6*
SCALES
.1-0.Introduction: Drawings of small objects can be prepared of the same size as the . o b j e c t st h e y r e p r e s e n tA . - j 5 0 m m l o n g p e n c i lm a y b e s h o w n b y a d r a w i n t of 5.0 rr.l.mlength. Drawings drawn Jf the same size as the objects, ar; ..i _ c a l l e d f u l l - s i z e d r a w i n g s .T h e o r d i n a r y f u l l _ s i z es c a l e sa r e u s e d f o r such drawings.A scale is defined as the ratio of the rineardimensionsof erement o f t h e o b j e c t .a s r e p r e s e n t e di n a d r a w i n gt o t h e a c t u a ld i m e n s i o n so f the same elemeni of the obiect itself.
1-1. Scales: ..1191generallyused {or generalengineering drawingsare snown . th," in table4-1 [SP: 46(19S8)]. TABTE4.1 Reducingscales
Enlargingscales
1..2 'l:2O l:200 1 : 2000 50:1 5:
Full size scales
I
1:5 1 :50 1:500 1 : 5000 20t1 2 t1
1 : 1 : 1 : 1 : 10:
10 100 1000 '10000 1
1:1
A l l t h e s e s c a l e sa r e u s u a l l y3 0 0 m m l o n g a n d s u b _ d i v i d e d throughout their .lengths. The scale is indicated on thj drawing at a suitatlle"place thg title.. The compiete designationof a scale c"onsistsof word scale :glr t o t i o w e d b y t h e r a t i o , i . e . s c a l e I : 1 o r s c a l e ,f u l l s i z e . ' a , - " y n o t b e a l w a y s . p o s s i b lteo p r e p a r ef u l l _ s i z ed r a w i n g s .T h e y are, -theretorei , drawn proportionately smaller or larger. When drawings are drawn smaller than the actual .size of the objects (as in case of bui-ldings, bridges, large machines etc.) the scale used is said to be a reducing scZle (1 :5). Drawings of small machrneparts, mathematicalinstruments/ watches etc. are made larger than their real size. These are said to be drawn on an enlargingsca/e (5 : 1). The scalescan be expressedin the following three ways: ( i ) E n g i n e e r ' ss c a l e : l n t h i s c a s e ,t h e r e l a t i o nb e t w e e n the dimension on the drawingand the actualdimensionof the object is mentionednumerically in the style as 10 mm = 5 m erc. (ii) Graphical scale: The scale is drawn on the drawing itself. As the . drawing becomes old, the engineer,sscale may shrink and may not give a c c u r a t er e s u l l s .
5 2 E n g i n e e r i n BD r a w i n g
However, such is not the case with graphical scale because if the d r a w i n g s h r i n k s ,t h e s c a l e w i l l a l s o s h r i n k . H e n c e , t h e g r a p h i c a ls c a l e i s c o m m o n l y u s e d i n s u r v e ym a p s . (iii) Representativefraction: The ratio of the length of the drawing to the actual length of the object representedis calied the Representative F r a c t i o n( i . e . R .F . ) . Length of the drawing R.F. _ Actual length of object W h e n a l c m l o n g line in a drawing represents1 metre length of the object, the R.F.is equal .
H
:
-'
,H
*,,
-
r0o1
und rhe scale of rhe
dra.rvingwill be 1 : 100 or,
full slze. Ihe R.F.of a drawing is greater ,*a than unity when it is drawn on an enlargingsca/e. For example, when a 2 m m l o n g e d g e o f a n o b j e c t i s s h o w n i n a d r a w i n gb y a l i n e 1 c m l o n g , 1cm 10mm - 5. Such a drawing is said to be drawn the R.h is , _n-' Z mon scale 5 :1 or five times full-size.
4-2. Scaleson drawings: W h e n a n u n u s u a ls c a l ei s u s e d , i t i s c o n s t r u c t e do n t h e d r a w i n gs h e e t . T o c o n s t r u c ta s c a l et h e f o l l o w i n gi n f o r m a t i o ni s e s s e n t i a l : ( i ) T h e R . F .o f t h e s c a l e . ( i i ) T h e u n i t s w h i c h i t m u s t r e p r e s e n t ,f o r e x a m p l e ,m i l l i r n e t r e sa n d centimetres,or feet and inches etc. ( i i i ) T h e m a x i m u ml e n g t hw h i c h i t m u s t s h o w . T h e l e n g t ho f t h e s c a l ei s d e t e r m i n e db y t h e f o r m u l a : Length of the scale = R.F.x maximum length required to be measured. It may not be always possible to draw as long a scale as to measure the longest length in the drawing. The scale is therefore drawn 15 cm to 30 cm long, Ionger lengths being measuredby marking them off in parts.
4-3. Typesof scales: T h e s c a l e su s e d i n p r a c t i c ea r e c l a s s i f i e da s u n d e r : ( i ) P l a i ns c a l e s (i i ) D i a g o n a I scales (i i i ) C o m p a r a t i v es c a l e s (iv) Vernier scales (v) Scaleof chords. ( i ) P l a i n s c a l e s :A p l a i n s c a l e c o n s i s t so f a l i n e d i v i d e d i n t o s u i t a b l e n u m b e ro f e q u a l p a r t so r u n i t s ,t h e f i r s t o f w h i c h i s s u b - d i v i d e di n t o s m a l l e r p a r t s . P l a i ns c a l e sr e p r e s e net i t h e rt w o u n i t s o r a u n i t a n d i t s s u b - d i v i s i o n . In every scale, ( i ) T h e z e r o s h o u l d b e p l a c e da t t h e e n d o f t h e f i r s t m a i n d i v i s i o n , i . e . b e t w e e nt h e u n i t a n d i t s s u b - d i v i s i o n s .
Scales 53
( i i ) F r o m t h e z e r o m a r k , t h e u n i t s s h o u l d b e n u m b e r e dt o t h e r i s h t a n d i t s s u b - d i v i s i o ntso t h e i e f t . ( i i i ) T h e n a m e s o f t h e u n i t s a n d t h e s u b - d i v i s i o n ss h o u l d b e s t a t e d c l e a r l yb e l o w o r a t t h e r e s p e c t i v ee n d s . ( i v ) T h e n a m e o f t h e s c a l e ( e . g . s c a l e ,j : 1 0 ) o r i t s R . F .s h o u l d b e m e n t i o n e db e l o w t h e s c a l e . Problem a'1. ffig. 4-1)t Constructa scaleof 1 :4 to show centrmetres and long enough to measureupto 5 decimettes.
CENTIMETRES
DECIMETRES R . F .= % Ftc.4_.1
(i)
D e t e r m i n e R . F .o f t h e s c a l e . H e r e i t i s ] .
(ii) Determine length of the scale.
4
Lengthof the scale= R.F.x maximumtength= = .12.5cm. ] x 5 dm 1 2 . 5 (iii) Draw a line cm long and divide it into 5 equal divisions, each representing1 dm. (iv) Mark 0 at the end of the first divisionand 1,2,3 and 4 at the e n d o f e a c h s u b s e q u e ndt i v i s i o nt o i t s r i g h t . ( v ) D i v i d et h e f i r s t d i v i s i o ni n t o ' 1 0e q u a ls u b - d i v i s i o n se,a c hr e p r e s e n t i n g I Cm.
(vi) Mark cms to the left of 0 as shown in the figure. To distinguish the divisions clearly, show the scale as a rectangleof small width (about 3 mm) insteadof only a /ine. Draw the divisio;-lines s h o w i n g d e c i m e t r e st h r o u g h o u tt h e w i d t h o i t h e s c a l e .D r a w t h e l i n e s f o r t h e s u b - d i v i s i o n s l i g h t l ys h o r t e ra s s h o w n . D r a w t h i c k a n d d a r k h o r i z o n t a l l i n 6 s i n t h e m i d d l e o f a l l a l t e r n a t ed i v i s i o n sa n d s u b - d i v i s i o n sT. h i s h e l o s i n t a k i n g m e a s u r e m e n t sB. e l o w t h e s c a l e ,p r i n t D E C I M E T R EoSn t h e r i g h t h a n d s i d e , C E N T I M E T R EoS n t h e l e f t - h a n ds i d e , a n d t h e R . F .i n t h e m i d d l e . To set-off any distance,say 3.7 dm, place one leg of the divider on 3 dm mark and the other on 7 cm mark. The distance between the ends of the two legs will represent3.7 dm. Problem 4-2. (tig. 4-2).: Draw a scale of 1 :60 to show metres and decimetresand long enough to measureupto 6 metres.
METRES ,I
Ftc.4-2
60
54
Engineeting Drar'ving
_1. (i) Determine R.F.oi the scale, here R.F.- 60 ( i i ) D e t e r m i n el e n g t ho f L h es c a l e . 'I 1 L e n g t ho f t h e s c a l e: 6 0 - * 6 . = 1 0 m e t r e : 1 0 c m ' ( i i i ) D r a w a l i n e 1 0 c m l o n g a n d d i v i d ei t i n t o 6 e q u a lp a r t s ' ( i v ) D i v i d e t h e f i r s t p a r t i n t o 1 0 e q u a l d i v i s i o n sa n d c o m p l e t et h e s c a l e a s s h o w n .T h e l e n g t h3 . 7 m e t r e si s s h o w n o n t h e s c a l e ' = 'l foot to shot\ Problem 4-3. (fi8. 4-g)t Constructa scaleof 1'5 inches inchesand long enough to measurcupto 4 feet' 2 FT.t0 rN -.
Frc.4-3
R . F .= %
= I Determine R.F.of the scale. R.F.= *@.,, t.nu; ( i i ) D r a w a l i n e ,1 . 5 x 4 = 6 i n c h e sl o n g . one foot' iiii) oiuia" it into four equal parts, each part representing representinS e a c h p a r t s , iiui oiuio" the first division into 12 equal T he distance 4 1 p r o b l e m i n 1 " . C o m p l e t et h e s c a l ea s e x p l a i n e d 2 ' - 1 Q " i s s h o w n m e a s u r e di n t h e f i g u r e . : to read yards Probtem 4-a. (fig. 4-4)t Constructa scaleof R'F AO and feet, and long enough to measure upta 5 yards' (i)
3Y.rF.
Ftc.4-4
R.F. =
-L
60
(i) Length of thescale= R.F.x max.length: fr " S ya = 3 inches' : iro ( i i ) D r a w a l i n e 3 i n c h e sl o n g a n d d i v i d e i t i n t o 5 e q u a l p a r t s ' ( i i i ) D i v i d e t h e f i r s t p a r t i n t o 3 e q u a ld i v i s i o n s . (iv) Mark the scale as shown in the figure. 1 = to show miles Probtem 4-5. (fig. 4'5): Constructa scaleof R.F *a* and furlongs and long enough to measurcupto 6 miles'
MILES I
FlG.4-5
84480
Scales
(i) Lengthof the scale=
*ir'
.6:
Ss
11'
;I-r +* u o u m i l e s = 4 ; . I
( i i ) D r a w u t i n " 4 i l o n g a n d d i v i d ei t i n r o 6 e q u a lp a r t s .D i v i d et h e f i r s t p a r t i n t o B e q u a l d i v i s i o n sa n d c o m p l e t et h e s c a l ea s s h o w n . T h e d i s t a n c e4 m i l e sa n d 3 f u r l o n g si s s h o w nm e a s u r e di n t h e f i g u r e . D^iagonalscales:A diagonalscaleis used when very minute distances ,(ii) such as 0.1 mm etc. are to be accuratelymeasuredor when measuremenfs are. required in three units; {or example, dm, cm and mm, or yard, foot and inch. S m a l l d i v i s i o n so f s h o r t l i n e s a r e o b t a i n e db y t h e p r i n c i p l eo f d i a g o n a l d i v i s i o n ,a s e x p l a i n e db e l o w . P r i n c i p l eo f d i a g o n a ls c a l e :T o o b t a i n d i v i s i o n so f a g i v e ns h o r t l i n e AB 1 i n m u l t i p l e so f i r s l e n g t h ,e . g . 0 . 1 A B ,O . Z A B ,0 . 3 A B e t c . ( f i g . 4 _ 6 ) . 1O ( i ) A t o n e . e n d ,s a y 8 , d r a w a l i n e p e r p e n d i c u l a r to
it, srep-off equaidivisions of 1_, l,!""q _ten anyild length, srartingfrom
B and ending at C. ( i i ) N u m b e rt h e d i v i s i o n - p o i sn,t9 , 8 , 7 , . . . . . 1a s shown. r i i i ) J o i nA w i r h c . ( i v ) T h r o u g ht h e p o i n t sj , 2 e r c .d r a w l i n e sp a r a l l e l to A B a n d c u t t i n gA C a t 1 ' , 2 , e t c . t t i s e v i d e n tt h a t t r i a n g l e s1 ' 1 C, Z ' 2 C . . . A B C a r e s i m i l a r . S i n c eC 5 : 0 . 5 8 C ,t h e l i n e 5 , 5 = 0 . 5 4 8 . S i m i l a r l y ,1 ' 1 : O . 1 A 8 , 2 , 2 = 0 . 2 4 8 e t c . Thus, each horizontal line below ,48 becomes progressivelyshorter in length by AA giuing lengthsin fr m u l t i p l e so f 0 . 1 4 8 -
A
6
i['"f",F '.f, " \ s'r 4
7 6 5
3 I
Frc.
Problem 4-6. (fig. 4-7\: Constructa diagonalscaleof 3 :2OO i.e. 1 :66 1 showing metres, decimetres and centimetrcs and to measure upto 6 meluel.
Lengthof the scale:
1
3ft
x 6 m = 9cm.
|,/,' url FI ull
=i r!l
ul DECIMETRES
n.r'=,1
240
Ftc.4-7
METRES
56
Enginee.ing Drawing
(i) Draw a line AB 9 cm long and divideit into 6 equalparts'Each will show a metre. (ii) Divide the first part A0 into 10 equal divisions'each showing d e c i m e t r eo r 0 . 1 m . along.it' 10 equal (iii) ' ' At A erect a perpendicularand step-off. lengttr,ending at D. Completethe rectangleABCD' oi "nyperpendicularsat metre-divisionsO, 1,2' 3 and 4' (iv) Erect (v) Draw horizontal lines through the division-pointson AD'
alongA0' viz' the.point iuil lo'n o with the end of the first division parallelto Pointsi.e.8, 7, 6 etc' drawlines iuiil inrougf,the remaining 1 dm or 0'1 m' Eachhorizontal! """ P^:l:: In a OFF,FF represents
in lengthbv-01FEtl':' I" :"f l:":YT ji p-gt.*;y'Oi *ini'shes m or 9 cm' !q"?r"i"b.s'ri"nai"fr"t"n* 6.srt dt = 0'9 dmor 0'09 1 cm or.0'0r- m 1nd o 1 :":b". T3t"'j:o Any lengthbetween. of 4.56metres,i'e' 4. m' 5 d,T."! I Aitr"n." tflir'i.|"1'f."tfo* " at 9 .yh"':,th",l-"'!.*l il.l"-:gt.;--T legof the.divider ;;;;"; ani the other leg at P wherethe cm o ir'"""gri iff;;;.;;l through 5 dm meetsthe same horizontal' R'F' Probfem 4'7. (Iig. 4-8\ Constructa diagonalscaleof metres' show metresand long enoughto measureupto 500
1 4000
I
4000
Frc. 4-B
Lengthof the scale :
1 4000
x500m=6
1
metre= 12.5cm.
(i) Draw a line 12.5cm long and divideit into 5 equalparts'Each will show I00 metres. (ii) Divide the first part into ten equal divisions'Eachdivision show 10 metres. and on it' step-off1 (iii) At the left-handend, erect a perpendicular . equaldivisionsof anYlength' and completethe scaleas shownin problem (iv) Drawthe rectangle The distancebetweenpointsA and I shows374 metres' :2'5' Probtem a-8. (fig. 4-9)t Draw a diagonalscale of .'l ce 20 upto cen;metres and milliietres and long enoughto measure
Scales 57 'l
L e n g t ho f t h e s c a l e -
20 cm - 8 cm. ,., " ( i ) D r a w a l i n e B c m l o n g a n d d i v i d e i t i n t o 4 e q u a l p a r t s . E a c hp a r t w i l l r e p r e s e n at l e n g t ho f 5 c m . ( i i ) D i v i d et h e f i r s tp a r ti n t o 5 e q u a ld i v i s i o n sE. a c hd i v i s i o nw i l l s h o w . l c m . ( i i i ) A t t h e l e f t - h a n de n d o f t h e l i n e , d r a w a v e r t i c a l l i n e a n d o n i t s t e p - o f f1 0 e q u a l d i v i s i o n so f a n y l e n g t h . C o m p l e t et h e s c a l ea s e x p l a i n e di n p r o b l e m4 - 6 . T h e d i s t a n c eb e t w e e n p o i n t sC a n d D s h o w s 1 3 . 4 c m .
url FI srl
EI
tl :t
I
CENTIMETRES
Frc.4-9 Problem 4-9. (fig. 4-10)t Constructa diagonatscaleof R.F. yards, feet and inches and to measure upto 4 yards.
