to drift steadily upwards at constant mach number, angle of attack and RPM, as mass decreases, such that it was always at the optimum altitude.

EN-ROUTE 71. C. In order to achieve maximum efficiency aircraft must operate at the altitude producing lowest specific fuel ,,, at the altitude where the consumption )SFC) and best TAS:Drag ratio. For jet aircraft this is at 1.32 V thrust at 85% to 95% RPM is equal to the drag force. But as flight progresses, reducing fuel load reduces aircraft mass. To maintain constant altitude it is necessary to reduce angle of attacic in order to reduce lift. But this also reduces drag, and so it is also necessary to reduce engine RPM in order to reduce thrust. But reducing angle of attack reduces TAS:drag ratio, whilst reducing RPk4 increases SFC. Both of these effects reduce efficiency. In order to regain optimum efficiency it is necessary to increase altitude such that drag at 1.32 V ,,, once again matches thrust at 85% to 95% RPM. This means that the optimum altitude gradually increases throughout cruising flight. Maximum possible efficiency would be achieved by empioying a cruise climb technique by allowing the aircraft to drift steadily upwards at constant mach number, angle of attack and RPM, as mass decreases, such that it was always at the optimum altitude. This gradual increase in altitude is however generally unacceptable for ATC purposes, which requires aircraft to remain at their authorised altitudes. Operators therefore generally employ the slightly less efficient technique of step climbing. In this technique an aircraft attains an initial altitude approximately 1000 ft above its optimum. It then cruises at 1.32 VIM, maintaining constant altitude by gradually reducing RPM until it is 3000 ft below the new optimum altitude. The aircraft is then climbed to 1000 ft above the new optimum, and the process is repeated. In this way the aircraft is kept at or as close to its optimum altitude, as ATC considerations will permit.

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EN-ROUTE 72. c A typical drag curve is illustrated at the right. From this it can be seen that drag decreases with increasing speed up to V ,, then increases at higher speeds. For all aircraft types, unless restricted by high speed buffet or structural strength limitations, the maximum steady speed is that at which the thrust available is just equal to the drag force. Maximum speed therefore requires the use of maximum thrust in order to accelerate the aircraft until increasing drag is equal to thrust. VIMD Maximum speed therefore requires maximumthrust incurs Maximum speed speed maximum drag. It should be noted however that had this question referred to aircraft configuration, then maximum speed would require maximum thrust and minimum drag configuration. This would be with flaps, spoilers and landing gear retracted.

EN-ROUTE 73. a. Although for any given aircraft configuration and load factor, the high speed buffet boundary changes only with altitude and temperature, the low speed buffet boundary is sensitive to changes in aircraft weight and C of G position. The effect of increasing weight is to increase the low speed stalling speed and hence the buffet boundary. Forward movement of the C of G increases the nose down pitching moment of the aircraft. To maintain level flight this nose down moment must be balanced by a tail down force from the tailplane. For level flight, the sum of upward forces must equal the sum of downward forces, so the lift produced by the wings must be increased to equal the sum of the aircraft weight and tailplane down force. This has the effect of increasing the low speed stalling speed. The resulting increase in the low speed buffet boundary, causes the buffet boundaries to converge. The effect of options c and d is to increase the high speed buffet margin, causing the boundaries to diverge.

EN-ROUTE 74. d. Increasing bank angle causes load factor to increase, making the effective weight of the aircraft greater. This increases stalling speed, thereby causing the buffet boundaries to converge. It should also be noted that rearward movement of the C of P has the same effect as forward movement of the C of G described in answer en-route 73 above, but the effect is likely to be less than that of increasing bank angle..

EN-ROUTE 75. C . The buffet margins are the speed ranges available between the actual speed of an aircraft and the speeds corresponding to the boundaries at which low speed and high speed buffet commence. So the margins decrease if the boundaries converge and increase if the boundaries diverge. As altitude increases, the TAS equating to the low speed buffet boundary increases, whilst that equating to the high speed buffet boundary decreases. This causes the two boundaries to converge. The overall effect is that the margin from any given aircraft speed to the low and high speed buffet boundaries both decrease as altitude increases.

EN-ROUTE 76. a. The maximum operating speed of a modern sub-sonic jet aircraft is limited by the effects of air loads on the structure and the buffeting and high speed stall that occl_rrat transonic speeds. These limiting conditions ,, ,V , is the maximum allowable CAS. M, is the limiting mach number. The air are termed VMo and.M loads acting on an aircraft are proportional to W pV2 where r is air density and V is TAS. Because air density reduces with increasing altitude, the air loads at any given CAS are greatest at low altitudes. VMo is therefore the limiting factor at low altitudes.

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EN-ROUTE 77. b. The maximum operating speed of a modern Crossover altitude sub-sonic jet aircraft is limited by the effects of air loads on the structure and the buffeting and high speed stall that occur at transonic speeds. These limiting conditions MMo VMoare termed VMoand.M ,, ,V , is the maximum allowable CAS. ,M , is the limiting mach number. Because the local speed of sound decreases with increasing ALT altitude, climbing at a constant mach number results in CAS decreasing. The CAS equating to MMotherefore decreases with increasing altitude. At a certain altitude, termed the crossover altitude VMoand MMorepresent the same value of CAS. At lower altitudes,V , is the limiting factor and at higher altitudes MMo is the limiting factor. The overall effect is illustrated in the diagram at the right.

