Cpl Navigation

CPL NAVIGATION

CPL Navigation CPL DOC 8 Revision 1/1/2001

FLIGHT TRAINING COLLEGE Version 7

INDEX CPL NAVIGATION 1. The Earth 2. Charts 3. Relative Velocity 4. Solar System & Time 5. Navigation Computer 6. Plotting

01 25 55 59 83 95

Annex A Annex B

137 149

Sample Exams Answers to Questions

Copyright  2001 Flight Training College of Africa All Rights Reserved. No part of this manual may be reproduced in any manner whatsoever including electronic, photographic, photocopying, facsimile, or stored in a retrieval system, without the prior permission of Flight Training College of Africa.

CPL Navigation CPL DOC 8 Revision 1/1/2001


CHAPTER 1 THE EARTH The earth is not a perfect sphere, there is a slight bulge at the Equator and a flattening at the Poles. The earth's shape is described as an oblate spheroid. The polar diameter is 6860.5 nm which is 23.2 nm shorter than the average equatorial diameter of 6883.7 nm. This gives a compression ratio of 1/2967 which for all practical purposes can be ignored. Cartographers and Inertial Navigation systems will take the true shape of the earth into account.

PARALLELS OF LATITUDE Parallels of Latitude are small circles that are parallel to the Equator. They lie in a 090 and 270 Rhumb Line direction as they cut all Meridians at 90. LATITUDE The Latitude of a point is the arc of a Meridian from the Equator to the point. It is expressed in degrees and minutes North or South of the Equator. It can be presented in the following forms. N 27:30

27:30 N

2730'N

3525'45"S

CPL NAVIGATION CPL DOC8 Revision 1/1/2001

35:25:45S


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LONGITUDE The Longitude of a point is the shorter arc of the Equator measured East or West from the Greenwich Meridian. It can be presented in the following forms. E032:15

3215' E

32:15 E

6524'W


6524'38"W

65:24:38 W


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FINDING PLACES WITH LAT/LONG Example 1 At 26:34 S / 26:16.5 E what do you find? Did you find a town?

Example 2 The lat/long of the township of Hartbeesfontein is?



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GREAT CIRCLE (GC)

A Great Circle is a circle drawn on the surface of a sphere whose centre and radius are those of the sphere itself. A Great Circle divides the sphere into two halves. The Equator is a Great Circle dividing the earth into the Northern and Southern Hemispheres. On a flat surface the shortest distance between TWO points is a straight line. On a sphere the shortest distance between two points is the shorter arc of a Great Circle drawn through the two points. To fly from Europe to the West Coast of America the shortest distance is of course a Great Circle which usually takes the least time and fuel used. A Great Circle cuts all Meridians at different angles. RHUMB LINE (RL) A Rhumb Line is a curved line drawn on the surface of the earth which cuts all Meridians at the same angle. An aircraft steering a constant heading of 065(T) with zero wind will be flying a Rhumb Line. MERIDIANS Meridians are Great semi-circles that join the North and South Poles. Every Great Circle passing through the poles forms a Meridian and its Anti-Meridian. All Meridians indicate True North or 000(T) and 180(T). As Meridians have a constant direction they are Rhumb Lines as well as Great Circles. EQUATOR The Equator cuts all Meridians at 90 providing a True East-West or 090(T) and 270(T) erection. As the Equator cuts all Meridians at 90 it is a Rhumb Line as well as a Great Circle.



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SMALL CIRCLE A Small Circle is a circle drawn on a sphere whose centre and radius are not those of the sphere itself.

DIRECTION True North True North is the direction of the Meridian passing through a position. True Direction Aircraft Heading or Track is measured clockwise from True North. It is usually expressed in degrees and decimals of a degree, e.g. 092(T) 107.25GC 265.37 RL Magnetic North Magnetic North is the direction in the horizontal plane indicated by a freely suspended magnet influenced by the earth's magnetic field only. Variation Variation is the angular difference between True North and Magnetic North

Magnetic Direction (M)

Aircraft Magnetic Heading or Magnetic Track is measured clockwise from Magnetic North, which is sometimes referred to as the Magnetic Meridian, e.g. 100(M)



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Compass North (C) Compass North is the direction indicated by the compass needle in an aircraft. Magnetic Fields in the aircraft will attract the compass needle away from Magnetic North causing Compass Deviation. Deviation The angular difference between Compass North and Magnetic North.

Deviation is Westerly when Compass North is to the West of Magnetic North Deviation is Easterly when Compass North is to the East of Magnetic North

DEVIATION EAST COMPASS LEAST

DEVIATION WEST COMPASS BEST

Heading l00(C) Dev+4e 104(M)

Heading 100(C) Dev-3w 096(M)

Deviation West is Negative (-) Deviation East is Positive (+) Deviation is a correction to Compass Heading to give Magnetic Heading



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CONVERGENCE AND CONVERSION ANGLE CONVERGENCE Meridians are Semi Great Circles joining the North and South Poles. They are parallel at the Equator. As the meridians leave the Equator either Northwards or Southwards they converge and meet at the Poles.

Convergence is defined as the angle of inclination between two selected meridians measured at a given Latitude.

Considering the two meridians shown above, one at 20W and the other at 20E. The Change of Longitude (Ch. Long) or Difference in Longitude (D Long) between the two meridians is 40. At the Equator (Latitude 0) they are parallel, the angle of convergence is 0. At the Poles (Latitude 90) they meet, and the angle of convergence is the Difference of Longitude, 40.

At any intermediate Latitude the angle of inclination between the same two meridians will between 0 and 40 depending on the Latitude. This is a sine relationship, convergence varies as Sine of the Mean Latitude. Convergence also varies as the Change of Longitude between the two meridians. The greater the Ch. Long, the greater the convergence. Convergence = Ch. Long x Sine Mean Latitude CPL NAVIGATION CPL DOC8 Revision 1/1/2001


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Example 1.

A B

Calculate the value of Convergence between A (N 45:25 E 025:36) and B (N 37:53 E042:17). N 45:25 N 37:53 N 41:39 Mean Latitude Convergence

= = = =

E 025:36 E 042:17 16:41 Change of Longitude Ch. Long 1641' 16.6833 11.0874

x Sin Mean Latitude x Sin 41 39' x Sin 41.65

NOTE Both Mean Latitude and Change of Longitude must be changed into decimal notation.

THE MERIDIANS CONVERGE TOWARDS THE NEARER POLE



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CONVERGENCE = CHANGE OF LONGITUDE x SIN MEAN LATITUDE CONVERGENCE = DIFFERENCE BETWEEN INITIAL AND FINAL GC TRACKS

Example 1

A and B are in the same hemisphere The Great Circle Track from A to B is 062 The Great Circle Track from B to A is 278 (a) In which hemisphere are A and B? (b) What is the value of Convergence between A and B?

Example 2

C and D are in the same hemisphere The Great Circle bearing of D from C is 136 (brg of D measured at C) The Great Circle bearing of C from D is 262 (brg of C measured at D) (a) In which hemisphere are C and D? (b) What is the value of Convergence between C and D?



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CONVERSION ANGLE (CA) CONVERSION ANGLE = DIFFERENCE BETWEEN GREAT CIRCLE AND RHUMB LINE Conversion Angle (CA) is used to change Great Circle bearings and tracks into Rhumb Line bearings and tracks or vice versa.

THE GREAT CIRCLE IS ALWAYS NEARER THE POLE THE RHUMB LINE IS ALWAYS NEARER THE EQUATOR

CONVERSION ANGLE

= ½ CONVERGENCE

CONVERGENCE

= TWICE CONVERSION ANGLE

CONVERGENCE

= CHANGE OF LONGITUDE x SIN MEAN LATITUDE

CONVERSION ANGLE

= ½ CHANGE OF LONGITUDE x SIN MEAN LATITUDE

CONVERSION ANGLE

= DIFFERENCE BETWEEN GREAT CIRCLE AND RHUMB LINE

CONVERGENCE

= DIFFERENCE BETWEEN INITIAL AND FINAL GREAT CIRCLES



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The Rhumb Line is a constant direction. If the Rhumb Line track from A to B is 100º, then the Rhumb Line track from B to A is 280º. You can always take the reciprocal of a Rhumb Line, NEVER A GC. Initial GC track A to B is 080° GC, initial GC track B to A is 300° GC (Conv. angle 20°) Example 3

The Great Circle bearing of A from B is 255 GC The Rhumb Line bearing of B from A is 084 RL

Example 4

The Great Circle bearing of X from Y is 072 GC The Rhumb Line bearing of Y from X is 259 RL What is the great circle bearing of Y from X?



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Example 5



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DISTANCE Kilometre (KM.) A Kilometre is 1/10 000 th. part of the average distance from the Equator to either Pole. It is generally accepted to equal 3280 feet. Statute Mile (SM) Defined in British law as 5280 feet. Nautical Mile (NM) A Nautical Mile is defined as the distance on the surface of the earth of one minute of arc at the centre of the earth. As the earth is not a perfect sphere the distance is variable. At the Equator I NM is 6046.4 feet

At the pole 1 NM -Is 6078 feet

For navigation purposes the Standard Nautical Mile is 6080 feet (South Africa and UK) ICAO 1 NM = 1852 metres or 6076.1 feet Most navigational electronic calculators use I NM = 6076.1 feet. To answer questions in the CAA examinations any of the following may be used :1 NM Conversion Factors

= 6080 feet or 1852 metres

1 Foot = 12 inches 1 Inch = 2.54 Centimetres

As one minute of arc is 1 NM, then Great Circle distance along a Meridian can be calculated. One minute of Latitude is 1 NM and 1Degree of Latitude is 60 NM. The Great Circle distance from N75:30 E065:45 to N82:15 W114:15 is:As W114:15 is the anti-meridian of E065:45 the Great Circle distance is along a Meridian over the Pole where 1 of Latitude equals 1 nm. N 75:30 to the Pole Pole to N 82:15

= = = = =

1430' change of Latitude (14=x 60 = 840 nm+30 nm) 870nm 745' change of Latitude (7 x 60 = 420nm + 45nm) 465nm + 870nm 1335 nm

CHANGE OF LONGITUDE (CH. LONG) or DEPARTURE DISTANCE Departure is the distance in Nautical Miles along a parallel of Latitude in an East-West direction. At the Equator, two meridians (5W and 5E) have a change of Longitude of 10 of arc. As the Equator is a Great Circle, 10 of arc equals 600 nautical miles. As Latitude increases, either to the North or to the South, the meridians converge, and the distance between them decreases, until they meet at the Poles where the distance between them is zero. Departure (nm) = ch long (mins) x cos mean lat: CPL NAVIGATION CPL DOC8 Revision 1/1/2001


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The departure between any 2 points is thus a function of their latitudes and the change of longitude, and the relationship is given by Where mean lat = lat A + lat B 2 E 032:45 E 021:15 11:30Ch.Long

W 067:25 Both East or West SUBTRACT E 027:30 One East & One West ADD 94:55 Ch. Long

DEPARTURE = CHANGE of LONGITUDE (in minutes) x COSINE LATITUDE Example 1

The distance from A (N 20:10 E 005:00) to B (N 20:10 W 005:00) is :-

Departure

= = = =

Ch. Long x cos Lat 10 x 60 x cos2010' 600 x cos 20.1667 563.2163 nm

Example 2

An aircraft leaves A (E 012:30) and flies along the parallel of S 29:30 in an Easterly direction. After flying 1050 nm its Longitude is :Departure = Ch. Long x cos Lat 1050nm = Ch. Long xcos2930' Ch Long = 1050 nm cos 29.5 = 1206.4 60 = 20 06' 24" Easterly +12 30' = E 032 36' 24"



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Example 3

An aircraft in the Northern Hemisphere flies around the world in an Easterly direction at an average groundspeed of 515 Kts in 14 hours. The Latitude at which the aircraft flew was :-

Departure GS 515 x 14 Hrs. 7210 21600

= =

Ch. Long cos Latitude 360 x 60 x cos Lat

=

cos Lat = 70 30’ N

DISTANCE ALONG A PARALLEL OF LATITUDE IS DEPARTURE DISTANCE ALONG A MERIDIAN IS CHANGE OF LATITUDE

As a Meridian is a Great Circle, then the arc of Change of Latitude can be converted into nautical miles. Example 4

The shortest distance from A (N 78:15 W 027:13) in B (N82:30 E 152:47) is :As E 152:47 is the anti-meridian of W 027:13, A to B is the arc of a Great Circle. N 78:15 to the North Pole North Pole to N 82:30

19° x 60 Example 5

= =

11:45 Change of Latitude 7:30 Change of Latitude _____ 19:15 Change of Latitude

= 1140nm + 15 minutes = 1155nm shortest (GC) distance A to B

An aircraft departs A (N 25:13 W017:25) and flies a track of 090°(T) at GS 360 for 1 hour 35 minutes. Then the aircraft flies a track of l80° (T) for I hour 55 minutes and arrives at position; Departure = Ch. Long x cos Latitude N 25:13 W017:25; Track 180° Change of Latitude Departure = Ch. Long x cos Latitude

Departure cos Lat = Ch. Long

GS360 x 1:35 cos 25:13 = 630 minutes of Longitude = 10°30-East of W 017:25 = W006:55 GS360 x 1:55 = 690nm =11°30 t= N 13:43 Track 180° = S Change of Latitude Old Latitude N 25:13 - 11:30 Position =

= N13:43

N 1343’ W 00655’



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RADIO BEARINGS

VHF D/F VERY HIGH FREQUENCY - DIRECTION FINDING

VDF

Major airports in South Africa have a VDF service, it is usually on the Approach frequency and will provide radio bearings to aircraft on request. The aircraft transmits on the appropriate frequency and direction finding equipment at the airport will sense the direction of the incoming radio wave. The bearing will be passed to the aircraft in Q-code form. Q CODE

QTE QDR QUJ QDM

TRUE bearing FROM the VDF station MAGNETIC bearing FROM the VDF station TRUE track TO the VDF station MAGNETIC track TO the VDF station

Take the shortest route to change one bearing to another QTE ± 180 = QUJ QDR ± 180 = QDM

VOR

QDM

± Variation

=

QUJ

QDR

± Variation

=

QTE

VOR Radials are Magnetic bearings from the VOR RMI Readings are Magnetic tracks to the VOR

= QDR = QDM

RMI BEARINGS (VOR & ADF) Usually termed RMI READING which is

QDM



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ADF BEARINGS ADF Relative bearings are measured from the Fore and Aft axis of the aircraft. ADF Relative bearings must be converted into True Bearings (QTE) before they can be plotted on a chart. RELATIVE BEARING + TRUE HEADING = QUJ  180 = QTE MAGNETIC VARIATION AT THE AIRCRAFT IS ALWAYS USED WITH ADF BEARINGS Lets demonstrate it to you.. ADF bearing Heading (T) + QUJ  QTE

095 Relative 057 152 (T) TO NDB 180 332 (T) FROM NDB

ADF bearing Heading (T) QUJ Subtract QUJ  QTE


200 Relative 318 518 360 158 (T) TO NDB 180 338 (T) FROM NDB


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QUESTIONS 1.

The Initial Great Circle track from A to B is 067 The Initial Great Circle track from B to A is 263 The Rhumb Line track from A to B is:(a) 059 (b) 075 (c) 083

2.

The Initial Great Circle track from C to D is 097 The Initial Great Circle track from D to C is 263 The Rhumb Line track from D to C is :(a) 256  (b) 262 (c) 270

3.

The Great Circle bearing of A from B is 255 The Rhumb Line bearing of B from A is 084 The Great Circle bearing of B from A is :(a) 093 (b) 096 (c) 099

4.

The Great Circle bearing of X from Y is 072 The Rhumb Line bearing of Y from X is 259 The Great Circle bearing of Y from X is :(a) 262 (b) 266 (c) 270

5.

The initial great circle track from A (S 30:45 E 045:15) to B(S 30:45 E 062:38) is (a) 085.6 (b) 094.4 (c) 098.9

6.

The initial great circle track from A (S 28:30 W 015:15) to B is 099 If A and B are on the same parallel of latitude the longitude of B is :(a) W 003:37 (b) E 012:15 (c) E 022:28



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7.

The latitude where the value of convergency is half the value of convergency at 60 N is (a) 30 00’ N (b) 27 52’ N (c) 25 39’ N

8.

The Initial Great Circle Track from A (S 27:30 E 017:45) to B (S 27:30 E 029:15)is: (a) 092.65 GC (b) 087.35 GC (c) 095.31 GC

9.

A and B are in the Northern Hemisphere. The Great Circle bearing of B from A is 068. If Conversion angle is 6 the Great Circle bearing of A from B is :(a) 242 GC (b) 254 GC (c) 260 GC

10.

A and B are in the Southern Hemisphere. The Great Circle bearing of A from B is 245 If Conversion Angle is 7 the Great Circle bearing of B from A is :(a) 079 GC (b) 065 GC (c) 051 GC

11.

A and B are in the same hemisphere. The Rhumb Line bearing of A from B is 100 The Great Circle bearing of B from A is 275 The Great Circle bearing of A from B is :(a) 105 GC (b) 100 GC (c) 095 GC

12.

A and B are in the same hemisphere. A bears 080 GC from B B bears 255 GC from A The Rhumb Line Track from B to A is :(a) 077.5 RL (b) 082.5 RL (c) 085.0 RL



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13.

A and B are in the same hemisphere. The bearing of B from A is 143 GC The bearing of A from B is 319 GC The Rhumb Line Track from A to B is :(a) 141 RL (b) 145 RL (c) 147 RL

14.

The Latitude where the Convergency between two meridians is twice the value of their Convergency at 20 N is :(a) N 42:45 (b) N 43:10 (c) N 43:16

15.

A and B are in the same hemisphere. The Great Circle bearing of A from B is 080 The Great Circle bearing of B from A is 270 The Rhumb Line Track from B to A is :(a) 075RL (b) 080RL (c) 085RL

16.

A and B are in the same hemisphere. The Great Circle bearing of A from B is 290 The Great Circle bearing of B from A is 118 The Rhumb Line bearing of A from B is :(a) 294RL (b) 298RL (c) 302RL

17.

A and B are in the same hemisphere. The bearing of A from B is 280 GC The bearing of B from A is 095 RL The Great Circle bearing of B from A is (a) 090GC (b) 100GC (c) 105GC



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18.

Two positions on the same parallel of Latitude are in the Northern Hemisphere. A at 171E and B at 173 W have a 8 angle of Convergency between them. The Initial Great Circle Track from A to B is :(a) 086GC (b) 090GC (c) 094GC

19.

Two positions on the same parallel of Latitude are in the Northern Hemisphere. A at 171 E and B at 173 W have a 8 angle of Convergency between them. The Latitude of A is :(a) 25N (b) 30N (c) 35N

20.

The position of A is S 30:00 W 010:00. Position B is on the same parallel of Latitude. The Initial Great Circle Track from A to B is 256. The Longitude of B is :(a) 56W (b) 61W (c) 66W

21.

The position of A is N 42:13 W 158:24. Position B is on the same parallel of Latitude. The Great Circle bearing of B from A is 278. The Longitude of B is :(a) E 175:13 (b) E 182:13 (c) E 177:47

22.

The distance from A (S 27:43 W 005:15) to B(S 27:43 E 018:29) is :(a) 703 nm (b) 1261 nm (c) 1452 nm

23.

An aircraft departs C (N 45:17 E 025:52) on a track of 270° (T) and arrives at D after a flight of 456 nm. The Longitude of D is :(a) E 015:04 (b) E 016:32 (c) E 017:25

24.

An aircraft in the Southern Hemisphere flies around the world in 16 hours 35 minutes at GS 478. The Latitude at which the aircraft flew was :(a) S 68:28 (b) S 68:47 (c) S 69:12



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25.

An aircraft leaves ORLANDO. FLORIDA (N 28:32 W 081 :20) at 08:20 Z, GS 375. The ETA at TENERIFE. CANARIES (N 28:32 W 016:16) is :(a) 17:29 Z (b) 17:44 Z (c) 17:57 Z

26.

An aircraft departed ABU ‘DHABI (N 24:23 E 054:44) at 06:52 Z and arrived at KARACHI (N 24:23 E 067:11) at 08:17 Z. The average groundspeed for the flight was :(a) 470 Kts (b) 480 Kts (c) 490 Kts

27.

An aircraft leaves X (N 57:42 E 030:15) on a Rhumb Line track of 270° (T) After flying 1037 nm the Longitude of the aircraft is :(a) W 002:06 (b) W 003:38 (c) W 004:29

28.

An aircraft flies 425 nm along the parallel of Latitude N 46:52. The change of Longitude is (a) 10°51’ (b) 10°36’ (c) 10°22’

29.

The length of one nautical mile is :(a) constant (b) maximum at the poles (c) maximum at the equator

30.

The shortest distance from A (N 75:39 E 123:17) to B(N 78:27 W 056:43) is :(a) 1554 nm (b) 1672 nm (c) 1739 nm

31.

An aircraft leaves X (S 34:58 E 018:24) at 06:30 Z, track 360° (T), GS 300 Kts. At 07:55 Z the aircraft turns right onto track 090° (T). The longitude of the aircraft at 09:05 is (a) E 024:30 (b) E 025:00 (c) E 025:30



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32.

The Latitude where the distance between two meridians is half the distance between the same two meridians at 25° N is :(a) N 62:08 (b) N 62:31 (c) N 63:03

33.

An aircraft leaves P (N 32:27 E 027:56) at 17:30 Z. track 270° (T), GS 390 Kts. At E 021:00 it flies due South and passes abeam of Q(N 20:20 E 021:56) at 20:12 Z. The groundspeed on the second leg was :(a) 384 Kts (b) 394 Kts (c) 404 Kts

34.