Lengthof the scale=
+
"
1
32
showing
4 yd = tB yd = ^ 1 " 2
YARDS
Frc.4-.10 (i) Drawa lineN
+!
bng.
I
32
( i i ) D i v i d e i t i n t o 4 e q u a l p a r t s t o s h o w y a r d s .D i v i d e t h e f i r s t p a r t A 0 i n t o 3 e q u a l d i v i s i o n ss h o w i n gf e e t . (iii) At A, erect a perpendicularand step-off along it, 12 equal divisions o f a n y l e n g t h , e n d i n g a t D . C o m p l e t et h e s c a l e a s e x p l a i n e di n problem 4-6. To show a distanceol 1 yard,2 feet and Z inches, place one leg of the . dividerat P, where the horizontalthrough7,,meetsthe veriicalfrom I y:ardand t h e o t h e r l e g a t Q w h e r e t h e d i a g o n atlh r o u g h2 , m e e t st h e s a m eh o i i z o n t a l .
Ingineeting Drawing
Problem4'10. (tig. 4-1'l)tDraw a scaleof full-size'showingfr uPto 5 inches' andto measure
i"n
INCHES
TENTH5
F r c .4 - 1 1 parts Eachpart Draw a line AB 5" long and divide it into five equal will show one inch. d i v i s i o nw i l l ( i i ) S u b - d i v i d et h e f i r s t P a r t i n t o . 1 0 e q u a l d i v i s i o n s 'E a c h 1 m e a s ur e 1 O I n c h ' equal ( i i i ) A t A , d r a w a p e r p e n d i c u l a tr o A B a n d o n i t ' s t e p - o f f t e n d i v i s i o n so f a n y l e n g t h ,e n d i n ga t D ' in (iv) Draw the reciangle,48CD and complete the scale as explained p r o b l e m4 - 6 ' T h e l i n e Q P s h o w s 2 ' 6 8 i n c h e s ' sq m' The length Probfem 4-11. (fig. 4-12)'.The areaof a field is 50,000 and B cm respectively' ana tii iieuatn of tie field, on the map is 10 cm Mark the length m.etre' one upto '""d can i ai^gonal scalewhich Z.rtiirr, scale? ii zss ,"rr" o-n the sca/e.What is the R'F o{ the (i)
100
o
100
zur
METRES
I
R.F.='566'
rrc.4-12 The areaof the field : 50,000sq m' : 80 cmr' The area of the field on the map : 10 cm x B cm
Scales 59
.- - :625 = -50000 sq m. OU I cm =25m. .1 cm 1 Now representativefraction 25 m 2500
..
lsqcm
..
Lengrh of rhescate=
s00-x100: zllo , ltST
2ocm.
Take 20 cm length and divide it into 5 equal parts. Complete the scale as shown tn fig. 4-12-
(iii). Comparative scales: Scales having same representative fraction . but graduated to read different units are callJd comparaiivesca/es.A drawing drawn with a. scale readinginch units can be iead in metric units bi meansot a metric comparativescale,constructedwith the same representative f r a c t i o n . C o m p a r a t i v es c a l e sm a y b e p l a i n s c a l e so r d i a g o n a ls c a l e sa n d may be constructed separatelyor one above the other. Problem a42, l{ig. 4-13(i) and fig. a-13(ii)l: A drawins is drawn in .?
inch units to a sca/e futt tit". Draw the scale showing I f, inch divisions and.to measure upto 15 inches. Construct a comparativi scale showing centimetresand millimetres,.and to read upto 40 centimetres. (i) lnch scale: Lengrhof the scale
= :
3
= 4s u x ts ;
-5 5
;
+l 'rl (rl EI
inches.
C o n s t r u c t h e d i a g o n a ls c a l ea s shown in fig. a-13(i).
F r c .J - t 3 ( t )
INCHES
I
(ii) Comparative scale:
Ei
[l
Lengthof the scale E I =i 3 =; x40 EI o
:
15 cm.
CENTIMETRES
C o n s t r u c tl h e d i a g o n a ls c a l ea s shown in fig. a-l3(ii).
F l c . 4 - . i3 ( l )
The line PQ on the inch scale shows a length equal ro 1i +. e q u i v a l e n tw , h e n m e a s u r e do n t h e c o m p a r a t i v es c a l ei s 2 8 . 9 c m . Problem 4.13. (fig. 4-14): Draw comparativescalesof R.F.: readupto BO kilometrcsand do yersts.7 verst = 1.067 km. VERSTS
KILOIYETRES Ftc. 4-14
.:_ 485000
Its
Engineering Drawing
x B0 x 1000 x 100 : 16 5 cm'
L e n g t ho f k i l o m e t r e, . u t " = ; u j * o L e n g t ho f v e r s t s c a l e :
x B 0 x 1 . 0 6 7x 1 0 0 0 x 1 0 0 : 1 7 ' 6 c m ' *arrrt Draw the two scalesone above the other as shown in the figure' Problem 4-14. (fig. 4-15)t On a road map, a scaleof miles is shown' On measuringfrom this scale,a distanceof 25 miles is shown by a line 10 cm lonT. Construct this scale to read miles and to measure upto 40 miles. Constructa comparative scale, attachedto this scale,to read kilometres upto 60 kilometres. 1 mi[e : 1.609 km. KILOMETRES
MI L E S F t c .4 - 1 5 (i)
Scaleof miles:
1-0 " 4 -0 = 16 cm. 2s D r a w a l i n e 1 6 c m l o n g a n d c o n s t r u c ta p l a i n s c a l et o s h o w m i l e s ' . L e n g t ho f t h e s c a l e
( i i ) Scaleof kilometres:
RF -
10
2 s ' l . o o t t l o o ox l o o 402250 1
x 60 x 1000 x 100 : 14'9 cm' ^O;so Constiuct the plain scale 14.9 cm long, above the scale of miles and attachedto it, to read kilometres. Problem 4-15. (fig. 4-16): The distancebetween Bombay and Poona is 78o km. A passen1erirain covers this distancein 6 hours. Construct a plain Length of the scale =
scaleto measuretime upto a singleminute. TheR.F.of the scale't mk' Find the distance covered by the train in 36 minutesS p e e do f t h e t r a i n -
1' a Rn
30 km/hour.
or= Frc.4-16
1 200000
Scales 61
( i ) D i s t a n c es c a j e( k i l o m e t r e s c a l e ) : Length of the scaie = R.F.x rnnyipym distance x 30 x looo x .too = 15 cm. 2o** ( i i ) T i m e s c a l e( m i n u t es c a l e ) : Speed of the train = 30 km/hour. =
i . e . 3 0 k m i s c o v e r e di n 6 0 m i n u t e s . A s l e n g t h o f t h e s c a l e o f 1 5 c m r e p r e s e n t s3 0 k m , 6 0 m i n u t e s w h i c h i s t h e t i m e r e q u i r e dt o c o v e r 3 0 k m , c a n b e r e p r e s e n t e d on the same length of the scale. (iii) Draw a line.15 cm Iong and divide it into 6 equal parts. Each part represents 5 km for thb distance scale and -10 minutes for the time scale. ( i v ) D i v i d e t h e f i r s t p a r t o f t h e d i s t a n c es c a l ea n d t h e t i m e s c a l ei n t o 5 a n d 1 0 e q u a lp a r t s r e s p e c t i v e l yC.o m p j e t et h e s c a l e sa s s h o w n .T h e d i s t a n c ec o v e r e di n 3 6 m i n u t e si s s h o w n o n t h e s c a l e . Probfem 4.16. (fig. 4-17): On a Russianmap, a scaleof verstsis shown. ^ On measuring it with a metric scale, 150 yelsts are found to measure.lS cm. Construct comparative scalesfor the two units to measure upto 200 versts and 200 km respectively.'lverct = 7.067 km. (i) Sca/e of verst: Length of the scale -
L5-l-200 = 20 cm. 150 D r a w a l i n e 2 0 c m l o n g a n d c o n s t r u c ta p l a i n s c a l et o s h o w v e r s t s . (ii) Sca/eof kilometres:
.
R'F= rso rr#* rooo , ro = c"r*' " .10 =
Lengthof the scale
x 200 x i 000 x : 12.4 cm. #* Construct the plainscalei2.4 cm long,abovethe scaleof verstsand attachedto it, to readkilometres(fig. 4-17).
'
-
1 160900
VERST
ttc. 4-17 (iv) Vernier scales: Vernier scales, like diagonal scales, are used to read to a very small unit with great accuracy."A vernier scale consists of a lwo. ?.arts..-. .primary scale and a vernier. fhe primary scale is a plain s c a l ef u l l y d i v i d e d i n t o m i n o r d i v i s i o n s . As it would be difiicuit to sub-dividethe minor divisions in the ordinary way, it is done with the help of the vernier. The graduations on the vernier are derived from those on the primary scale.
6 2 E n g i n e € r i n gD r a w i n g
P r i n c i p l e o f v e r n i e r : F i g . 4 - 1 8 s h o w s a p a r t o f a p l a i n s c a l ei n w h i c h t h e l e n g t f iA 0 r e p r e s e n t s1 0 c m . l f w e d i v i d eA 0 i n t o t e n e q u a l p a r t s ,e a c h ourt *iil represent 1 cm. lt would not be easy to divide each of these p a r t s i n t o t e n e q u a l d i v i s i o n st o g e t m e a s u r e m e n t isn m i l l i m e t r e s ' Now, if we take a length B0 equal to 1 0 + 1 : 1 1 s u c h e q u a l p a r t s ,t h u s r e p r e s e n t i n1g1 c m , a n d d i v i d ei t i n t o t e n e q' u a l d i v i s i o n s ,e a c h o f t h e s e d i v i s i o n s 11
w i l l r e p r e s e n t, , n = 1 1 c m o r 1 1 m m . The difference between one part of A0 CENTIMETRES a n d o n e d i v i s i o no f B 0 w i l l b e e q u a l -l F I C . 4 - 1B 1.1 - 1.0 = 0.1 cm or mm. Similarly, w i l l b e 0.2 cm or 2 mm. the difference belween two parts o{ e a c h o Ihe upper sca/e 80 is the vernier. The c o m b i n a t i o n f t h e P l a i n s c a l e a n d t h e v e r n i e ri s t h e v e r n i e rs c a l e . n u n i t s i s d i v i d e di n t o n e q u a l p a r t s ' I n g e n e r a l ,i f a l i n e r e p r e s e n t i n g !n = 1 u n i t B u t , i f a l i n e e q u a lt o n + l o f t h e s e u n i t s e a c hp a r t w i l l s h o w i s t a k e na n d t h e n d i v i i e d i n t o n e q u a lp a r t s ,e a c ho f t h e s ep a r t sw i l l b e e q u a l 1 n+1 = ti units The difference between one such part and one 1 + ,o ; !# = , " i t S i m i l a r l yt,h e d i f f e r e n c e f o r m e r p a r t w i l l b e e q u a la I i between two parts from each will be j
unit.
Least count of a vernier: lt is the differenceof 1 primary scale d i v i s i o na n d I v e r n i e rs c a l ed i v i s i o n .l t i s d e n o t e db y Z C ' Z C = 1 p r i m a r y s c a t ed i v i s i o n- 1 v e r n i e r s c a l ed i v i s i o n 'T h e v e r n i e r s c a l e sa r e c l a s s i f i e da s u n d e r : (i) Forward vernier: ln this case, the length of one division of the vernier scale is smallerthan the length of one division of the primary scale. The vernier divisions are marked in the same d i r e c t i o na s t h a t o f t h e m a i n s c a l e ' (ii) Backwardvernier: The length of each division of vernier scale is greater .than the length of each division. of the primary stale' T h e n u m b e r i n gi s d o n e i n t h e o p p o s i t e d i r e c t i o n a s t h a t o f t h e prima!'yscale. = to 'eud Probfem 4-17. (fig. 4-1g\t Draw a vernier scaleof R'F ]; centimetresupto 4 metres and on it, show lengthsrepresenting2'39 m and 0.91m. CENTIMETRES
DECIMETRES Length of the scale
F r c .4 - 1 9 1
4 x 100 : 16 cm.
Scal€s 53
(i) Draw u cm.long and divide it inro 4 equal parts to show 16 metres.lil". Divide each of tl]ese partsinto l0 equalpartsto sholv decjmetres. (ii) To construct a vernier, take .11 parts o f d m l e n g t h a n d d i v j d e i t r n t o . 1 0 e q u a l p a r t s . E a c h o f t h e s e parts will show a length of 1.1 dm or 11 cm. a l e n g t hr e p r e s e n t i n 2 g. 3 9 m , p l a c eo n e l e g o f t h e d i v i d e r ^ .t A^ ' o on -.:ujr." a 99 cm mark and the othei leg at B on t.+ m m " a r k r. h e l e n l t h AB wilf show 2.39 metres (0.99 + j.; : 2.3g). S i m i l a r l y ,t h e l e n g t h ,C D s h o w s 0 . 9 . 1m e t r e ( O . B + 0 . . 1 1= 0 . 9 1 ) . T h e n e c e s s i t yo f d i v i d i n gt h e p l a i n s c a l ei n t o m i n o r d i v i s i o n st h r o u g h o u t its length is quite evidentlrom the above measurements. -1S. (fig. 4-2O)t Construct a full_sizevernier scaleof inches a_.n d,t].obf"s h o w o n i t l e n g t h s3 . 6 7 " , 1 . 5 4 " a n d 0 . 4 8 " .
Ftc.4-20 ( i ) D r a w a p l a i n f u l l - s i z es c a l e4 , , l o n g a n d d i v i d ei t f u l l y t o s h o w 0 . 1 , , l en g t hs . ( i i ) C o n s t r u c ta v e r n i e ro f l e n g t he q u a l t o 1 0 + j : j l p a r t sa n d d i v i d e i t i n t o 1 0 e q u a l p a r t s .E a c ho f t h e s e p a r t s w i l l t f f ; q = o.rr ". T h e l i n e A B s h o w s a l e n g t ho f 3 . 6 7 , ,( O . 7 7 " + 2 . g , : E . O Z , , )S. i m i t a r l y , , l ]n . " t . C D a n d P Q s h o w l e n - g t h so i 1 . s 4 " ( 0 . 4 4 " + 1.1i = 1.s4,,) an() 0 . 4 8 " ( 0 . S 8 "- O . 4 " : 0 . 4 8 , , ) r e s p e c r r v e t y . Probfem a-19, (lig. 4-21): Constructa vernier scaleof R.F.= { to ,"ud ov inchesand to measureupto 15 yards.
Ftc. 4-21 .
Length ot the scate _
|
8b
I
ts yd =
?
;
?"
yd = d4
(i)
Draw the plain scale Of; ring and divide ir f u l l y r o snow yards and feet. (ii) To construct the vernier, take a length of 12 + 1 = 1 3 f e e t - d i v i s i o n sa n d d i v i d e i t i n t o 1 2 e q i a t p a r t s . - f a c f , part will 1? represent tt or 1,_1". ;2 LinesA8, CD and pe show ^respectivelylengths representing4 y d l f r 9 i n ( 9 ' - 9 " + 4 ' ) ,6 y d 2
f t 3 i n ( 3 ' _ 3 ,+, . r i , l a n & o y a i t i ) - i ^V , - 7" - s").