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EN-ROUTE 78. b. The local speed of sound decreases with decreasing air temperature, as altitude increases up to 36000 feet, Climbing at constant mach number therefore causes IAS to decrease. But lift is proportional to IAS multiplied by C, and C, is proportional to angle of attack. So if weight remains constant in a constant mach climb, the angle of attack must be increased to maintain the balance between lift and weight.

EN-ROUTE 79, a. At the absolute ceiling there is only one speed at which an aircraft can maintain straight and level flight. , For both jet and propeller aircraft this speed is V., For both aircraft types, the CAS value of V, and ,V remain constant at all altitudes. In the case of jet aircraft V, is.,V, For propeller aircraft V, is less than ,V , and because the two remain constant with changes in attitude, they do not converge at the absolute , and propeller aircraft must fly at ceiling. At their absolute ceilings jet aircraft must therefore fly at ,V VxpRop,in order to maintain straight and level flight. It should be noted that this means that a propeller , at its absolute ceiling, because it would have insufficient power available to aircraft cannot fly at ,V do so.

EN-ROUTE 80. b. The buffet margins are the speed ranges available between the actual speed of an aircraft and the speeds corresponding to the boundaries at which low speed and high speed buffet commence. So the margins decrease if the boundaries converge. As altitude increases the TAS equating to the low speed buffet boundary increases, whilst that equating to the high speed buffet boundary decreases. This causes the two boundaries to converge. The overall effect is that the margin from any given aircraft speed to the low and high speed buffet boundaries both decrease as altitude increases. But increasing weight increase the stalling speed and hence increases the low speed boundary at any given altitude. Increasing weight also increases the required angle of attack at any speed. This increases the acceleration of air over the wings, and hence decreases the value of the critical mach number, McRIT.This decrease in McRITdecreases the high speed buffet boundary at any given altitude. So increasing weight moves the low and high speed buffet boundaries closer together, decreasing the buffet margins.

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EN-ROUTE 81. C. The lift generated by an aircraft is equal to CL%pV2S,where: C, is the coefficient of lift. p is the air density. V is the true airspeed (TAS). For straight and level flight, lift must be equal to weight. So as speed increases at any given weight, the CL must be reduced to offset the increasing V2, such that lift remains constant. But CL is proportional to angle of attack, so as speed increases, the angle of attack must decrease. The best 1ift:drag ratio occurs at the and jet aircraft maximum range speed is approximately 1.32 .V ,,, So the minimum drag speed (V,), angle of attack of a jet aircraft at maximum range speed must be lower than that for best 1ift:drag ratio.

EN-ROUTE 82. C . As a jet aircraft cruises, its weight gradually decreases due to the burning of fuel. In order to maintain constant altitude it will therefore be necessary to decrease the lift force to match the decreasing weight. This is achieved by decreasing angle of attack. But this also decreases drag, so if thrust were to remain constant, the aircraft would accelerate. Thrust must therefore be gradually decreased in order to maintain constant speed (IAS).

EN-ROUTE 83. C . % gradient = (ROD Iground speed) x 100%. For small angles of climb or descent in still air, the flight path is approximately horizontal so % gradient is approximately equal to the (RODITAS) x 100%. The 5% gradient specified in this question represents a small descent angle so this problem can be solved using the following equation: % gradient = (RODITAS) x 100% This can be rearranged to give ROD = TAS x % gradient. So ROD = 140 Kts x 5% which gives ROD = 7 Kts. But the options are all specified in fpm, so 7 Kts must be converted into fpm. Although 'I Kt = 101.3 fpm, an approximation of 1 Kt'= 100 fpm is frequently used. Using 1 Kt = 101.3 fpm would give ROD = 709.1 fpm, whilst using 1 Kt = 100 fpm would give 700 fpm, which is option c in this question.

EN-ROUTE 84. b. Ground speed when descending in still air can be calculated using the following equation: Still air ground speed = TAS x Cosine of the descent gradient. But for very small gradients the Cosine is very close to one, so for small gradients ground speed is approximately equal to TAS. When descending at small % gradients into a headwind, the groundspeed i s reduced by an amount approximately equal to the speed of the headwind. So when descending into a headwind the groundspeed is approximately equal to the TAS minus the headwind. So with in this question the approximate ground speed = 140 Kts - 20 kts which is 120-Kts. Comparing this approximation with the options reveals that option b 119.8 Kts is the most appropriate.

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It should be noted that had the options been closer together (ie 117.8, 118.8, 119.8 120.8) or the gradient very large, then it would have been necessary to use the more accurate figure taking into account the actual values of the Cosine of the gradient.

EN-ROUTE 85. d. % gradient = (ROD Iground speed) x 100%. Ground speeci when descending in still air can be calculated using the following equation: Still air ground speed = TAS x Cosine of the descent gradient. But for very small gradients the Cosine is very close to one, so for small gradients ground speed is approximately equal to TAS. So for such gradients the % gradient is approximately equal to the RODITAS x 100%. So in this question the approximated ground speed in still air = 140 Kts. And gradient = 5% = (ROD1140 Kts) x 100%. This can be rearranged to give ROD = 5% x 140 Kts / 100%. ... ..Equation 1 gut when descending into a headwind the groundspeed is reduced by an amount equal to the speed of the headwind. So when descending into a headwind the groundspeed is approximately equal to the TAS minus the headwind. So in this case groundspeed in the headwind is approximately = 140 Kts - 20 Kts = 120 kts. And new gradient = ROD I120 Kts. Which gives ROD = New gradient x 120 kts I100%............ Equation 2. But headwinds do not affect the ROD?so ROD is the same in equations 1 and 2 above. So the two equations are equal. So New gradient x 120 Kts 1100% = 5% x 140 Kts I100% Simplifying and rearranging this equation gives New gradient = 5% x 1401120 which = 5.83% which is option d.