An aircraft departs X (S 27:34 W 034:15) at 09:00 Z, track 090°(T). GS 455 Kts. At W 015:00 it flies due North at GS 422 Kts. The ETA abeam of Y (S 19:35 W 013:45) is: (a) 12:23 (b) 12:33 (c) 12:43

35.

Aircraft A and aircraft B depart the same position (60ºN 150ºW). Aircraft A flies a track of 090ºT. Aircraft B flies directly north to the pole then a track of 180ºT so as to intercept aircraft A. The groundspeeds are the same for A and B. At what longitude will aircraft A and B intercept each other? (a) 90E (b) 30W (c) 30E



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CHAPTER 2 CHART PROJECTION THEORY The original problem of map making is still with us even in the 21st century, how can you represent the curved surface of the earth on a flat piece of paper without distortion??? The answer is IT CANNOT BE DONE!! It’s the same as trying to flatten out a Orange peel, it too cannot be done. Charts which are produced by conic projections are used widely in aviation – mainly because conic projections “ 1. 2. 3. 4.

preserve true shapes preserve angular relationships (called conformal or orthomorphic) have a reasonably constant scale over the whole chart show great circle as straight lines..

Lets now look at the chart projections and properties that we as pilots are interested in: ORTHOMORPHISM Orthomorphism means true shape. In theory a cartographer starts with a 'reduced earth' which is the earth reduced by the required scale. The 'reduced earth' is a true undistorted representation of the earth. Details, such as Parallels of Latitude, Meridians and topographical features are 'projected' from the reduced earth onto a cylinder (Mercator's Projection), a cone (Lambert's Projection) or a flat sheet of paper (Polar Stereographic Projection). The ideal chart would possess the following features.     

Scale, both correct and constant Bearings correct Shapes correctly shown Areas correctly shown Parallels of Latitude and Meridians will intersect at 90

Unfortunately to reproduce a spherical surface on a flat sheet of paper is impossible. Distortions will occur. Only one of the above features can be shown correctly. If shapes and areas are approximately correct to enable map reading, then slight distortions can be tolerated. Bearings and scale must be correct, but we cannot have both.



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The 1 nm square on the reduced earth is correct, the diagonal of a square is 45 and bearings are correct. The 1 nm square of the reduced earth projected onto a cylinder becomes a rectangle. Bearings are no longer correct. The scale has been expanded in the North/South direction to a greater degree than the East/West case. To overcome this problem the scale expansion North/South is reduced mathematically to equal the scale expansion East/West. The rectangle becomes a square and the diagonal is 45 Bearings are now correct. Meridians and Parallels of Latitude intersect at 90 Scale is expanded, but by the same amount in all directions over short distances. Shapes and areas are approximately correct and the chart is orthomorphic. On the Mercator, Lambert and Polar Stereographic charts the Parallels of Latitude are adjusted in the above manner. Bearings are correct but the scale is variable. SCALE Scale is the ratio of a line drawn on a chart to the corresponding distance on the surface of the earth. Statement In Words

1 inch equals 40 nm

Usually found on radio facility charts. 1 inch on the chart equals 40 nm.

Graduated Scale Line 0 10 20 30 40 50 60 70 80 90 100 1_____1_____1_____1_____!_____1_____1_____1_____1______1



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Representative Fraction 1 1000 000

or 1/1 000 000

or

1:1 000 000

 1 Unit on the chart equals 1 000 000 units on the earth  1 Centimetre on the chart equals 1 000 000 centimetres on the earth .  1 Inch on the chart equals 1 000 000 inches on the earth SCALE FACTOR Due to the inherent difficulty of presenting a spherical object (the earth) on a flat sheet of paper. there is no such thing as a constant scale chart. Scale expansion or contraction will occur. Usually scale will be correct at a certain Latitude but expands else where. For example :Mercator Chart

Scale 1:1 000 000 at the Equator What is the scale at 40N with a Scale factor of 1.3054 1 _______ 1 000 000

x Scale factor 1.3054

Scale at 40N = 1: 766 049

Example 1

A chart has a scale of 1:2 500 000. How many nautical miles are represented by 4 cm on the chart? Chart Length (CL) = ________________ Earth Distance (ED)

Scale

1 ________ 2 500 000

4 cm ______ ED

ED = 2 500 000 x 4 cms 2 500 000 x 4 cms ______________ = 53.96nm 2.54 x 12 x 6080 Example 2

Divide by 2.54 = Inches Divide by 12 = Feet Divide by 6080 = Nautical Mile;

32 centimetres on a chart represents 468 nm. The scale of the chart is : CL

Scale = ED

32 cms _________________________ 468 nm x 6080 x 12 x 2.54


=

1 _______ 2 710 282


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Example 3

The scale of a chart is 1: 3 500 000. The length of a line that represents 105 nm is :-

CL Scale = ___ ED

=

1 ________ = 3 500 000

CL __________________________ 105 nm x 6080 x 12 x 2.54

= 3 500 000 x CL = 105 nm x 6080 x 12 x 2.54 105 nm x 6080 x 12 x 2.54 CL = _____________________ 3 500 000 Example 4

= 5.56 cms

Chart A has a scale of 1:2 500 000 Chart B has a scale of 1:1 750 000

Which chart has the larger scale? 1 Chart B has the larger scale ___ 2

1  ___ 4

The smaller denominator is the larger scale (half a cake is larger than quarter of a cake)



Page 28

LAMBERT CONFORMAL CONIC CHART The Lambert's chart was developed from the Simple Conic chart. Simple Conic A cone is placed over a reduced earth so it is tangential to a selected parallel of latitude. The apex of the cone is above the pole. A light source at the centre of the reduced earth projects details onto the cone. The cone is opened to give a simple conic projection.



Page 29

The scale is correct at the parallel of tangency (45N) and expands north and south of 45N. Due to the scale expansion the chart is not suitable for navigation.

The Meridians are straight lines converging on the nearer pole and the value of convergence is constant throughout the chart. Parallels of Latitude are arcs of circles radius the Pole.



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SIMPLE CONIC CONVERGENCE When the cone is opened, 360 of Longitude is represented by the angular extent of the chart which is 254.5584. The angular extent of the chart is controlled by the latitude chosen to be the parallel of tangency. Angular extent of the chart 254.5584 ______________________________ Change of Longitude 360

= 0.7071 Constant of the Cone or 'n' factor

Two Meridians 1 apart have a convergence 0.7071 and this is called the: CHART CONVERGENCE FACTOR (CCF) Parallel of Tangency 45 Sine 45 = 0.7071 = CCF = Constant of the Cone = 'n' factor

LAMBERT CONFORMAL CONIC CHART The Lambert's chart is based on the simple conic and is produced mathematically from it. Firstly, the scale is reduced throughout the chart. Since scale on the simple conic is correct only on the parallel of tangency and expands either side, the reduction will give two Standard Parallels (SP) on which scale is correct, one on either side of the simple conic parallel of tangency which is renamed the Parallel of Origin (// 0). Further mathematical modification is applied by adjusting the radius of the parallels of latitude to produce an orthomorphic projection. The above can be shown be lowering the simple conic cone so that it cuts the earth at the two Standard Parallels instead of the original parallel of tangency of the simple conic. Lambert's Chart Properties PARALLELS OF LATITUDE

Arcs of circles, radius the Pole, unequally spaced.

MERIDIANS

Straight lines converging towards the nearer Pole

SCALE

Correct at the two Standard Parallels Expands outside the Standard Parallels Contracts between the Standard Parallels Scale variation throughout 1:1 000 000 and 1:500 000 charts is negligible and can be considered constant if the band of Latitude projected is small and the Standard Parallels are positioned according to the one sixth rule. That is one sixth of that Latitude band from the top and bottom of the chart. Charts of the North Atlantic with a scale of 1:5 600 000 have a marked scale variation and care must be taken when measuring distances.

RHUMB LINES

Curves concave to the Pole and convex to the Equator.

GREAT CIRCLES

A straight line joining two positions on the Parallel of Origin curves slightly concave to the Parallel of Origin.



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CONVERGENCE

Chart Convergence Chart Convergence Chart Convergence Chart Convergence

Constant throughout the chart Correct at the Parallel of Origin Ch. Long x sin Parallel of Origin Ch. Long x CCF (Chart Convergence Factor) Ch. Long x 'n' Ch. Long x Constant of the Cone

SHAPES and AREAS

Slight distortion

CHART FIT

Charts of the same scale and Standard Parallels will fit N/S and E/W. Charts with different SP will not fit.



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Lambert's Chart - Tracks For all practical purposes the Great Circle is a straight line.

The Rhumb Line track is parallel to the mean Great Circle track at the Mid Meridian between two positions The difference between the Great Circle and the Rhumb Line is : Chart Conversion Angle (CCA) The difference between the Initial Great Circle track and the Final Great Circle track is Chart Convergence (CC) NB: For examination purposes Unless otherwise stated in a question, the Great Circle is taken to be the straight line and Chart Convergence (CC) is used.  Where a question asks for 'the most accurate value of the Great Circle' or 'the true Great Circle' then Earth Convergence (EC) is used.  The Parallel of Origin of a Lamberts chart is mid way between the two Standard Parallels  If the Standard Parallels (SP) are 20S and 40S, then the Parallel of Origin (// 0) is 30S  If one SP is 20S and the // 0 is 30S - Then the other SP is 40S  Chart Convergence (CC) = Change of Longitude x sine Parallel of Origin  Chart Convergence (CC) = Change of Longitude x Chart Convergence Factor  Sine Parallel of Origin = Chart Convergence Factor (CCF)  If a statement regarding convergence is given :-



Page 33

e.g. a Lamberts chart has a chart convergence of 5 between the meridians of 10E and 20E) then the Parallel of Origin can be calculated (CC 5 = ch. long 10 x sin 30) and the CCF = 0.5 As convergence is proportional to the CCF, convergence between any two meridians is easily found. Lets look at some problems that can arise in the examination..

Q1.

On a North Hemisphere Lamberts chart (SP 20N & 40 N) the initial GC track from A (10E) to B (42E) is 065. The GC track at B is: CC 16

A 65

65

B

42E

10E CC

= Ch Long X sin // O = 32 X sin 30 = 16 65 + 16  = 81 GC at B Q2.

The Chart Convergency factor of a Lamberts chart is .5. The Great Circle track from C (20N 10E) to D (45N 30 W) measures 316 at C. The Rhumb Line Track from C to D is: D GC C

RL CC CC CC CCA Q3

= = = =

CH Long X CCF 40 X 0.5 20 10

Track C to D CCA Track C TO D

316 GC 10 306

The CCF of a Lambert's chart is 0.5 If one Standard Parallel (SP) is 25S then the Latitude of the other Standard Parallel is :The Parallel of Origin (// 0) is midway between the two Standard Parallels CCF 0.5 = sin//0 = 30S SP25S Parallel of Origin 30S Other SP35S



Page 34

Lambert's Chart

Plotting Radio Bearings

Radio bearings are Great Circles. Straight Lines on a Lambert's chart are Great Circles and plotting radio bearings is simple. VOR & VDF VOR and VDF bearings are determined at the station, that is measured from the Meridian of the station. VOR Radial (QDR) VOR RMI reading

Correct for VOR station Variation only and plot from VOR Meridian RMI reading is a QDM VOR Variation 180 = QTE Plot QTE from VOR Meridian (do not apply compass deviation)

VDF

QDM = Station Variation 180 = QTE QDR Station Variation = QTE QUJ ± 180 = QTE Plot QTE from VDF station Meridian

Plotting ADF Bearings A problem arises with the plotting of ADF bearings due to the bearing being measured at the aircraft's Meridian and plotted from the NDB's Meridian which differ by the amount of Convergence between the two positions.

ADF QUJ ± 180 = Bearing to plot from Aircraft Meridian paralleled through NDB ADF QUJ ± 180 ± CC = Bearing to plot from NDB Meridian



Page 35



Page 36

Scale Problems Lambert's scale 1:2 500 000, SP20 S and 40S. The scale is correct at the two Standard Parallels Scale 20S = Scale at 40S Some problems that may arise.. Example 1

A Lambert's chart has Standard Parallels of 30N and 50 N The Rhumb Line distance from A (50N 30E)to B (50N 10E) is 13.75 inches. The scale at 30N is :-

CL 13.75 inches Scale = __ = ________________________________ ED 20 Ch. Long x 60 x cos 50 x 6080 x 12 (Departure in nm) Example 2

=

1 ________ 4 092 898

On a Lambert's chart the Standard Parallel of 35S measures 58.4 cms. The other Standard Parallel measures 43.9 cms. The Latitude of the second Standard Parallel is :-

CL 58.4 cms CL 43.9 cms Scale at 35S = _________________ Scale at 2nd SP = _______________ ED Ch. Long x cos 35 ED Ch.Long x cos Lat The scales are equal. As CH. Long is the same in both equations it disappears 58.4 cms ___________ cos 35

43.9 cms ____________ cos Lat

=

cos lat = 0.6158 lat

= (.6158)COS-1

lat

= 52S



Page 37

Lambert's Questions 1.

The convergence between the meridians of 37E and 59E at 30S on a Lamberts chart (Standard Parallels 29S and 41S) is :(a) 11.0 (b) 11.8 (c) 12.6

2.

Chart convergency on a Lamberts chart between the meridians of 10E and 10W is 12. If one standard parallel is S 30:20 the other standard parallel is at :(a) S 40:48 (b) S 41:52 (c) S 43:24

3.

On a Lamberts chart the standard parallel of 32N measures 77.5 cms. The other standard parallel measures 72 cms. The latitude of the second standard parallel is :(a) 37N (b) 38N (c) 39N

4.

A Lamberts chart has standard parallel of 20S and 40S. The rhumb line distance from A (20S 15E) to B (20S 37E) is 76.62 cms. The scale of the chart at 40S is :(a) 1:3 000 000 (b) 1:3 500 000 (c) 1:4 000 000

5.

A Lamberts chart has standard parallels of 25S and 45S. The rhumb line distance from X (45S 17W) to Y (45S 05E) is 80.5 cms. The rhumb line distance from 25S 17W to 25S 05E is :(a) 97 cms (b) 100 cms (c) 103 cms

6.

A Lamberts chart has standard parallels of 30N and 50N. The initial great circle track from A (32N 63W) to B (45N 15W) is 056(T). The longitude at which the great circle track becomes 090(T) is :(a) W 012:37 (b) W 010:06 (c) W 008:28



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7.

A northern hemisphere Lamberts chart has a CCF of 0.57. A straight line from A (27W) to C (32E) passes through B (09E). The initial great circle track at A is 062(T). The great circle track at B is :(a) 082.5 (b) 087.2 (c) 093.8

8.

An aircraft heading 321(T) in the northern hemisphere receives an ADF bearing of 094 relative. If chart convergency between the aircraft and the NDB is 5 the bearing to plot from the NDB on a Lamberts Chart is :(a) 230 (b) 235  (c) 240

9.

An aircraft heading 112(T) in the southern hemisphere receives an ADF bearing of 156 relative. If chart convergency between the aircraft and the NDB is 4 the bearing to plot from the NDB on a Lamberts Chart is :(a) 084 (b) 092 (c) 096

10.

A Lambert's chart has a chart convergency of 12.04 between the Meridians of 70E and 84E. If One Standard Parallel is 64N, the Latitude of the other SP is :(a) N59:19 (b) N57:15 (c) N54:38

11.

On a Lambert's chart, a straight line from A (45N 045E) to B (47N 035E) cuts the 40E Meridian at 75. Convergency between A and B is 8. The Rhumb Line track from A to B is :(a) 255 (b) 270 (c) 285

12.

On a Lambert's chart, Convergency between A(50S 005E)and B (52S 015E) is 8. The Great Circle track from A to C (50 S 020 E) is :(a) 084  (b) 090 (c) 096



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13.

A bearing obtained from a NDB is 273 relative. Aircraft heading 330 (T). d.long between aircraft and NDB is 14, mean Latitude 26S. Parallel of Origin 30S. The bearing to plot on a Lambert's chart is :(a) 063 (b) 070 (c) 077

14.

Two Meridians. 175W and 171E, in the Northern Hemisphere, have a convergency of 7 on a Lambert's chart. The GC track from X (169E) to Y (175W) is 080(T) at X. The Rhumb Line track from Y to X is :(a) 256 (b) 260 (c) 264

15.

A Northern Hemisphere Lambert's chart has a CCF of 0.75. The straight line from A (20E) to C (60E) passes through B (044E). The direction of the track at A is 100 (T). The Great Circle track at B is :(a) 109 (b) 115  (c) 118

16.

A Lambert's chart has Standard Parallels of 24S and 46S. Position X (46S 045W) and Y (46S 025W) are plotted on the chart. The Great Circle bearing of Y from X is :(a) 095.74 (b) 096.94 (c) 097.19

17.

The chart convergency factor of a Lambert's chart is 0.50. A straight line is drawn from X (20N 010E) to Y (40N 030W) and measures 305 (T) at X. The RL track from X to Y is :(a) 295 (b) 305 (c) 315

18.

An aircraft heading 173 (T) obtains a relative bearing of 313 from a NDB. Convergency between aircraft and NDB is 14. What bearings would you plot from the NDB on a Lambert's chart in: 1. Northern Hemisphere? (a) 292 (b) 306 (c) 320

2. Southern Hemisphere? (a) 292 (b) 306 (c) 320



Page 40

19.

Positions X and Y are in the Southern Hemisphere. An aircraft at X. heading 240 (T). obtains a relative bearing of 198 from a NDB at Y. When plotted as a straight line from the NDB on a Lambert's chart, the bearing measures 250 (T). The RL track from X to Y is :(a) 074 (b) 078 (c) 082

20.

The Standard Parallels on a Lambert's chart are 10N and 30 N. A straight line is drawn from Position A (30 N 160W) to B (10 N 150 E) on a Mercator and a Lambert's chart. The straight line cuts the 180 Meridian at 244 on the Mercator. The GC tracks from A to B on the Lambert's chart at 160 W and 170 E are :-. 160 W (a) 249.67 (b) 252.55 (c) 254.05

21.

170 E (a) 241.41 (b) 242.29 (c) 243.84

Positions X (45 N 010 W) and Y (20 N 060 E) are joined by a straight lines on a Mercator and Lambert's chart. Lambert's: SPs 20 N and 45 N. Mercator track X to Y is 124 (T) at 40 E. The Longitude that the track on the Lambert's chart equals 124 (T) is :(a) 23E (b) 25E (c) 27E

22.

On a Northern Hemisphere Lambert's chart the Initial Great Circle track from A to B is 082(T). An aircraft leaves A steering a constant heading of 082(T) in zero wind conditions. The aircraft will pass :(a) North of B (b) Overhead B (c) South of B

23.

A Lambert's chart has Standard Parallels of 15S and 35S. The difference between Chart Convergency and Earth Convergency of Meridians 22 apart at 34S is :(a) 2 (b) 3 (c) 4

24.

A Lambert's chart has Standard Parallels of 25N and 45N. A straight line is drawn on the chart from X (42N 15W) to Y (43N 15E). The true Great Circle between X and Y will be :(a) to the North of the straight line (b) the same as the straight line (c) to the South of the straight line



Page 41

25.

Parallels of Latitude on a Lambert's chart are :(a) Parallel straight lines unequally spaced (b) Arcs of circles equally spaced (c) Arcs of circles unequally spaced

26.

Lambert's charts of the same scale and Standard Parallels will fit :(a) N/S only (b) E/W only (c) both N/S and E/W

27.

A Lambert's chart has a scale of 1:2 500 000. The chart length of 1 of Latitude varies as follows :Latitude 70N 65N 60N 55N 50N 45N

Chart length of 1 of Latitude 4.471 cms 4.441 cms 4.422 cms 4.441 cms 4.471 cms 4.516 cms

The Standard Parallels are :-

(a) 54N & 66N (b) 55N & 67N (c) 56N & 68N



Page 42

MERCATOR CHART Before the advent of Inertial Navigation, and GPS computers aircraft flew constant headings. They flew Rhumb Lines. The Mercator chart was constructed so that Rhumb Lines are straight lines and the headings flown were easily plotted.

A cylinder is positioned over the reduced earth tangential to the Equator. A light source at the centre of the reduced earth projects details of the reduced earth onto the cylinder and we have a Geometric Cylindrical Projection. After adjusting the Parallels of Latitude so that the scale expansion North/South equals the scale expansion East/West it becomes a Mercator chart.



Page 43

MERCATOR CHART PROPERTIES POINT OF PROJECTION

Centre of the reduced earth

POINT OF TANGENCY

Equator

PARALLELS OF LATITUDE

Parallel straight lines, unequally spaced

MERIDIANS

Parallel straight lines, equally spaced

CONVERGENCY

Constant Value Zero Correct at the Equator

SCALE

Correct at the Equator Expands as the secant of the Latitude Straight Lines

RHUMB LINES

GREAT CIRCLES

Complex curves towards the nearer Pole Convex to the Pole, Concave to the Equator



Page 44

SHAPES & AREAS

Approximately correct, excellent between 12N and 12S becoming distorted with increasing Latitude. The chart has a limit of 70N and 70S.

CHART FIT

Charts of the same equatorial scale will fit N/S. diagonally.

USES

Plotting and Met charts Topographical maps between 12N and 12S

ADVANTAGES

Rhumb Lines are straight lines - plotting easy

DISADVANTAGES

Great Circles (radio bearings) are complex curves. Great care must be taken measuring distances due to rapidly changing scale.