64
Engineering DtatYing
used to set out or measure (v) Scale of chords: The scale of chords'is It is basedon the lengths ungtu'*tl"n' J pt"it"tt"' L ""t "ti'li!1", on the same arc and rs measured of chords of different angles below' shown as constructed (fig' 4-22)' (i) Draw a lineAB of any length
( i i ) A t B ' e r e c td P e r p e n d i c u l a r ' t* t""t*' describean^arcAC cutting ff:11:"l:: (iii) With S AC ) subtendsan chord "' (or the arc'4C ut u pt'iii c in""' the angle of 90' at the centre 8' may be done ( i v ) D i v i d eA C i n t o n i n e e q u a l p a r t s 'T h i s p1:'j:Jjtu*'"g ( a ) b y d i v i d i n gt h e a r c A C i n t o t h r e e " e q u a l and then A8' radius and C u"' *ittt ZentresA and three equal parts by (b) by dividing each o{ these parts into t h e nine equal parts iach of t"'itoa t'i"i';;i''"; c e n t r e8 ' ' u [ t " n a t a n a n g l eo f 1 0 ' a t t h e arc to the straiSht line (v) franster each divisi"on-pointfrom the and radii equal to chords A8-procluced'by taking A as centre A -10, A -20 etc' rectanqle below AD' The (vi) ComPlete the scale by drawing a d i v i s i o n s o b t a i n e d a r e u n e q U a | / o e c r e a s i n "fgr ogm r a' dAl atlol y far o m lvident'that the distance to D' ; ;;';"i; t o t h e l e n g t ho f t l r e : - 1 : 1 1 C i v i s i o n - p o i not n t h e s c a l ei s e q u a l be noreo b v i t a t l h e c e n r r eB l t m a v tt th;;I;i;'t;;renJed A 8 ' the radius tn"t tni lnoJ 4-60 is equal to ten equal i ' e e a c h d i v i s i o nd i v i d e di n t o T h e s c a l er n u , * ' u " ' d i v i d e d ' of 5' d e g r e e sa r e s h o w n i n m u l t i p l e s o a r t s t o s h o w d e g r e e s l' n t h e f i g u r e '
F$' 4-23 Ftc.4-22 anglesof 47' and 125' by means P r o b l e m 4 ' 2 0 . ( f i 8 . 4 - 2 3 ) : Construct of the scale of chords' (i) Draw any line AB and radius equal to.0-60 (ii) With any Point P on it as centre point R' at"* an arc cutting AP at a (from the scaleo{ iorltl, to 0-47 (chord of 47') cut ( i i i ) W i t h R a s c e n t r e and radius equal the arc at a Point T. ZRPT = 47"' ( i v ) D r a w a l i n e j o i n i ng P with T fhen
Scales 65
As_the scaleof chordsgivesanglesupto only 90", angleof 125. may be set-offin two partsviz. 60' * 65" or 90" + 35. as shownin the figure. It may also be obtainedby setting-offa chordSe of 55" (180. - j25.) from the right. Problem4-21.(fig. 4-24\2Measurcthegiven angle PQRby meansof the scaleof chords. (i) With Q as centreand radiusequalto 0-60,drawan arc cuttingpe at A and RQ at 8. (ii) Takethe chord lengthAB and applyit to the scaleof chordswhich shows the angle to be of 37". Ftc.4-24
ErerciseslV 1. Fill-upthe blanksin the followingsentences, usingappropriatewords,' selectedfrom thosegivenin the brackets: (a) The ratio of the length of the drawing of the object.to the actuallengthof the object is called (resultingfraction, representativefigure, representative fraction). (b) When the drawing is drawn of the same size as that of the object,the scaleusedis _ (diagonalscale,full-sizescale, vernierscale). (c) For drawingsof small instruments,watchesetc. ___ scale is alwaysused(reducing,full-size,enlarging). (d) Drawingsof buildingsare drawn using _.__ (full-sizescale, reducingscale,scaleof chords). (e) The R.F.in caseof (b), (c) and (d) abovewould be, and _ respectively(equalto 1, less than 1, greaterthan 1). (f) When measurements are requiredin three units__ scaleis used (diagonal,plain,comparative). (g) The scale of chords is used to set out or measure (chords,lines,angles). Answercto Ex. (1): a-3,b-2, c-3, d-2, e-1, 3 and 2, 11, g-3. 2 . Constructa scaleof 1:5 to show decimetresand centim€tresand to readupto 1 metre.Showthe lengthof 7.Gdm on it. 3 . Constructa scaleof 1.5 cm = 1 dm to readupto 1 metreand show on it a lengthof 0.6 metre. Draw a diagonalscaleof R.F.= 1fu, showingmetres,decimetresand centimetres,and to measureupto S metres.Show the lengthof 3.69 metreson it. 5 . Draw a scaleof 1:50 showingmetresand decimetres, and to measure upto 8 metres.
66
E n g i n e e r i n gD r a w i n g
s l e n g t ho f 4 m e t r e s .E x t e n dt h i s l i n e t o 6 . A 3.2 cm long line representa 25 metres and show on it units of metre and 5 measure lengihs upto m m e t r e s .S h o w t h e l e n g t ho f 1 7 e t r e so n t h i s l i n e . Construct a diagonal scale of R.F. = ar*1 a read upto 1 kilometre and to read metres on it. Show a length of 653 metres on it. B . O n a m a p , t h e d i s t a n c eb e t w e e n t w o p o i n t s i s 1 4 c m ' T h e r e a l d i s t a n c eb e t w e e nt h e m i s 2 0 k m . D r a w a d i a g o n a ls c a l e o f t h i s m a p to read kilometres and hectametres,and to measure upto 25 km. S h o w a d i s t a n c eo f 1 7 . 6 k m o n t h i s s c a l e . 9 . A n a r e a o f 1 4 4 s q c m o n a m a p r e p r e s e n t sa n a r e a o f 3 6 s q k m o n t h e f i e l d . F i n d t h e R . F .o f t h e s c a l ef o r t h l s m a p a n d d r a w a d i a g o n a l scale to show kilometres,hectametresand decametresand to measure u p t o 1 0 k i l o m e t r e s .I n d i c a t eo n t h e s c a l ea d i s t a n c eo f 7 k i l o m e t r e s , 5 hectametresand 6 decametres. 1 0 . C o n s t r u c tt h e f o l l o w i n g s c a l e sa n d s h o w below each, its R.F.and the u n i t s w h i c h i t s d i v i s i o n sr e p r e s e n t : 7.
: I foot, to measureupto 5 feet and showing feet 1 (a) Scale i "t and inches.. (b) Scale of "7 = 1 yard, to measure upto 10 yards and showing yards and feet. = 1 m i l e , t o m e a s u r eu p t o 4 m i l e s a n d s h o w i n g ( c ) S c a l eo f 1 ? milesand furlongs. 11. Construct a scale of 1" = 1 foot to read upto 6 feet and show on it, 4' - 7" length. ' 1 2 . T h e R . F .o f a s c a l e s h o w i n g m i l e s , f u r l o n g s a n d c h a i n s ' t u o * f u Draw a scale to read upto 5 miles and show on it, the length representing3m5f3ch. .1 'l Draw a 4" long diagonalscaleof " = " and show on it, the length of I J. 2 . ' 1 4 "a n d 3 . 7 9 " . 1n The points are 1 4 . The distance between two points on a map is 5 ;' actually 20 miles apart. Construct a diagonal scale of the map, s h o w i n gm i l e s a n d f u r l o n g sa n d t o r e a d u p t o 2 5 m i l e s . I J. C o n s t r u c t c o m p a r a t i v ed i a g o n a l s c a l e so f m e t r e s a n d y a r d s h a v i n g to show upto 400 metres. 1 metre : 1.0936 yards' 2200 1 6 . D e f i n e s i m p l e a n d c o m p a r a t i v es c a l e s .W h a t i s t h e R ' F o f a s c a l e w h i c h m e a i u r e s2 . 5 i n c h e s t o a m i l e ? D r a w a c o m p a r a t i v es c a l e o f k i l o m e t r e st o r e a d u p t o 1 0 k m . 1 m i l e = 1 6 0 9 k m . , 0 km are found to equal 1 7 . O n a m a p s h o w i n ga s c a l e o f k i l o m e t r e s 6 R . F ? C o n s t r u c ta c o m p a r a t i v es c a l e o f w h a i i s t h e 4.5 inches. '1093.6 = E n g l i s hm i l e s . 1 k m Yards. R.F.=
Scales 67
1 8 . On a Russianmap/ a scale of versts is shown. On measuringit rvith a m e t r i c s c a l e , 1 2 0 v e r s t s a r e f o u n d t o m e a s u r e1 0 c m . C o n s t r u c t comparativescalesfor the two units to measure upto 150 vers{s and 1 5 0 k m r e s p e c l i v e l y1. v e r s t : 1 . 0 6 7 k m . 1 9 . Prepare a scale of knots comparativeto a scale of j cm = 5 km, A s s u m e s u i t a b l el e n g t h s .1 k n o t = 1 . 8 5 k m . 20. Draw a full-size vernier scale to read -T,^a -f4
it lensrhs * t !r,
z slo ^a 2]o.
l;
lengthsand mark on
2 1 . C o n s t r u c ta s c a l eo f * . r . = t o s h o w d e c i m e t r e sa n d c e n t i m e t r e s + a n d b y a v e r n i e rt o r e a d m i i l i m e t r e st,o m e a s u r eu p t o 4 d e c i m e t r e s . 2 2 . C o n s t r u c ta v e r n i e rs c a l et o s h o w y a r d s ,t h e R . F .b e i n g ] . n n . S h o wt n e d i s t a n c er e p r e s e n t i n 2 g f u r l o n g s9 9 y a r d s . 2 3 . C o n s t r u c t a s c a l e o f c h o r d s s h o w i n g 5 " d i v i s i o n sa n d r v i t h i t s a i d s e t - o f fa n g l e so f 2 5 ' , 4 0 ' , 5 5 ' a n d 1 3 d . . 2 4 . D r a w a t r i a n g l eh a v i n gs i d e sB c m , 9 c m a n d 1 0 c m l o n g r e s p e c t i v e l and measure its angleswith the aid of a scale of chords. The distance between Vadodaraand Surat is 130 km. A trarn covers t h i s d i s t a n c ei n 2 . 5 h o u r s .C o n s t r u c a t p l a i n s c a l et o m e a s u r et i m e u p t o a s i n g l e m i n u t e . T h e R . F .o f t h e s c a l e , , ,*bb'. covered by the train in 45 minutes.
Find rhe disrance
2 6 . O n a b u i l d i n gp l a n , a l i n e 2 0 c m l o n g r e p r e s e n t sa d i s t a n c eo f 1 0 m . D e v i s e a d i a g o n a ls c a l e f o r t h e p l a n t o r e a d u p t o 1 2 m , s h o r v i n e metres, decimetresand centimetres.Show on your scale the lengths 6 . 4 8 m a n d 1 ' 1 . 1 4m . 2 7 . A room of 1728 m3 volumeis shownby a cube of 216 cm3 volume. F i n d R . F .a n d c o n s t r u c ta p l a i n s c a l et o m e a s u r eu p t o 4 2 m . M a r k a distance of 22 m on the scare.
28. A n o l d p l a n w a s d r a w n t o a s c a l e o { 1 c m = 2 4 n . l t h a s s h r u n K so that actual length of 100 m at site now works out to 96 rn ds pei scale on plan. Find out the shrinkagefactor and the corrected R.F.of t h e p l a n . ( H i n t : S h r i n k a g ef a c t o r = p r e s e n t l e n g t h o n s c a l e / o r i g i n a Iength on scale.) 2 9 . The actual length of 500 m is representedby a line of 15 cm on a drawing. Construct a vernier scale to read upto 600 m. MJrk on the scale a length of 549 m.
30. Draw a vernier scale of R.F.= 5 to read I ..
und I cm ancl tc measure upto 5 cm. Mark on the scale distinces of 2.112" cm
3 1 . A car is running at a speed of 50 km/hour..Constructa diagonal scale t o s h o w 1 k m b y 3 c m a n d t o m e a s u r eu p t o 6 k m . M a r k j l s o o n t h e s c a l e t h e d i s t a n c ec o v e r e db y t h e c a r i n 5 m i n u t e s2 8 s e c o n d s .
GEOMETRICA
5
CONSTR,UCTIO
5-0. Introduction: I n t h i s c h a p t e r ,w e s h a l l d e a lw i t h p r o b l e m so n g e o m e t r i c a l w h i c h a r e m o s t l y b a s e do n p l a n eg e o m e t r ya n d w h i c h a r e v e r y e s s e n t i a.l . h e y a r e d e s c r i b e da s u n d e r : t l r e p r e p a r a t i o no f e n g i n e e r i n gd r a w i n g s T 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
1.1.Specialmethodsof d B i s e c t i na g line p e r p e n d i c u l a r s regularpolygons To draw inscr polygons 12. Regular To draw parallel ines circles To dividea line 1 3 . T o d r a w r e g u l a rf i g u r e su s i T o b i s e c ta n a n g l e T-squareand set-squares T o t r i s e c ta n a n g l e 14. To draw tangents To find the centre of an arc T o c o n s t r u c te q u i l a t e r atlr i a n g l e s 1 5 . L e n g t h so f a r c s 1 6 . C i r c l e sa n d l i n e si n c o n t a c T o c o n s t r u c ts q u a r e s 1 7 . l n s c r i b e dc i r c l e s . T o c o n s t r u c tr e g u l a rp o l y S o n s
5-l . Bisectinga line: Problem 5-1, Io bisect a given straight/ine (fig. 5-1). (i) Let AB be the given line. With centre A and radius greater than A8, drarv arcs on both sides of A8. (ii) With centre I and the same radius,draw arcs intersecting p r e v i o u sa r c s a t C a n d D . ( i i i ) D r a w a l i n e j o i n i n g C a n d D a n d c u t t i n gA B a t F . 1
Then/E:EB=;AB.
Further,CD bisectsAB at right angles. Problem 5-2. To bisecta given arc (fig. 5-2). Let AB be the arc drawn with centre O. Adopt the same method a-< s h o w n i n p r o b t e m5 - 1 . T h e b i s e c t o rC D , i f p r o d u c e d ,w i l l p a s sthrough the centre O.
Ceometrical Construclion 69
3-2,To draw perpendiculars: Problem 5-3. To draw a perpendicular to a given line from a point within it (fis. 5-3).
(o)
I Frc.5-3
It @)
(a) When the point is nearthe middle of the line. Let48 be the given line and p the point in it. ( i ) W i t h P a s c e n t r e a n d a n y c o n v e n i e n tr a d i u s R 1 , draw an arc c u t t i n gA B a t C a n d D . ( i i ) W i t h a n y r a d i u sR 2 g r e a t e rt h a n R 1 a n d c e n t r e s C and D, draw arcs intersectingeach other at O. (iii) Draw a linejoiningp and O. Then PO is the required perpendkular. (b) When the point is nearan end of the line. LetAB be the given line and p the point in it. Method I: (i) With any point O as centreand radius equal to Op, draw an arc greater than the semi_circle,cutting Ag at C. ( i i ) D r a w a l i n ej o i n i n gC a n d O , a n d p r o d u c e it to cut the arc at e. Draw a linejoining p and e. T h e n P Q i s t h e r e q u i r e dp e r p e n d i c u l a r . Method II: ( i ) W i t h P a s c e n t r e a n d a n y c o n v e n i e n tr a d i u s , draw an arc cutting /B at C. (ii) With the same radiuscut (from the arc) two equal divisions CD and Df. (iii) Again with the same radius and centres D and E, draw arcs i n t e r s e c t i n ge a c h o t h e r a t e . D r a w a l i n e j o i n i n g p and e. p e Then i s t h e r e q u i r e dp e r p e n d i c u l a r . Probfem S-4. To draw a perpendicularto a given line from a point outside it (fig. 5-4).
7 C E n g i n e e r i n SD r a w i n S
( a ) W h e n t h e p o i n t i s n e a r e rt h e c e n t r et h a n t h e e n d o f t h e l i n e . LetAB be the given line and P the point, (i) With centre P and any convenient radius, draw an arc cutting AB at C and D. ( i i ) W i t h a n y r a d i u sg r e a t e rt h a n h a l f C D a n d c e n t r e sC a n d D . (iii) Draw the arcs intersectingeach other at E. (iv) Drirw a line joining P and E and cutting AB at Q. T h e n P Q i s t h e r e q u i r e dp e r p e n d i c u l a r .
(b)
(o)
Flc.5-4
(b) When the point is nearer the end than the centre of the line. L e t A B b e L h eg i v e n l i n e a n d P t h e p o i n t . (i) With centre A and radius equal to AP, draw an arc [F cutting AB or AB-produced,at C. (ii) With centreC and radiusequalto CP,draw an arc cutting FFat Dg B at Q. ( i i i ) D r a w a l i n e . j o i n i n gP a n d D a n d i n t e r s e c t i n A T h e n P Q i s t h e r e q u i r e dp e r p e n d i c u l a r .