EN-ROUTE 86. a. The requirements governing this matter are specified in JAR OPS 1.540 (b) which states that, "The altitude at which the aircraft may be assumed to be cruising may not exceed that at which the ROC is 300 fpm with all engines operating at maximum continuous power."

EN-ROUTE 87. b. CAP 698 (March 2000 edition) page 27 paragraph 3.1, states that "All obstacles in the accountability area must be cleared by a vertical interval of 50 ft."

EN-ROUTE 88. b. In order to achieve maximum efficiency commercial aircraft cruise at an altitude at which the power produced with all engines operating at their maximum efficiency RPM, will be just sufficient to maintain cruising speed. This RPM is very close to the maximum continuous power setting. The failure of an engine in the cruise will therefore cause power available to be less than power required and so the aircraft will loose altitude. The rate of drift down will be equal to the power deficiency divided by the mass, so the greater the power deficiency the greater will be the drift down rate. Increasing the number of engines will decrease the proportion of total power that is produced by each. This in turn will reduce the power deficiency in the case of a single engine failure. The failure of one of two engines

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results in a 50% power reduction,whereas the failure of one of four engines results in only a 25% reduction. Increasing the number of engines therefore decreases the drift down rate in the event of engine failure.

EN-ROUTE 89. b. In order to achieve maximum efficiency, commercial aircraft cruise at an altitude at which the power produced with all engines operating at their maximum efficiency RPM, will be just sufficient to maintain cruising speed. This RPM is very close to the maximum continuous power setting. The failure of an engine in the cruise will therefore cause power available to be less than power required and so the aircraft will loose altitude. The rate of drift down will be equal to the power deficiency divided by the mass, so the smaller the mass of the aircraft, the lower will be the rate of descent in the drift down. As altitude decreases, the increasing air density will increase power available and decrease the power required. This will gradually reduce the power deficiency and hence the drift down rate. The drift down will continue to the stabilising altitude, at which the power available is equal to the power required. Because decreasing mass decreases power required at any altitude it will increase the stabilising altitude. Reducing mass by dumping fuel will therefore decrease drift down rate and increase stabilising altitude. It should be noted that this answer is based upon the assumption that drift down occurs at the best drift down speed, rather than at the best glide speed VIMD.The rate of descent in a glide is not related to aircraft mass.

EN-ROUTE 90. b. JAR OPS 1.500 (d) and JAR OPS 1.505 (b) specify that "This semi-width is increased to 10 nm, if navigational accuracy is unable to satisfy a 95% containment level."

EN-ROUTE 91. d. JAR OPS 1.500 (c) specifies that "The drift down path from cruising altitude to the aerodrome of intended landing must clear all terrain and obstacles by a minimum vertical interval of 2000 ft."

EN-ROUTE 92. d. JAR OPS 1.500 (a) specifies that "The net flight path must have a positive gradient at 1500 ft above the aerodrome at the landing is assumed to be made."

EN-ROUTE 93. b. JAR OPS 1.500 (b) specifies that "The net flight path must have a positive gradient at 1000 ft above all terrain and obstructions in the obstacle accountability area."

EN-ROUTE 94. a. As altitude increases, the power available decreases and the power required increases. The maximum rate of climb at any altitude is equal to excess power (power available minus power required), divided by mass. The absolute ceiling for all aircraft types is that at which power available is equal to power required. Under these circumstances maximum ROC is zero. It should be noted that the gross ceiling is that at which ROC = 500 fpm for jets and 100 fpm for piston aircraft. And the net ceiling is that at which ROC = 750 fpm for jets and 150 fpm for piston aircraft. None of the options in this question therefore equate to the gross or net ceiling for any aircraft types.

EN-ROUTE 95. a. A number of factors limit the maximum altitude at which an aircraft can be operated. The aerodynamic ceiling is the altitude at which the low speed and high speed buffet boundaries cross, Flight at this altitude

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is possible only at one speed and flight at any speed at any higher altitude will induce buffet. The absolute ceiling is that at which the power available is equal to the power required and maximum possible rate of climb is zero. It is therefore the maximum altitude that can be achieved using all of the available power. Steady flight above this altitude is not possible because insufficient power is available to sustain it. A further potential limitation is the maximum altitude at which the cabin structure can sustain the loads exerted upon it by internal pressurisation. But if aircraft were to operate at any of these altitudes there would be a constant danger of buffet, structural failure or loss of altitude. In order to ensure that flights can be conducted safely, a lower maximum operating altitude is specified for normal operations. At this altitude an aircraft can attain a load factor of 1.3 without buffet, has sufficient structural strength to prevent failure and has sufficient power to maintain arrange of speeds.

EN-ROUTE 96. a. This question refers to a class A aircraft. JAR OPS 1.500 (c) specifies that for class A aircraft, "The drift down path from cruising altitude to the aerodrome of intended landing must clear all terrain and obstacles by a minimum vertical interval of 2000 ft." It should be noted that option b is incorrect because gradient will never be positive in a drift down. And options c and d are incorrect because net performance is always less than gross.