E/W and


Page 45

SCALE Scale is correct at the Equator and expands North and South as the secant of the Latitude. Every Parallel of Latitude has its own scale. (SF x cos LAT) Equator 5S 10S 30S 60S

1:2 000 000 1:1 992 389 1:1 969 615 1:1 732 051 1:1 000 000

Great care must be taken when measuring distances on a Mercator chart due to the variable scale. Use the Latitude scale at the mid point between the two positions. SCALE PROBLEMS Scale problems are easily solved by use of ABBA

SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A Example 1

The scale of a Mercator chart is l:2500000 at 15S. 15S = A What is the scale at 45N? 45N = B

SCALE DENOMINATOR A x COS B = SCALE DENOMINATOR B x COS A

Example 2

2 500 000 x cos 45

=

Scale B x cos 15

2 500 000 x cos 45 cos 15

=

1 830 127 Scale at 45N 1:1 830 127

The scale of a Mercator chart is 1:3 500 000 at 10N At what Latitude is the scale 1:2 500 000?

SCALE DENOMINATOR A x COS B

10N = A Lat X = B

=

SCALE DENOMINATOR B x COS A

3 500 000 x cos X

=

2 500 000 x cos 10

cos X

= =

2 500 000 x cos 10 3 500 00 0:7034

=

(0.7034)cos -1

=

45 17'49" N/S

X



Page 46

Example 3

The Meridian spacing on a Mercator chart is 2.7 cms. The scale at 30S is If ABBA cannot solve the problem, then revert to:-

CL Scale = __ ED

=

=

2.7 cms _______________________________ 1 Long x 60 x cos 30 x 6080 x 12 x 2.54 2.7 9 629 426

(scale is 1/xxxxxx)

= 1:3 566 454 PLOTTING RADIO BEARINGS VDF & VOR Radio bearings are Great Circle bearings. They have to be converted into Rhumb Line bearings by applying Conversion Angle before they can be plotted. Both VDF and VOR bearings are measured at the station, thus station variation must be applied. Conversion angle is also applied where the bearing was measured, that is the VDF or VOR station. VDF & VOR

APPLY STATION VARIATION

BEARING VARIATION QDM 100 10 W QDR 258 16 W RMI Reading 113 15E RMI Reading 088 11 W QDR 129 20 E QDM 285 14 W QUJ 131 24 E Answers on next page

APPLY CA TO QTE

HEMISPHERE NORTHERN SOUTHERN NORTHERN SOUTHERN NORTHERN SOUTHERN NORTHERN


CA 2 3 4 2 3 4 1

PLOT RL


Page 47

NOTE VOR RMI readings are QDM's. Apply VOR station variation, but not compass deviation.

Answers to VOR/VDF bearings QDM 100 QDR 258 QDM 113 QDM 088 QDR 129 QDM 285 QUJ 131

VAR 10 W VAR 16 W VAR 15 E VAR 11 W VAR 20 E VAR 14 W

QUJ 090

180

QUJ 128 QUJ 077

180 180

QUJ 271

180 180

GC QTE 270 GC QTE 242 GC QTE 308 GC QTE 257 GC QTE 149 GC QTE 091 GC QTE 311

N CA -2 S CA +3 N CA -4 S CA +2 N CA +3 S CA -4 N CA -1

RL QTE 268 RL QTE 245 RL QTE 304 RL QTE 259 RL QTE 152 RL QTE 087 RL QTE 310

PLOTTING ADF/NDB BEARINGS ADF bearings are presented to the pilot by either a RELATIVE BEARING INDICATOR (RBI) or by a RADIO MAGNETIC INDICATOR (RMI). RELATIVE BEARING INDICATOR (RBI) ADF bearings are measured clockwise from the fore and aft axis of the aircraft and are termed RELATIVE BEARINGS, that is relative to the aircraft's fore and aft axis. ADF Relative bearings must be converted into True Bearings (QTE) before they can be plotted on a chart,

RELATIVE BEARING + TRUE HEADING = GC QUJ ± CA = RL QUJ ± 180 = RL QTE

NB

The GC QUJ must be converted into a RL QUJ before the reciprocal is taken. The reciprocal of a Rhumb Line can always be taken, never the reciprocal of a Great Circle

MAGNETIC VARIATION AT THE AIRCRAFT IS ALWAYS USED WITH ADF BEARINGS



Page 48

RADIO MAGNETIC INDICATOR (RMI) The RMI is a remote gyro compass on which radio bearings (both ADF and VOR) are shown. As it is a compass, the heading index is heading compass and it may suffer from deviation, for which a correction must be made to ADF bearings but not VOR bearings. The sharp end of the pointers are referred to as RMI readings or QDM. The opposite or blunt end of the needle will be a QDR.

ADF QDM ± Deviation ± Aircraft Variation = GC QUJ ± CA RL QUJ ± 180= RL QTE

ADF BEARINGS HEADING 016 (C) 065 (C) 225 (C) 345 (C)

DEVIATION 2W 3E 4W 2E

VARIATION 13 E 18 W 21 W 19 W

BEARING 071 REL 214 REL 069 REL 123 RMI


HEMISPHERE NORTHERN SOUTHERN NORTHERN SOUTHERN

CA 4 5 3 2


Page 49

Hdg 016 ( C ) Dev 2W Hdg 014 (M) Var 13 E Hdg 027 (T) ADF 071 Rel QUJ 098 GC CA + 4 N QUJ 102 RL  180 QTE 282 RL

Hdg 065 ( C ) Dev 3E Hdg 068 (M) Var 18 W Hdg 050 (T) ADF 214 Rel QUJ 264 GC CA +5 S QUJ 269 RL 180 QTE 089 RL

Hdg 225 ( C ) Dev 4 W Hdg 221 (M) Var 21 W Hdg 200 (T) ADF 069 Rel QUJ 269 GC CA - 3 N QUJ 266 RL 180 QTE 086 RL

Station West of ACFT

RMI 123 Dev 2E QDM 125 Var 19 W QUJ 106 GC CA -2 S QUJ 104 RL ± 180 QTE 284 RL

Station East of ACFT

NB VOR & VDF

APPLY STATION VARIATION DO NOT APPLY DEVIATION APPLY CA TO QTE

RMI READING

= QDM

ADF NDB

APPLY AIRCRAFT DEVIATION & VARIATION GC QUJ  CA = RL QUJ  180 = RL QTE



Page 50

Mercator Questions

1.

A Mercator chart has a scale of 1:2 500 000 at 20N.

The scale at 50N is:-

(a) 1:1 710 100 (b) 1:1 835 000 (c) 1:1 972 060 2.

A Mercator chart has a scale of 1:2 000 000 at 20N. The latitude where the scale would be 1:1 500 000 is :(a) N 44:11 (b) N 45:11 (c) N 46:1 I

3.

The distance between meridians on a Mercator chart 1 apart is 2.7 centimetres The scale of the chart at 33S is :(a) 1:3 247 500 (b) 1:3 453 800 (c) 1:3 694 400

4.

An aircraft at position S 22:35 E 029:45 obtains an ADF QDM of 095. If variation is 14W and conversion angle 2 the bearing to plot on a Mercator chart is :(a) 263 (b) 261 (c) 259

5.

An aircraft obtains a QDM of 275 from a VHF D/F station in the Southern Hemisphere. Aircraft Variation is 16W and VHF D/F station Variation is 18W. lf Convergency between the aircraft and the VHF D/F station is 6 the bearing to be plotted on a Mercator chart is :(a) 080 (b) 074 (c) 071

6.

An aircraft at position S 23:15 E 027:32 (variation 15W) heading 267(T) obtains a relative bearing of 346 Relative from a NDB at S 25:27 E 019:32 (variation 17W). CA 2. The bearing to plot on a Mercator chart is :(a) 060 (b) 071 (c) 075



Page 51

7.

On a Mercator chart the distance between two meridians 5º apart is 156 millimetres. The scale of the chart at 32N is :(a) 1:3 022 286 (b) 1:3 127 654 (c) 1:3 226 327

8.

Two straight lines of equal length are drawn East/West on a Mercator chart. One at 15S and the other at 45N. (a) The line at 15S represents a greater distance than the line at 45N (b) The line at 45N represents a greater distance than the line at 15S (c) Both line represent the same distance

9.

Two straight lines representing 200 nm are drawn East/West on a Mercator chart. One at 22S and the other at 48N. (a) The line at 22S is longer than the line at 48N (b) The line at 48N is longer than the line at 22S (c) The lines are of equal length

10.

The Meridian spacing on a Mercator chart is 3 cms. The Latitude where the scale is 1:2 500 000 is (a) 42 45' (b) 45 15' (c) 47 35'

11.

The Meridians on a Mercator chart are 5,75 cms apart. The ratio of nautical miles to the centimetre at 52N is :(a) 6.42 (b) 7.53 (c) 8.64

12.

The Parallels of Latitude on a Mercator chart are :(a) Parallel straight lines equally spaced (b) Parallel straight lines unequally spaced. (c) Arcs of circles radius the nearer Pole

13.

The Meridians on a Mercator chart are :(a) Parallel straight lines equally spaced (b) Parallel straight lines unequally spaced. (c) Straight lines converging on the nearer Pole

14.

A Great Circle on a Mercator chart is :(a) A curve concave to the nearer Pole (b) A curve concave to the Equator (c) A curve convex to the Equator



Page 52

15.

Mercator charts of the same scale at the Equator will fit :(a) N/S only (b) E/W only (c) in all directions



Page 53



Page 54

CHAPTER 3 RELATIVE VELOCITY Relative Velocity is the comparison of aircraft speeds or the speed of one aircraft relative to another. The calculations can be broken down into 3 main areas: Aircraft meeting Aircraft overtaking Speed adjustment Meeting

Aircraft A GS 240 Kts

Aircraft B GS 300 Kts SPEED OF CLOSING 540 Kts

Overtaking





Aircraft A GS 340 Kts Aircraft A being faster will overtake Aircraft B

Aircraft B GS 250 Kts SPEED OF CLOSING 90 Kts

 Aircraft A GS 220 Kts

 Aircraft B GS 290 Kts

Aircraft B being faster than Aircraft A will fly further ahead of B SPEED OF OPENING 70 Kts POSITION BOTH AIRCRAFT AT THE SAME TIME



Page 55

Example 1. Aircraft A is overhead NDB PY at 0900 Z enroute to VOR CN. GS 240 Kts Aircraft B is overhead VOR CN at 0920 Z enroute to NDB PY. GS 300 Kts Distance PY to CN is 1150 nm

1150 nm

PY

CN

556nm 595nm

Note: Times have been rounded off to the nearest minute Example 2

Aircraft A. GS 180 Kts, passes overhead X at 1200 Z bound for Y Aircraft B, GS 270 Kts, passes overhead X at 1225 Z bound for Y At what time will aircraft B overtake aircraft A?

Y

X B



Page 56

Example 3

Two aircraft at the same Flight Level following the same route are approaching a VOR. Aircraft A, GS 350 Kts, is 260 nm from the VOR at 0800 Z. Aircraft B, GS 450 Kts, is 390 nm from the VOR at 0750 Z. At what time must aircraft B reduce to GS 350 in order to :(a) (b)

ensure a 50 nm separation at the VOR? ensure a 5 minute separation at the VOR?

As aircraft B reduces speed to the same speed as aircraft A it is a 'speed of closing' problem. If aircraft B reduces speed to a different speed than aircraft A it is a 'delay' problem. 390 nms

315nms

B 0750

260 nms

B 0800

A 0800 5 mins @ 350 kts =29 nms

(a)

Speed of closing Distance to close (55-50) Time to close Reduce speed at

Example 4

100 kts 5 nm 3 mins 0803Z

(b)

Speed of closing Distance to close (55-29) Time to close Reduce speed at

100 kts 26 nms 16 mins 0816

An aircraft, GS 450 Kts, estimates overhead 'Delta' at 0915 Z. ATC requests the aircraft to cross 'Delta' at 0920 Z. To accomplish this the aircraft reduces speed to 390 Kts at time :-

Delay x Old GS x New GS Distance = ______________________ = Difference in GS x 60

5 x 450 x 390 ___________ 60 x 60

=

243.75 nm

GS 450

Dist 243.75 nm

Time 32½ mins ETA 0915 - 32½ mins

= 0842½

GS 390

Dist 243.75 nm

Time 37½ mins ETA 0920 - 37½ mins

= 0842½'

Alternative solution

DISTANCE = SPEED x TIME

At the point where speed is reduced, the aircraft is 'D nm' from Delta. At GS 450

D = 450 x T

As D is common, then

At GS 450 At GS 390

At GS 390 450 T 450 T 450 T-390T 60 T T

ETA 0915 - T 32½ mins ETA 0920 - (T+5) 37½ mins


D = 390 x (T+5) = 390 (T + 5) = 390 T + 1950 = 1950 = 1950 = 32½ mins = 0842½ = 0842½ FLIGHT TRAINING COLLEGE Version 7

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Example 5

A/c A, GS 180 Kts passes over NDB PB 5 minutes ahead of a/c B A/c B. GS 260 Kts. passes over VOR CPL 8 minutes ahead of a/c A. The distance from NDB PB to VOR CPL is :-

As aircraft B overtakes aircraft A. the times are added. Delay x Old GS x New GS Distance = ______________________ = Difference in GS x 60 Example 6

13 x 180 x 260 ______________ = 126.75 nm 80x60

Aircraft A. GS 250 Kts, passes NDB DN 14 minutes ahead of aircraft B. GS 315 Kts.

Aircraft A then passes VOR PON 5 minutes ahead of aircraft B The distance from NDB DN to VOR PON is :As aircraft B does not overtake aircraft A the times are subtracted Delay x Old OS x New GS 9 x 250 x 315 Distance = _____________________ = _______________ Difference in GS x 60 65 x 60

= 181.73 nm

QUESTIONS Q1.

An aircraft is at FL 330, at M.85 with IOAT -35ºC W/V 250/45. The aircraft Estimates overhead PWV at 1011Z, track to PWV is 61ºT the local variation is 21ºE. ATC requests that the aircraft change GS to 200 so as to arrive overhead PWV at 1020Z. At what time must the aircraft be slowed down. a. b. c. d.

Q2.

Aircraft A is overhead PNV at 1010Z FL 240 enroute to JWV, GS 250kts. Aircraft B is overhead JWV at 1020Z FL 330 enroute to PNV, GS 420 KTS. Distance PNV to JWV is 998 NMS. What time will they cross? a. b. c. d.

Q3.

1015Z 1005Z 1001Z 1008Z

11:49Z 11:43Z 11:46Z 11:52Z

An aircraft flying at FL120, IAS 200 knots, temperature –5° C, wind component +30 knots. At a position 100 nm from the next reporting point the aircraft is ordered to delay arrival by 5 minutes. The immediate reduction in IAS to comply with this order is: a. 50 kt b. 16 kt c. 32 kt d. 41 kt



Page 58

CHAPTER 4 THE SOLAR SYSTEM & TIME The measurement of the passage of time is based upon observations of events occurring at regular intervals. The two repetitive events which most influence life on Earth are the rotation of the Earth on its axis. Causing day and night, and the movement of the Earth in its orbit around the Sun, causing the seasons. THE EARTH’S ORBIT The orbit of a planet around the Sun conforms with Kepler’s Laws of Planetary Motion which state :-

1.

The orbit of a planet is an ellipse, with the Sun at one of the foci.

2.

The line joining the planet to the Sun, known as the radius vector, sweeps out equal areas in equal in equal intervals of time.

A

Y SYC

SAX

C X

In the above sketch the planet (P) moves anticlockwise in its orbit and is at its closest position to the Sun at position A which is called PERIHELION. At Perihelion the Earth is about 91½ million miles from the Sun and occurs on January 4. At position C the planet is furthest from the Sun and is known as APHELION. At Aphelion the Earth is about 94½ million miles from the Sun and occurs on July 4. The mean distance of the Earth from the Sun is about 93 million miles. According to Kepler’s Law the radius vector sweeps out equal areas in equal intervals of time. If the area SAX equals the area SYC then as the distance AX is greater than the distance CY and the orbital speed of the planet is faster at Perihelion than at Aphelion. The orbital speed of the Earth is variable.



Page 59

The Earth completes one orbit around the Sun in about 365.25 days. The plane of the orbit is called the plane of the Ecliptic, and the N/S axis of the Earth is inclined to this plane at an angle of 66½. The plane of The Ecliptic is at an angle of 23½ º to the Earth’s Equator and this angle is known as the obliquity of the ecliptic. THE SEASONS One effect of the tilt of the Earth’s axis is the annual cycle of seasons. As the Earth moves around the Sun, on or near 23rd of December the North Pole is inclined away from the Sun, which is vertically above Latitude 23½°N. This is known as winter solstice and is midwinter in the Northern Hemisphere and midsummer in the Southern Hemisphere. As the Earth travels around its orbit, being a gyro. Its axis will always point in the same direction relative to space and will reach a point at the summer solstice, on or about 22 nd June, when the Sun is vertically overhead Latitude 23½°N. It is then midsummer in the Northern Hemisphere and midwinter in the Southern Hemisphere. Between these dates the Sun. will be overhead the Equator. These events occur on 21st March which is the spring or vernal equinox, and 23rd September which is the autumn equinox.

Jan 4 Mar 21 Jun 22 July 4 Sep 23 Dec 23

Perihelion Vernal or Spring Equinox Summer Solstice Aphelion Autumn Equinox Winter Solstice

Sun 91½ million miles Sun overhead Equator Declination 00:N/S Sun overhead Tropic of Cancer Declination 23½°N Sun 94½ million miles Sun overhead Equator Declination 00:N/S Sun overhead Topic of Capricorn 23½°S

The seasons apply to the Northern Hemisphere and reversed in the Southern Hemisphere.



Page 60

MEASUREMENT OF TIME – THE DAY The rotation of the Earth on its axis is used as a basis for the measurement of the length of a day. The length of time taken for the Earth to complete one revolution on its axis can be found by taking the time between two successive transits of a fixed point in space over a particular meridian. Sidereal Day (23 hours 56 minutes 4 seconds) As stars are at immense distances from the Earth, they can be considered to be at infinity and rays of light from stars can be considered parallel regardless of the position of the Earth in its orbit round the Sun. The time interval between two successive transits of a star or a fixed point in space over a meridian is called a SIDEREAL DAY and is constant at 23 hours 56 minutes and 4 seconds.

Apparent Solar Day The time interval between two successive transits of the True Sun over a meridian is an Apparent Solar Day. The Sun and a star are in transit overhead a meridian. After 23 hours 56 minutes and 4 seconds the star is in transit For a second time (a Sidereal Day), rays of light from a star being parallel. Due to the Earth’s orbital speed (approximately 58 000 Kts) it has moved some 1 400 000 nm along its orbit and the Earth has to rotate ‘X’ degrees before the Sun is in transit for a second time. This of course takes time thus an Apparent Solar Day is always longer than a Sidereal Day. An average of 365 Apparent Solar Days is taken and termed a Mean Solar Day which is 24 hours. Mean Solar Day The 24 hour day is based on the Mean Sun. When the Mean Sun is overhead a meridian it is 12:00 Local Mean Time (LMT). Each and every meridian has its own LMT. THE EQUATION OF TIME (E)

The equation of time is the time difference between the apparent solar day and the mean solar day and is of varying duration.



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The Siderial Day

Because of the relative proximity of the earth to the sun, attempts to measure the length of the day (one revolution of the earth) are contaminated by the movement of the earth in its orbit relative to the sun. To solve this problem, a fixed point in space is chosen which is so enormously distant that the movement of the earth in its orbit relative to this point is basically zero. This point in space is called the Siderial point or the first point of Aries. The Siderial day then, is defined as two successive transits of the Siderial point at the same meridian. The Siderial day is of constant duration : 23 hours 56 mins 4 seconds. The Earth rotates on its axis from West to East. It is more convenient to imagine the Earth stationary with the Sun rising in the East and setting in the West. At the Greenwich Meridian the sun is rising at 06:00 LMT. At 90ºE the sun is overhead at 12:00 LMT. At 180ºE/W the sun is setting at 18:00 LMT. At 90ºW it is midnight 24:00 LMT on the 5th LD Local Date or 00:00 LMT on the 6 th LD. The Local Date changes at midnight and also at the International Date Line. ARC TO TIME The Earth rotates through 360 in 24 hours. 90 in 6 hours, or 15° per hour, there is a direct relationship between Longitude and LMT. The Conversion of Arc to Time table is available in the Navigation Tables booklet provided in the examination. The first six columns are degrees of Longitude on the left with the corresponding time in hours and minutes on the right.

10º

0:40

15º

1:00

134º

8:56

314º

20:56

The right hand column gives the time equivalent for minutes of Longitude. 10' Long

40 seconds

131º 16'E Arc to Time

16' Long 131º = 8:44

1 minute 04 seconds

16’ long = 1 minute 04 seconds 131 º16’ = 08:45:04



Page 62

UNIVERSAL CO-ORDINATED TIME (UTC) UTC is the LMT at the Greenwich Meridian and is used as the standard reference from time keeping for aviation. UTC is the same as GMT (Greenwich Mean Time). CONVERSION OF LMT TO UTC

LONGITUDE EAST

UTC LEAST

LONGITUDE WEST

UTC BEST

(Tables used for these following questions start on page 65) Example 1.

At position A (N 45:05 E 065:30) it is 13:15 LMT on 23rd March. The UTC at this position is :A E 065:30 Arc to Time A

13:15 LMT 23 March 4:22 08:53 UTC 23 March

Longitude East - UTC Least UTC must be an earlier time than LMT

Example 2.