5-3. To draw Parallellines: Problem 5-5. To draw a line through a given point, parallel to a given straight line (fi8. 5-5). Let AB be the given line and P the Point. With centre P and any convenient radius, draw an arc cD cutting AB at E. (ii) With centre E and the same radius,draw an arc cuttin:.r''B3t F (iii) With centre f and radius equal to FP, draw an arc to cut C:-)at Q.
(i)
(iv) Draw a straight line through P and Q. Then this i:: the r e q u i r e dl i n e .
GeometricalConstruction
71
Frc.5-6 Probfem 5-6. To.draw a line parallel to and at a given distance from a given straight /ine (fig. 5-6). Let AB be the given line and R the given distance. (i) Mark points.p and e on AB, as far apart as convenient. ( i i ) W i t h p a n d e . a s c e n t r e sa n d r a d i u s e q u a lt o R , d r a w a r c s o n t h e samesideof AB. (iii) Draw the line CD, just touching the two arcs. CD is the required l i ne .
5-4. To divide a line: Probfem 5-7(a). To divide a given straight line into any number of e q u a l p a r t st f i g . 5 -Z ( i ) ] . Let AB be the given line to be divided into say, seven equal parts. (i) Draw a line AC of any length inclined at some convenient angle to AB (preferablyan acute angle). (ii) From A and along.AC, cut-off with a divider seven equal divisions o f a n y c o n v e n i e n tl e n g t h . (iii) Draw a line joiningB and 7. (iv) With the aid of two set-squares,draw lines through i, 2,3 etc., p a r a l l e l t o B Z ( p r o b l e m .1 _ 4 , C h . 1 ) i n t e r s e c t i n g " AB at points 1 ' , 2 ' , 3 ' e t c . , t h u s d i v i d i n gi t i n t o s e v e ne q u a l ;arts.
Ftc.5-7(i)
7 2 E n g i n e e r i n BD r a w i n g
Problem 5-7(b). To divide a Siven suai9htline into unequal Partstfig.5-7(ii)1. L e t A B b e t h e g i v e n l i n e t o b e d i v i d e d i n t o u n e q u a l parts say 1 1111
6 ' a , i , a a n dt '
(i) Draw a line AB of given length, say, 120 mm' ( i i ) E r e c t p e r p e n d i c u l aA r D and BC at the ends A and B' Complete r e c t a n g lua r A B C D . at f. ( i i i ) J o i n d i a g o n a l sA C a n d 8 D i n t e r s e c t i n g (iv) Draw perpendicularfrom E on AB as shown' 1 (v) Then 47 = , AB. ( v i ) J o i n D a n d F . T h e l i n e F D intersectsthe diagonal AC at C. Drop 1 perpendicularfrom C to AB. T h e n A H = a A B . 1
( v i i )S i m i l a r l ym a k e c o n s t r u c t i o nin figure,for obtaining t 1
and a AB as shown
o
Frc.5-7(ii)
5-5. To bisect an angle: Problem 5-8. Io bisect a given angle (fig' 5-8)'
Frc.5-B
1
A8, ; A B
Ceometrical Construction 73
L e t , 4 8 Cb e t h e g i v e n a n g l e . ( i ) W i t h I a s c e n t r e a n d a n y r a d i u s ,d r a w a n a r c c u t t i n g A B a t D ano 5L ar t. ( i i ) W i t h c e n t r e s D a n d I a n d t h e s a m e o r a n y c o n v e n i e n tr a d i u s , draw arcs intersectingeach other at F. (iii) Draw a line joining I and F. 8f bisects the angle ABC, i.e. ZABF : z FBC. Problem 5-9. To dnw a line inclined to a given line at an angle equal to a given angle (fig. 5-9).
Frc.5-9 Let PQ be ihe given line and AO8 the given angle. (i) With O as centre and any radius, draw an arc cutting OA at C and OB at D. (ii) With the same radius and centre P/ draw an arc EF cutting PQ at F. (iii) With F as centre and radius equal to CD, draw an arc cutting rne arc tf ar L. ( i v ) F r o m P , d r a w a l i n e p a s s i n gt h r o u g h C . T h i s i s t h e r e q u i r e dl i n e .
5-6. To trisect an angle: Problem 5-10. Io tfisect a given right angle (fig. 5-10). L e t , A B Cb e t h e g i v e n r i g h t a n g l e . (i)
With centre I and any radius, draw an arc cutting AB at D and BC at E. (ii) With the same radius and centres D and [, draw arcs cutting the arc DE at points Q and P. ( i i i ) D r a w l i n e s j o i n i n g I w i t h P a n d Q . B P a n d 8 Q r r i s e c rr h e r i g h t a n g l eA B C . rhus, ZABP = ZPBQ = ZQBC =
Flc. 5-10
]
z,oec.
F l c .5 - 1 1
-:
E cs i ne e r i n g D . a w i n t
5-7. To find the centre of an arc: Probfem 5-11. To find the centreof a given arc (fig. 5-.li). Let AB be the given arc. (i) ln AB, draw two chords CD and EF of any lengths. ( i i ) D r a w p e r p e n d i c u l a rb i s e c t o r s o { C D a n d f F i n t e r s e c t i n g e a c h o t h e r a t O . T h e n O i s t h e r e q u i r e dc e n t r e . Probfem 5.72. To draw an arc of a given radius, touching a given straight line and passingthrough a given point (fig. 5-12). Let,48 be the givenline,P the point and R the radius. (i) Draw a line CD parallel to and at a distance equal to R from AB ( P r o b l e m5 - 6 ) . (ii) With P as centre and radiusequal to R, draw an arc cutting CD at o. ( i i i ) W i t h O a s c e n t r e ,d r a w t h e r e q u i r e da r c .
Ftc. 5-12
F r c .S - 1 3
Problem 5-13. Io draw an arc of a given radius touching two given stnight lines at right anglesto each other (fig. 5-13). L e tA B a n d A C b e t h e g i v e n l i n e sa n d R t h e g i v e n r a d i u s . (i) With centre A and radius equal to R, draw arcs cutting AB at p and AC at Q. (ii) With P and Q as centresand the same radius, draw arcs intersecting . each other at O. ( i i i ) W i t h O a s c e n t r ea n d r a d i u se q u a l t o R , d r a w t h e r e q u i r e da r c . Probfem 5-14. To draw an arc of a given radius touching two given gtraight lines which make any angle between them tfig. S-14(a) and fig. s-1a(b)1.
Frc.5-14(a)
Frc.5-14(b)
Ceometrical Construction 75
L e tA B a n d A C b e t h e g i v e n l i n e sa n d R t h e g i v e n r a d i u s . ( i ) D r a w a l i n e P Q p a r a l l e tl o a n d a t a d i s t a n c ee q u a l t o R f r o m A B . ( i i ) S i m i l a r l y ,d r a w a l i n e [ f p a r a l l e lt o a n d a t a d i s t a n c ee q u a l t o R from AC, intersectingPQ at O. ( i i i ) W i t h O a s c e n t r ea n d r a d i u se q u a l t o R , d r a w t h e r e q u i r e da r c . Probfem 5"15. Io draw an arc of a given radius touching a given arc and a given straighLline. C a s e I ; [ f i g . 5 - 1 5 ( a ) ] :L e t A B b e t h e g i v e n l i n e , C D r h e g i v e n a r c d r a w n w i t h c e n t r eO a n d r a d i u se q u a l t o R . , ,a n d R , t h e g i v e n r a d i u s . (i) With O as centre and radius equal to (R1- R2), draw an arc FF. ( i i ) D r a w a l i n e p a r a l l e lt o a n d a t . a d i s t a n c ee q u a l t o R 2 f r o m A B and intersectin6 gf at a point P. ( i i i ) W i t h P a s c e n t r ea n d r a d i u se q u a lt o R 2 , d r a w t h e r e q u i r e da r c .
F r c .s - 1 5 ( a )
F r c . 5 - 15 ( b )
Case ,r: tfig. 5-15(b)l: Let AB be the given line, CD the given arc drawn with centre O and radius equal to R.,,and R2 the given radius. ( i ) W i t h O a s c e n t r ea n d r a d i u se q u a lt o ( R 1 * R 2 ) ,d r a w a n a r c E F . ( i i ) D r a w a l i n e p a r a l l e lt o a n d a t a d i s t a n c ee q u a l t o R 2 f r o m A B a n d intersectingff at a point P. ( i i i ) W i t h P a s c e n t r ea n d r a d i u se q u a lt o R 2 , d r a w t h e r e q u i r e da i c . Problem 5"16. Io draw an arc of a given radius touching two given . - |bl, a f c s ( T r g5 Let AB be the given arc drawn with centre O and radius equal to R1; CD the arc drawn with centreP and radiusequal to R2, and R3 the eiven radius. Case ,: ( i ) With O as centre and radius equal to (R1 + R3), draw an arc EF. ( i i ) With O as centre and radius equal to (R2 + R3), draw an arc intersectingfF at a point Q. (i i i ) With Q as centre and radius equal to R3, draw the required arc. Case II: ( i ) With O as centre and radius equal to (R1- R3), draw an arc ff. ( i i ) W i t h P a s c e n t r e a n d r a d i u s e q u a l t o ( R 2 + R 3 ) , d r . a wa n a r c intersectingEF at a point Q. (i i i ) With Q as centre and radius equal to R3, draw the required arc.
75 tngineering Drawin8
Case III: (i) With O as centre and radius equal to (R3- R1), draw an arc EF. (ii) With P as centre and radius equal to (R3 - R2). draw an arc intersecting fF at a point Q. ( i i i ) W i t h Q a s c e n t r ea n d r a d i u se q u a l t o R 3 , d r a w t h e r e q u i r e da r c .
T]]IO Ftc.5-16
F t C .S - 1 7
Problem 5-17. To draw an arc passingthrough three given points not in a straight line (fig. 5-17). Let A, I and C be the given points. ( i ) D r a w l i n e sj o i n i n g I w i t h A a n d C . ( i i ) D r a w p e r p e n d i c u l a rb i s e c t o r s o f , 4 8 a n d B C i n t e r s e c t i n ge a c h other at a point O. (iii) With O as centre and radius equal to OA or OB or OC, draw the r e q u i r e da r c . Problem 5.18. Io draw a continuous curve of circular arcs passing through any number of given points not in a straight/ine (fig. 5-18).
: r c .5 - 1 8 LeIA, B, C, D and f be the givenpoints. (i) Draw linesjoining.,4with 8, I with C, C with D etc. (ii) Draw perpendicular bisectorso{ AB and 8C intersecting at O. (iii) With O as centreand radiusequalto OA, draw an arc ABC. (iv) Draw a line joining O and C.
Ceom etaical.Construction 77
( v ) D r a w t h e p e r p e n d i c u l a rb i s e c t o r o f C D i n t e r s e c t i n eO C o r O C produced, at P. (vi) With P as centre and radius equal to pC, draw an arc CD. ( v i i ) R e p e a tt h e s a m e c o n s t r u c t i o n N . ote that the centreof the arc is a t t h e i n t e r s e c t i o no f t h e p e r p e n d i c u l abr i s e c t o ra n d t h e l i n e , o r t h e l i n e - p r o d u c e dj ,o i n i n g t h e p r e v i o u sc e n t r e w i t h t h e l a s t p o i n t o f t h e p r e v i o u sa r c .
5-8. To constructequilateraltriangles: Problem 5-1g. To construct an equilateral of the side (fie.5-19). ""*^*-.,
(o) f rc. 5-19
,*
-rr"
(b)
(a) With T-squareand set-squareonly. (i) With the T-square,draw a line AB of given length. ( i i ) W i t h 3 0 ' - 6 0 ' s e t - s q u a r ea n d T - s q u a r e ,d r a w a l i n e t h r o u g h A m a k i n g 6 0 . a n g l ew i t h 4 8 . ( i i i ) S i m i l a r l y ,t h r o u g h B , d r a w a l i n e m a k i n g t h e s a m e a n g l e with AB and intersectingthe first Iine at C. T h e n , 4 8 Ci s t h e r e q u i r e dt r i a n g l e . (b) With the aid of a compass. (i) With centres A and I and radius equal to AB, draw arcs intersectingeach other at C. ( i i ) D r a w l i n e sj o i n i n g C w i t h A a n d B . T h e n A B C i s t h e r e q u i r e dt r i a n g l e . Probfem 5-2O. To construct an equilateraltriangle of a given altitude (fig. s-20). (a) With T-squareand set-squareonly. (i) With the T-square,draw a line AB of any length. (ii) From a point P in AB, draw with a set-square,the vertical pe equal to the given altitude. ( i i i ) W i t h T - s q u a r ea n d 3 0 . - 6 0 . s e t - s q u a r ed, r a w l i n e s t h r o u g h e o n b o t h s i d e s o f a n d m a k i n g 3 0 . a n g l e sw i t h p e a n d c u t t i n g ,48 in R and f. T h e n Q R f i s t h e r e q u i r e dt r i a n g l e .
7B Engin€ering Drawing
(o)
(b) Flc.5-20
t6t With lhe aid of a compass. (i) Draw a lineAB of any length. ( i i ) A t a n y p o i n t P i n A B , d r a w t h e p e r p e n d i c u l aPr Q e q u a l t o t h e g i v e n a l t i t u d e( P r o b l e m5 - 3 ) . (iii) With centre Q and any radius,draw an arc intersectingPQ at C. (iv) With centre C and the same radius, draw arcs cutting the first arc at F and F. (v) Draw bisectorsof CEand CFto intersectAB at R and f respectively. T h e n Q R f i s t h e r e q u i r e dt r i a n g l e .
5-9. To €onstructsquares: Problem 5-21. To construct a square, Iengthof a side given (fi8. 5-21). 'a) With T-squareand set-squareonly. With the T-square,draw a line AB equal to the given length. (ii) At A and 8, draw verticalsz\f and 8F. (iii) Draw a line AC inclined at 45" to AB, cutting 8F at C' (iv) Draw a line BD inclined at 45" to AB, cutting AE at D. (i)
( v ) D r a w a l i n e j o i n i n gC w i t h D . Then ABCD is the required square.
(b)
(o)
Ftc.5-21
Ceometli.alConstruction
79
(b) With the aid of a compass (i)
Draw a line A8 equal to the given length.
(ii) At A, draw a line AE perpendicularto AB. ( i i i ) W i t h c e n t r eA a n d r a d i u sA B , . d r a wa n a r c c u t t i n t A E a t D . (iv) With centres I and D and the same radius, oraw arcs intersecting at C. (v) Draw lines joining C with B and D. fhen ABCD is the required square.
5-10. To construct regular polygons: Problem 5.22, To construct a regularpolygon, given the length of its side. Let the number of sides of the polygon be seven (i.e. heptagon). Method I: (fig. s-22\: (i)
Draw a line AB equal to the given length. (ii) With centre .,4and radius AB, draw a semi-circlegp. ( i i i ) W i t h a d i v i d e r .d i v i d e t h e s e m i - c i r c l ei n t o s e v e n e q u a l o a r t s i ( s a m ea s t h e n u m b e r o f s i d e s ) .N u m b e r t h e d i v i s i o n - o o i n i a ss 1, 2, etc. starting from p. ( i v ) D r a w a l i n e j o i n i n g A w i t h r h e s e c o n dd i v i s i o n - p d i n2t .
Y (o)
(b) Ftc, S_22
(a) /nscribe circle method. (i)
D r a w p e r p e n d i c u l a rb i s e c t o r s o l A 2 each other at O.
and Ag intersecting
(ii) With centre O and radius OA. describe a circle. (iii) With radius AB and starting from B, cut the circle at points, c, D.....2. (iv) Draw lines BC, CD etc. thus completingthe required heptagon.