EN-ROUTE 97. a. IWaximum range is achieved by flying such that the ratio of ground speed to fuel consumption is maximised. The effect of a headwind would be to reduce the ground speed at any given TAS, thereby reducing range. Although increasing TAS in a headwind will regain some of the lost groundspeed, it will also increase the power required and hence fuel consumption rate. The still air range cannot therefore be fully recovered by increasing TAS in a headwind. It is however possible to regain some of the lost range by increasing TAS by an amount somewhat less than the speed of the headwind. It should be noted that this method is applicable only to low performance aircraft. This is because cruising speed for high performance aircraft is close to McRITand any increase in TAS will cause a disproportionate increase in drag.

EN-ROUTE 98. C. The stratosphere is that layer of the atmosphere extending from 160000 ft down to the tropopause at 36000 ft. Ambient temperature gradually decreases with decreasing altitude throughout the upper part of the stratosphere down to 85000 ft, then remains constant down to 36000 ft. The local speed of sound (LSS) is proportional to temperature. So when descending through the stratosphere down to 85000 ft temperature and LSS decrease. And when descending in the stratosphere between 85000 ft and 36000 ft the temperature and LSS remain constant. Mach number is the ratio of TAS:LSS. So descending at constant mach number from the top of the stratosphere to 85000 ft causes TAS to decrease and from 85000 ft to 36000 ft causes TAS to remain constant. But few if any current commercizl aircraft types operate at or above 85000 ft so option c (TAS remains constant) is the most appropriate for this question.

EN-ROU'TE 99. b. Mach number is the ratio of TAS:LSS. LSS is proportional to temperature, which in turn is proportional to altitude. So cruising at constant mach number and constant flight level (FL) means cruising at constant TAS. Specific air range (SAR) is equal to the distance flown for each unit of fuel burned. To achieve s e that they achieve best TAS:drag ratio and minimum specific fuel maximum SAR, aircraft must c r ~ ~ i such consumption (SFC). Best TAS:drag ratio occurs within a limited range of C, TAS and altitude. Minimum

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En-Route

SFC occurs within a limited engine RPM range. If an aircraft is cruising at its optimum altitude (giving maximum SAR), any increase of decrease in any of the above factors will decrease SAR. As flight progresses, the burning of fuel will reduce tne mass of the aircraft, so lift must be gradually reduced to maintain constant FL. This requires a reduction in TAS, C,, or both. Each of these changes will require a change in RPM which will reduce SAR. Of the options offered in this question, only option b (C, must decrease) is true.

EN-ROUTE 100. b. Maximum possible range in a jet aircraft is achieved at the altitude at which the drag at 1.32 V ,,, is equal to the thrust and maximum efficiency engine RPM (within the 85% to 95% range). But drag is proportional to lift, so as the burning of fuel gradually reduces aircraft weight, lift must be reduced to maintain constant altitude. But this would also reduce drag, causing the aircraft to accelerate beyond its optimum speed. If however RPM is reduced such the thrust matches the reducing drag in order to maintain speed, then the RPM will move out of its optimum range. The overall effect of these factors is that optimum cruising altitude increases as flight progresses. If flight continues at constant altitude, then efficiency and hence range will gradually worsen. Maximum range is achieved by employing a cruise climb technique by allowing the aircraft to drift steadily upwards at constant mach number, angle of attack and RPM as mass decreases, such that it is always at the optimum altitude. This gradual increase in altitude is however generally unacceptable for ATC purposes, which requires aircraft to remain at their authorised altitudes. Operators therefore generally employ the slightly less efficient technique of step climbing. In this technique ,,, the aircraft attains an initial altitude approximately 1000 ft above its optimum. It ther; cruises at 1.32 ,V maintaining constant altitude by gradually reducing RPM until it is 3000 ft below the new optimum altitude. The aircraft is then climbed to 1000 ft above the new optimum, and the process is repeated. In this way the aircraft is kept at or as close to its optimum altitude, as ATC considerations will permit. Comparing the three methods described above, range at constant altitude is tess than in a step climb, which is less than in a cruise climb. The best option is therefore b, ( (1) c (2) c (3) ).

EN-ROUTE 101. a. In straight and level constant speed flight, thrust required equals drag and lift equals weight. Drag and lift can be calculated using the following formulae:

D = C, K pV2S and L = C, K pV2S If weight is reduced whilst maintaining constant speed, the lift must be reduced to maintain level flight. If this is achieved by reducing speed, the effect will also inciude a change in the drag force. The fact that the aircraft in this question is cruising, indicates that its speed is greater than .,V, Decreasing speed will therefore reduce drag. This in turn will reduce the thrust required to maintain the lower IAS.

EN-ROUTE 102. a. The maximum operating altitude of a JAR certificated passenger aircraft is that altitude at which it can just attain a positive load factor of 1.3 before the onset of buffet. Load factors greater than 1.3 when flying at this altitude will cause buffeting.