The time is 06:45 UTC on 21st May GD (Greenwich Date). At position B (S 28:37 W 092:20) the LMT is :B W 092:20 Arc to Time B

06:45:00 UTC 21°May GD° 6:09:20 00:35:40 LMT 21" May LD

Longitude West - UTC Best UTC must be a later time than LMT

Example 3.

If the UTC is 15:30 on the 22nd June GD and the LMT at position X is 09:45 on 22nd June, LD the Longitude of X is :15:30 UTC 22nd June 09:45 LMT 22nd June

Time difference

Example 4.

5:45 Time to Arc = W 086° 15' Longitude

An aircraft departs C (N 45:35 E 010:15) at 15:30 LMT on 15th May LD. Flight time to D (42:37 E 135:45) is 11 hours 18 minutes. The ETA in LMT is :-

C E 010:15 Arc to Time C Flight Time D E 135:45 Arc to Time D

ETD ETD ETA ETA ETA

15:30 LMT 15th May LD 0:41 14:49 UTC 15th May GD 11:18 26:07 UTC 15 th May GD 02:07 UTC 16 th May GD 9:03 11:10 LMT 16 th May LD

NOTE: In flight the time standard is UTC. always work in UTC. CPL NAVIGATION CPL DOC8 Revision 1/1/2001


Page 63

LOCAL STANDARD TIME As every Meridian has a different LMT, LMT is not suitable for civil time keeping. Durban has a different LMT to Johannesburg. Each country has its own standard time factor which is applied to UTC to give local standard time. Standard Time tables appear on page 67 onwards. For GMT (Greenwich Mean Time) read UTC. List 1 Mainly countries with Easterly Longitude (including Spain & Portugal which are Westerly Long.) List 2 Countries normally keeping GMT or UTC. List 3 Countries with Westerly Longitude Apply Standard Times in the same manner as LMT (Long East - UTC Least & Long West UTC Best) or apply as given at the top of each list. Ignore summer time. INTERNATIONAL DATE LINE The International Date Line roughly follows the 180 E/W meridian, with some divergences to accommodate certain groups of South Sea Islands and regions of Eastern Siberia. GOING EAST 1 DAY LEAST (lose a day) GOING WEST 1 DAY BEST (gain a day)



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Page 69

TIME QUESTIONS 1.

An aircraft departs X(S 23:46E 023:45) at 21:00Z on 3 July and arrives at Y(S 29:13 W 066:30) at 05:58 LMT on 4 July. If the distance from X to Y is 4732 nm the average groundspeed was :(a) 353 Kts (b) 376 Kts (c) 396 Kts

2.

An aircraft departs Ascension (S 07:58 W 014:30) at 22:15 LMT on 15th June LD enroute for Johannesburg (S 26:08 E 028:15). If the flight time is 9 hours 14 minuses the ETA FAJS is :(a) 08:31 SAST (b) 10:20 SAST (c) 10:27 SAST

3.

An aircraft departs Perth (S31:57 E 115:57) at 0930 LMT on a flight to Mauritius (S20:26 E 057:40). Distance 3207 nm, average groundspeed 427 kts. The LMT of arrival at Mauritius is :(a) 13:00 (b) 13:07 (c) 13:14

4.

At 08:15 LMT on 19 th Sept Local Date an aircraft leaves A (N27:00 E 035:15). After a flight of 7 hours 27 minutes the aircraft arrives at B (N32:00 W 028:45). The LMT of arrival at B is :(a) 11:26 (b) 13:21 (c) 15:16

5.

An aircraft leaves Prestwick (N55:00 W 005:00) at 1 115 LMT on 23rd June LD. Flight time to San Francisco (N37:30 W 122:00) is 11 hours 15 minutes. The LMT of arrival at San Francisco is :(a) 14:02 (b) 14:22 (c) 14:42

6.

An aircraft is to fly from Wellington. New Zealand (S 41:10 E 174:45) to Tahiti. The Standard Time Factor New Zealand is 12 hours) Arrival at Tahiti (S 17:29 W 149:29) must not be later than Sunset 1823 LMT on 5th March LD. If the flight time is 5 hours, the latest local mean time and date at which the aircraft must leave Wellington is :(a) 10:55 5 th (b) 11:00 6 th (c) 11:05 5 th



Page 70

7.

An aircraft arrived at Vancouver (N49:00 W 123:15) at 1057 LMT on Jan 18 after a flight of 8 hours 10 mins. It departed X at 2000 LMT on Jan 18. The Longitude of X is :(a) 135°E (b) 137°E (c) 139°E

8.

An aircraft flies 1952 kilometres on a track of 270° (T) along a parallel of Latitude. If the LMT of arrival at the destination is the same as the LMT of departure and the flight time is two hours, the parallel of Latitude which the aircraft followed was :(a) 52:37 N/S (b) 54:09 N/S (c) 56:15 N/S

9.

An aircraft leaves Tokyo (N36:00 E 139:45) at 2100 Standard time on Oct 12 (Standard time factor Japan 9 hours). After a 11 hour flight it arrives at San Francisco (N37:30 W 122:00) The LMT of arrival at San Francisco is:(a) 14:32 (b) 14:42 (c) 14:52

10.

An aircraft arrived at A (168°W) at 22:08 Standard Time on 2nd July. Standard Time Factor 11 Hr. If the flight time from B (174°E) was 7 hours 6 minutes the LMT of departure from B was :(a) 14:26 2 nd (b) 14:36 2 nd (c) 13:38 3 rd

11.

An aircraft heading 090° (T) crosses the International Date Line at 0600 UTC on 6th May GD The local date :(a) changes from 6 th May to the 5 th May (b) changes from 5 th May to the 6 th May (c) remains 6 th May

12.

An aircraft heading 090(T) crosses the International Date Line at 1200 UTC on 6 May GD. The GMT (UTC) date is :-

th

(a) changes from 6 th May to the 5 th May (b) changes from the 5 th May to the 6 th May (c) remains 6 th May 13.

An aircraft departs New York at 13:00 LST on 28 th February 1992 on a 8 hour 30 minute flight to Frankfurt, Germany. The LST of arrival at Frankfurt is :(a) 03:30 LST 29 th Feb (b) 03:30 LST 1st Mar (c) 02:30 LST 1 stMar



Page 71

SUNRISE & SUNSET The time at which the upper limb of the sun is coincident with the horizon. Corrections have been made for atmospheric refraction.

CIVIL TWILIGHT The time at which the upper limb of the sun is 6 below the horizon. The degree of illumination at the beginning of morning and end of evening twilight (in good conditions and in the absence of other illumination) is such that the brightest stars are just visible, and terrestrial objects can be easily distinguished. The rising and setting tables on pages 74 onwards give the UTC at the Greenwich Meridian. For all places of the same latitude the corresponding phenomena will occur at approximately the same LMT and this will be approximately the UTC tabulated. since UTC = LMT at Greenwich. To obtain the UTC of a phenomenon at a particular position the longitude must be convened into time and applied to the time extracted from the tables. A further correction for Standard Time Factor must be made for LST (Local Standard Time). Sunrise LMT Longitude (Arc to Time) = Sunrise UTC ± Standard Time Factor = Sunrise LST In high latitudes the Sun may be above or below the horizon all day, or civil twilight may last all night. The following symbols indicate the occurrence of these conditions. Sun remains continuous above the horizon Sun remains continuously below the horizon Twilight lasts all night, the Sun is less than 6 below the horizon and it is always lighter than at the beginning or end of morning or evening civil twilight.



Page 72

Example 1. The Standard Time of Sunrise at Amsterdam. Holland (N 52:18 E 005:15) on 27 January is:

January 54N 52N

26 0757 0749

(1) (1)

27 0756 0748

(2) (2)

28 0754 0746

(1) (1)

29 0753 0745

NOTE Making the time difference between dates symmetrical (1 - 2 – 1, 1 - 0 - 1 etc.) will give a maximum error of 20 seconds which can be ignored. Interpolation for Latitude, 8 minutes time difference for 2 of Latitude 8 minutes ÷ 2 x 0018’ (5218’N – 52N) = 1 min (to nearest minute) = Sunrise 0749 LMT Sunrise Amsterdam E 005:15 Arc to Time Sunrise Standard Time Factor Holland Sunrise

07:49 LMT 00:21 07:28 UTC +1:00 08:28 LST



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Page 81

SUNRISE - SUNSET - TWILIGHT QUESTIONS 1.

The Standard Time of Sunrise at Madrid, Spain (N 40:30 W 003:24) on l5 April is: (a) (b) (c)

2.

The Standard Time of Sunset at Paris, France (N 48:40 E 002:00) on 7 May is: (a) (b) (c)

3.

19 minutes 29 minutes 39 minutes

Find the ST of sunset at Leningrad Russia CIS (59:46N 030:20E) on local date 22 January. (a) (b) (c)

7.

0938 LMT 1251 LMT 1236 LMT

The duration of evening civil twilight at Cape Town (S 33:58 E 018:36) on 6 January is: (a) (b) (c)

6.

38 minutes 48 minutes 58 minutes

An aircraft is to fly from A (S 12:00 E 022:15) to B (S 10:00 W 063:45). If the Flight time is 11 hours 15 minutes the latest LMT that the aircraft may depart from A to arrive at B 15 minutes before sunset on 27 February is: (a) (b) (c)

5.

2009 LST 2015 LST 2031 LST

The duration of morning civil twilight at N 55:45 W 010:15 on 3rd June is: (a) (b) (c)

4.

0430 LST 0636 LST 0536 LST

16h 49m ST 09h 36m ST 15h 49m ST

Give the GMT of sunrise at Auckland NZ (36:50S 174:48E) on local date 3 February. (a) (b) (c)

17h 38m 2 Feb 17h 38m 3 Feb 17h 38m 4 Feb



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CHAPTER 5 Navigational computer VECTOR TRIANGLE (TRIANGLE OF VELOCITIES) Navigation plotting is based around the Vector Triangle which comprises of three vectors.

NOTE All directions are TRUE DIRECTIONS (measured from TRUE NORTH) The length of each Vector is the value for ONE HOUR. (TAS 240 = 240 nm)(W/V 340/30 = 30 nm) The AIR VECTOR (TAS & True Heading) has one arrow and is called the AIR PLOT. The GROUND VECTOR (True Track & Groundspeed) has two arrows and is called the TRACK PLOT. The W/V has three arrows. W/V 340/30 is the direction from which the wind blows at 30 Kts. In the above sketch the Drift angle is 7 Right. The Wind blows from the Air Vector to the Ground Vector. The units cannot be interchanged. The Air Vector is TAS and True Heading only (never TAS & Track) If four of the six values are known, the other two can be calculated. CPL NAVIGATION CPL DOC8 Revision 1/1/2001


Page 83

NAVIGATIONAL COMPUTER Prior to flight the Heading and GS must be known as well as the fuel required for the flight and the time intervals between enroute points. This can all be found by using simple calculations from the flight computer or Whiz Wheel. There are many wide and varied versions of the Whiz Wheel, but basically they can all do the same thing in the same way. There are two methods of working with the wind side: TAS under the grommet (center) wind down. or GS under the grommet (center) wind up – Jeppesen method. The first method can solve all 3 common triangle of velocity problems, method 2 can only solve 2. Therefore method one will be used in this chapter. In this method the wind is plotted down from the grommet. WIND EXAMPLES Example 1 HDG TAS W/V

330º 150kts 040/25

Find: Track made good The groundspeed Solution: Step1 Plot wind down, then set HDG 330º under index on top. Step2 Read off the drift 10º left, the TRK is therefore 320º Step3 Read off the GS 144 kts Example 2 Required Track 150 TAS 100kt W/V 360/30 Find the Hdg and GS Step 1 CPL NAVIGATION CPL DOC8 Revision 1/1/2001


Page 84

Place the W/V on the plotting disk: Step 2 Move the circular scale to have the Track under the Index mark. Step 3 The drift is noted to be 7ºR, adjust the disk so that 143º (150º-7) is under the index. Observe that the drift has changed and is now 8ºR. Futher adjust the disk until the difference between the required track and the HDG under the index equals the drift. Note that if this is done correctly HDG 141 is under the index and the drift will be 9ºR. The TRK will equal 150º which is what we require. Now read off the GS at the end of the wind vector 125 kts. Example 3 If TAS is 174kt, Track is 290, the wind velocity is 240/40. Find the Heading and GS. ANSWER: Approx 280, 145kt Example 4 The in-flight type of problem…finding wind You know the following figures, find out the Wind Velocity. HDG TAS TRK GS

138 120kts 146 144kts Step 1 Place the HDG under index and TAS in the middle on the whiz wheel:



Page 85

Step 2 (diagram to the right) Now in your head work out the drift, and its found to be 8 right, so now draw in a straight line along the 8 right drift. The GS is 144kt, draw a line along the 144 line so as to intersect the 8º drift line. Draw a line from the grommet to the intersection of 144kt and 8 drift and you have drawn in the wind vector.

Step 3 (diagram below) Rotate the grid until the wind vector blows straight down. Under the Index mark you can read the direction, in this case its 360, and from the 120kts under the grommet at the beginning of the wind vector to its tail is the strength of the wind, in this case its 30kt. So the answer is 360/30



Page 86

THE CALCULATOR SIDE OF THE COMPUTER

This side of the computer can do many weird and wonderful calculations, but we are only concerned with the GS/Dist/Time and the Fuel Qty/Fuel Flow/Time problems. In order to make things simple we shall use the whiz wheel in the same manner as you would a electronic calculator, in that we use the following methods for the equations: DISTANCE

TIME

GS Fuel Qtty

TIME

Fuel flow

Example 1 If the aircraft has a GS of 154kts, and the Distance for the leg it 77nm, what is the EET for the leg?

Answer = 30minutes (found under the Arrow head)



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Example 2 At a GS of 147kt, how far will you travel in 11minutes?

ANSWER 27nm

Example 3 A leg is 25 minutes long, and the fuel flow is 32 lph, what is the fuel burn for this leg of the flight?

ANSWER = 13.3 litres, say 14 litres.. Example 4 If you burn 24 litres per hour, and the duration of the leg is 88minutes, what will be the fuel burn?

ANSWER = 35 litres



Page 88

MULTIPLE DRIFT W/V USING THE WHIZ WHEEL Given:-

TAS 190Kts

Heading 040° + Drift 8° Right Heading 085° + Drift 8½° Right Heading 355° + Drift 1° Right

Method: 1. 2. 3. 4. 5.

Set TAS 190 Kts at the CENTRE. Set HEADING 040° against TRUE INDEX, draw 8° RIGHT DRIFT LINE. Set HEADING 085° against TRUE INDEX, draw 8½° RIGHT DRIFT LINE. Set HEADING 355° against TRUE INDEX, draw 1° RIGHT DRIFT LINE. Place the intersection of the THREE DRIFT LINES on the CENTRE LINE below the CENTRE CIRCLE. Read off WIND DIRECTION 348° against the TRUE INDEX. Read off WIND SPEED 30 Kts along the CENTRE LINE

6. 7. 2

4

3

5

1



Page 89

TRACK & GROUNDSPEED W/V Given

DOPPLER W/V

Heading 126°(T) TAS 156Kts Doppler Drift 10° Right or Track 136° Doppler GS 142 Kts

Method: 1. 2. 3. 4. 5.

Set HEADING 126° against TRUE INDEX Set TAS 155 Kts at the centre circle Draw 10° Right DRIFT LINE. Draw arc of GROUNDSPEED 142 Kts. Position the intersection of the DRIFT and GROUNDSPEED lines BELOW the CENTRE CIRCLE. Read off WIND DIRECTION 070° against the TRUE INDEX. Read off WIND SPEED 30 Kts along the CENTRE LINE

6. 7.

MULTIPLE DRIFT W/V PRACTICE PROBLEMS TAS 230 Kts Heading 195° Heading 257° Heading 332°

Drift 7° Right Drift 6° Right Drift 2° Left

W/V 135/30

TAS 200 Kts Heading 045° Heading 090° Heading 340°

Drift 10° Right Drift 6° Right Drift 5° Right

W/V 313/32

DOPPLER W/V PRACTICE PROBLEMS Heading 045° 225° 352°

TAS 240 300 420

Drift 10 Right 7° Left 12° Right

Groundspeed 275 285 465

W/V 282/57 289/39 242/103

The DOPPLER DRIFT may be given on one heading and the DOPPLER GROUNDSPEED on another. In this case the W/V can only be solved by the manual nav computer. Given:

1000 z 1012 Z

Heading 055° (T) Heading 010° (T)

TAS 250 Kts Doppler Drift 10° Right Doppler GS 235 Kts

Method 1. Set TAS 250 Kts at CENTRE 2. Set HEADING 055° at TRUE INDEX 3. Draw 10 Right DRIFT LINE 4. Set HEADING 010° at TRUE INDEX 5. Draw arc of GROUNDSPEED 235 Kts 6. Position the intersection of the DRIFT and GROUNDSPEED lines BELOW the CENTRE CIRCLE. 7. Read off WIND DIRECTION 303° against the TRUE INDEX. 8. Read off WIND SPEED 50 Kts along the CENTRE LINE.



Page 90

Given:

1800 Z Heading 120° (T) TAS 200 Kts Doppler Drift 12 Left 1812 Z Heading 055° (T) Doppler GS 250 Kts 1825 Z The Groundspeed on Heading 335° is :-

Method: 1. 2. 3.

Calculate W/V 232/50 as above Set Heading 335° at TRUE INDEX and TAS 200 Kts at CENTRE Read off DRIFT 13° Right and GROUNDSPEED 218 Kts

MEAN W/V The following winds are forecast for a climb to cruising altitude:045/25 080/45 120/55 The mean W/V for the climb is :Method: Select a vacant area on the chart and start from the intersection of a Meridian and Parallel of Latitude. This can best be done on normal ruled paper using the lines as reference and a suitable scale. Draw the three wind vectors to scale, from head to tail. Join the end (tail) of the third wind vector to the starting point (head) and measure the wind direction. Measure the length of the vector and divide by the number of W/V’s to give the wind speed. The above method is used to calculate the mean W/V at cruising altitude when several W/V are given for a route. 091º/36.7



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ELECTRONIC NAVIGATION COMPUTERS TO CALCULATE TRACK & GROUNDSPEED (GS) ON THE PATHFINDER

Given

Heading 090°

TAS240Kts

W/V 010/60

Method: 1.

Select WIND function

2.

Enter HEADING 090°

3.

Enter WIND SPEED 60 Kts as GROUNDSPEED

4.

Enter TAS 240Kts

5.

Enter WIND DIRECTION 010° as CRS (TRACK)

6.

Computed W/V 104/237

7.

WIND DIRECTION 104° is the TRACK

8.

WIND SPEED 237 Kts is the GROUNDSPEED

TO CALCULATE MEAN W/V ON PATHFINDER/SPORTY: Method: USE

Req TAS function

1.

Enter 1st W/V in

W Dir W Spd

2.

Enter 2nd W/V in

Crs GS

Answer Hdg = Wind Direction TAS = Wind Spd 3.

Enter 3st W/V in

] remember this ]

W Dir W Spd

Enter Hdg (from 2) in Crs Enter TAS (from 2) in GS ANSWER Mean W/V Hdg = Mean Wind Direction TAS = Mean Wind Speed (÷ by number of winds)



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Mean wind can also be computed on the back of the Flight Computer using the square grid. Each wind is plotted separately and the resultant wind speed must be divided by the number of winds. TAS CALCULATION

Given:

RAS 140 Kts. Pressure Altitude 8000 feet. OAT

+20°C

Using the AIRSPEED WINDOW set pressure Altitude 8000 feet against OAT +20°C Against RAS 140 on the INSIDE SCALE, read off TAS 164 Kts on the OUTSIDE SCALE.



Page 93

AIRSPEED COMPRESSIBILITY CORRECTION ELECTRONIC CALCULATORS Electronic calculators correct for the compressibility error at high speeds. Given OAT or Corrected OAT use PLANNED TAS Given Indicated OAT use ACTUAL TAS MANUAL FLIGHT COMPUTER Given: Pressure Altitude 20 000 feet

OAT -23C

RAS 320kts

Using the AIRSPEED WINDOW set Pressure Altitude 20 000 feet against OAT -23C Against RAS 320 on the INSIDE SCALE Read off TAS 440 kts on the OUTSIDE SCALE. RAS 320 and TAS 440 kts are too high due to compressibility, use the correction factor from the table below.

ALTERNATIVE METHOD (EQUIVALENT AIR SPEED EAS) RAS 320 x 0.97 = EAS 310.4 Using the AIRSPEED WINDOW set Pressure Altitude 20 000 feet against OAT -23°C Against EAS 310.4 on the INSIDE SCALE Read off TAS 427 kts on the OUTSIDE SCALE. ELECTRONIC CALCULATORS If given OAT or COAT use PLAN TAS or PLAN MACH If given IOAT use ACT TAS or ACT MACH



Page 94

CHAPTER 6 LAMBERT'S PLOTTING CHART The chart used for the South African Commercial Pilot's plotting examination is the Lambert's Conformal Conic of Southern Africa. Scale 1:5 000 000 with Standard Parallels of S 20:20 and S 33:40. The Chart Convergency Factor is 0.45 which is the sine of the Parallel of Origin S 27:00. Straight lines drawn on the chart are considered to be GREAT CIRCLES for all practical purposes, which is the prime advantage of the chart, especially when plotting radio bearings. The other main advantage of the chart is that Great Circle tracks can be flown which are shorter than Rhumb Line tracks. This is useful when using Great Circle navigation systems such as INS, and GPS . The main disadvantage of the chart is when Rhumb Line navigation is used (flying constant headings or tracks). This is overcome by splitting the Great Circle track into short segments of 200 to 300 nautical miles or perhaps 5º of Longitude and using the MID MERIDIAN technique. MEASUREMENT OF DISTANCES USE THE VERTICAL LATITUDE SCALE ONLY TO MEASURE DISTANCES IN NAUTICAL MILES 1° OF LATITUDE = 60 NM

1° OF LATITUDE HAS 12 INCREMENTS OF 5 NM EACH

Errors in distance measurement can easily occur. It is suggested that every distance is measured twice as a check. Each VOR has a feather indicating Magnetic North. There is a warning on the chart that these feathers should not be used for plotting purposes due to possible inaccuracies.