B0 Engineering Drawing
(b) Arc method. (i) With centre I and radius A8, draw an arc cutting the line A 6 - p r o d u c e da t C (ii) With centre C and the same radius, draw an arc cutting t h e l i n e A 5 - p r o d u c e da t D . ( i i i ) F i n d p o i n t s f a n d F i n t h e s a m em a n n e r . (iv) Draw lines 8C, CD etc. and complete the heptagon' Method II: ffig. 5-23(i)l: Ceneralmethod for drawing any polYson:
(a) (b) (c)
(d)
( i ) D r a w a l i n e A B e q u a lt o t h e g i v e n l e n g t h . ( i i ) A t 8 , d r a w a l i n e 8 P p e r p e n d i c u l aar n d e q u a l t o A B ' ( i i i ) D r a w a l i n e . i o i n i n gA w i t h P . (iv) With centre I and radius AB, draw the quadrantAP' ( v ) D r a w t h e p e r p e n d i c u l a rb i s e c t o r o f A B t o i n t e r s e c t t h e s t r a i g h tl i n e A P i n 4 a n d t h e a r c A P i n 6 . A square of a side equal to AB can be inscribedin the circle d r a w n w i t h c e n t r e4 a n d r a d i u s4 4 ' A r e g u l a r h e x a g o no f a s i d e e q u a l t o A B c a n b e i n s c r i b e d i n the circle drawn with centre 6 and radius 46. T h e m i d - p o i n t5 o f t h e l i n e 4 - 6 i s t h e c e n t r eo f t h e c i r c l e o f t h e r a d i u s A i i n w h i c h a r e g u l a r p e n t a g o no f a s i d e e q u a l t o A B can be inscribed. To locate centre 7 for the regular heptaSon of side AB, step-off a d i v i s i o n6 - 7 e q u a l t o t h e d i v i s i o n5 - 6 . ( i ) W i t h c e n t r e 7 a n d r a d i u se q u a l t o 4 7 , d r a w a c i r c l e ' ( i i ) S t a r t i n gf r o m 8 , c u t i t i n s e v e n e q u a l d i v i s i o n sw i t h r a d i u s
equal to AB. (iii) Draw lines 8C, CD etc and complete the heptaSon' Regularpolygons of any number of sides can be drawn by this method'
I Frc.5-23(i)
+ Frc.5-23(ii)
Frc.s-23(iii)
ceometricalConstructionBl Alternative method ltig. 5-23(ii)l: (i) On AB as diameter,describea semi-circle. (ii) With either A or B as centre and AB as radius, describe an arc o n t h e s a m es i d e a s t h e s e m i - c i r c l e . ( i i i ) D r a w a p e r p e n d i c u l a rb i s e c t o r o f A B c u t t i n g t h e s e m i - c i r c l e at point 4 and the arc at point 6. ( i v ) O b t a i n p o i n t s 5 , 7 , 8 e t c . a s e x p l a i n e di n m e t h o d , l t . Fig. 5-23(iii) shows a square, a regular pentagont a regular hexagon and a regularoctagon, all constructedon AB as a common side.
5-11.Specialmethods of drawing regular polygons: Problem 5-23, To construct a pentagon, Iength of a side given. Method I: (tig. 5-24): (i)
Draw a line AB equal to the given length.
( i i ) W i t h c e n t r eA a n d r a d i u sA 8 , d e s c r i b ea c i r c l e - l . (iii) With centre I and the same radius, describe a circle-2 cuttting circle-l at C and D. (iv) With centre C and the same radius, draw an arc to cut circle-1 and circle-2 at f and F respectivelv. (v) Draw a perpendicularbisectorof the line AB to cut the arc fF at C. (vi) Draw a line fC and produce it to cut circle-2 at p. (vii) Draw a line FC and produce it to cut circle-j at R. (viii) With P and R as centres and AB as radius, draw arcs intersecting each other at Q. (ix) Draw lines 8P, PQ, QR and RA, thus completingthe pentagon.
X Ftc.5-24
Ftc.5-25
82
Engineering Dlawing
Method II: (fig.5-25): (i)
D r a w a l i n e A B e q u a lt o t h e g i v e n l e n g t h .
( i i ) B i s e c tA B i n a p o i n t C . ( i i i ) D r a w a l i n e B D p e r p e n d i c u l aar n d e q u a l t o A B . (iv) With centre C and radius CD, draw an arc to intersect the line AB-producedat E. (v)
Then AF is the length of the diagonalof the pentagon.
(vi) Therefore, with centre A and radius AB, draw an arc intersectingthe arc drawn with centre I and radius Af at R. (vii) Again with centre A and radius AE, draw an arc intersecting the arc drawn with centre I and radius AB at P (viii) With centres A and I and radius AE, draw arcs intersecting each other at Q. (ix) Draw lines 8P, PQ, QR and RA, thus completingthe pentagon. Probfem 5-24. To constructa hexagon,length of a side given (fig. 5-26). (a) With T-squareand 30"-60" set-squareonly. (i)
Draw a line AB equal to the given length.
( i i ) F r o n iA , d r a w l i n e s A 1 a n d / 2 m a k i n g 6 0 ' a n d 1 2 0 " a n g l e s respectivelywith A8. (iii) From 8, draw lines 83 and 84 making 60' and 120' angles respectivelywith AB. (iv) From O the point of intersectionof 41 and 83, draw a line parallelto AB and intersectingA2 at F and 84 at C. (v)
From F, draw a line parallelto 8C and intersecting83 at [.
(vi) From C, draw a line parallelto,4F and intersectingA1 at D. ( v i i ) D r a w a l i n e j o i n i n gE a n d D . fhen ABCDEFis the required hexagon.
ut
o
(o)
Ftc.5-26
G€ometrical Codstruclion 83
(b) With the aid of a compass. (i)
With any point O as centre and radius equal to the given l e n g t h ,d r a w a c i r c l e . (ii) With the same radius and startingfrom any point on the circle, set-off six divisions on the circle. ( i i i ) J o i n t h e d i v i s i o n - p o i n t si n p r o p e r s e q u e n c ea n d c o m p l e t e the hexagon. Problem 5-25. To inscribe a rcgular octagon in a given square (fig. 5-27). (i)
Draw the given square ABCD. (ii) Draw diagonalsAC and BD inter.sectingeach other a1 O. (iii) With centreA and radius,4O, draw an arc cuttingAB at 2 and AD at 7. (iv) Similarly,with centres B, C and D and the same radius, draw arcs and obtain points 1, 3, 4 etc. as sholvn. D r a w l i n e s 2 - 3 , 4 - 5 , 6 - 7 a n d 8 - - 1 ,t h u s c o r r , p l e t i n gt h e o c t a g o n .
Ftc.5-28
5-12. Regularpolygonsinscribedin circles: Problem 5-26. To inscribe a regular polygon of any number of sides, say 5, in a given circle (fig. 5-28). (i) With centre O/ draw the given circle. (ii) Draw a diameter AB and divide it into five equal parts (same number of parts as the number of sides) and number tltenr a5 snown, (iii) With centres A and I and radius A8, draw arcs intersecting each other at P. (iv) Draw a line PZ and produce it to meet the circle at C. Then /C is the length of the side of the pentagon.
8 4 E n g i n e e r i n gD r a w i n g
(v) Starting from C, step-off on the circle, divisions CD, DE etc. equal to ,4C. ( v i ) D r a w l i n e s C D , D F e t c . , t h u s c o m p l e t i n gt h e p e n t a g o n . Problem 5-27. Toinscribea squarein a given circle (fig. 5-29). (i) With centre O/ draw the given circle. (ii) Draw diametersAB and CD perpendicular A to each other. (iii) Draw lines AC, CB, BD and DA, thus c o m p l e t i n gt h e s q u a r e . Problem 5-28, To insuibe a regular pentagon in a given circle (fig. 5-30). (i) With centre O, draw the given circle. Frc.5-29 ( i i ) D r a w d i a m e t e r sA B a n d C D p e r p e n d i c u l at ro each other. ( i i i ) B i s e c tA O i n a p o i n t P . W i t h c e n t r e P a n d radius PC, draw a:. a r c c u t t i n gO B i n Q . (iv) With centreC and radiusCQ, draw an arc cuttingthe circle in E and.. (v) With centresi and F and the same radius, draw arcs cuttin: the circle in C and H respectively. (vi) Draw lines CE, EC, CH, HF and fC, thus completing the requirec pentagon.
)
Frc.5-30
(o)
Ftc.5-31
Problem5-29,To inscribe a regular hexagon in a given
(b) ( f i g . 5 - 3 1: .
Applythe samem e t h o da s s h o w n i n P r o b l e m5 - 2 4 ( b ) . Note;(a) When two sides of the hexagon are required to be horizontal the s t a r t i n g p o i n t f o r s t e p p i n g - o f fe q u a l d i v i s i o n s s h o u l d b e o n a n e n c of the horizontal diameter. (b) lf they are to be vertical, the starting point should be on an end o:' the vertical diameter In either case, to avoid inaccuracy, the points should be joined with the aid of T-square and 30"-6O" sef-sguare.
ceometrical Construction 85
Problem 5-30. Io lnscribe a regular heptagonin a given circle (fig. S-32). (i) With centre O, draw the given circle. (ii) Draw a diameter Ag. With centre A and radius AO, dralv an arc cutting the circle at E and F. ( i i i ) D r a w a l i n e f F , c u t t i n gA O i n G . Then FC or FC is the length of the side of the heptagon, Jherefore,.from any point on the circle, say ,4, step_off divisions equal . to fC, around the circle. Join the division_poinisand obtain the heptagon.
Flc.5-32
Ftc.5_33 Problem 5-31. To inscribea regular octagonin a given circle(fig. 5-33). (i) With centre O, draw the given circle. (ii) Draw diametersAB and CD at right angles to each other. (iii) Draw diametersEF and CH bisecting angles,4OC and COB. (iv) Draw lines Af, 6C etc. and complete the octagon.
i.13. To draw regular figures using T_squareand set-squares: Problem S-32. To describean equilateralt angle about a given circle (fig. 5-3a).
Ftc.5-34 (i) With centre O, draw the given circle. ( i i ) D r a w a v e r t i c a lr a d i u sO 4 .
86
Engin€ering Dtawing
( i i i ) D r a w r a d i i O B a n d O C w i t h a 3 0 ' - 6 0 ' s e t - s q u a r e ,s u c h t h a t /AOB=ZAOC=12o"' (iv) At A, I and C, draw tangents to the circle, i'e' a. horizontal line EF through A, and lines FC and Cf through B and C respectivel\ with a 30'-60" set-square. Tlren fFC is the required triangle. Problem g-33. To draw a squareabout a given circle (fig' 5-35)' (i) W'th centre O, describethe given circle. (ii) Draw diametersAB and CD at right anglesto each other as shown' (iii) At A and B, draw vertical lines, and at C and D, draw horizonta fines intersectingat E, F, C and H. EFCH is the required square. Problem 5-34. To describe a regular hexagon about a Siven circle Ifits.5-36(d)1. ' (i) With centre O, draw the given circle. ( i i ) D r a w h o r i z o n t a ld i a m e t e rA 8 , a n d d i a m e t e r sC D a n d E F m a k i n g 6 0 ' a r r g l ew i t h A B . (iii) uraw tangents at all the six ends, i.e' verticals at A and B' a n d l i n e i w i t h a 3 0 ' - 6 0 ' s e t - s q u a r ea t t h e r e m a i n i n gp o i n t s at 1, 2,....-6. intersecting A hexagonwith two sides horizontal can be drawn by drawing a vertical diaireter AB and the other lines as shown in fig 5-36(b)'
(o)
Frc.s-36
(b)
l-tL.
J-J
/
Problem 5-35. Io describea regulat octaSonabout a Siven circle (fig. s-37). (i) With centre O, describethe given circie. (ii) Draw a horizontal diameter A8, a vertical diameter CD and diametersEF and CH at 45" to the first two. ( i i i ) D r a w t a n g e n t s a t t h e e i g h t p o i n t s A , 8 . . . . H i n t e r s e c t t n go n e 'l a n o t h e ra t , 2 . . ' . . A .T h e n 1 , 2 . . . . . 8i s t h e r e q u i r e do c t a g o n '
Ceometrical Conslruction 87
5-14.To draw tangents: Problem 5-36. To draw a tangent to a given circle at any point on tr (fig. 5-38). (i) With centre O, draw the given circle and mark a point p on it. ( i i ) D r a w a l i n e j o i n i n gO a n d p . (iii) ProduceOP to Q so that pe = Op. (iv) With centres O and e and with any convenient radius, draw arcs intersecting each other at R. Frc.5-38 ( v ) D r a w a l i n e t h r o u g hp a n d R . T h e n t h i s l i n e i s t h e r e q u i r e d tangent. Problem 5.37. To draw a tangent to a given circle from any pornt outside it. ( i ) W i t h c e n t r eO , d r a w t h e g i v e n c i r c l e ( f i g . 5 _ 3 9 ) . ( i i ) M a r k a p o i n t p o u t s i d ei t . (iii) Draw a linejoiningO and p. (iv) With OP as diameter,draw a semi-circlecutting the given circle at R. Draw a line through p and R. Then this line is-the required tangent. line through p and R1 is the other tangentwhich ,The can be drawn from the same point. \ ,,,r,
f r tcc. . s5--3399 F
F r c .5 _ 4 0 ( a ) F t c . s-40(b) Frc. 5_40(b) Problem 5-38(a),.Io draw a tangent to a given arc of inaccessible centre at any point on it ftig.5_40(a)1. Let AB be the given arc and p the point on it. (i) With centre P and any radius, draw arcs cutting the arc AB at C and D. Draw EF,the bisector of the arc CD. lt will pass through p. (ii) Through P. draw a line RS perpendicularto EF. RS'is rhe req'uired tangent. . Probfem 5-38(b). Io draw a tarigentto a given circle and parallelto a Biven line tfig. 5-a0(b)1. The circle with centre O and the line Ag are given. (i) From O, draw a line perpendicularto Ag and c u t t i n gt h e c i r c l e a t a point P or Q. (ii) Through P or Q, draw the required tangent CD or C1 D1 ( P r o b l e m5 - 3 6 ) .
Drawing Engineering Problem 5-39. Io draw a common tangent to two given circles of e q u a l r a d i i ( fi g . 5 - 4 1 ) . Draw the given circles with centres O and P. (al Externaltangents[fig. 5-41(a)]: ( i ) D r a w a l i n e j o i n i n gO a n d P . (ii) At O and P, erect perpendicularsto OP on the same side oi it and intersectingthe circles at A and 8. ( i i i ) D r a w a l i n e t h r o u g hA a n d 8 . T h i s l i n e i s t h e r e q u i r e dt a n g e n t ' At 81 is the other tangent. \
Ftc.5-41(a)
Ftc. 5-41 (b)
(b\ lnternaltangents:tfig. 5-a1(biJ: (i) Drawa linejoiningO and P. ( i i ) B i s e c t O P i n R . D r a w a s e m i - c i r c l ew i t h O R a s d i a m e t e rt c cut the circle at 4. ( i i i ) W i t h c e n t r e R a n d r a d i u s R A , d r a w a n a r c t o i n t e r s e c tt h e other circle on the other side of OP at 8. ( i v ) D r a w a l i n e t h r o u g h/ a n d B . T h i s l i n e i s t h e r e q u i r e dt a n g e n t ' The other tangent through A1 and B1 can also be similarly drawn Probfem 5-4O. To draw a common tanlent to two given circles o; unequal radii tfig. 5-42(a) and fig. s-42(b)1. Draw the given circles with centres O and P, and radii R1 and Rrespectively,of which R1 is greater than R2. (a) Externaltangentslfig. 5-42(a)l: ( i ) D r a w a l i n e j o i n i n g c e n t r e sO a n d P . (ii) With centre O and radius equal to (R1 - R2), draw a circle' ( i i i ) F r o m P , d r a w a t a n g e n tP f t o t h i s c i r c l e ( P r o b l e m5 - 3 7 ) . (iv) Draw a line OI and produce it to cut the outer circle at A' (v) Through P, draw a line PB parallel to OA, on the same side of OP and cutting the circle at 8. ( v i ) D r a w a l i n e t h r o u g hA a n d 8 . T h e n t h i s l i n e i s t h e r e q u i r e d tangent. T h e o t h e r s i m i l a rt a n g e n tw i i l p a s st h r o u g hA 1 a n d 8 1 .
Ceometrical Const.uction 89
Ftc.5-42(a)
Frc.5-42(b)
(b) lnternal tangentslfig. 5-42(b)l: ( i ) D r a w a l i n e j o i n i n g t h e c e n t r e sO a n d p ( i i ) W i t h c e n t r eO a n d r a d i u se q u a l t o ( R 1 * R 2 ) ,d r a w a c i r c l e . ( i i i ) F r o m P , d r a w a l i n e p I t a n g e n tt o t h i s c i r c l e . (iv) Draw a line Of cutting the circle at A. (V) Through P, draw a line pB parallel to O,4, on the other side of OP and cutting the circle at g. ( v i ) D r a w a l i n e t h r o u g hA a n d L T h e n t h i s l i n e i s t h e r e q u i r e d t a n g e n t .T h e s e c o n dt a n g e n tw i l l p a s st h r o u g hA 1 a n d 9 1 .
5-15. Lengthsof arcs: Probfem S-41. To determine the length of a given arc (lig. S-43).