LANDING

LAND I.c. The LDR is the landing distance required. It is proportional to the horizontal distance from the point at which a landing aircraft reaches screen height to the point at which it is brought to a halt. This distance is then multiplied by a factor which is determined by aircraft class and type, to produce the LDR. LDR is proportional to landing speed, such that reducing landing speed reduces LDR. Increasing flap angle decreases stalling speed, which in turn decreases landing speed and hence decreases LDR. But to carry out a go-around the aircraft must climb to regain height. Go-around performance is therefore dependent upon climb performance. Climb performance is in turn dependent upon excess thrust and excess power. But increasing flap angle increases drag, which increases the power required. The increased drag decreases excess thrust, whilst the increased power required decreases excess power. So increasing flap angle decreases climb performance and hence decreases go-around performance. The overall effect of increasing landing flap angle from 35 degrees to 45 degrees is therefore that LDR is decreased and go-around performance is degraded.

LAND 2. b. The LDR is the landing distance required. It is proportional to the horizontal distance from the threshold to the point at which the aircraft is brought to a halt. This definition is based on the assumption that the aircraft reaches screen height at the threshold. This distance is then multiplied by a factor which is determined by aircraft class and type, to produce the LDR. LDR is proportional to landing speed, such that increasing landing speed increases LDR. Landing speed is proportional to stalling speed, which increases with increasing aircraft mass. So increasing mass increases LDR. The ratio of TAS to IAS increases with decreasing air density. But stalling speeds and landing speeds are related to IAS. This means that as density decreases, th6 TAS equating to any given landing speed increases. But the distance required to stop the aircraft after touch down is related to it ground speed which is proportional to TAS. So as density decreases, the ground speed and hence LDR increase. If the runway has a downhill slope , the weight of the aircraft will tend to accelerate it down the slope. This increases the distance required to bring it to a halt. So LDR will be greatest with a high mass, low density and downhill runway slope.

LAND 3. b. Landing distance will not be affected unless the depth of snow is sufficient to make the runway either wet or contaminated. The specific gravity of dry snow is less than 0.35, so 5 mm of such snow produces a water equivalent depth of 1.75 mm or less. Although this is insufficient to qualify as contamination, it does qualify as a wet runway. So 5 mm of dry snow will increase landing distances compared to a dry runway.

LAND 4. b. The landing distance available (LDA) must be at least equal to the landing distance required (LDR). But landing distance required decreases with decreasing aircraft mass. As a flight progresses, the burning of fuel reduces aircraft mass and hence reduces landing distance required. The greatest landing distance

424

Landing

will be required if the flight is aborted shortly after take-off and the aircraft returns to the departure field. So the landing distance available at the destination field can be less than that at the departure field.

LAND 5. a. JAR OPS 1.520 states that if meteorological conditions actual report or forecast for the time of arrival indicates that the runway will be wet, then the dry runway LDR must be multiplied by a factor of 115%. This constitutes a 15% increase.

LAND 6. c. JAR OPS 1.515 states that the dry runway LDR for a class A turbojet must not exceed 60% of the LDA.

LAND 7. a. JAR OPS 1.515 states that the dry runway the landing distance used by a class A turbojet must not exceed 60% of the LDA. The landing distance must therefore be multiplied by 1.67 to calculate the LDR. This is the minimum acceptable LDA.

LAND 8. c. JAR OPS 1.515 states that the dry runway landing distance used by a class A turbo-prop aircraft must not exceed 70% of the LDA. This means that the landing distance used mclst be multiplked by a safety factor of 143% to calculate the LDR. This is the minimum acceptable LDA.

LAND 9. a. JAR OPS 1.515 states that the dry runway landing distance used by a class A turbo-prop aircraft must not exceed 70% of the LDA. This means that the landing distance must be multiplied by a safety factor of 143% to calculate the LDR. This is the minimum acceptable LDA.

LAND 10. d. JAR OPS 1.515 states that the dry runway landing distance used by a class A turbo-prop aircraft must not exceed 70% of the LDA. This means that a safety margin of 30% of LDA is required over the distance that will actually be used.

LAND 11. b. The coefficient of friction of a grass runway is considerably lower than that of concrete or asphalt surfaces. This reduces the friction between the tyres and the surface, and hence increases the LDR. Although very long grass will increase rolling resistance, this effect is much less significant than the reduction in friction.

LAND 12. c. For smali % gradients in still air the ground a rsasonable approximation can be calculated using the following equation:

% Gradient = (ROD ITAS ) x 100% Which can be rearranged to give ROD = (TAS x % gradient) I100% But both TAS and ROD must be in the same units. This can be achieved by muliiplying the TAS in Kts by 100 fpm/Kt. In this case we have ROD = (150 Kts x 100 fpmIKt x -5%) I100% = -750 fpm This problem can now be solved using a ROD of -750 fprn in the following equation:

Landing

425

% Gradient in 20 Kts headwind = (ROD lground speed ) 1100% Where, for small % gradients the ground speed = TAS + tailwind In this case

% gradient in 20 Kt headwind = (-750 fpm /(I50 + 20 Kts x 100 fpmlKt)) / 100% Which is % gradient = (750113000) 1 100% = -4.44% This result illustrates the fact that a tailwind decreases descent gradient.

LAND 13. c. Increasing ambient temperature decreases air density. This increases the TAS at any given IAS and decreases the thrust produced by the engines. LDR is proportional to ground speed, which is in turn proportional to TAS. But landing speed is related to IAS. So as temperature increases, the TAS equating to landing speed increases, causing the LDR to increase. The weight, altitude, temperature (WAT) limit is related to the ratio of thrust available and aircraft mass. The WAT limit represents the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient after taking-off in any given corr~binationof altitude and ambient temperature. By decreasing the thrust available, an increase in ambient temperature will decrease the WAT limited take-off mass. So increasing ambient temperature will increase LDR and decrease WAT limit?