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MID MERIDIAN TECHNIQUE - GREAT CIRCLE NAVIGATION Draw the straight line GREAT CIRCLE TRACK from A to B. Select the nearest Meridian to the mid point along the track, this is the MID MERIDIAN WHERE THE MEAN GREAT CIRCLE TRACK IS MEASURED.

The usual exam question is -the MEAN MAGNETIC HEADING from A to B is:  Measure the MEAN TRUE TRACK: at the MID MERIDIAN.  Using the TAS and W/V calculate the mean TRUE HEADING.  Apply MAGNETIC VARIATION at the MID MERIDIAN. Other exam questions are :- the INITIAL or FINAL MAGNETIC TRACK or HEADING from A to B is:      

Measure the INITIAL TRUE TRACK: at the nearest MERIDIAN to A. Using the TAS and W/V calculate the INITIAL TRUE HEADING. Apply MAGNETIC VARIATION of the first ISOGONAL along TRACK. Measure the FINAL TRUE TRACK at the last MERIDIAN before B. Using the TAS and W/V calculate the FINAL TRUE HEADING. Apply MAGNETIC VARIATION of the last ISOGONAL along TRACK.

NAVIGATION LOG A Navigation Log is supplied for the exam. Its use is optional and it is not inspected by the examiner. Use of the log does help in answering questions. CPL NAVIGATION CPL DOC8 Revision 1/1/2001


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BASIC PLOTTING SYMBOLS Several lines drawn on a chart become meaningless after a short while. A Navigation Plot shows the history of a flight. It is a legal document, even if it is in pencil. Every line must have ARROWS denoting what the line represents. Every symbol must have a time. ONLY TRUE VALUES MAY BE PLOTTED - NEVER MAGNETIC AIR VECTOR or TRUE HEADING The direction that the aircraft: is steering GROUND VECTOR or TRUE TRACK The track or path of the aircraft over the ground WIND VELOCITY W/V The direction FROM WHICH THE WIND BLOWS (270°) FIX A POSITIVE GROUND POSITION The aircraft was over the position shown at 1015 Z



1015

AIR POSITION The Air Position of the Aircraft at 1320 Z

+

1320



1550

DR POSITION The calculated or assumed position of the aircraft at 1550 Z RADIO POSITION LINE A bearing from a VOR. VDF or NDB at 1608 Z The aircraft is somewhere along the position line

1608

TRANSFERRED POSITION LINE The 1608 Z Position Line transferred to a Position Line at 1622 Z to give a fix

1622

-

AIR PLOT - THE PATH OF THE AIRCRAFT RELATIVE TO THE AIR

-

TRACK PLOT - THE PATH OF THE AIRCRAFT OVER THE GROUND

ZERO WIND CONDITIONS If an aircraft is flying in zero wind conditions navigation would be easy because HEADING = TRACK & TAS = GROUNDSPEED (GS) Unfortunately most of the time there is wind (occasionally 200 Kts or more). CPL NAVIGATION CPL DOC8 Revision 1/1/2001


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FLIGHT PLAN Plotting starts from a known point of departure and the initial flight conditions are taken from a flight plan. Navigation from the point of departure to destination will consist of fixing the aircraft's position, and if off track, calculating a new heading and ETA for the destination. TRACK PLOT An aircraft is to fly from A to B as per flight plan. Some time later, after maintaining a constant heading a fix is obtained at C.

This is the simple I in 60 rule application that was covered in Navigation General Track Error  __________ = 60

Distance Off 9 run _______________ Distance Run 108 nm

Track Error 5 Alter Heading 5 Right to parallel Track

Track Error  ___________ = 60

Distance Off 9 nm ________________ Distance To Go 135 nm

 = 4 Alter Heading a further 4 Right to B

The procedure is effective and accurate for short range navigation if the angles are small and no alteration of heading has been made between A and C. To make life easier and no calculation required try the following :TMG

The Track Error is actually Drift Error when the forecast W/V is in error which is of course usual. The alteration of heading is made without knowledge of the actual W/V. If the angles are large then the actual W/V must be found and used to alter heading for B.



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TRACK AND GROUNDSPEED W/V The Vector Triangle or Triangle of Velocities consists of three vectors, namely Drift Angle

True Heading & TAS W/V True TRK & GS

knowing any four of the six values, the other two can be calculated. Example 1. TAS 240 Kts Heading 035(T) The W/V is By computer WV 260/47 Kts

Track 042(T) GS 275 Kt

Sector Track Distance A to B 105124nm B to C 157 102nm

Drift 8Rt

Example 2. TAS 215 215

Time 31 mins

Assuming the W/V remains constant the heading to steer and the elapsed time from B to C is? Firstly calculate the W V TAS 215 Kts Heading 097(T) Then B to C Heading 158(T)

Track 105 GS 255 Kt

GS 240 Kt Time 24 min

W/V 333/40

Example 3. An aircraft passes overhead VOR PON at 0915 Z maintaining Radial 123 Heading 132(M) TAS 220 Kt, Variation 18W. At 0935 Z PON DME indicates 85 nm. The W/V is? Heading 114(T)

TAS 220 Kts Track 105(T) GS 255 Kts


W/V 243/51 kts


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TRACK & GROUNDSPEED W/V An aircraft is overhead A at 0900 Z. Heading 082(T), TAS 200 Kts. A fix is obtained at 0945 Z. Calculate the mean W/V affecting the aircraft.

Heading 082(T) TAS 200 Kts Measure Track Made Good 090 (at Mid-Meridian) Measure Distance A to fix 180 nm in 45 minutes. Calculate GS 240 Using the Pathfinder or Nav. Computer enter Heading, TAS. Track & GS Calculate W/V 304/50

TRACK PLOT

Calculate the Track & GS W/V from 0900 to 0945. Draw a line from A to the fix (this is the TMG, Track Made Good) and extend for 12 minutes at GS 240, that is 48 nm to give a DR position at 0957. From the 0957 DR position draw the new track to B and calculate the new heading and ETA using the W/V found.



Page 100

AIR PLOT With the knowledge of the aircraft's TRUE HEADING, TAS and W/V, an AIR PLOT starting from the point of departure enables a pilot to calculate the position of the aircraft at any time regardless of the number of alterations of heading made.

+

AIR POSITION (The position of the aircraft in zero wind conditions) DR POSITION (The ground position of the aircraft)

1000 1030

Overhead A. Heading 090(T), TAS240kts, W/V 360/40 kts. Alter Heading for B

1. 2.

From A plot Heading 090 (T) Plot the 1030 Air Position 120 nm along the Air Vector (30 minutes of TAS 240). This is the position of the aircraft in ZERO W/V conditions, but the W/V is 360/40. From the Air Position, plot the Wind Vector DOWNWIND. The Wind is blowing FROM 360 The length of the vector is proportional to time, 30 minutes of 40 kts = 20 nm. This is the DR DEAD RECKONING (GROUND) position of the aircraft at 1030. Join the DR position to B, measure the track and distance. Compute with TAS and W/V. Determine new Heading and ETA for B.

3.

4. 5. 6.

0900 Overhead X. Heading 090(T), TAS I80 kts, W/V 360/40 0930 Alter Heading 135(T) 0950 Alter heading for Y

X



Page 101

NOTE

The AIR PLOT has been running from 2100 to 2224 (I hour 24 minutes). If the W/V is 330/30 the length of the WIND VECTOR will be 42 nm (30 kts for 1 hour 24 minutes).



Page 102

AIR PLOT W/V

The information required for an Air Plot is readily available. that is TRUE HEADING and TAS. 1000 Overhead A, Heading 080(T). TAS 240 kts. 1030 Overhead B. 1030 AIR POSITION

Plot the Heading 080(T) from A using the nearest meridian. This is the AIR VECTOR. Along the AIR VECTOR plot the AIR DISTANCE flown 120 nm (TAS 240 kts for 30 minutes).

This is the AIR POSITION at 1030, and would be the position of the aircraft in zero wind conditions. But the aircraft is overhead B. the reason being the WIND VELOCITY. Join the AIR POSITION to the FIX. This is the W/V.

Measure the WIND DIRECTION 330 from the nearest meridian. The wind is blowing from 330. Measure the WIND VECTOR 25 nm. the Wind has affected the aircraft for 30 minutes

which gives a WIND SPEED of 50 kts. W/V 330/50.



Page 103

Further example of AIR PLOT W/V 1315 1355 1430

Overhead C, Heading 080(T), TAS 180 kts. Alter Heading 120(T) Overhead D, the W/V is:1355 AIR POSITION

From C plot Heading 080(T) Plot the 1355 AIR POSITION (TAS 180 kts for 40 minutes = 120 nm) From the 1355 AIR POSITION plot the new heading 120(T) using the nearest meridian. Plot the 1430 AIR POSITION (TAS 180 kts for 35 minutes = 105 nm) Join the 1430 AIR POSITION to the FIX. Measure the WIND DIRECTION 090 Measure the WIND VECTOR 56 nm in 75 minutes = Wind SPEED 45 kts.


W/V 090/45


Page 104

RADIO BEARINGS Q CODE

QTE QDR QUJ QDM

TRUE bearing FROM the station MAGNETIC bearing FROM the station TRUE track TO the station MAGNETIC track TO the station

Take the shortest route to change one bearing to another

QDM

 Variation

180 QDR

QUJ 180

Variation

QTE

Take the shortest route to change one bearing to another

VDF VOR

Apply station Variation, calculate QTE and plot VOR Radials are Magnetic bearings RMI Readings are Magnetic tracks to the VOR Apply station Variation, calculate QTE and plot

QDR QDM

RMI BEARINGS (VOR & ADF) Usually termed RMI READING which is


QDM


Page 105

ADF BEARINGS ADF Relative bearings are measured from the Fore and Aft axis of the aircraft. ADF Relative bearings must be converted into True Bearings (QTE) before they can be plotted on a chart. RELATIVE BEARING + TRUE HEADING = QUJ  180 = QTE MAGNETIC VARIATION AT THE AIRCRAFT IS ALWAYS USED WITH ADF BEARINGS CHART CONVERGENCE IS NOT APPLIED TO ADF BEARINGS ON THE SA CHART

ADF bearing Heading (T)+ QUJ  QTE

095 Relative 057 152 (T) TO NDB 180 332 (T) FROM NDB

ADF bearing Heading (T) QUJ Subtract QUJ  QTE


200 Relative 318 518 360 158 (T) TO NDB 180 338 (T) FROM NDB


Page 106

USE OF SINGLE POSITION LINES GROUNDSPEED CHECK & REVISED ETA A Position Line at right angles (80) to track may be used as a Groundspeed check 0800 Overhead A en-route to B, Distance 166 nm, 0824 QTEC180

Measure the distance from A to the position line. Calculate Groundspeed Measure the distance from the position line to B and calculate time and ETA. THIS PROCEDURE IS VERY IMPORTANT AT ALL TIMES TRACK CHECK A bearing from a radio facility that the aircraft has overflown or departed from (Back bearing) is an indication of the aircraft's track or TMG (Track Made Good) 1205 1220

Overhead ABC, Heading 090(T) NDB ABC bears 172 Relative 172 Relative + Heading 090 = QUJ 262 - 180 = QTE 082



Page 107

DRIFT During the 1939 - 1945 war and for some years thereafter most aircraft were equipped with drift sights. They were optical devices that protruded through the side of the aircraft and had a grid that could be aligned with objects on the ground tracking below the aircraft. An accurate drift angle could be measured. Although drifts sights are obsolete, drift may be given in an exam question. 1315 1330 1335

Overhead X, Heading 090(T) Drift 8 Right QTE CP360

x

CP The drift is applied to the Heading to give a Track Made Good and is used as a position line to give a Fix at 1335. Note that the drift position line at 1330 is not transferred but extended. MULTI - POSITION LINE FIX Two or more position lines may be used to construct a fix. The ideal situation is that two position lines are obtained at the same time, preferably at 90 to each other. 0956 JSVVOR/DME Radial l06 D105nm

JSV



Page 108

TRANSFER OF POSITION LINES A position line is usually a bearing of the aircraft from a radio facility. If the radio station were moved along a track parallel to the aircraft's track and at the same groundspeed. the bearing of the aircraft from the radio station would remain constant

QTE 340. Track 090. Groundspeed 240 kts.

The aircraft at position A at 0900 Z obtains a QTE of 340 from VDF station X. At 0912 the aircraft will have flown 48 nm along track to position B. If the VDF station is imagined to travel from X to Y at the same speed as the aircraft, then XY is equal and parallel to AB and the line joining Y to B will be an imaginary position line parallel to AX. The distance AB or XY is known as the run. The line BY drawn through the aircraft's position at 0912 is known as a transferred position line and has two arrows and no time. In practice 48 nm is measured forward from where the 0900 position line cuts the track and the position line redrawn through this point.



Page 109

THREE POSITION LINE FIX (RUNNING FIX)

From 1200 to 1230, 120 nm in 30 minutes - GS 240 Kts. Transfer the 1223 position line 17 minutes at GS 240 Kts = 68 nm along track Transfer the 1230 position line 10 minutes at GS 240 kts = 40 nm along track NOTE: If a GS check cannot be made the DR groundspeed may be used.

TRANSFER OF CIRCULAR (DME) POSITION LINES

0810

Overhead A, Track 090(T), Groundspeed 180 kts

0830

JSV DME Range 58 nm

0842

JSV DME Range 125 nm

The transfer of DME circular position lines is achieved by the track & GS method of moving the DME station along a line parallel to the aircraft's track at the aircraft's GS and re-plotting the original range from the transferred DME position.



Page 110

THREE POSITION LINE FIX (RUNNING FIX) NO POINT OF ORIGIN The following radials are obtained from TIMBUKTU VOR TIM Aircraft track 045(T), Groundspeed I50kts. 0815 VOR TIM 288 0824 VOR TIM 321 0835 VOR TIM 348

TIM As there is no fix from which to start the plot, the 045 track is drawn to cut the three position lines. It is parallel to the actual track of the aircraft. A groundspeed check cannot be carried out so the DR groundspeed is used. Transfer the 0815 P/L 20 mins of GS 180 = 60 nm along track Transfer the 0824 P/L 11 mins of GS 180 = 33 nm along track



Page 111

CONSTANT SPEED (RAS) CLIMB This is the normal climb technique used by commercial aircraft. Navigation wise the most important parameter required for a climb to cruising altitude is the mean climb TAS. This will occur at the mid-point of the climb in TIME and not ALTITUDE. The mid-point of the climb occurs at approximately two-thirds of the climb for both piston and turbine aircraft. Example 1. Climbing from Sea Level to FL 240 at RAS 175 kts and a mean rate of climb of 800 feet per minute. Temperature is ISA + 10C 24 000 feet x 2/3 = ISA at Sea Level 16000ft x 2C/1000 ISA at FL 160 ISA + 10 OAT By computer

16 000 feet +15C -32 -17C +10 - 7C

FL160 OAT-7C RAS 175 kts

TAS 227 kts

Example 2. Climbing from 6000 feet to FL 270 at RAS 225 and a mean rate of climb of 1250 feet/minute. Temperature Deviation ISA +13C 27 000 feet 6 000 feet 21 000 feet x 2/3 = 14 000 feet + 6 000 feet (Initial climb altitude) 20 000 feet Mean Climb Altitude ISA at Sea Level + 15C 20000ft x 2C/1000 - 40 ISA at FL 200 - 25C ISA + 13 + 13 OAT - 12C By computer FL200 OAT-12C RAS 225kts

TAS 311 kts

The above calculations were made with an electronic calculator which corrects for compressibility. If a manual navigation computer is used the compressibility correction must be made According to the table below.



Page 112

THE DESCENT A CONSTANT RATE OF DESCENT is normally used. The temperature and W/V at the mid altitude of the descent are used to calculate TAS and Groundspeed. Example: An aircraft cruising at FL 370 at GS 495 obtains a fix at 1000 Z which gives a distance of 230 nm to go to destination. Descent details:

Rate of Descent 2000 feet per minute Mean Descent GS 360 kts

Plan a descent to arrive overhead the destination at FL 90. The Navigation Log is useful for descent calculations.

AVERAGE WIND COMPONENTS  The average WC is the difference between the average TAS and the average GS.  The average TAS is calculated by the Total Air Nautical Miles flown divided by the Total Time.  The average GS is calculated by the Total Ground Nautical Miles flown divided by the Total Time. GNM ANM Sector TAS WC GS DIST TIME DIST Average TAS A to B 150 -10 140 280 2:00 300 780 ANM4:50 = 161 Kts B to C 160 -20 140 210 1:30 240 C to D 180 -30 150 200 1:20 240 Average GS 690 GNM4:50=143 Kts

690

4:50

780 Average WC 18 Kt HW

ANM can also be written NAM nautical air miles



Page 113

GENERAL NAVIGATION QUESTIONS 1.

An aircraft is flying from A to B, distance 212 nm. After flying 135 nm the aircraft is 9 nm Left of track. The alteration of heading to fly to B is a) b) c)

2.

An aircraft is flying from C to D. track 270(T). distance 345 nm. After flying 135 nm the aircraft is 14 nm Left of track. The new track to D is a) b) c)

3.

WV 184/42 WV 261/42 WV 004/42

X to Y, Distance 123 nm. FL 130, OAT -7C. Headwind Component 25 Kts. The required RAS to fly from X to Y in 42 minutes is a) b) c)

7.

257(M) 262(M) 267(M)

At 0900 Z an aircraft is heading 137(T) to maintain Radial 151º from VOR CDX. TAS 220 Kts Variation 22W. At 0919 Z DME CDX indicates 78 nm The W/V is a) b) c)

6.

257(M) 228(M) 235(M)

An aircraft is heading 087(M) to maintain a track of 062(T). Variation 20W. The heading to fly the reciprocal track is a) b) c)

5.

274(T) 276(T) 280(T)

An aircraft leaves F heading 041(M) in order to maintain a track of 045(M). After flying 220 nm the aircraft is 11 nm Right of track. The heading to steer to return to F is: a) b) c)

4.

8 Right 11 Right 14 Right

160 Kts 165 Kts 171 Kts

An aircraft flies from P to Q. distance 372 nm in 1 hour 50 minutes at TAS 180 Kts. In order to fly the return trip in 1 hour 45 minutes the TAS should be a) b) c)

215 Kts 225 Kts 235 Kts



Page 114

8.

Sector Track Distance X to Y 177 153 nm Y to Z 127 109 nm

TAS 205 205

Drift 5L

Time 51 mins

Assuming the W/V remains constant the heading to steer from Y to Z is a) b) c) 9.

An aircraft is heading 315(M) to maintain a track of 287(T). Variation 20W. The heading to fly the reciprocal track is a) b) c)

10.

025(M)/56 257(M)/56 057(M)16

An aircraft heading 115(M) TAS 260 Kts. obtains the following readings from a VOR/DME 0743 Z Radial 253 0748 Z Radial 208 DME 21 nm 0753 Z Radial 163 The W/V is? a) b) c)

12.

119(M) 127(M) 135(M)

At 0900 Z an aircraft is overhead NDB NL heading 145(M), TAS 350 Kts. At 0920 Z the aircraft is overhead NDB CD. 125 nm from NDB NL and the ADF tuned to NL reads 172 Relative. The W/V is (a) (b) (c)

11.

130 135 140

022(M)/25 Kts 039(M)/32 Kts 057(M)/16 Kts

After flying on a heading of 040(C) for 2 hours at GS 150 Kts. a pilot discovers that he has not made allowance for Variation 6E and Deviation 2W The aircraft is off track by a) b) c) d)

20 nm to the right 20 nm to the left 10 nm to the right 10 nm to the left



Page 115

PLOTTING CHART HEADING & ETA QUESTIONS 1.

An aeroplane departs CAPE TOWN CTV (S 33:58 E 018:36) on a direct track to DURBAN DNV(S 29:46 E 031:00), TAS210Kts, W/V 310/45. The mean magnetic heading to steer is :(a) 080 (M) (b) 087 (M) (c) 094 (M)

2.

An aeroplane departs CAPE TOWN CTV (S33:58 E 018:36) at 1700 Z on a direct track to MARGATE MG (S 30:52 E 030:20), TAS 220 Kts, W/V 300/55. The ETA MARGATE is :(a) 1928 Z (b) 1934 Z (c) 1940 Z

3.

An aeroplane departs LADYSMITH LYV (S 28:33 E 029:45) on a direct track to SUTHERLAND SLV (S32:24 E 020:39), TAS 285 Kts, W/V 320/45. The mean magnetic heading to steer is :(a) 274 (M) (b) 281 (M) (c) 286 (M)

4.

An aeroplane departs MASERU MZV (S 29:46 E 027:34) at 1615Z on a direct track to CAPE TOWN (S 33:58 E 018:36), TAS180, W/V 300/50. The ETA abeam VICTORIA WEST VWV (S31:24 E 023:08) is:(a) 1750 Z (b) 1757 Z (c) 1804 Z

5.