Ftc.5-43 Let AB be the given arc drawn with centre O. (i) At A, draw a tangent to the arc. (ii) Draw the chord AB and produce it beyondA to a point C such that AC = ; AB. (iii) With cenrre C and radius equal to CB, describe an arc cutting the tangent at D. ( i v ) T h e n t h e l e n g t h A D i s a p p r o x i m a l e l yequal to the length of the arc AB-
9 0 E n g i n e e r i n gD r a w i n g
This method is satisfactoty for arcs which subtend at the centre, angles smaller than 60". Problem 5-42. To determine the tength of the circumference of a given circle ([ig. 5-44). Let the circle with centre O be given. (i)
D r a w a d i a m e t e rA B .
( i i ) A t A , d r a w a t a n g e n tA C e q u a lt o 3 t i m e s A B . ( i i i ) D r a w a r a d i u sO D m a k i n ga n a n g l eo f 3 0 . w i t h O B . ( i v ) F r o m D , d r a w a l i n e D E p e r p e n d i c u l at ro O B . ( v ) D r a w a l i n e j o i n i n g f a n d C . T h e n f C i s a p p r o x i m a t e l ye q u a l i n l e n g t ht o t h e c i r c u m f e r e n coef t h e c i r c l e .
ttc.5-44
Flc.5-45
5-16. Circlesand lines in contact: Probfem 5-43. To draw a circle passingthrough a given point and tangent to a given line at a given point on it (fig. 5-a5). A p o i n t P a n d a l i n e A B w i t h a p o i n t e i n i t a r e g i v e n .A t e , d r a w a line perpendicularto ,'48. (i)
Draw a line joiningP and e.
(ii) Draw a perpendicularbisector of pe to intersectthe perpendicular from Q at O. ( i i i ) W i t h c e n t r eO a n d r a d i u sO p o r O e , d r a w t h e r e q u i r e dc i r c l e . Problem 5-44. To draw a circle passingthrough a given point and touchin7 a given circle at a given point on it (fig. S-ag. A p o i n t P , a c i r c l e w i t h c e n t r eA a n d a p o i n t e o n t h e c i r c l e a r e g i v e n . (i) Draw a linejoiningP and Q. (ii) Draw.a perpendicularbisector o{ pe, to intersect the line through AandQatO. ( i i i ) W i t h c e n t r eO a n d r a d i u sO p , d r a w t h e r e q u i r e dc i r c l e . T h e r e q u i r e d c i r c l e i n c l u d e st h e g i v e n c i r c l e w h e n t h e p o i n t i s i n a p o s i t i o ns u c h a s P ' .
GeometricalConsiruclion
F C. t c .5-4b 5-46
91
Ftc. 5_47
Problem 5.45, To draw a circle to touch a given line and a given circle at a given point on it (fig. 5-47). A i i n e A 8 , a c i r c l e w i t h c e n t r eC a n d a p o i n t p o n t h e c i r c l e a r e g i v e n . From P, draw a tangent to the circle intersectingAB in D. (a) Draw a bisector ol ZPDB, to intersect the line throueh C and p a t O . W i t h c e n t r e O a n d r a d i u s O p , ' d r a w t h e r e o u i r e dc i r c l e . (b) Draw a bisector ol IPDA to meet the line throush C and p aI O'. Then O'is the centre of anothercircle which will include the given circle within it. Probfem 5-46. To draw a circle to touch a given circle and a given line at a given point on it (fig. 5-a8). A circle with centre C and a I i n e A B w i t h d p o i n t P i n i t a r e g i v e n . Through C, draw a line p e r p e n d i c u l a rt o A B d n d c u l t i n g t h e c i r c l ei n E o r F . (a) Draw a line joiningP and F a n d i n t e r s e c t i n gt h e circleat C. At P, drawa perpendicular to ,,48intersectingthe line throughC and C at O. With centre O and radius OP, d r a w t h e r e q u i r e dc i r c l e .
Frc.5-48
(b) Draw a line through P and f and obtain centre O, for another c i r c l e i n t h e s a m e m a n n e r .l t w i l l i n c l u d et h e g i v e n c i r c l e w i t h i n i t . Probfem 5-47. To draw a circle touching two given circles,one of them at a given point on it [fig. 5-49(a) and fig. 5-49(b)]. Circles with centres A and B, and a point p on the circle A are given
tfig.s-ae(a)1.
92
E n g i n e e r i n gD r a w i n g
( i ) D r a w a l i n e j o i n i n gA a n d P . ( i i ) T h r o u g h8 , d r a w a l i n e p a r a l l etl o , 4 P a n d i n t e r s e c t i ntgh e c i r c l ei n C . ( i i i ) D r a w a l i n e P C a n d p r o d u c e i t ( i f n e c e s s a r y )t o c u t t h e c i r c l e ( w i t h c e n t r e- B ) i n D . ( i v ) D r a w a i i n e t h r o u g hD a n d I t o i n t e r s e c t . 4 P o r A P - p r o d u c e da, t O . W i t h c e n t r eO a n d r a d i u sO P , d r a w t h e r e q u i r e dc i r c l e .
Frc.5-49(a)
F t c . s-49(b)
F i g . 5 - 4 9 ( b ) s h o w s c i r c l e - ( - 1w) h i c h i n c l u d e sone of the given circles, a n d c i r c l e - ( 2 )w h i c h i n c l u d e sb o t h o f t h e m .
5 . 1 7 . I n s c r i b e dc i r c l e s : Problem 5-48. To insuibe a circle in a given triangle (fig. 5-50). Let ABC be the triangle. (i) Bisect any two angles by lines intersectingeach other at O. ( i i ) D r a w a p e r p e n d i c u l a frr o m O t o a n y o n e s i d e o f t h e t r i a n g l e , meeting it at P. ( i i i ) W i t h c e n t r eO a n d r a d i u sO P , d e s c r i b et h e r e q u i r e dc i r c l e . Problem 5-49. To draw a circle touching three lines inclined to each other but not fotming a triangle (fig. 5-51). Let AB, BC and AD be the given lines. (i)
Draw bisectorsof the two angles intersectingeach other at O. ( i i ) F r o m O , d r a w a p e r p e n d i c u l atro a n y o n e l i n e i n t e r s e c t i n gi t a t P . (iii) With centre O and radius OP, draw the required circle. Protrlem 5-50. Io inscribea circle in a regularpolygon of any number of sides, saya pentagon (fig. 5-52).
Geometrical Construction 93
LeI ABCDEbe the pentagon. (i)
Bisect any two angles by lines intersectingeach other a1 O. (ii) From O, draw a perpendicularto any one side of the pentagon cutting it at P. ( i i i ) W i t h c e n t r eO a n d r a d i u sO p , d r a w t h e r e q u i r e dc i r c l e .
Frc.5-50
Ftc.5-51
Flc.5-52 Problem 5-51. To draw in a regular polygon, the same number of equal circles as the sides of the polygon, -circlei .each circle touching one side of the polygon and two of the oihe; 1fig. s_sr). (i)
Let ABCD be the given square.
( i i ) D r a w b i s e c t o r so f a l l t h e a n g l e so f t h e s q u a r e .T h e y will meet at O , t h u s d i v i d i n g t h e s q u a r ei n t o f o u r e q u a l t r i a n g l e s . I n e a c h t r i a n g l ei n s c r i b ea c i r c l e ( p r o b l e m5 - 4 g ) . E a c hc i r c l e w i l l t o u c h a s i d e o f t h e s q u a r ea n d t w o o t h e r c i r c l e s a s r e q u i r e d . F i g . 5 - 5 4 s h o w s f i v e e q u a l c i r c l e s i n s c r i b e di n a r e g u l a rp e n t a g o n in ., the same manner-
rrL. i-i.t
FllCc.. 55_-5544 F Probfem 5-52. To draw in a regularpolygon, the same number -potygon,'"each of equal circles as the sldes of th.e circle touchrng two adjacent sides of the polygon and two oi tne otne, circles (fig. s_is).
94
E n g i n e e r i n gD r a w i n g
LeI ABCDEFbe the given hexagon. ( i ) D r a w t h e p e r p e n d i c u l a rb i s e c t o r so f a l l s i d e s o f t h e h e x a g o n . T h e y r v i l l m e e t a t O a n d w i l l d i v i d e t h e h e x a g o ni n t o s i x e q u a l q u a dr i l a t e r a ! s . ( i i ) I n s c r i b ea c i r c l e i n e a c h q u a d r i l a t e r aal s s h o w n i n c a s e o f / 1 0 2 a n d a s e x p l a i n e db e l o w . (i i i ) Bisect any two adjacent angles with bisectors intersecting each other at P. (iv) From P, draw a perpendicularto any one side of the quadrilateral, meeting it at Q. With centre P and radius PQ, draw one of the r e q u i r e dc i r c l e s . ( v ) D r a w o t h e r c i r c l e si n t h e s a m e m a n n e r .
F trcc..5 - 5 55
Ftc.5-56
Problem 5-53. Io draw in a given regular hexagon, three equal circles, each touchlng one side and two other circles (fig. 5-.56). (i)
Draw the given hexagon.
(ii) Draw perpendicularbisectorsof its 1wo alternatesides, to intersect e a c h o t h e r a t O a n d t o m e e t t h e m i d d l e s i d e p r o d u c e do n b o t h .i sides at and 2. ( i i i ) I n s c r i b ea c i r c l e i n t r i a n g l eO . 1 2 . S i m i l a r l y ,d r a w t h e o t h e r t w o r e q u i r e dc i r c l e s . Problem 5-54. To draw in a given circle,any number o{ equal circles,say four, each touching the given circle and two of the other circles(fi1. 5-57). (i)
Divide the given circle into four equal parts by diametersAB and CD.
( i i ) D r a w a t a n g e n t t o t h e c i r c l e a t D . D r a w l i n e s b i s e c t i n gZ A O D and ZBOD and meeting the tangent at 1 and 2. ( i i i ) I n s c r i b ea c i r c l e i n t h e t r i a n g l eO 1 2 . Draw the other circles in the same manner.The centresfor the remaining c i r c l e s m a y a l s o b e d e t e r m i n e db y d r a w i n g a c i r c l e w i t h c e n t r e O a n d r a d i u s O P t o c u t t h e d i a m e t e r sa t t h e r e q u i r e dp o i n t s .
Geometrical Construction 9S
Frc.5-57
Frc.5-5B
Problem 5.55. To draw outside a given regularpolygon, the same "the numbe.r of eq,ual circles the srdes of polyion, iaci'circle touching , .as one side and two of the other circies (fig. S_!A). (i) Let ABCDE be the given pentagon. (ii) Draw bisectors of .two adjacent angles, say ZA and ZB, and produce them outside the pentagon. (iii) Draw a circle touching the extended bisectors and the side Ag ( P r o b l e m5 - 4 9 ) . O b t a i n t h e o t h e r f o u r r e q u i r e dc i r c l e si n t h e same manner. Problem 5-56. Io draw outside a given circle any number of equa! circles, say six, each touching the given circle and two otier crrcles (fig. S_i9). (i) Draw the given circle and describe a regular hexagon about it. (ii) Draw the required six equal circles outsidl the hexigon as shown i n t h e p r e v i o u sp r o b l e m .
Frc.5-59
Ftc.5-60 P r o b l e m 5 - 5 7 . T o d r a w a c i r c l e t o u c h i n g t w o c o n v e r g i n gl i n e s a n d passingthrough a given point between them (flg. 5_60). (i) LinesAB and CD, and the point p are given. ( i i ) P r o d u c e l i n e s A B a n d C D t o i n t e r s e a ta t a p o i n t f. Draw the bisector EF of .AEC.
Engineering Drawing
(iii) Mark any point Q on tF and from it draw a perpendicularQR on AB' (iv) With Q as centre and QR as radius draw a circle which will touch the line CD also' ( v ) D r a w a l i n e j o i n i n g P w i t h F , c u t t i n gt h e c i r c l e a t a p o i n t G ' (vi) Draw the line QC. E Fa t a p o i n t O ' ( v i i ) F r o m P , d r a w a l i n e p a r a i l e tl o Q C i n t e r s e c t i n g (viii) From O, draw a perpendicularON to either AB or CD' ( i x ) W i t h O a s c e n t r ea n d O N a s r a d i u s ,d r a w t h e r e q u i r e dc i r c l e ' P r o b l e m 5 - 5 8 . I o d r a w t w o c i r c l e st o u c h i n g e a c h o t h e r a n d t w o convetginglines, the smallercircle being of given radius (fig' 5-61)'
Flc.5-61 ( i ) L i n e sA B a n d C D a n d r a d i u sR o f t h e s m a l l e rc i r c l e a r e g i v e n ' ( i i ) P r o d u c e l i n e s A B a n d C D t o i n t e r s e c ta t a p o i n t f ' D r a w t h e bisector EF oI ZAEC. (iii) Draw a line parallel to and at distanceR from AB to intersect Ef in a point Q. ( i v ) D r a w t h e p e r p e n d i c u l aQr P o n t h e l i n e A B . (v) With Q as centre and QP as radius, draw the smaller circle' ( v i ) M a r k p o i n t s f a n d N a t w h i c h t h e c i r c l ec u t s f F ' ( v i i ) D r a w t h e l i n e j o i n i n g f w i t h P . D r a w a l i n e N S p a r a l l e lt o IP intersectingAB in the Point S. F r o m S , d r a w t h e p e r p e n d i c u l at ro A B c u t t i n gE F i n t h e p o i n t O W i t h O as centre and OS as radius, draw the required circle. O ' i s t h e c e n t r e o f s m a l l e rc i r c l e , o b t a i n e d i n t h e s a m e m a n n e r , t o u c h i n gt h e t w o S i v e nl i n e s a n d t h e S i v e nc i r c l e .
E x e r c i s eVs 1 . D r a w a l i n e 1 2 5 m m l o n g a n d q u a d r i s e citt . 2 . Draw a line AB 80 mm long and divide it into {ive parts, one of them 20 mm long and the remaining each 15 mm long, by the method of bisection. 3. With centre O and radius equal to 50 mm, draw two arcs of any lengths on opposite sides of O. Bisect the two arcs and produce the b i s e c [ o r st i l l t h e v m e e t .
Ceom€lrical Construction 97
4.
Draw a line AB 75 mm long. At B, erect a perpendiculargC j 00 mm long. Draw a line joining A and C, and measure its length. C o n s t r u c ta s q u a r eo n e a c h l i n e a s a s i d e .
5.
Draw a line PQ 1OOmm long. At any point O in it near its cenrre,erecr a perpendicularOA 65 mm long. ThroughA, draw a line parallelto pe.
6 . M a r k a n y p o i n t O . D r a w a l i n e A B , s u c h t h a t i t s s h o r t e s td i s t a n c e from O is 50 mm. 7 . Construct a rectangleof sides 65 mm and 40 mm long. B. Draw a line A8 75 mm long. Mark a point C, 65 mm from A and 90 mm from 8. Join C with / and B. Through the poinrs A, B and C, draw l i n e s ( i ) p e r p e n d i c u l aar n d ( i i ) p a r a l l e tl o t h e i r o p p o s i t el i n e s . 9 , C o n s t r u c ta s q u a r e o f 7 5 m m s i d e . D r a w t h e d i a g o n a l si n t e r s e c t i n g a t O . F r o m O , d r a w l i n e s p e r p e n d i c u l at ro t h e s i d e so f t h e s q u a r e . 1 0 . D r a w a c i r c l e o f 5 0 m m r a d i u s .D i v i d e i t ( i ) i n t o B e q u a l p a r t s b y c o n t i n u e db i s e c t i o na n d ( i i ) i n t o 1 2 e q u a l p a r t s b y b i s e c t i o no f a l i n e and trisection of a right angle methods. 11. Draw two lines AB and AC making an angle of 25.. Draw a circle o f 2 5 m m r a d i u st o u c h i n gt h e m . 12. Construct a right angle pQR. Describea circle of 20 mm radius touching the sides PQ and eR. 1 3 . D r a w a l i n e A B o f a n y l e n g t h . M a r k a p o i n t O a t a d i s t a n c eo f 25 mm from AB. With O as centre, draw a circle of 40 mnr d i a m e t e r .D e s c r i b ea n o t h e r c i r c l e ( i ) o f 2 0 m m r a d i u s ,t o u c h i n g t h e circfe and AB; (ii) of 35 mm radius, touching AB and the circle, and i n c l u d i n gt h e c i r c l e w i t h i n i t . 14. Draw two circles of 20 mm and 30 mm radii respectivelywitlt centres 6 5 m m a p a r t . ( i ) D e s c r i b ea t h i r d c i r c l e o f S 0 m m r a d i u s t o u c h i n g t h e t w o c i r c l e s a n d ( a ) o u t s i d e t h e m ; ( b ) i n c l u d i n g2 0 m m c i r c l c ; ( c ) i n c l u d i n g3 0 m m c i r c l e . ( i i ) D e s c r i b ea c i r c l e o f 7 5 m n r r a d i u s , t o u c h i n g b o t h c i r c l e s a n d i n c l u d i n gb o t h o f t h e m w i t h i n i t . 1 5 . M a r k p o i n t s A a n d B , 5 0 m m a p a r t .M a r k a t h i r d p o i n t 7 5 m m f r o m b o t h , 4 a n d 8 . D e s c r i b ea c i r c l e p a s s i n gt h r o u g ht h e t h r e e p o i n t s . 1 6 . D r a w t h e m a c h i n eh a n d l e s h o w n i n f i e . 5 - 6 2 . A l l d i m e n s i o n s. r r e i n millimetres.