LAND 14. a. The coefficient of friction of a grass runway is considerably lower than that of concrete or asphalt surfaces. This reduces the friction between the tyres and the surface, and hence increases the LDR. But LDA must be at least equal to LDR, so a grass runway will reduce the maximum mass that can be landed using any given runway LDA. The LDR is related to aircraft ground speed, such that increasing ground speed increases LDR. A tailwind increases ground speed at any given IAS, and hence increases LDR. A tailwind will therefore reduce the maximum mass that can be landed using any given LDA. The most limiting condition will therefore be when landing on a grass runway with a strong tailwind.

LAND 15. b. Decreasing ambient temperature increases air density. This decreases the TAS at any given IAS and increases the thrust produced by the engines. LDR is proportional to ground speed, which is in turn proportional to TAS. But landing speed is related to IAS. So as temperature decreases, the TAS equating to landing speed decreases, causing the LDR to decrease. The weight, altitude, temperature (WAT) iimit is related io the ratio of thrust available and aircraft mass. The WAT limit represents the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient after taking-off in any given combination of altitude and ambient temperature. By increasing the thrust available, a decrease in ambient temperature will increase the WAT limited take-off mass. Decreasing ambient temperature will decrease LDR and increase WAT limit?

LAND 16. b. JAR regulations defines the following speeds: a. VAT,is the speed to be achieved at screen height when landing with 1 engine inoperative. b. VAT, is the speed to be achieved at screen height when landing with all engines operating,

426

Landing

c. ,,V , is the minimum speed at which it is possible to maintain control on the ground, in the take-off configuration, following a critical engine failure. , is the minimum speed at which it is possible to maintain control in the air, in the take-off d. ,,V configuration, following a critical engine failure.

LAND 17. d. The LDR is the landing distance required. It is proportionai to the horizontal distance from the point at which a landing aircraft reaches screen height to the point at which it is brought to a halt. This distance is then multiplied by a factor which is determined by aircraft class and type, to produce the LDR. LDR is proportional to landing speed such that reducing landing speed reduces LDR. Increasingflap angle decreases stalling speed, which in turn decreases landing speed and hence decreases LDR. But to carry out a go-around the aircraft must climb to regain height. Go-around performance is therefore dependent upon climb performance. Climb performance is in turn dependent upon excess thrust and excess power. But increasing flap angle increases drag, which increases the power required. The increased drag decreases excess thrust, whilst the increased power required decreases excess power. So increasing flap angle decreases climb performance and hence decreases go-around performance. The overall effect of increasing flap angle is therefore that LDR is decreased and go-around perforniance will be degraded.

LAND 18. c. 0.5 mm of wet snow is not sufficient to affect performance, so a runway in this condition is considered to be dry. LDR will not therefore be affected.

LAND 19. b. JAR OPS 1.520 states that for a class A turbo-prop aircraft the landing distance used must not exceed 61% of the LDA. This means that if the LDA is multiplied by a factor of 0.61, the result will be the maximum landing distance that may be used. The wording of this question is somewhat dubious in that it asks by what the landing distance must be factorised. It should ideally have asked "what is the maximum % of the LDA that the landing distance used may be ...." Given the option available however, only option b, 0.61 can be correct.

LAND 20. c. When there is a danger of aquaplaning it is necessary to achieve a positive contact between the tyres and surface as soon as possible. Also, because breaking effectiveness is likely to be reduced, it is necessary to apply the brakes as soon as possible. In many cases the aquaplaning speed Vii, is very low so waiting until this speed is reached before applying the brakes would take up too much of the runway. A

LAND 21. a. The time at which a decision might be made to go to an alternative rather than a main destination cannot be predicted. It is not therefore possible to predict that the landing mass at the alternative field will differ from that at the main destination. The landing distance required at the alternative field is therefore the same as that at the main destination, provided all other factors such as surface type, slope, condition and winds are the same.

LAND 22. b. Landing distance will not be affected unless the depth of snow is sufficient to make the runway either wet or contaminated. The specific gravity of dry snow is less than 0.35, so 5 mm of such snow produces a water equivalent depth of 1.75 mm or less. Although this is insufficient to qualify as contamination, it

Landing

427

does qualify as a wet runway. So 5 rnm of dry snow will increase landing distances compared to a dry runway.

LAND 23. c. JAR OPS 1.520 states that the landing distance used by a class a turbojet using a wet runway must not be more than 52% of the LDA. So the maximum allowable landing distance used can be calculated by multiplying the LDA by a factor of 0.52.

LAND 24. a. JAR OPS 1.520 states that the landing distance used by a turbojet aircraft landing on a wet runway must not be more than 52% of the LDA. So minimum acceptable LDA must be 1.92 of the landing distance used (LD). So minimum acceptable LDA is found by increasing Ld by 92%

LAND 25. b. ,,V , is the maximum IAS at which the brakes of an aircraft are capable of bringing it to a stop, within the remaining ASDA, following a decision to abort a take-off. The ability of the brakes to stop an aircraft is related to the ground speed rather than to the IAS. But increasing tailwind velocity will increase ground speed at any given IAS. So increasing tailwind decreases VMBE.