An aeroplane departs CAPE TOWN CTV (S 33:58 E 018:36) on a direct track to EAST LONDON ELV(S 33:02 E 027:53), TAS180Kts, W/V 330/45. The initial magnetic heading to steer is :(a) 089 (M) (b) 094 (M) (c) 099 (M)

6.

An aeroplane departs GEORGE GGV (S 34:00 E 022:22) at 0730 Z on a direct track to KEETMANSHOOP KTV(S 26:35 E 018:05), TAS180Kts, W/V 225/45 The ETA at the WINDHOEK FIR Boundary (S 27:30) is :(a) 0948 Z (b) 0957 Z (c) 1008 Z



Page 116

7.

An aeroplane departs MMABATHO MMV (S 25:51 E 025:30) at 1330 Z on a direct track to KEETMANSHOOP KTV(S 26:35 E 018:05), TAS 195Kts, W/V 255/40. The ETA at the WINDHOEK FIR Boundary (E 020:00) is :(a) 1526 Z (b) 1536 Z (c) 1546 Z

8.

An aeroplane departs CAPE TOWN CTV (S 33:58 E 018:36) on a direct track to DURBAN DNV (S 29:46 E 031:00). The true track of the aircraft just before reaching DURBAN is :(a) 060 (T) (b) 063 (T) (c) 066 (T)

9.

An aeroplane departs ALEXANDER BAY ABV (S 28:34 E 016:30) on a direct track to LADYSMITH LYV (S28:33 E 029:45), FL 130, OAT-5C,RAS 175 Kts. W/V 345/50, Compass Deviation +2E. The mean compass heading to steer is :(a) 100 (C) (b) 097 (C) (c) 093 (C)



Page 117

PLOTTING QUESTIONS 1.

1000 Z

Overhead BLOEMFONTEIN BLV(S 29:06 E 026:20) on a direct track to SUTHERLAND SLV(S 32:24 E 020:40)

1100 Z

Overhead VICTORIA WEST VWV (S31:23 E 023:08)

The alteration of heading at 1100 Z to reach SUTHERLAND is :(a) 8 Right (b) 13 Right (c) 19 Right

2.

0930 Z

Overhead NEW HANOVER NHV (S 29:21 E 030:33) on a direct track to GEORGE GGV (S 34:00 E 022:22)

1102 Z

Overhead COOKHOUSE CH (S 32:46 E 025:46)

The alteration of heading at 1102 Z to reach GEORGE is :(a) (b) (c)

3.

9 Right 15 Right 22 Right

1300 Z

Overhead VICTORIA WEST VWV (S 31:24 E 023:09) en route to MARGATE MG (S 30:53 030:19)

1350 Z

Overhead BURGERSDORP BDV (S 30:59 E 026:16)

The alteration of heading at 1350 Z to reach MARGATE is :(a) 10 Right (b) 15 Right (c) 20 Right

4.

1147 Z

Overhead KEETMANSHOOP KTV (S 26:34 E 018:08), FL 210, OAT -20C, RAS 180 Kts, Heading 107 (M).

1312 Z

Overhead MMABATHO MMV (S 25:51 E025:31)

The mean W/V since 1147 is:(a) 137/30 (b) 317/40 (c) 218/43



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5.

0915 Z

Overhead KIMBERLEY KMV (S 28:49 E 024:46) Heading 281 (M), FL 200. OAT -16C, RAS 207.

1059 Z

Overhead ALEXANDERBAY ABV (S28:34 E 16:31)

The mean W/V since 0915 is :(a) 332/57 (b) 215/59 (c) 319/53

6.

1343 Z

Overhead WOLFKOP WKP (S30:13 E 017:11) Heading 122 (M), FL230, OAT-23C, RAS162Kts.

1525 Z

Overhead COOKHOUSE CH (S 32:46 E 025:45)


7.

0900 Z

Overhead ULCO UCV (S 28:24 E 024:20) Heading 097 (T). FL 230. OAT -17C. RAS 195 Kts.

0955 Z

Overhead LADYSMITH LYV (S 28:32 E 029:45)

1007 Z

Alter Heading for DURBAN DNV (S 29:46 E 031:00)

The mean magnetic heading to steer at 1007 Z is :(a) 181 (M) (b) 201 (M) (c) 221 (M)

8.

1520 Z

Overhead VICTORIA WEST VWV (S31:24 E 023:07) Heading 055 (T), FL170. OAT-10C, RAS 160Kts.

1624 Z

Overhead MASERU MZV (S 29:26 E 027:34)

1636 Z

Alter Heading for HEIDELBERG HGV (S 26:43 E 028:16)




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9.

1235 Z

Overhead BISHO BOV (S 32:55 E 027:20), Heading 261 (T), FL 200, OAT -8C, RAS 175 Kts.

1349 Z

Overhead GEORGE GGV (S 34:00 E 022:22)

1401 Z

Alter Heading for CAPE TOWN CTV (S 33:58 E 018:36)

The ETA CAPE TOWN is :(a) 1429 Z (b) 1437 Z (c) 1445 Z

10.

2110 Z

Overhead UPINGTON UPV (S 28:24 E 021:15) FL 330, OAT -42C, Mach 0.82, Heading 066 (M)

2139 Z

Overhead MMABATHO MMV (S 25:51 E 024:30)

2145 Z

Alter Heading for HOEDSPRUIT HS (S 24:22 E 03 1:02)

The ETA HOEDSPRUIT is :(a) 2212 Z (b) 2220 Z (c) 2227 Z

11.

1522 Z

Overhead PORT ELIZABETH PEV (S 33:57 E 025:35' Heading 032 (T). TAS 275 Kts. W/V 230/50.

1607 Z

Alter heading for MARGATE MG (S 30:53 E 030:20)


12.

1100 Z

Overhead KIMBERLY KMV (S 28:48 E 024:46) Heading 008 (T), TAS310Kts. W/V 235/50.

1125 Z

Alter heading for JOHANNESBURG JSV (S 26:09 E 028:14)

The ETA at JOHANNESBURG is :(a) 1151 Z (b) 1201 Z (c) 1211 Z



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13.

1315Z

Overhead DURBAN DNV (S 29:56 E 030:00) Heading 355 (T), TAS320Kts, W/V 310/45.

1403 Z

Alter heading for MAPUTO VMA (S 25:55 E 032:35)

The mean magnetic heading to steer for MAPUTO at 1403 is :(a) 092 (M) (b) 102 (M) (c) 112 (M)

14.

0900 Z

Overhead ALEXANDER BAY ABV (S 28:34 E 016:30) Heading 042 (T), TAS 285 Kts, W/V 300/40.

0942 Z

Alter heading for UPINGTON UPV (S 28:24 E 021:16)

The ETA at UPINGTON is :(a) 1012 Z (b) 1019 Z (c) 1427 Z

15.

0627 Z

Overhead CAPE TOWN CTV (S 33:58 E 018:36) Heading 067 (M), TAS300Kts. W/V 270/45.

0703 Z

Alter heading 127 (M).

0725 Z

Alter heading for JAGERSFONTEIN JF (S 29:47 E 025:25)


16.

2200 Z

Overhead LADYSMITH LYV (S 28:32 E 029:44) Heading 282 (M), TAS 300 Kts, W/V 010/55.

2235 Z

Alter heading 324 (M).

2310Z

Alter heading for UPINGTON UPV (S 28:24 E 021:16).

The ETA at UPINGTON is :(a) 2328 Z (b) 2334 Z (c) 2342 Z



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17.

2135 Z

Overhead MMABATHO MMV (S 25:51 E 025:32) Heading 290 (M), TAS 285 Kts, W/V 225/50.

2213 Z

Alter heading 335 (M)

2238 Z

Alter heading for DORDABIS DSV (S 22:50 E 017:55)


18.

1530 Z

Overhead ORAPA OP (S 21:18 E 025:18) Heading 055 (M), TAS 240 Kts, W/V 060/45.

1624 Z

Alter heading 119(M)

1714 Z

Alter heading for BEIRA VBR (S19:47 E 034:55)

The ETA at BEIRA is :(a) 1817 Z (b) 1827 Z (c) 1837 Z

19.

0500 Z

Overhead NIEUWOUDVILLE NVV (S 31:21 E 019:02) Heading 347 (T), TAS 195 Kts.

0542 Z

Alter heading 061 (T)

0621 Z

Overhead UPINGTON UPV (S 28:23 E 021:17)

The mean W/V since 0500 Z is :(a) W/V 295/30 (b) W/V 320/27 (c) W/V 345/25 20.

2213 Z

Overhead GEORGE GGV (S 34:00 E 022:22) Heading 337 (T), TAS 175 Kts.

2248 Z


2321 Z

Overhead VICTORIA WEST VWV (S 31:24 E 023:08)

The mean W/V since 2213Z is :(a) W/V 280/39 (b) W/V 300/44 (c) W/V 320/50 CPL NAVIGATION CPL DOC8 Revision 1/1/2001


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21.

0747 Z

Overhead BISHO BOV (S 32:55 E 027:15) Heading 033 (T), TAS 240 Kts.

0837 Z


0857 Z

Overhead LADYSMITH LYV (S 28:33 E 029:46)

The mean W/V since 0747 is :(a) W/V 225/42 (b) W/V 250/35 (c) W/V 282/38

22.

1522 Z

Overhead GABORONE GBV (S 24:32 E 025:56) Heading 337 (M), TAS 220 Kts.

1556 Z

Alter heading 012 (M), TAS 240 Kts.

1622 Z

Overhead ORAPA OR (S 21:16 E 024:18)

The mean W/V since 1522 is :(a) W/V 258/45 (b) W/V 283 40 (c) W/V 327 46

23.

1132 Z

Overhead LIVINGSTONE VLR (S 17:46 E 025:49) Heading 113 (M), TAS 245 Kts.

1216 Z


1303 Z

Overhead CHIREDZI CZ (S 21:00 E 031:30)

The mean W/V since 1132 Z is :(a) W/V 288/40 (b) W/V 265/35 (c) W/V 231/30



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24.

1410 Z

Overhead SKUKUZA SZ (S 25:00 E 031:35) Heading 013 (M), TAS 310 Kts.

1429 Z


1506 Z

Overhead VILANCULOS VI (S 21:58 E 035:18)

The mean W/V since 1410 Z is :(a) W/V 270/50 (b) W/V 290/43 (c) W/V 315/42

25.

1410 Z

Overhead UPINGTON UPV (S 28:24 E 021:16) on a direct track to KROONSTAD KS (S 27:41 E 027:16) Heading 109 (M), TAS 180kt

1522 Z

VRYBURG NDB VB (S 26:59 E 024:42) bears 265 Relative

The revised ETA for KROONSTAD is :(a) 1602 Z (b) 1607 Z (c) 1612 Z

26.

0415 Z

Overhead BUGERSDORP BDV (S 30:59 E 026:17) enroute to GEORGE GGV (S 34:00 E 022:22). Heading 259 (M), TAS 210

0446 Z

COOKHOUSE NDB CH (S 32:46 E 025:46) bears 259 Relative

The revised ETA for GEORGE is :(a) 0536 Z (b) 0539 Z (c) 0544 Z

27.

1652 Z

Overhead EAST LONDON ELV (S 33:02 E 027:53) enroute to LADYSMITH LYV (S 28:33 E 029:45), Heading 013 (T), TAS190

1736 Z

MARGATE NDB MG (S 30:53 E 030:20) bears 098 Relative

The revised ETA for LADYSMITH is :(a) 1801 Z (b) 1808 Z (c) 1817 Z



Page 124

28.

1000 Z

Overhead HOTAZEL HZ (S 27:15 E 022:56) TAS 170 to. On a direct track to LADYSMITH LYV (S 28:33 E 029:44)

1057 Z

Overhead WELKOM WM (S 28:00 E 026:42) Increase TAS 195 Kts

The revised ETA for LADYSMITH is :(a) 1130 Z (b) 1138 Z (c) 1146 Z

29.

1605 Z

Overhead PORT SHEPSTON EZTZ (S 30:44 E 030:25) TAS 275 kts. On a direct track to PORT ELIZABETH PEV (S 33:59 E 025:37)

1657 Z

Overhead BISHO BOV (S 32:55 E 027:16) Reduce TAS 235 kts.

The revised ETA for PORT ELIZABETH is :(a) 1721 Z (b) 1729 Z (c) 1735 Z

30.

1725Z

Overhead ORAPA OR (S 21:18 E 025:18) TAS 230 kts. On a direct track to CHIREDZI (S 2l:00 E 031:30)

1752 Z

Overhead FRANCISTOWN FT (S 21:13 E 027:29) Increase TAS 255 kt;

The revised ETA for CHIREDZI is :(a) 1839 Z (b) 1846 Z (c) 1852 Z

31.

2100 Z

Overhead GABORONE GBV (S 24:32 E 025:55) on a direct track to THORNHILL VTL (S 19:26 E 029:53) TAS 300 Kts.

2139 Z

GREEFSWALD VOR GWV (S 22:15 E 029:26) Radial 272

2152 Z

GREEFSWALD VOR GWV (S 22:15 E 029:26) Radial 317

The ETA for THORNHILL is :(a) 2221 Z (b) 2228 Z (c) 2235 Z



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32

2100 Z

Overhead ALEXANDER BAY ABV (S 28:34 E 016:29) enroute to BURGERSDORP BDV (S 30:59 E 026:18). TAS 300 Kts.

2142 Z

UPINGTON VOR UPV (S 28:32 E 021:46) Radial 218

2158 Z

UPINGTON VOR UPV Radial 163

The position of the aircraft at 2158Z is :(a) S 29:50 E 022:26 (b) S 30:12 E 022:41 (c) S 29:30 E 022:05

33.

1515Z

Overhead CAPETOWN CTV (S 33:58 E 018:35) enroute to BURGERSDORP BDV (S 30:59 E 026: IS) TAS 240Kts.

1550 Z

GEORGE VOR GGV (S 34:00 E 022:22) QDM 179

1605 Z

GEORGE VOR GGV QDM 233

The position of the aircraft at 1605 Z is :-

34.

(a) S 32:58 (b) S 32:40 (c) S 32:07

E 023:05 E 023:15 E 023:31

1617Z

Overhead KEETMANSHOOP KTV (S 26:33 E 018:06) on a direct track to LADYSMITH LYV (S 28:32 E 029:44), GS 255 Kts.

1722 Z

UPINGTON UPV (S 28:24 E 021:15) DME124nm

1747 Z

KIMBERLEY KMV (S 28:48 E 024:46) DME71nm

The position of the aircraft at 1747 Z is :(a) S 27:36 E 024:33 (b) S 27:40 E 025:11 (c) S 27:58 E 025:43



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35.

0900 Z

Overhead CAPE TOWN CTV (S 33:58 E 018:35) enroute to BURGERSDORP BDV (S 30:59 E 026:18) GS 240 Kts.

0927 Z

GEORGE GGV (S 34:00 E 022:22) DME 105 nm

0959 Z

GEORGE GGV DME 105 nm


36.

2119Z

Overhead UPINGTON UPV (S 28:24 E 021:16) Heading 136 (M), TAS 215 Kts.

2135 Z

Observed drift 7 Left

2203 Z

KIMBERLEY KMV (S 28:48 E 024:46) Radial 226


37.

1000 Z

Overhead MASERU MZV (S 29:26 E 027:34) TAS 350 Kts, Heading 030 (T), Drift 11 Right.

1025 Z

Alter Heading 335 (T). Groundspeed 325 Kts.


38.

1215 Z

Overhead LADYSMITH LYV (S 28:24 E 029:44) TAS 300 Kts, Heading 270 (T), Drift 8 Left.

1250 Z

Alter Heading 325 (T), Groundspeed 270 Kts.

The mean W/V is :(a) 276/62 (b) 347/40 (c) 008/44



Page 127

39.

0625 Z

Overhead KIMBERLEY KMV (S 28:48 E 024:46) TAS 325 Kts, Heading 135 (T), Drift 7 Right.

0630 Z


The Groundspeed on a Heading of 020 (T) will be :(a) 265 kts (b) 285 kts (c) 305 kts

40.

0900 Z

Overhead POMFRET PF (S 25:51 E 023:31) on a direct track to VICTORIA WEST VWV (S 31:24 E 023:08)

0940 Z 0948 Z 0958 Z

KIMBERLEY KMV (S 28:47 E 024:46) Radial 320 KIMBERLEY KMV Radial 293 KIMBERLEY KMV Radial 255


41.

1121 Z

Overhead NIEUWOUDTVILLE NVV (S 31:22 E 019:03)on a direct track to KEETMANSHOOP KTV (S 26:34 E 018:06)

1153 Z 1205 Z 1216 Z

ALEXANDER BAY ABV (S 28:33 E 016:30) RMI reading 303 ALEXANDER BAY ABV RMI reading 279 ALEXANDER BAY ABV RMI reading 257


42.

The following VOR Radials were obtained from NIEUWOUDTVILLE NVV (S 31:21 E 019:02). Aircraft track 025 (T), GS 180Kts. 0900 Z 0912 Z 0927 Z

NVV VOR Radial 169 NVV VOR Radial 124 NVV VOR Radial 083

The position of the aircraft at 0927 Z is :(a) S 31:00 E 019:52 (b) S 30:50 E 020:13 (c) S 30:39 E 020:38 CPL NAVIGATION CPL DOC8 Revision 1/1/2001


Page 128

43.

The following VOR Radials were obtained from BURGERSDORP BDV (S 30:59 E 026:18) Aircraft track 337 (T), GS 240 Kts. 1124 Z 1140 Z 1156 Z

BDV Radial 221 BDV Radial 269 BDV Radial 316

The position of the aircraft at 1156Z is :(a) S 30:40 (b) S 30:30 (c) S 30:20 44.

E 025:30 E 025:07 E 024:45

The following VOR Radials were obtained from MMABATHO VOR MMV (S 25:51 E 025:32), Aircraft track 352(T), GS 350 Kts. 1342 Z 1352 Z 1406 Z

MMV Radial 237 MMV Radial 273 MMV Radial 318


45.

The following W/V are forecast for a climb to cruising altitude :270/25

250/35

210/50

The mean climb W/V is :(a) 250/37 (b) 240/40 (c) 235/33

46.

The following W/V are forecast for a climb to cruising altitude :045/27

070/40

100/55

The mean climb W/V is :(a) 080/38 (b) 085/45 (c) 070/43



Page 129

47.

The following W/V are forecast for a flight from DURBAN to CAPE TOWN The first third of the route W/V 280/30 The second third of the route W/V 315/45 The last third of the route W/V 350/65 The mean W/V for the route is :(a) 307/44 (b) 330/50 (c) 325/42

48.

An aeroplane is overhead FAJS at 6000 ft climbing to FL 190 at a mean rate of climb of 750 feet per minute. RAS 165 Kts. Temperature deviation ISA +12C The mean climb TAS is :(a) 200 Kts (b) 210Kts (c) 220 Kts

49.

An aeroplane is overhead LANSERIA at 5000 feet climbing to FL 180 at a mean rate of climb of 600 feet per minute. Temperature deviation ISA +15C. RAS 172 Kts. The mean climb TAS is :(a) 197 Kts (b) 207 Kts (c) 217 Kts

50.

An aircraft is overhead FABL at 5500 feet climbing to FL 210 at a mean rate of climb of 800 feet per minute, RAS 185 Kts, Temperature deviation ISA +13C. The mean climb TAS is :(a) 220 Kts (b) 230 Kts (c) 240 Kts

51.

An aircraft is overhead FAKM at 5000 feet climbing to FL 230 at a mean rate of climb of 1200 feet per minute. Mean climb TAS 230 Kts, mean climb W/V 225/30, Track 135 (T). The distance of the TOC position from FAKM is :(a) 52 nm (b) 57 nm (c) 62 nm



Page 130

52.

1315 Z Overhead PORT St JOHNS enroute to DURBAN, distance 128 nm FL190, GS235Kts. A descent is planned to arrive overhead DURBAN at 4000 feet. Mean rate of descent 750 feet per minute, mean descent GS 195 Kts. The latest time to commence descent is :(a) 1331 Z (b) 1334 Z (c) 1337 Z

53.

1420 Z

Fix JSV Radial 127, JSV DME 195 nm, FL 330, track 307 (M), GS 455 Kts A descent is planned to arrive overhead JSV at FL 100 Mean rate of descent 2000 feet per minute, mean descent GS 425 Kts The latest time to commence descent is :(a) 1432 Z (b) 1435 Z (c) 1438 Z

54.

Heading 120 (T)

TAS 350 Kts

W/V 045/60

The groundspeed is :(a) 330 Kts (b) 340 Kts (c) 350 Kts 55.

Heading 270 (T)

TAS 420 Kts

W/V 190/75

The groundspeed is :(a) 400 Kts (b) 414 Kts (c) 428 Kts

56.

An aeroplane is flying from SUTHERLAND SLV (S 32:24 E 020:40) to KIMBERLEY KMV (S 28:48 E 024:47). Heading 036T. When the aeroplane is abeam of VICTORIA WEST NDB VW (S 31:23 E 023:08) the ADF will indicate :(a) 081 Relative (b) 090 Relative (c) 098 Relative



Page 131

57.

An aeroplane is flying from KIMBERLEY KLMV (S 28:48 E 024:47) to EAST LONDON ELV (S 33:02 E 027:53), Heading 156 (T). When the aeroplane is abeam of COOKHOUSE NDB CH (S 32:46 E 025:45) the ADF will indicate :(a) 082 Relative (b) 090 Relative (c) 098 Relative

58.

Overhead GROOTFONTEIN GF (S 19:36 E 018:03) on a direct track to BULAWAYO VBU (S 20:05 E 028:42) TAS 240 kts, W/V 015/45. The VOR Radial to maintain the outbound track from GROOTFONTEIN is (a) Radial 096 (b) Radial 104 (c) Radial 107

59.