Flc.5-62 1''
The distance between the centres of two circles of 65 rnrn and 9 0 m m d i a m e t e r si s 1 2 0 m m . D r a w a n i n t e r n a l a n d a n e x t e r n a l common tangent to the two circles.
98
Engineeling Drawing
1 8 . Draw a circle with centre O and radius equal to 30 mm. From a point P, 75 mm from O, draw a line joining P and O, and produce it to cut the circle at Q. From P and Q draw tangents to the circle. 1 9 . Two shafts carry pulleys of 90 cm and .135cm diametersrespectivelyThe distance between their centres is 270 cm. Draw the arrangement showing the two pulleys connected by (i) direct belt (ii) crossed belt T a k e1 m m : 2 cm. 20. An arc AB drawn with 50 mm radius subtendsan angle of 45' at the centre. Determine approximatelythe length of A8. 2 1 . Determine the length of the circumferenceof a 75 mm diameter circle2 2 . .A point P is 25 mm from a lineA8. Q is a point in AB and is 50 mm from P. Draw a circle passing through P and touching AB at Q. 2 3 . The centre O of a circle of 30 mm diameter is 25 mm from a line A8. Draw a circle (i) to touch the given circle and the line AB at a point P, 50 mm from O; (ii) to touch AB and the given circle at a point Q/ 20 mm from AB. 2 4 . Construct an equilateral triangle ABC of 40 mm side. Construct a square/ a regular pentagonand a regular hexagonon its sides ,48, 8C and CA respectively. 2 5 . Two circles of 40 mm and 50 mm diameters have their centres 60 mm apart. Draw a circle to touch both circles and (i) to includ€ the bigger circle, the point of contact on it being 75 mm from the centre of the other circle; (ii) to include both the circles, the point of contact being the same as in (i). 2 6 . Constructa regularpentagonof 30 mm side by three different methods 2 7 . On a line AB 40 mm long, construct a regular heptagon by two different methods. Construct a regular octagon of 40 mm side. Inscribe another octagoo with its corners on the mid-points of the sides of the first octagon2 9 . Constructthe following regularpolygonsin circles of 100 mm diameter, using a different method in each case: (i) Pentagon(ii) Heptagon. 3 0 . Draw the following regularfigures,the distancebetweentheir opposite sides being 75 mm: (i) Square; (ii) Hexagon; (iii) Octagon. Jt. Construct a regular octagon in a square of 75 mm side. 32. Describe a regular pentagon about a circle.of 100 mm diameter. .t J.
Constructa trianglehavingsides25 mm,30 mm and 40 m m long Draw three circles,each touchingone of the sides and the two sides Droduced.
34. lnscribea circle in a trianglehavingsides50 mm, 65 mm and Z5 mm lon& 3 5 . Construct a regular heptagonof 25 mm side and inscribea circle in iL 36. Construct a regular hexagonof 40 mm side and draw in it, six equd circles, each touching one side of the hexagonand two other circler Construct a square of 50 mm side and draw in it, four equal circles, each touching two adiacentsides and two other circles. 3 8 . In a regular octagon of 40 mm side, draw four equal circles, each touching one side of the octagon and two other circles.
Geometrical Construction gg
3 9 . D r a w a c i r c l e o f 1 2 5 m m d i a m e t e ra n d d r a w i n i t , f i v e e o u a l circles, each touching the given circle and two other circtes. 4 0 . C o n s t r u c t a s q u a r e o f 2 5 m m s i d e . D r a w o u t s i d e i t f o u r e q u a l. ' circles, each touching a side of the square and two other circles. 4 1 . O u t s i d e a c i r c l e o f 2 5 m m d i a m e t e r ,d r a w f i v e e q u a l c i r c l e s ,e a c h t o u c h i n gt h e g i v e n c i r c l e a n d t w o o t h e r c i r c l e s . 42. fwo lines convergeto a point making an angle of 30. between them. Draw three circlesto touch both these lines,the middle circle b e i n g o f 2 5 m m r a d i u sa n d t o u c h i n gt h e o t h e r t w o c i r c l e s . 43. Two lines convergeto a point making an angle of 30. between them. A p o i n t P i s b e t w e e nt h e s el i n e s l 5 m m f r o m o n e l i n e a n d 2 5 m m f r o m the other. Draw a circle to touch both the lines and pass through p. 44. Draw a series of four circles, each touching the preceding circle and two converging lines which make an angle of 256 between them. T a k e t h e r a d i u s o f t h e s m a l l e s tc i r c l e a s 1 0 m m . 45. A vertical straight line AB is at a distanceoI 90 mm from the centre of a circle of 75 mm diameter. A straight line pe passes through the centre of the circle and makes an ansle of 60. with the verticil. Draw circles having their centres on pd and to touch the straight l i n e A B a n d t h e c i r c l e . M e a s u r et h e r a d i u s o f e a c h c i r c l e . 4 6 . D f t w a s e m i - c i r c l eo f 1 2 5 m m d i a m e t e ra n d i n s c r i b ei n i t t h e l a r e e s t equilateral triangle having a corner at the centre. The semi-circle is t h e d e v e l o p m e n to f a c o n e a n d t h e t r i a n g l e t h a t o f a l i n e o n i t s s u r f a c e .D r a w L h e p r o j e c t i o n so f t h e c o n e i e s t i n g o n i t s b a s e o n t h e g r o u n d s h o w i n gt h e l i n e i n b o t h v i e w s . !ii-r't 4 7 . C o n s t r u c ta l e v e r a s s h o w n i n f i s . 5 - 6 3 . / lr".,! I
Frc.5-63
CENTRALLiBRAR
i;';l;.:,t-f#4
1 0 0 E n g i n e e r i n BD r a w i n g
48. Construct a special spanneras shown in fig. S-6a. (Hint: The method of drawinq curve is shown on left-hand side. The s a m e i s t o b e f o l l o w e i o n t h e r i g h t - h a n ds i d e . T h i s c u r v e i s k n o w n a s o g l e e c ur v e .)
Ftc.5-64 49. Two shafts, 120 cm apart are connected by flat belt. The flat belt pulleys of 30 cm diameter and 60 cm diameter are fixed on the shafts. Draw the arrangementand determine approximatelylength of the belt. 50. Draw plan-view of a hexagonal nut of 20 mm using standard d i m e ns i ons . 5 1. ( i ) D r a w a n u m b e r ' 8 ' o f h e i g h t 1 0 5 m m a n d 1 5 m m t h i c k . ( i i ) D r a w a n a l p h a b e t' S ' o f h e i g h t 1 0 5 m m a n d 15 m m t h i c k .
CURVES USED IN ENGINEERING PN,ACTICE
6 6-0. Introduction:
The profile of number of objects consists of various types of curves. This chapter deals with various types of curves which are comrnonty u s e d i n e n g i n e e r i n gp r a c t i c ea s s h o w n b e l o w : 1. 2. 3. 4. 5. 6.
Conic sections Cycloidal curves lnvolute Evolutes Spirals Helix.
We shall now discussthe above in details with reference to their c o n s t r u c t i o na n d a p p l i c a t i o n s .
6-1. Conic sections: The sections obtained by the intersectionof a r i g h t c i r c u l a r cone by a plane in different positions relativeto the axis of the cone are called conics. Refer to fig. 6-1.
6
\9
F r c . 61-
F r c . 6 _ j( i )
F r c . 6 - 1( i i )
F r c .6 - 1 ( i i i ) W h e n t h e s e c t i o n p l a n e i s i n c l i n e dt o t h e a x i s a n d c u t s all the generators on one side of the apex, the section is an ellipse tfig.6-1(i)1. ( i i ) W h e n t h e s e c t i o n p l a n e i s i n c l i n e dt o t h e a x i s a n d i s p a r a l l e lt o _ o n e o f t h e g e n e r a l o r st,h e s e c t i o ni s a p a r a b o l a tfig. 6_1{ii}J. ( i i i ) W h e n t h e . s e c r i o np l a n e c u t s b o t h t h e p a r t s of ite double .one o n o n e s i d e o f t h e a x i s ,t h e s e c t i o ni s a h y p e r b o l a tfig. 6_1(iii)1. (i)
1 0 2 E n g i n e e r i n gD r a w i n g
T h e c o n i c m a y b e d e f i n e da s t h e l o c u s o f a p o i n t m o v i n g i n a . p l a n ei r ' t such a way that ihe ratio of its distancesfrorn a fixed point .and a {ixed straight line is always constant. The fixed point is called the focus and the fixed line, the direcfrix. distance of the point from the focus is cailed eccentrici1 distanceof the Point from the directrix 1 for a n d i s d e n o t e d b y e . l t i s a l w a y s l e s s t h a n 1 f o r e l l i p s e ,e. q u a l t o = 1 for e < for ellipse, 1 e purabolaand greaier than 1 for hyperbola i.e. parabolaand e > 1 for hYPerbola. T h e I i n e p a s s i n g t h r o u g h t h e f o c u s a n d p e r p e n d i c u l a rt o t h e is directrix is catied thJ axis. T6e point at which the conic cuts its axis called the vertex. The ratio
6-1-1. ElliPse: , onuments' U s e o f e l l i p t i c a lc u r v e si s m a d e i n a r c h e s ,b r i d g e s , . d a m sm can be ellipse an rnun-hol"r, glands and stuffing-bnxesetc. Mathematically \ur) 2 v2 = 'l . Here a' and 'b are half the length oi * describedby equation L, m a j o r a n d m i n o r a x e so f t h e e l l i p s ea n d x a n d y c o - o r d i n a t e s ' (a) General method of construction of an ellipse: Probfem 6-1. (fiC. 6-2): To construct an ellipse when the distanceoi 'j' the focusfrom the directrixis equalto 50 mm and eccentticityis
Directrix and focus Frc.6-2 ( i ) D r a w a n y v e r t i c a ll i n e A B a s d i r e c t r i x . (ii) At any point C in it, draw the axis' (iii) Mark a focus F on the axis such that CF = 50 mm' ( i v ) D i v i d e C F i n t o 5 e q u a ld i v i s i o n s . (v) Mark the vertex V on the third division-pointfrom C'
Curves Used in Engineering practice 103 .t/E2
lhus, eccentricity,= = lC i. " (vi) A.scalemay now be constructed on the axis(asexplained below), will directlygive the distanc", in if,. Lquired ratio. yhi:h ... .(vii) At Y.rdraw a perpendicular VE equalto yF. Draw a line joining L
ano t.
I n u s , l n t n a n g t eC- .VvLE, v F 2 : ia ie i ( v i i i )M a r k a n y . p o i n t . t o n t h e axis and through it, draw a perpendicularto meet CF_producedat 1,. (ix) With centre F and radius equal to 1-1,, draw arcs to intersect the perpendicularthrough 1 ai points p1 and p1,. T h e s ea r e t h e p o i n t s o n t h e e l l i p s e ,b e c a u s e t h e d i s t a n c eo f p 1 f r o m 1-1' - vr A B i s e q u a lt o C 1 , p 1 F = 1 - 1 ' =vc=i' ?. "^A S i m i l a r l y ,m a r k .p o i n t s 2 , 3 e t c . o n t h e a x i s and obtain points ^ P2 and P2', p3 and p3, eic. (x) Draw the ellipse through these points. lt is a closed curve having two foci and two directrices. ( b ) C o n s t r u c t i o no f e l l i p s e b y o t h e r m e t h o d s : Ellipse is also defined as a curve traced out by a point, moving in the same plane as and in such a way that the sum of itr' dirtuna", from two fixed points is always the same. (i) Each of the two fixed points is called the focus. ( i i ) T h e l i n e p a s s i n gt h r o u g h t h e two foci and terminated by the curve, is called the major axis. ( i i i ) T h e . l i n e b i s e c t i n gt h e m a j o r a x i s a t r i g h r a n s l e s and terminated b y t h e c u r v e ,i s c a l l e dt h e m i n o r a x i s Conjugate axes: Ihese axes are called conjugate axeswhen they are parallelto the tangentsdrawn at their extremitiis.
Conjugateaxes Frc.6-2(i)
104 Engincering Drawing
ln fig. 6-2(i), AB is the major axis, CD the minor axis and Ft and Fz are the foci. The foci are equidistantfrom the centre O. The points A, P, C eIc. are on the curve and hence, according to the definition, ( A F 1+ A F ) = ( P F j + P F ) : ( C F 1 + C F 2 ) e t a . (AF1+ AFz\ -- AB. But (PFr + PF2) = A8, the major axis. .'. Therefore, the sum of the distancesof any point on the curve from the two foci is equal Io the maior axis. Again, (CF1 + CF2\ : AB. CF1 = 67;. But 1 CFt =CFz:tAB. :. Hence, the distanceof the ends of the minot axis frcm the foci is equal to half the maior axis. Problem 6-2. To construct an ellipse, given the maior and minor axes' The ellipse is drawn by, first determining a number of points through which it is known to pass and then, drawing a smooth curve through them, either freehand or with a french curve. Largerthe number of points, more accuratethe curve will be. Method I:
Arcsof circles method (fig. 6-3).
Draw a line AB equal to the major axis and a line CD equal to the minor axis, bisectingeach other at right angles at O. (ii) With centre C and radius equal to half AB (i.e' AO ) draw arcs cutting AB at F1 and f2, the foci of the ellipse.
(i)
( i i i ) M a r k a n u m b e ro f p o i n t s 1 , 2 , 3 e t c . o n 4 8 . (iv) With centres Fl and F2 and radius equal to A1, draw arcs on both sides of AB.
Arc of circle method Frc.6-3
Curv€s Used in Engineering Practice 10S
(v) With same centres and radius equal to 81, draw arcs intersecting the previous arcs at four points marked pt. (vi) Similarly,with radii 42 and 82, A3 and 83 etc. obtain more points. (vii) Draw a smooth curve throughthese points.This curve is the required e l l i os e . Method Il:
Concentric circles method (fig. 6-a).
C o n c e n t r i cc i r c l e m e t h o d =tc. 6-4 (i)
Draw the major axis,48 and the minor axis CD intersectingeach other at O.
(ii) With centre O and diametersAB and CD respectively,draw two circles. ( i i i ) D i v i d et h e m a j o r - a x i s - c i r cilnet o a n u m b e ro f e q u a ld i v i s i o n ss, a y 1 2 and mark points1, 2 etc. as shown. (iv) Draw lines joining these points with the centre O and cutting the minor-axis-circa l et p o i n t s 1 ' , Z ' e t c (v) Through point 1 on the major-axis-circle,draw a line parallel to CD, the minor axis. (vi) Throughpoint 1'on the minor-axis-circle d ,r a w a l i n e p a r a l l e lt o A8, the major axis. The point Pl, where these two lines intersect i s o n t h e r e q u i r e de l l i p s e . (vii) Repeat the construction through all t h e p o i n t s . D r a w t h e e l l i p s e through A, P1, P2,.. etc. Method IIr:
Loop of the thread method (fig. 6-5).
This is practical applicationof the first method. (i) Draw the two axes AB and CD intersectingat O. Locate the foci F1 and F2.
'106 Engineering Drawing
( i i ) I n s e r t . ap i n a t e a c hf o c u s - p o i nat n d t i e a p i e c eo f t h r e a di n t h e f o r m or a roop around th.epins, in such a way that the pencil point w h 9 L p l a c e d i n . t h e l o o p ( k e e p i n gt h e t h r e a d t i g h t ) , i s j u s t o n t h e eno L ot the mtnor axis. (iii) Move. the pencil around the foci, maintaining an even tension in the thread throughout and obtain the ellipsel I t i s e v i d e n tt h a t p F l + p F 2 : 6 p , + C F , e t c .