Climb and approach gradients are equal to the RODIground speed. The take-off climb and approach speeds are specified in terms of IAS. But a tailwind increases ground speed at any given IAS. So increasing tailwind velocity decreases both climb and approach gradients. In the event of a decision to reject a landing it is necessary for the aircraft to regain height and go-around. The climb limited landing mass is the maximum mass at which such an aircraft can achieve the minimum acceptable climb gradient. The field length limited landing mass is the maximum mass at which an aircraft can land safely given the prevailing runway length, wind conditions, flap position, and pressure altitude. Although a headwind or tailwind might be expected to affect both the climb limited mass and field length limited mass, the effect on the latter is the greater of the two. So wind effects are only included in the field length limited mass calculations. This is illustrated in the relevant CAP 698 graphs, where wind is included in the field length limit mass graph, but not in the climb limited mass graph.

LAND 26. a. A wet runway will decrease the coefficient of friction between the tyres and the surface. This in turn will increase the distance required to bring a landing aircraft to a halt. JAR OPS therefore requires that the LDR for a turbojet landing on a dry surface is increased by 15% when the surface becomes wet.

LAND 27. b. The landing distance available (LDA) is the distance from the threshold to the end of the runway. It is not therefore affected by whether or not the surface becomes damp, wet of contaminated. The LDA remains constant.

LAND 28. c. A landing aircraft must cross the threshold at VAT.This is 1.3 V.,

So if V, is 120Kts, VATwill be 156 Kts.

LAND 29. b. JAR OPS 1.515 states that the dry runway LDR for a class A turbojet must not exceed 60% of the LDA. This means that a safety margin of 40% of LDA over LDR is required.

428

Landing

LAND 30. d. If an approach to land is aborted, the aircraft must climb to regain height in order to conduct a safe goaround. The purpose of specifying an approach climb limit speed is to ensure that such an aircraft has adequate climb performance to achieve at least the minimum acceptable climb gradient with one engine out in a go-around situation. This will not however guarantee obstacle clearance, as this depends upon the location and height of the obstacles.

LAND 31. b. Although a headwind will not affect the gradient at which an aircraft flies through the air, it will cause the mass of air through which it is flying, to be pushed in the opposite direction to flight. This will reduce the ground speed, whilst leaving rate of descent unchanged. But gradient is rate of descent divided by ground speed, so a headwind will increase descent gradient. So if an aircraft established in an approach to land, suddenly encounters a headwind, its descent gradient will increase. In order to regain the correct gradient it will be necessary to increase TAS. This will increase rate of descent and ground speed. Provided TAS is increased by the appropriate amount, the correct gradient is restored.

LAND 32. d. The reason for specifying a minimum acceptable landing climb gradient is to ensure that an aircraft can regain height to conduct a go-around should the landing be aborted. JAR OPS 1.525 defines the requirements for the landing climb gradient. For class B aircraft the minimum acceptable value is 2.5% with all engines operating, and 0.75% with one engine inoperative.

LAND 33. c. The reason for specifying a minimum acceptable landing climb gradient is to ensure that an aircraft can regain height to conduct a go-around should the landing be aborted. JAR OPS 1.525 defines the requirements for the landing climb gradient. For class B aircraft the minimum acceptabie value is 2.5% with all engines operating, and 0.75% with one engine inoperative.

LAND 34. d. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.550 specifies that this must not be more than 70% of the LDA for a class B aircraft on a hard dry surface.

LAND 35. b. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.550 specifies that this must not be more than 70% of the LDA for a class B aircraft on a hard dry surface. This means that LDA must be equal to the LD plus a safety factor of 30% of the LD.

LAND 36. d. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.550 specifies that this must not be more than 70% of the LDA for a class B aircraft on a hard dry surface. This type of problem can be solved using the inverse function of a pocket calculator. That is to say, if maximum LD is 0.70 of LDA, then minimum LDA is 1/0.70, or 1.428 of LD.

Landing

429

LAND 37. c. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.550 specifies that this must not be more than 58.3% of the LDA for a class B aircraft on a dry grass.

LAND 38. b. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.550 specifies that this must not be more than 58.3% of the LDA for a class B aircraft on a dry grass. This type of problem can be solved using the inverse function of a pocket calculator. That is to say, if maximum LD is 0.583 of LDA, then minimum LDA is 110.583, or 1.715 of LD. So the minimum acceptable LDA must be equal LD plus a further safety factor of 71.5% of the LD.

LAND 39. a. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.550 specifies that this must not be more than 58.3% of the LDA for a class B aircraft on a dry grass. This type of problem can be solved using the inverse function of a pocket calculator. That is to say, if maximum LD is 0.583 of LDA, then minimum LDA is 110.583, or 1.715 of LD.

LAND 40. b. The LDR is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt, multiplied by a various factors to take account of variables. JAR OPS 1.555 specifies that for a class B aircraft landing on a wet hard surface, the LDR must be 115% of that for a dry hard surface.

LAND 41. b. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.555 specifies that this must not be more than 50.7% of the LDA for a class B aircraft on wet grass.

LAND 42. a. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought n of the LDA for a class B aircraft to a hait. JAR OPS 1.555 specifies that this must not be m ~ r e ~ t h a50.7% on wet grass. This type of problem can be solved using the inverse function of a pocket calculator. That is to say, if maximum LD is 0.507 of LDA, then minimum LDA is 110.507, or 1.973 of LD. So the minimum acceptable LDA is equal to the LD plus a safety factor of 97.3% of the LD.