An aircraft is overhead CAPE TOWN CTV (S 33:58 E 018:36) on a direct track to PORT ELIZABETH PEV (S 33:57 E 025:35) The Radial from the VICTORIA WEST VOR VWV (S 31:24 E 023:08) that will be received when crossing the PORT ELIZABETH FIR BOUNDARY at 24 E is :(a) Radial 164 (b) Radial 188 (c) Radial 185

60.

0930 Z

Overhead SUTHERLAND SLV (S 32:24 E 020:40), TAS 260Kts.

1030 Z

Overhead KIMBERLEY KMV (S 28:48 E 024:47)

If the aircraft returns to SUTHERLAND at 1030 Z the ETA is :(a) 1130 Z (b) 1142 Z (c) 1154 Z

61.

1515 Z

Overhead GEORGE GGV (S 34:00 E 022:22),TAS 180 Kts.

1615 Z

Overhead VICTORIA WEST VWV (S31:24 E 023:08).

If the aircraft returns to GEORGE at 1615Z the ETA is :(a) 1704 Z (b) 1710 Z (c) 1715 Z CPL NAVIGATION CPL DOC8 Revision 1/1/2001


Page 132

62.

2114 Z

Overhead EAST LONDON ELV (S 33:02 E 027:53) TAS 220 Kts.

2214 Z

Overhead DURBAN DNV (S 29:56 E 031:00)

If the aircraft returns to EAST LONDON at 2214 Z the ETA is :(a) 2314 Z (b) 2322 Z (c) 2330 Z

63.

1300 Z 1305 Z 1310 Z

Heading 050 (T) Heading 120 (T) Heading 190 (T)

Drift 9 Right Drift 2 Left Drift 11 Left

TAS 200 Kts

The W/V is :(a) 288/41 (b) 310/37 (c) 257/43

64.

1000 Z

Overhead BURGERSDORP BDV (S 30:58 E 026:18) Heading 068 (T), TAS 150 kts, Drift 7 Left.

1024 Z 1035 Z 1048 Z

MASERU VOR MZV (S 29:25 E 027:35) Radial 200 MASERU VOR MZV (S 29:25 E 027:35) Radial 165 MASERU VOR MZV (S 29:25 E 027:35) Radial 128

From the 1048 Z position alter heading for RICHARDS BAY RB (S 28:46 E 032:05) The ETA RICHARDS BAY is :(a) 1143 Z (b) 1150 Z (c) 1157 Z

65.

0900 Z

Overhead GEORGE GGV (S 34:00 E 022:20) Heading 045 (T), TAS 210 kts, Drift 8 Right.

0918 Z 0934 Z 0950 Z

PORT ELIZABETH VOR PEV (S 33:58 E 025:36) Radial 315 PORT ELIZABETH VOR PEV (S 33:58 E 025:36) Radial 348 PORT ELIZABETH VOR PEV (S 33:58 E 025:36) Radial 020

From the 0950 Z position alter heading for NEW HANOVER VOR NHV (S 29:21 E 030:31) The mean magnetic heading to steer for NEW HANOVER is :(a) 065 (M) (b) 072 (M) (c) 081 (M)



Page 133

66.

1600 Z

Overhead GEORGE GGV(S 34:00 E 022:22) Heading 040 (T), TAS 190 Kts, Drift 12 Right.

1700 Z

Alter heading 220 (T), Drift 10 Left.

1800 Z

Alter heading for GEORGE

The ETA for GEORGE is :(a) 1816 Z (b) 1824 Z (c) 1832 Z

67.

1200 Z

Overhead CAPE TOWN CTV (S 33:58 E 018:36) TAS 360, Heading 060 (T).

1300 Z

Alter Heading 240 (T).

1420 Z

Overhead CAPE TOWN CTV

The mean W/V since 1200 is :(a) Light and Variable (b) 240/50 (c) 240/120

68.

1000

Overhead MASERU MZV (S 29:26 E 027:34). TAS 300 Kts, Heading 030 (T), Drift 11 Left.

1032



69.

1500 Z

Overhead EAST LONDON ELV (S 33:02 E 027:53) Heading 180 (T), TAS 480 kts.

1600 Z

Alter Heading 360 (T)

1648 Z

Overhead PORT ALFRED PA (S 33:35 E 026:52)

The mean W/V since 1500 Z is :(a) 142/45 (b) 163/55 (c) 185/48 CPL NAVIGATION CPL DOC8 Revision 1/1/2001


Page 134

70.

2000 Z

Overhead GEORGE GGV (S 34:00 E 022:21) Heading 180 (T), TAS 420

2100 Z

Alter Heading 360 (T)

2212 Z

Overhead PLETTENBERG BAY PY (S 34:05 E 023:19)

The mean W/V since 2000 Z is :(a) 275/50 (b) 300/55 (c) 330/45

71.

An aircraft flying at FL 270, heading 026ºM, drift 10ºR, groundspeed 360 kts. At 1203 the relative bearing of the an island was 330º. At 1209 the relative bearing of the same island 270º. If the variation is 10ºW, what is the bearing and approximate range to plot to the island from the aircraft at 1209? a. b. c.

106T – 30nm 286T – 30nm 296T – 36nm



Page 135



Page 136

ANNEX A SAMPLE EXAMS



Page 137

PAPER 1 1.

The great circle bearing of X from Y is 072º The rhumb line bearing of Y from X is 259º The great circle bearing of Y from X is: a) b) c)

2.

The initial great circle track from A (S28:30 W015:15) to B is 099º. If A and B are on the same parallel of latitude the longitude of B is: a) b) c)

3.

170 nm 180 nm 190 nm

A radio facility chart has a scale of 1:3 750 000. If two reporting points are 17 cms apart on the chart the actual distance is: a) b) c)

7.

08:37 Z 08:47 Z 08:57 Z

On a chart a line 58 centimeters in length represents 365 statute miles. On the same chart a line 13 inches in length would represent: a) b) c)

6.

1554 nm 1672 nm 1739 nm

An aircraft leaves MOCAMEDES (S15:19 E012:10) at 0627 Z. GS 375 Kts. The ETA at LUSAKA (S15:19 E028:25) is; a) b) c)

5.

W003:37 E 012:15 E 022:28

The shortest distance from A (N75:39 E123:17) to B (N78:27 W056:43) is: a) b) c)

4.

262º 266º 270º

344 nm 364 nm 384 nm

A Mercator chart has a scale of 1:2 500 000 at 20ºN. The scale at 50ºN is: a) b) c)

1:1 710 010 1:1 835 000 1:1 972 060



Page 138

8.

An aircraft heading 031º ( C), Var 20ºW, Deviation –2º, obtains an ADF bearing of 104º relative from an NDB in the Southern hemisphere. If the CA is 1º the bearing to plot on a Mercator Chart is: a) b) c)

9.

On a Lamberts chart the standard parallel of 32ºN measures 77.5 cms. The other standard parallel measures 72 cms. The latitude of the second standard parallel is; a) b) c)

10.

292º 293º 294º

37ºN 38ºN 39ºN

A Lamberts chart has standard parallels of 30ºN and 50ºN. The initial great circle track from A (32ºN 63ºW) to B (45ºN 15ºW) is 056º (T). The longitude at which the great circle track becomes 090º (T) is: a) b) c)

11.

An aircraft heading 112º (T) in the Southern Hemisphere receives an ADF bearing of 156º relative. If chart convergency between the aircraft and the NDB is 4º the bearing to plot from the NDB on a Lamberts Chart is: a) b) c)

12.

W 012:37 W 010:06 W 008:28

084º 092º 096º

Aircraft A is overhead WY at 1127 Z, groundspeed 235 Kts. Aircraft B is overhead WY at 1139 Z, groundspeed 320 Kts. If both aircraft are following the same route B will overtake A at: a) b) c)

13.

1212 Z 1222 Z 1232 Z

Aircraft A, GS 245 Kts reports overhead CD 6 minutes ahead of aircraft B. Aircraft B, GS 375 Kts reports overhead HP 8 minutes ahead of aircraft A. The distance from CD to HP is: a) 165 nm b) 175 nm c) 185 nm



Page 139

14.

An aircraft estimates VOR CPL at 0830 Z flying at GS 360 Kts. ATC requires the aircraft to cross CPL at 0835 Z. In order to comply with the requests the aircraft reduces GS to 315 Kts at: a) b) c)

15.

A to B distance 413 nm, FL 150, OAT +5ºC, tailwind 35 Kts. The required RAS to fly from A to B in 1 hour 32 Minutes is: a) b) c)

16.

9º right 12º right 15º right

At 1015 Z an aircraft is overhead VOR CAR heading 195º (T) to maintain the radial 208º outbound, variation 21ºW, TAS 245 Kts. At 1032 Z the CAR DME indicates 78 nm. The W/V is: a) b) c)

20.

252º (M) 262º (M) 272º (M)

An aircraft is flying from A to B, distance 266 nm. At 111 nm from A, a fix is obtained 13 nm left of track. The alteration of heading to arrive overhead B is: a) b) c)

19.

205 Kts 220 Kts 235 Kts

An aircraft is heading 072º (M) in order to maintain a track of 062º (T) Variation 20ºW. The heading to fly the reciprocal track is: a) b) c)

18.

180 Kts 190 Kts 200 Kts.

An aircraft flies from A to B 372 nm in 1 hour 50 mins at TAS 180 Kts. In order to fly from B to A in 1 hour 45 mins the TAS should be; a) b) c)

17.

0750 Z 0755 Z 0800 Z

321/47 347/53 012/38

The duration of morning civil twilight at N47:45 E015:37 on May 18th is: a) b) c)

32 mins 37 mins 42 mins



Page 140

21.

The standard time of sunset at Lisbon, Portugal (N 38:42 W 009:30) on June 3rd is: a) b) c)

22.

TAS 200 Kts Heading 095º (T) The groundspeed is: a) b) c)

23.

24.

1857 1957 2057 W/V 010/45 Kts

191 kts 201 kts 211 kts

An aircraft is overhead FAJS at 6500 ft climbing to FL 230 at a mean rate of climb of 850 ft/min. RAS 175 kts, Temperature deviation ISA +13ºC. The mean climb TAS is: a) b) c)

223 kts 228 kts 233 kts

1732 1751

Overhead BARBERTON BT (S 25:45 E 031:30) TAS 180 kts. Overhead MHLUME HL (S 26:50 E 031:00) TAS 205 kts.

The ETA at DURBAN DNV (S 29:55 E 031:00) is:

25.

a) b) c)

1840 1845 1850

1127

Overhead UPINGTON UPV (S 28:25 E 021:16) TAS 195 kts, on a direct track to BISHP BOV (S 32:55 E 027:16)

1235

VICTORIA WEST VOR VWV (S 31:24 E 023:10) Radial 063º

The ETA at BISHO is:

26.

a) a) b)

1335 1342 1351

1508

Overhead CAPE TOWN CTV ( S33:58 E 018:35) Heading 031º (T) TAS 240 kts, W/V 260/55 Alter heading 115º (T) The mean heading magnetic to steer to reach PORT ELIZABETH (S 33:58 E 025:37) is:

1553 1637

a) b) c)

184º (M) 194º (M) 204º (M)



Page 141

27.

The following RMI readings were obtained from the LADYSMITH VOR LYV (S 28:33 E 029:45) Aircraft track 078º (M) GS 300 kts. 0937 0946 0957

LYV VOR RMI reading 118º LYV VOR RMI reading 155º LYV VOR RMI reading 208º

The position of the aircraft at 0957 is:

28.

a) b) c)

S27:00 E030:04 S27:28 E029:58 S26:39 E030:09

1115

Overhead EAST LONDON ELV (S 33:02 E 027:53) Heading 032º (M), TAS 260KT

1147

Alter heading 070º (M)

1218

Overhead DURBAN DNV (S 29:56 E 031:00)

The mean W/V since 1115 is:

29.

a) b) c)

265/55 317/48 003/60

1315

Overhead ALEXANDER BAY ABV (S 28:34 E 016:30) on a direct track to LANSERIA (S 25:58 E 027:55) TAS 180 kts, W/V 230/45

The ETA at LANSERIA is: a) b) c) 30.

1554 1604 1614

The following VOR radials were obtained from Nieuwoudtville NVV, (S 31:21, E 019:0E). Aircraft track 025(T), GS 180kts. 0900 Z NVV VOR radial 169 0912 Z NVV VOR radial 124 0927 Z NVV VOR radial 083 The position of the aircraft at 0927Z is: (a) (b) (c)

31:00S 019:52E 30:50S 020:13E 30:59S 020:38E



Page 142

PAPER 2 1.

The initial great circle track from A to B is 067º. The initial great circle track from B to A is 263º. The rhumb line track from A to B is: a) b) c)

2.

The position of A is N42:13 W158:24. B is on the same parallel of Latitude. The Great Circle bearing of B from A is 278º. The Longitude of B is: a) b) c)

3.

176 Kts 186 Kts 196 Kts

The latitude where the value of convergency is half the value of convergency at 60ºN is: a) b) c)

6.

E 000:33 E 001:38 W000:18

An aircraft departed George (S 33:59 E 022:23) at 0652Z and arrived at Port Elizabeth (S 33:59 E 025:36) at 0741Z. The average groundspeed for the flight was; a) b) c)

5.

E 177:47 W 182:31 W 177:47

An aircraft departs from C (S 23:07 W 005:13) on a track of 090º (T) RL and arrives at D (S 23:07) after a flight of 378 nm. The Longitude of D is: a) b) c)

4.

059º 075º 083º

30º 00’N 27º 52’N 25º 39’N

A Mercator chart has a scale of 1:2 000 000 at 20ºN. The latitude where the scale would be 1:1 500 000 is: a) b) c)

N44:11 N45:11 N46:11



Page 143

7.

Chart convergency on a Lamberts chart between the meridians of 10ºE and 10ºW is 12º. If one standard parallel is S 30:20 the other standard parallel is at: a) b) c)

8.

Aircraft A is overhead NDB DN at 0900 Z enroute to NDB PY, GS 245 Kts. Aircraft B is overhead NDB PY at 0920 Z enroute to NDB DN, GS 305 Kts. If the distance between DN and PY is 485 nm the aircraft will pass each other at: a) b) c)

9.

063º 070º 077º

A Lambert’s chart has Standard Parallels of 30ºS and 50ºS. The Rhumb Line distance from A (50ºS 010ºE) to B (50ºS 010ºW) is 13.75 inches. The scale at 30ºS is: a) b) c)

12.

218º 222º 224º

A bearing obtained from a NDB is 273º relative. Aircraft heading 330º (T), d.long between aircraft and NDB is 14º, mean Latitude 26ºS, Parallel of Origin 30ºS. The bearing to plot on a Lambert’s chart is: a) b) c)

11.

0942 Z 0953 Z 1004 Z

An aircraft in the Southern Hemisphere heading 180º ( C) has a relative bearing of 210º indicated on the ADF. Deviation 2º W, aircraft variation 12ºE, NDB variation 14ºE, convergency between the aircraft and the NDB is 4 degrees. The bearing to plot on a Mercator chart is: a) b) c)

10.

S 40:48 S 41:52 S 43:24

1:4 092 898 1:4 234 729 1:438 197

Aircraft A, GS 240 Kts is 262 nm from BT at 0915 Z. Aircraft B, GS 285 Kts is 298 nm from BT at 0922 Z. In order to ensure a 50 nm separation at BT, aircraft B must reduce its groundspeed to 240 Kts at: a) b) c)

0934 Z 0937 Z 0941 Z



Page 144

13.

The distance between meridians on a Mercator chart 1º apart is 2.7 centimetres. The scale of the chart at 33º S is: a) b) c)

14.

1:3 247 500 1:3 453 800 1:3 694 400

Sector A to B

Track 177º

Distance 153 nm

TAS 205 KTS

Drift 5ºL

Time 51 mins

The W/V affecting the aircraft is; a) b) c) 15.

What distance in cms would two fixes taken 20 minutes apart appear on a chart whose scale is 1:1 000 000 if the GS is 180 Kts? a) b) c)

16.

1615 LMT 1537 LMT 1452 LMT

The standard time of sunrise at Amsterdam (N 52:15 E 005:15) on 27 June is: a) b) c)

18.

11.12 cms 12.12 cms 13.67 cms

An aircraft leaves Tokyo (N36:00 E139:45) at 2100 Standard time on Oct 12 (Standard time factor Japan 9 hours). After a 11 hour flight it arrives at San Francisco (N 37:30 W 122:00) The LMT of arrival at San Francisco is: a) b) c)

17.

141/30 177/30 213/30

0320 0420 0520

Aircraft heading 115º (T) The groundspeed is: a) b) c)

TAS 185 KTS

W/V 200/40

186 Kts 177 Kts 196 Kts



Page 145

19.

20.

0845

Overhead Cape Town CTV (S 33:58 E 018:35) on a direct track to East London ELV (S 33:02 E 027:53) TAS 175 kts, W/V 300/35. The ETA at the Port Elizabeth FIR boundary 24ºE is:

a) b) c)

1005 1011 1017

0915

Overhead Sutherland SLV (S 32:24 E 020:40) TAS 265 kts, W/V 240/50, Heading 010º (M).

0945

Alter heading 305º (M).

1012

Alter heading for Alexander Bay ABV (S 28:34 E 016:31) The ETA at Alexander Bay is:

21.

a) b) c)

1034 1044 1054

1310

Overhead Victoria West VWV (S 31:24 E 023:09) Heading 355º (T), TAS 310,

1339

Alter heading 268º (T).

1356

Overhead Upington UPV (S28:24 E021:16) The mean W/V since 1310 is:

22.

a) b) c)

190/42 170/32 220/55

1427

Overhead Pomfret PF (S 25:51 E 023:30 on a direct track to Burgersdorp BDV (S 30:58 E 024:47).

1525

Overhead Kimberley KMV (S 28:49 E 024:47) The alteration of heading to reach Burgersdorp is:

a) b) c) 23.

5º left 11º left 17º left

The following W/V are forecast for a climb to cruising altitude: 180/30 225/45 260/60 The mean climb W/V is: a) b) c)

220/47 252/50 233/40



Page 146

24.

2213

Fix CPL VOR/DME Radial 287, DME 195nm. FL410, GS465 kts. Descent details – Mean rate of descent 2000 ft/min Mean descent GS 365 kts The latest time to commence descent to arrive overhead CPL at 7000 ft is: a) b) c)

25.

2225 2228 2231

An aircraft flying at GS 240 kts along track 204º (T) obtains the following bearings from Maputo VOR VMA (S 25:56 E 032:34); 1020 VMA QDM 141º 1031 VMA QDM 110º 1043 VMA QDM 085º The position of the aircraft at 1043 is:

26.

a) b) c)

S26:20 E031:22 S26:35 E030:42 S26:45 E030:11

2115

Overhead Port St Johns PJ (S 31:38 E 029:32) Heading 286º (M), TAS 250 kts. Overhead Victoria West VWV (S 31:23 E 023:08)

2243

The mean W/V since 2115 is;

27.

a) b) c)

337/57 262/48 225/41

1715

Overhead East London ELV (S 33:02 E 027:53) on a direct track to Maputo VMA (S25:56 E032:34), Heading 023º (T), GS 310 kts. Durban VOR/DME DNV (S 29:55 E 031:00) Range 65 nm. Durban VOR/DME DNV Range 65 nm.

1752 1811

The position of the aircraft at 1811 is:

28.

a) b) b)

S29:10 E030:05 S28:52 E030:44 S28:50 E031:15

1915

Overhead Pomfret NDB PF (S 25:51 E 023:30) on a direct track to Phalaborwa PHV (S 23:56 E 031:08) Heading 085º (M). Gabarone NDB GBE (S 24:33 E 025:55) bears 270º Relative.

1954

The ETA for Phalaborwa is: a) b) c)

2047 2057 2116



Page 147

29.

30.

1315 1355 1356

Overhead C, Hdg 080T, TAS 180kts, Track 077 Alter heading 120T, GS 144kts Overhead D, the Wind velocity is?

(a) (b) (c)

070/65 080/45 090/45

Find the wind velocity affecting the aircraft if: TAS 230kts Hdg 195 Drift 7 right Hdg 257 Drift 6 right Hdg 332 Drift 2 left (a) (b) (c)

135/30 144/24 125/25



Page 148

ANNEX B ANSWERS



Page 149

CHAPTER 1 1 2 3 4 5 6 7 8 9 10 11 12 13

B C A B B C C A C A A A A

14 15 16 17 18 19 20 21 22 23 24 25 26

B C A A A B C C B A A A B

27 28 29 30 31 32 33 34 35


A C B A B C C A B


Page 150

6.

7.

090RL 099GC CA9 CI8 C = Ch. Long sin Lat 18 = Ch.Longsin2830' 18 _______ = Ch. Long = 37.7233 = 37 43' East of W 015 15' sin2830' 37 43" Easterly Longitude E 022 28C = Ch. Long sin Lat C = 1 sin 60 = 0.866 - 2 = 0.433 = sin 25 39'



Page 151

9.

10.

11.

12.

13.



Page 152

14. C = Ch. Long sin Lat C = 1 X sin 20 - 0.342 x 2 - 0.684 = sin 43.1602 = 4309' 37" 15

16

17.



Page 153

18

19. C = Ch. Long sin Lat 8 = 16 sin Lat

8 16

= 0.5 = sin 30

20.

21

22.

23.