LAND 43. d. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.555 specifies that this must not be more than 50.7% of the LDA for a class B aircraft on wet grass. This type of problem can be solved using the inverse function of a pocket calculator. That is to say, if maximum LD is 0.507 of LDA, then minimum LDA is 110.507, or 1.973 of LD.

LAND 44. c. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.555 specifies that this must not be more than 61% of the LDA for a class B aircraft on a wet hard surface.

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Landing

LAND 45. a. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.555 specifies that this must not be more than 61OO/ of the LDA for a class 6 aircraft on a wet hard surface. This type of problem can be solved using the inverse function of a pocket calculator. That is to say, if maximum LD is 0.61 of LDA, then rr~inimumLDA is 110.61, or 1.639 of LD. This means that the minimum acceptable LDA is equal to the LD plus a safety factor of 63.9%.

LAND 46. c. The LD is the landing distance actually used from the threshold to the point at which the aircraft is brought to a halt. JAR OPS 1.555 specifies that this must not be more than 61% of the LDA for a class B aircraft on a wet-hard surface. This type of question can be solved by using the inverse function of a pocket calculator. That is to say if maximum LD is 0.61 of LDA, then LDA is 110.61, which is 1.64 of LD.

LAND 47. a. Although long grass causes a slight increase in rolling resistance, its effect in reducing the friction between the tyres and the surface is far greater. The overall effect of this is that wet grass increases the LDR.

LAND 48. a. Thrust reverse systems enable engine thrust to be used to retard the aircraft during the landing run. This decreases the LDR at any given mass or increases the mass that can be landed on any given LDA. So if the thrust reverse system is unserviceable the LDR will inc~easeand the maximum landing mass will decrease.

LAND 49. a. For class A aircraft, JAR regulations specify a threshold speed (VAT,) of 1.3 V, at a screen height of 50 feet, with all engines operating. The minimum threshold speed with one engine inoperative (VAT,) must be not less than VAT,. VMCAis the minimum speed at which control can be maintained in the air following a critical engine failure in the take-off configuration. It has no relevance to landing speeds.

LAND 50. b. The weight, altitude, temperature (WAT) limited landing mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient in the event of a decision to abort a landing and goaround again. JAR 25.121 specifies that this minimum gradient is 2.1% for a twin engine class A aircraft.

LAND 51. c. The weight, altitude, temperature (WAT) limited landing mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient in the event of a decision to abort a landing and goaround again. JAR 25.121 specifies that this minimum gradient is 2.4% for a three engine class A aircraft.

LAND 52. d. The weight, altitude, temperature (WAT) limited landing mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient in the event of a decision to abort a landing and goaround again. JAR 25.121 specifies that this minimum gradient is 2.7% for a four engine class A aircraft.

Landing

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LAND 53. d. The weight, altitude, temperature (WAT) limited !anding mass is the maximum mass at which an aircraft can achieve the minimum acceptable climb gradient in the event of a decision to abort a landing and goaround again. JAR 25.121 specifies that this minimum gradient is 2.1% for a twin engine class A aircraft. But this is increased to 2.5% or the published aerodrome minimum gradient, whichever is greater, when conducting instrument approaches with decision heights below 200 ft.

LAND 54. c. An Airbus A320 is a twin engine class a aircraft. JAR 25.121 specifies that this minimum gradient is 2.1% for a twin engine class A aircraft. But this is increased to 2.5% or the published aerodrome minimum gradient, whichever is greater, when conducting instrument approaches with decision heights below 200 ft. JAR 25.1.510 states that if this gradient cannot be achieved, the decision height must be increased to at least 200 ft.

LAND 55. a. When gusty or turbulent conditions require an increase in VAT,the LDR must be recalculated. The new LDR can be calculated for changes in VATusing the following equation: New LDR = Old LDR x (VAT2I(VREF9 7)2) This gives New LDR = 4000 m x ( 130' 1 (110 9 7)2) Which is New LDR = 4939 m.

LAND 56. b JAA regulations define a steep approach as one in the descent gradient is 4.5O or greater.

LAND 57. c. JAR OPS 1.515 states that if an operator is unable to comply with the forecast wind requirements at the destination field. but can comply with the still air requirements, the aircraft can still be despatched provided an alternative field is available at which all of the landing requirements can be met. I

,

I I I I

I

LAND 58. c. The measured landing distance is horizontal distance between an aircraft actually attaining screen height and the point at which it comes to a full stop when landing on a hard dry surface. This distance is then multiplied by suitable factors to take account of actual conditions when calculating the landing distances required (LDR) for any given field.

LAND 59. b. JAR OPS 1.550 states that for a class b aircraft using a wet grass surface, the LD must be not more than 50.7% of the LDA. This means that the LDA must be 1.97 times the LD. Questions of this type can be solved by using the inverse function on a pocket calculator. That is to say 11.507 is 1.97.

LAND 60. c. A declared safe area is an area free of obstructions immediately before the threshold. Its maximum permitted length is 90 metres, minimum width is twice that of the runway, and between 5% upslope and 2% downslope. It is required for short landings, when the screen is considered to be at the beginning of the declared safe area.

249497952-AircraftPerformance-Keith-Williams.pdf

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