W005:15 E 018:29 _______ 23:44 Ch. Long 14:24 mins Long,

Departure = Ch. Long x cos Latitude = 1424 x cos 27:43 = 1260.6 nm

Departure = Ch. Long x cos Latitude 456 nm _______ = 648 mins of Long = 10°48' West of E 025:52 - E015:04 cos 45:17



Page 154

24.

GS478 x 16:35 = 7926.8333 nm = Ch. Long 360° x cos Latitude 7926.8333 _________ 360° x 60

25.

= 0.367 = cos 68°28'

W 081:20 W 016:16 ________ 65:04 = 3904 mins of Long

Departure = Ch.Long x cos Latitude = 3904 x cos 28:32 = 3430 nm at GS375 = 9:09 = 08:20 ETA = 17:29

26. E 067:11 E 054:44 12:27

= Long 747 minutes of

Departure = Ch. Long x cos Latitude = 747 x cos 24:23 =

27.

ETA 08:17 ETD 06:52 1:25 Flight Time

680 nm in 1:25 = GS480Kts

Dep = Ch long x cos LAT 1037 nm = Ch.Long x cos Lat 57:42 1037 cos 57:42

= Ch long = 1941 mins Long = 32°21 West of E 030:15 = W 002:06

28.

Departure 425 nm = Ch. Long x cos 46:52 425 nm _________ cos 46:52

= 622 mins Long = 10° 22' Change of Longitude

29.

One nautical mile at the Pole One nautical mile at the Equator

30.

N 75:39 to North Pole North Pole to N 78:27

6108 feet 6046 feet

14:21 Change of Latitude 11:33 Change of Latitude 25:54 Change of Latitude = 1554 nm



Page 155

31.

32.

Departure

= Ch. Long x cos Latitude = 1 minute Longitude x cos 25 = 0.906308  2 = 0.453154

0.453154

= cos 6303' 14"

33.

34.



Page 156

CHAPTER 2 Mercator Questions 1 2 3

A B B

4 5 6

7 8 9

C B C

10 11 12

A A B

C A B

13 14 15

A B C

DETAILED ANSWERS 1.

ABBA

A

2 500 000 x cos 50 2 500 000 x cos 50 cos 20

20N

Scale 1:2 500 000

=

Scale B x cos 20

=

1710100

B 50N

Scale B 1:1 710 100

2. ABBA

A 20N Scale 1:2 000 000

2000000 x cos B cos B

B Scale 1:1 500 000

= 1 500 000 x cos20 = 1 500 000 x cos 20 2 000 000

= (0.7048)COS-1 = 45 11'21" NS

3.

S =

4.

CL __ ED

=

QDM VAR QUJ CA QUJ QTE

2.7 cms ___________________________ 1 x 60 x cos 33 x 6080 x 12 x 2.54

095 14W 081 GC -2 079 RL ± 180 259 RL

5.

QDM VAR QUJ QTE CA QTE

275 18W 257 GC ± 180 077 GC -3 074 RL

=

6.

1.3 453 802

HDG ADF QUJ QUJ CA QUJ QTE


267 (T) 346 613 GC -360 253 GC +2 255 RL ± 180 075 RL


Page 157

7.

S =

10.

CL __ ED

=

CL S = ___ ED

156 millimetres ___________________________ 5 x 60 x cos 32 x 6080 x 12 x 25.4 ( Departure )

1 __________ 2500000

2 500 000 x 3cms 2 500 000 x 3 cms ____________________ 1 x 60 x 6080 x 12 x 2.54

11.

CL Ratio = ___ ED

=

=

1:3 022 286

3cm = _____________________________ 1 x 60 x cos Lat x 6080 x 12 x 2.54 ( Departure ) = 1 x 60 x cos Lat x 6080 x 12 x 2.54

= Lat = (0.6745)cos –1 = 47 35' N/S

5.75 cms ____________ 1 x 60 x cos 52 ( Departure )

= 6.4243 nautical miles to the centimetre



Page 158

Lamberts chart questions

1 2 3 4 5

C C B A C

6 7 8 9 10

B A C B C

11 12 13 14 15

C C B C C

16 17 18(1) 18(2) 19

A A C A A

20 21 22 23 24

B B A B A

25 26 27

C C A

DETAILED ANSWERS

1.

SP 29S & 41S

Parallel of Origin 35S

CC = Ch. Long x sin//O CC = 22 x sin 35 CC = 12.62 2.

3.

CC = Ch.Long x sin//O 12 = 20 x sin//O 12 = 0.6 = sin 36 52 20 SP 30 20' Change of Lat 6 32"

77.5 cms _______________ Ch. Long x cos 32

4. Scale =

5.

= Scale

SP 30 20 Change of Latitude Parallel of Origin Change of Latitude SP

72 cms ________________ Ch. Long x X?X Lat

=

Latitude : 3801'N

76.62 cms 1 _______________________________ = ________ 22 x 60 x cos 20 x 6080 x 12 x 2.54 3 000 103

80.5 cms _______________ = Scale = Ch. Long x cos 45

CL ______________ Ch. Long x cos 25


632' 36 52' 6 32' 43=24'

at 20 & 40S

CL = 103.2 cms


Page 159



Page 160

10.

CC = 12.04 =

Ch. Long x sin. // O 14 x sin // O

12.04 SP 64 00' _____ .= (0.86)SIN-1 = 59 19' 14 4 41'

SP 64 00' Change of Latitude Parallel of Origin Change of Latitude SP

4 41' 59 19' 4 41' 54 38'

11.

12.



Page 161

13.

15.



Page 162

16.

17.

18.



Page 163

19.

20.

21.

On a Lambert's chart the Great Circle and Rhumb Line tracks are parallel at the mid meridian between any two positions. The mid meridian between 10W and 60E is 25E where the track is 124RL And 124GC



Page 164

22.

23.

SP 15S & 35S

// O 25S

Earth Convergency = Ch. Long 22 sin Lat 34 Chart Convergency = Ch. Long 22 sin // O 25

24.

EC = 12.3022 CC = 9.2976 Difference = 3.0046

The True Great Circle between two positions on the Parallel of Origin is a straight line. The True Great Circle between two positions North of the Parallel of Origin will be a slight curve concave to the Parallel of Origin or North of the straight line

CHAPTER 3 RELATIVE VELOCITY Q1

B

Q2

C

Q3

D



Page 165

CHAPTER 4 SUNRISE SUNSET TWILIGHT ANSWERS 1. B

2. B

3. C

4. C

5. B

6. A

7. A

DETAILED ANSWERS 1.

April 14 15 16 45N 0518 0516 0515 40N 0525 0523 0522 Time difference 7 minutes ÷ Sunrise W 003:24 Arc/Time Sunrise Spain ST Factor Sunrise

17 0513 0520 5 x 0030’ = 1 minute = 0522 LMT 05:22:00 LMT 13:36 05:35:36 UTC +1 06:35:36 LST

May 50N 45N

9 1929 1913

2. 6 1924 1910

7 1926 1911

Time difference Sunset 2E Arc/Time Sunset France ST Factor Sunset

8 1927 1912

15 minutes ÷ 5x 340’ = 11minutes= 1923 LMT 19:23 LMT 8 19:15 UTC +1 20:15 LST

3. Twilight N 55:45 June 1 56N Sunrise 0323 Twilight 0225 Difference 58 54N Sunrise 0335 Twilight 0244 Difference 51 Time difference 7 minutes ÷

2

3

58

59

51 52 2 x 145’ =

4 0320 0221 59 0333 0241 52 +6 minutes 58 minutes twilight

4 Sunset S 10:00 W 063:45 February 24 26 10S 1823 1823 Sunset W 063 :45 Arc/Time Sunset 15 mins before SS ETA Flight time ETD E 022:15 Arc/Time

18:22 4:15 22:37 15 22:22 11:15 11:07 1:29

27 1822

28 1822

LMT UTC UTC UTC ETD

12:36 LMT



Page 166

5.

30S

50S

Twilight S 33:58 January 5 Sunset 1905 Twilight 1933 Difference 28 Sunset Twilight Difference

1918 1947 29

6

7

28

27

8 1906 1933 27

29

1918 1947 29

29

Duration of evening civil twilight 29 minutes 6.

20 N60 1543 N59:46 N58 1556 1550 LMT -0201 30:20E 1349 UTC +0300 1649 ST

22 1548 1550 1600

23 1550

1 0519

3 0521 0517 0511

4 0522

7.

S35 S36:50 S40 0508 3 FEB 0517 -1139 174:48E 2 FEB 1738 UTC

1602

0512

Time Questions 1 2

A C

3 4

B A

5 6

C B

7 8

A B

9 10

C C

11 12 13

A C A

DETAILED ANSWERS

1.

Y ETA W 066:30 Arc to Time Y ETA X ETD Flight Time Distance Groundspeed


05:58 4:26 10:24 21:00 13:24 4732 nm 353 Kts

LMT 4 July UTC 4 July UTC 3 July


Page 167

2.

Ascension ETD W 014:30 Arc to Time Ascension ETD Flight Time FAJS ETA SA Standard Time Factor FAJS ETA

22:15 0:58 23:13 9:14 08:27 02:00 10:17

LMT 15 June UTC 15 June UTC 16 June SAST 16 June

3.

Distance 3207 nm Perth E 115:57 Arc to Time Perth Flight Time Mauritius E 057:40 Arc to Time Mauritius

GS 427 Kts Time 7:30:38 ETD 09:30 LMT 07:43:48 ETD 01:46:12 UTC 07:30:38 ETA 09:16:50 UTC 03:50:40 ETA 13:07:30 LMT

4.

A E 035:15 Arc to Time A Flight Time B W 028:45 B

ETD

Prestwick W 005° Arc to Time Prestwick Flight Time San Francisco W 112° Arc to Time San Francisco

ETD

5.

6.

7.

ETD ETA ETA

08:15 2:21 05:54 7:27 13:21 1:55 11:26

LMT 19thD UTC 19th GD UTC 19th GD LMT 19th LD LMT 23rd LD

ETA

11:15:00 :20 11:35:00 11:15 22:50:00 8:08 14:42-00

Tahiti Sunset ETA W 149:29 Arc to Time Tahiti ETA Flight Time Wellington ETD E 174:45 Arc to Time Wellington ETD ETD

18:23:00 9:57:56 28:20:56 5:00 23:20:56 11:39 34:59:56 10:59:56

LMT 5th LD

Vancouver ETA W 123:15 Arc to Time Vancouver ETA Flight Time X ETD X ETD Time Difference

10:57:00 8:13 19:10:00 8:10 11:00:00 20:00:00 9:00:00

ETD ETA


UTC 23 rd GD UTC 23 rd GD LMT 23 rd LD

UTC 5 th GD UTC 5 th GD LMT 5 th LD LMT 6 th LD LMT 18 th LD UTC 18 th GD UTC 18 th GD LMT 18 thLD = 135°E


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8.

If the LMT of arrival at a destination is the same as the LMT of departure then the aircraft has flown at the same speed as the sun. that is 15°S cos Latitude per hour. 1952 Kms in 2 hours = GS976Kms - GS527Kts Departure = Ch. Longitude x Cos Latitude 527 = 15° x 60 x Cos Latitude 527 ___ = (0.5856)COS-! = 54:09:28 N S 900

9.

Tokyo ETD Standard Time Factor Tokyo ETD Flight Time San Francisco ETA 122°W Arc to Time San Francisco ETA

10.

A 168W Standard Time Factor A Flight Time B 174°E Arc to Time B B

21:00 LST 9:00 12:00 UTC 11:00 23:00 UTC 8:08 14:52 LMT ETA ETA ETD ETD ETD

22:08 11:00 33:08 7:06 26:02 11:36 37:38 13:38

12 th LD 12 th GD 12 th GD 12 th LD

LST 2nd LD UTC 2nd GD UTC 2nd GD LMT 2nd LD LMT 3rd LD

11.

See sketch on page 4. An aircraft flying heading 090°(T) is passing from Easterly Longitude to Westerly Longitude. The local date will change from the 6 th to the 5th whilst the Greenwich date will remain the 6th

12.

At 1200 UTC all Longitudes will be 6th May. At 179°59' E the LMT will be 23:59:56 on 6th May At 179°59' W the LMT will be 00:00:04 on 6th May

13

New York St Factor New York New York Flight Time Frankfurt St Factor Germany Frankfurt

NOTE

ETD ETD ETA ETA ETA

13:00 5:00 18:00 8:30 26:30 02:30 +1 03:30

LST

28th Feb

UTC 28th Feb UTC 28th Feb UTC 29th Feb UTC 29th Feb

If the year is divisible by 4 it is a Leap Year 1992 ÷ 4 = 498 = Leap Year 1993 ÷ 4 = 498.25 not a Leap Year



Page 169

Chapter 6 Heading and ETA questions 1 2 3

A A A

4 5 6

B B A

7 8 9

A C C

FOR DETAILED ANSWERS SEE NAVIGATION LOG ON NEXT PAGE



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Plotting Questions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

B B A C B A B C C A C A A A C B A C A B B B

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

C C C C B B B A B A B B A C A C C A C B C A

45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66

C A C B C C B A B B B C A C C C A C A B B C

67 68 69 70 71

B A A C B

Detailed Answers 1. Method: Plot required track BLOEMFONTEIN to SUTHERLAND. Plot TMG (Track Made Good) BLOEMFONTEIN to VICTORIA WEST and extend. Join VICTORIA WEST to SUTHERLAND (new track). Place protractor over VICTORIA WEST with 360 aligned with the TMG. Measure the alteration of track, 13 Right. Alter Heading 13 Right



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Question 1



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Question 24



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Question 25

Question 26



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Question 27

Question 28



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Question 29



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Question 30



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37 1. 2. 3. 4. 5. 6. 7. 8.

Set the 1000 Z Heading 030 (T) against the True Index of the Nav. Computer. Set TAS 350 kts at the centre. Draw the 11 Right drift line. Set the 1025 Z Heading against the True Index. Draw an arc of Groundspeed 325 kts to intersect the drift line. Bring the intersection of the drift line and GS arc to the centre line below the centre. From the centre to the intersection is the Wind Velocity. Read off Wind Direction against the True Index 268 Read off Wind Speed 90 kts (length of vector). W/V 268/90

Question 38

Similar to the above W/V 008/44

Question 39 Similar to the above. After calculating the W/V set Heading 020(T) with TAS 325 kts at the centre. The OS 305 kts appears under the intersection of the drift line and GS arc. GS 305



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Question 48 19 000 feet 6 000 feet _________ 13 000 feet

x 2/3 =

8 667 feet +6 000 feet 14 667 feet mean climb altitude 14 667 feet @ 2/1000 feet

- 29 C colder than MSL + 15 C ISA at MSL - 14 C ISA at 14 667 feet Temperature Deviation ISA +12 C OAT - 2 C Altitude 14 667 feet

OAT-2C

RAS165kts

TAS210 kts

Question 49 18 000 feet 5 000 feet 13 000 feet x 2/3 = 8 66 7 feet + 5 000 feet 13 667 feet mean climb altitude 13 667 feet @ 2/1000 feet

Temperature Deviation

Altitude 13 667 feet

OAT+3C

- 27 C colder than MSL +15 C ISA at MSL ISA

- 12 C ISA at 13 667 feet +15 C + 3 C

RAS172kts


TAS 217 kts


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Question 50 21000 feet 5500 feet 15 500 feet x 2/3 = 10 333 feet + 5 500 feet 15 833 feet mean climb altitude 15 833 feet @ 2/1000 feet

Temperature Deviation

Altitude 15 833 feet

OAT - 4 C

ISA OAT

- 32 C colder than MSL + 15C ISA at MSL - 17C ISA at 14 667 feet + 13 C - 4 C

RAS 185 kts

TAS 240 kts

Question 51 23 000 feet - 5000 feet = 18 000 feet @ 1200 feet/minute = 15 minutes Mean Climb TAS 230 kts Track 135 (T) Mean Climb W/V 225 30 kts Mean Climb GS 228 kts for 15 minutes ~ 57 nm Question 52 GMT

OBSERVATION

GS

DIST

TIME

ETA

Place

1315

Port St. Johns

235

63

16

1331

TOD

195

65

20

1351

DEST

1331

TOD Top Of Descent

Question 53 GMT

OBSERVATION

GS

DIST

TIME

ETA

Place

1420

JSV DME 195nm

455

15

1435

TOD

1435

TOD Top Of Descent

425

113.5 195 81.5

11.5

1446.5

DEST



Page 214

Question 54 Using the Pathfinder or similar Nav. computer :GS 339 kts Select W/V Enter HDG 120 Enter Wind Direction 045 as CRS Enter TAS 350 kts Enter Wind Speed 60 kts as GS Compute W/V 130/339 Wind Direction 130 is the Track Wind Speed 339 kts is the GS Question 55 Using the Pathfinder or similar Nav. computer :GS 414 kts Select W/V Enter HDG 270 Enter Wind Direction 190 as CRS Enter TAS 420 kts Enter Wind Speed 75 kts as GS Compute W/V 280/414 Wind Direction 280 is the Track Wind Speed 414 kts is the GS Question 56 Measure QTE VWV when abeam = 314 - 180 = QUJ 134 Relative Bearing + Heading (T) Relative Bearing Relative Bearing

= QUJ = QUJ – Heading (T) = 134 - 036 = 098 Relative

Question 57 Measure QTE CH when abeam Relative Bearing Heading (T) Relative Bearing Relative Bearing

= 058 + 180 = QUJ 238 = QUJ = QUJ – Heading (T) = 238 - 156 = 082 Relative

Question 58 Initial Track from GROOTFONTEIN to BULAWAYO is 095 (T), Variation 12W, Radial 107 CPL NAVIGATION CPL DOC8 Revision 1/1/2001


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Question 59 The QTE of the point along track at 24 E is 164, VWV Variation 21W, Radial 185 Question 60 093 OZ Overhead SUTHERLAND SL TAS 260kts 1030 Z Overhead KIMBERLEY KMV Distance SL to KMV 304 nm in 1 Hour = GS 304 kts SL – KMV TAS 260 kts KMV – SL TAS 260 kts

Wind Component 44 kts Tailwind Wind Component 44 kts Headwind

KMV – SL GS 216 kts

Distance 304 nm

Time 1:24

GS 304 kts GS 216 kts ETA 1154Z

Question 61 1515 Z Overhead GEORGE GGV TAS 180 kts 1615Z Overhead VICTORIA WEST VWV Distance GGV to VWV 161 nm in I Hour – GS 161 kts GGV-VWV TAS 180 kts VWV – GGV TAS 180 kts VWV-GGV GS 199 kts

Wind Component 19 kts Headwind GS 161 kts’ Wind Component 19kts Tailwind GS 199 kts Distance 161 nm

Time 0:49

ETA 17:04 Z

Question 62 2114 Z Overhead EAST LONDON ELV TAS 220 kts 2214 Z Overhead DURBAN DNV Distance ELV to DNV 246 nm in 1 Hour = GS 246 kts ELV - DNV TAS 220 kts Wind Component 26 kts Tailwind DNV – ELV TAS 220 kts Wind Component 26 kts Headwind DNV – ELV GS 194 kts

Distance 246 nm

Time 1:16

GS 246 to GS 194 kts ETA 23:30 Z

Question 63

Multiple Drift wind by nav. Computer W/V 288/41



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Question 66 If an aeroplane, in zero wind conditions, flies a heading of 040(T) for one hour, and then flies the reciprocal heading 220(T) for one hour at the same TAS it will return at its point of departure. This question involves a strong wind, but with an Air Plot the Air Position will be overhead GEORGE at 1800 Z. The W/V is calculated by the multiple drift method. Heading 040 (T) Drift 12 Right W/V 333/40 kts

Heading 220 (T) Drift 10 Left

The aeroplane will be 2 hours of the W/V 333/40 DOWNWIND from GEORGE. That is on a bearing of 153 (T) at a distance of 80 nm from GEORGE. Track to GEORGE 333 (T)

Drift zero Heading 333 (T)

TAS 190 kts WC 40 kts HW GS 150 kts Distance 80 nm Time 0:32 ETA1832Z Question 67 Similar to the above question. 1200 Z Heading 060 (T) for I hour at TAS 360 kts. The Air Position is 360 nm from CTV. 1300 Z Heading 240 (T) for I hour 20 minutes. The Air Position is 480 nm on a bearing of 240 from the 1200 Air Position, that is 120 nm on a bearing of 240 from CTV and is the Wind Vector. 120 nm in 2 hours 20 minutes = Wind Speed 51 kts

W/V 240/51

Question 68

same as numbers 37 38 W/V 100/60



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Question 71 026M + 270 relative 296 QDM - 16W 286 QUJ The distance need not be worked out, but by construction will be 26 nms.



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CHAPTER 6 ANSWERS TO GENERAL QUESTIONS 1. B 2. C 3. C 4. A 5. B 6. B 7. C 8. B 9. A 10. B 11. C 12. A 11.



Page 221

PAPER 1 1.

B

2.

C

3.

A

4.

C

5.

B

6.

A

7.

A

8.

A

9.

B

10.

B

11.

B

12.

A

13.

A

14.

B

15.

A

16.

C

17.

C

18.

B

19.

A

20.

B

21.

C

22.

B

23.

C

24.

A

25.

C

26.

C

27.

B

28.

B

29.

B

30.

B



Page 222

PAPER 2

1.

B

17.

B

2.

A

18.

A

3.

B

19.

A

4.

C

20.

B

5.

C

21.

A

6.

B

22.

B

7.

C

23.

C

8.

C

24.

A

9.

A

25.

B

10.

B

26.

C

11.

A

27.

B

12.

C

28.

C

13.

B

29.

C

14.

C

30.

A

15.

A

16.

C



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Cpl Navigation

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