Loopof the threadmethod Frc.6-5 Method IV: Oblongmethod tfig.6-6(i)1.
Oblongmethod Ftc.6_6(i) (i)
Draw the two axesAB and CD intersectingeach other at O. (ii) Construct the oblong FfCH having its sides equal to the two axes. (iii) Divjde_thesemi-major-axis AO into a number of equal parts, say +, a n d A [ i n t o t h e s a m e n u m b e r o f e q u a l p a r t s , n u m b e r i n gt h e m from A as shown. (iv) Draw lines joining 1,, 2, and 3, with C. 'l (v) From.D, draw_lines .through , 2 and 3 intersectingC1,, C2' and C3, at points P1,P2 and P3 respectively.
Curves Used in Engineering Practice 107
( v i ) D r a w t h e c u r v e t h r o u g hA , P 1 . . . . . C l.t w i l l b e o n e q u a r t e r o f t h e e l l i ps e . ( v i i )C o m p l e t e t h e c u r v e b y t h e s a m e c o n s t r u c t i o n i n e a c h o f t h e t h r e e r e m a i n i n gq u a d r a n t s . As the curve is symmetricalabout the two axes/ points in the remaining q u a d r a n t s m a y b e l o c a t e d b y d r a w i n g p e r p e n d i c u l a r sa n d h o r i z c n t a l i Irom P1, P2 etc. and making each of them of equal length on both the sides of the two axes. For example, P2x : x P11and P2y = yPr, A n e l l i p s e c a n b e i n s c r i b e dw i t h i n a p a r a l l e l o g r a m by usingthe above methodas shown in fig.6-6(ii).
Frc.6-6(ii) L i n e sP Q a n d R S , . j o i n i n gt h e m i d - p o i n t so f r h e o p p o s i t es i d e s o f t h e parallelogramare called conjugateaxes. Method V: Trammelmethod (fig. 6-7).
.
Trammel method Ftc.6-7 ( i ) Draw the two axes AB and CD intersectingeach other at O. Along the edge of a strip of paper which may be used as a trammel, mark PQ equal to half the minor axis and pR equal to half the maior axis. ( i i ) Place the trammel so that R is on the minor axis CD and e on the major axis A8. Then P will be on the required ellipse. By moving the trammel to new positions,always keepingR on CD and Q on AB, obtain other points. Draw the ellipsethrough these points.
I08
E n s i n e e r i n BD r a w i n g
Normal and tangent to an ellipse: The normal to an ellipse at any p o i n t o n i t b i s e c t st h e a n g l e m a d e b y l i n e sj o i n i n g r h a t p o i n r w i i h r h e f o c i . T h e t a n g e n tt o t h e e l l i p s ea t a n y p o i n t i s p e r p e n d i c u l atro t h e n o r m a l at that point. Problem 6-3, tfig. 6-2(i)l: To draw a normal and a tangentto the eltipse at a point Q on it. Join Q with the foci F1 and F2. (i) Draw a line NM bisectingZ F1 eF2. NM is the normal to rhe ellipse. ( i i ) D r a w a l i n e S f t h r o u g h Q a n d p e r p e n d i c u l atro N M . S f i s t h e t a n g e n tt o t h e e l l i p s ea t t h e p o i n t e . Problem 6-4, (fig. 6-8); To draw a curve parallelto an ellipse and at dislance R from it. This may be drawn by lwo methods: (a) A large number of arcs of radius equal to the required distanceR, with centres on the ellipse, may be described. The curve drawn t o u c h i n g t h e s e a r c s w i l l b e p a r a l l e lt o t h e e l l i p s e . ( b ) I t m a y a l s o b e o b t a i n e db y d r a w i n g a n u m b e r o f n o r m a l st o t h e e l l i p s e . m a k i n g t h e m e q u a l t o t h e r e q u i r e d d i s t a n c eR a n d t h e n d r a w i n ga s m o o t hc u r v e t h r o u g ht h e i r e n d s .
f lc. 6-8 Frc. b-B
Ftc. 6-9
Problem 6-5. (fig. 6-9): To find the major axis and minor axis of an ellipse whose conjugate axesand angle between them are given. Conjugateaxes PQ and R5, and the angle oc between them are given. (i) Draw the two axes intersectingeach other at O. ( i i ) C o m p l e t e t h e p a r a l l e l o g r a ma n d i n s c r i b e t h e e l l i ps e i n i t a s describedin problem 6-2, method (iv). (iii) With O as centre and OR as radius, draw the semi-circle cutting t h e e l l i p s ea t a p o i n t F . (iv) Draw the line RE.
Cutves Used in Engineering Practice 109
( v ) T h r o u g h O d r a w a l i n e p a r a l l e lt o R E a n d c u t t i n g t h e e l l i p s e a t p o i n t sC a n d D . C D i s t h e m i n o r a x i s . ( v i ) T h r o u g hO , d r a w a l i n e p e r p e n d i c u l at ro C D a n d c u t t i n gt h e e l l i p s e at points A and B. ,48 is the maior axis. Probfem 6-6. (fig. 6-10): Io find the centre, maior axis and minor axisof a given ellipse. (i)
Draw any two chords 1-2 and 3-4 parallel to each orner. ( i i ) F . i n dt h e i r m i d - p o i n t sP a n d e , a n d d r a w a l i n e p a s s i n gt h r o u g h them, cutting the ellipse at points R and S. Bisect the line RSrtn t h e p o i n t O w h i c h i s t h e c e n t r eo f t h e e l l i p s e . With O as centre and any convenientradius, draw a circle cutting the .. ellipse in points E, F, C and H. Complete the rectangleEFCH.Throug-hO, d r a w a I i n e p a r a l l e lt o E F c u t t i n g t h e e l l i p s e i n p o i n t s A a n d B . A g a i n through O, draw a line parallel to FC cutting the ellipse at points C lnd D. AB and CD are respectivelythe major axis and the minor axis.
F r c .6 - 1 0
Directrixand focus Ftc.6-11
6-1-2. Parabola: . . Use of parabolic curves is made in arches, bridges, sound reflectors, l i g h t r e f l e c t o r se t c . M a t h e m a t i c a l l ya p a r a b o l ac a n b e d e s c r i b e db y a n e q u a t i o ny 2 = 4 a x o r x 2 : 4 a y . (a) General method of construction of a parabola: Problem 6-7. (fig. 6-11)t To construct a parabola,when the destance of the focus ftom the dircctrix is S0 mm. ( i ) D r a w t h e d i r e c t r i x , , 4a Bn d t h e a x i s C D . (ii) Mark focus f on CD, 50 mm from C. (iii) Bisect CF in y the vertex (becauseeccentricity : 1).
11o Engineering Dlawing
( i v ) M a r k a n u m b e ro f p o i n t s 1 , 2 , 3 them, draw perpendicularsto it.
etc.on the axisand through
(v) With centre F and radius equal to C1, draw arcs cutting the p e r p e n d i c u l atrh r o u g h 1 a t P 1 a n d P 1 ' . (vi) Similarly, locate points P2 and P2', P3 and P3' etc. on both the s i d e so f t h e a x i s . ( v i i )D r a w a s m o o t h c u r v e t h r o u g h t h e s e p o i n t s . T h i s c u r v e i s t h e required parabola. lt is an open curve. Problem 6-8. (fig. 6-12): To find the axisof a given parabola. Draw any two chords AB and CD across the parabola, parallel to each other and any distanceapart. (ii) Bisect AB and CD in points E and F respectivelyand draw a line C H p a s s i n gt h r o u g ht h e m . T h e l i n e C H w i l l b e p a r a l l e lt o t h e a x i s .
(i)
( i i i ) D r a w a c h o r d P Q , p e r p e n d i c u l atro C H . (iv) BisectPQ in the point O and through it draw a line XY parallelto CH Then XY is the required axis of the parabola.
F lrcc. .6 *- 1122
F t c . 6 - 13
Probfem 6-9. (fig. 6-13): To find the focus and the directrix of a parabola whose axis is given, (i) Mark any point P on the parabola and draw a perpendicular PA to the axis. Mark a point B on the axis such that BV = VA ( i i ) D r a w a l i n e j o i n i n gI w i t h P . (iii) Draw a perpendicularbisector FF o{ BP, intersectingthe axis at a p o i n tf . Then F is the focus of the parabola. (iv) Mark a point O on the axis such thai OV - VF. Through O, draw a l i n e C D p e r p e n d i c u l atro t h e a x i s . T h e n C D i s t h e d i r e c t r i xo f the parabola.
Curyds Used in En8ineerin8 Practic€ 111
(b) Construction of parabola by other methods: Method I: Rectanglemethod [lig. 6-14]. Problem 6-10. To construct a parabolagiven the base and the axis. (i) Draw the base AB. ('i) At its mid-point E, draw the axis FF at right angles to ,48. ( i i i ) C o n s t r u c ta r e c t a n g l e ,B C D , m a k i n gs i d e B C e q u a l t o F F . (iv) Divide AE and AD into the same number of eoual oarts and n a m e t h e m a s s h o w n ( s t a r t i n gf r o m A ) . ( v ) D r a w l i n e sj o i n i n g f w i t h p o i n t s 1 , 2 a n d 3 . T h r o u g h1 , , 2 , a n d 3 ' , d r a w p e r p e n d i c u l a r st o . , 4 B i n t e r s e c t i n gF 1 , F 2 a n d F 3 a t p o i n t s P1, P2 and P3 respectively. (vi) Draw a curve through A, P1,p2 etc. lt will be a half parabola. Repeat the same construction in the other half of the rectansle to complete the parabola. Or, locate the points by drawing lines through the points P1, P2 etc. parallel to the base and making eich of them of equal length on both the sides of EF, e.g. p1O - Op,t,. AB and EF are called the base and the axis respectivelyof the parabola.
Rectanglemethod Ftc.6-14
Ftc.6-15
Fig. 6-15 shows a paraboladrawn in a parallelogramby this method. Method II: Tangentmethod ({ig. 6-16). (i) Draw the base AB and the axis EF. (Theseare taken different from those in methodl.) (ii) ProduceEF to O so that fF : fO. (iii) Join O with / and 8. Divide lines O4 and OB into the same number of equal parts, say 8. ( i v ) M a r k t h e d i v i s i o n - p o i n tass s h o w n i n t h e f i g u r e .
1 1 2 E n g i n e e t i n gD r a w i n g
( v ) D r a w l i n e s j o i n i n g 1 with 1', 2 with 2' etc. Draw a curve starting from A and tangent t o l i n e s 1 - 1 ' , 2 - 2 ' e t c . T h i s c u r v e i s t h e required parabola.
Parallelogrammethod F r c .6 - 1 6
6-1-3.Hyperbola: Use of hyperbolicalcurves is made in cooling towers, water channelsetc. Rectangular hyperbola: lt is a curve traced out by a point moving in such a way that the product of its distancesfrom two fixed lines at riSht angles to each other is a constant. The fixed lines are called asyrnptotes. This curve graphically represents the Boyle's Law, viz. P x V = a constant. lt is also useful in design of water channels. General method of construction of a hyperbola: Mathematically,we can describea hyperbolaby ')t = 1 . F i 8 . 6 - 1 7a n d f i g . 6 - 1 8 . n Problem 6-11. (fig. 6-17\2 Consttuct a hyperbola, when the distance of the focus from the dircctrix is 65 mm and eccentricity is ;' (i)
Draw the directrix AB and the axis CD.
(ii) Mark the focus F on CD and 65 mm from C. (iii) Divide CF into 5 equal divisions and mark V the vertex, on the second division from C.
: eccentricit, Thus, # .
=
i
To constructthe scale{or the ratio I d.u* u line VF perpendicular to CD suchthat VE = VF.Join C wiii E.
CvE,# : # = t Thus, in triangle
C u r v e sU s e d i n E n g i n € e r i n gp r a c t i c e 1 1 3
(iv) Mark any point 1 on the axis and through it, draw a perpendicular to meel CE-producea dt i,. (v) With centre F and radius equal to 1-1,, dtaw arcs intersecting t h e p e r p e n d i c u l atrh r o u g h 1 a t p 1 a n d p 1 , .
Ftc.6-17 ( v i ) S i m i l a r l y ,m a r k a n u m b e r o f p o i n t s 2 , 3 P2 and P2', P3 and p3, etc.
elc.,and obtain points
( v i i ) D r a w t h e h y p e r b o l at h r o u g ht h e s e p o i n t s . P r o b f e m 6 - 1 2 . ( t i g . 6 - 1 8 ) : I o d r a w a h y p e r b o l aw h e n i t s f o c i a n d vertices are given, and to locate its asymplotes. (i) Draw a horizontal line as axis and on it, mark the siven foci F anC f | , a n d v e r t i c e sV a n d V l . ( i i ) M a r k a n y n u m b e ro f p o i n t s 1 , 2 , 3 e t c . t o t h e r i g h t o f F 1 . (iii) With F and F1 as centres and radius, say V2, draw four arcs. (iv) With the same centres and radius V12, draw four more arcs i n t e r s e c t i n gt h e f i r s t f o u r a r c s a t p o i n t s p 2 . T h e n t h e s e p o i n t s l i e on the hyperbola. (v) Repeat the irrocesswith the same centres and radii V1 and V.,1. V3 and V13 etc. Draw the required hyperbola through the p o i n t st h u s o b t a i n e d . (vi) With FF1as diameter,draw a circle. (vii) Through the vertices V and Vr, draw lines perpendicular to the axis, cutting the circle at four points A. From O, the centre o{ the circle, draw lines passing through points A. These Iines are the required asymptotes.
1 1 4 E n g i n e e r i n gD r a w i n g
F rrcc..6 --1188
F r c .6 - 1 9
Problem 6-13. (fig. 6-19): Io locate the directrix and asymPtotesof a hyperbola when its axis and foci are given. F r o m t h e f o c u s F 1 , d r a w a p e r p e n d i c u l a rt o t h e a x i s i n t e r s e c t i n g t h e h y p e r b o l aa t a p o i n t A . Draw a line joining A with the other focus F. Draw the bisecto: oI Z FAF|, cutting the axis at a point B. (ii) Through B, draw a line CD perpendicularto the axis. CD is the required directrix. (i i i ) With O the mid-point of FF1 as centre and OV1 as radius, drau a circle cutting CD at points f and C. (iv) L i n e sd r a w n f r o m O a n d p a s s i n gt h r o u g hE a n d C a r e t h e r e q u i r e i asymptotes. (v) The asymptotes will also pass through the points o{ intersection (R and S) between the circle of radius OF1 and the vertical through Vt.
(i)
P r o b l e m 6 - 1 4 . (fig. 6-20): To draw a rectangular hyperbola, given the position of a point P on it. B
Ftc.6-20
Curves Used in Engin€€ring practice .lt5
(i) (ii) (iii) (iv) (v) (vi)
Draw lines OA and OB at right anglesto each other. M a r k t h e p o s i t i o no f t h e p o i n t p . Through P, draw lines CD and EFparallelto OA and OB respectively. AlongPD, marka numberof points1, 2. 3 etc.not necessarily equidistant. D r a w l i n e s 0 1 , 0 2 e t c . c u t t i n gp F a t p o i n t s 1 , , 2 ' e \ c T h r o u g hp o i n t 1 , d r a w a l i n e p a r a l l e tl o O B a n d t h r o u g h1 , , d r a w a l i n e p a r a l l e lt o O A , i n t e r s e c t i n g e a c ho t h e r a t a p o i n t ? 1 . (vii) Obtain other points in the same manner. For locating the point, say p6, to the lefl of P, the line 06 should be extended to meet PE at 6'. Draw the h y p e r b o l at h r o u g h t h e p o i n t s P6, P, P1 etc. A hyperbola,through a given point situated between two lines making any angle between them, can similarly be drawn, as shown in lig,. 6-21,
Ftc. 6-21
6-l-4, Tangentsand normalsto conics: ( a ) G e n e r a lm e t h o d : T h e c o m m o n r u l e f o r d r a w i n gt a n g e n t sa n d n o r m a l s : When a tangent at any point on the curve is produced to meet the .. d i r e c t r i x ,t h e l i n e j o i n i n g t h e f o c u s w i t h t h i s m e e t i n gp o i n t w i l l b e a t r i g h t a n g l e st o t h e l i n e j o i n i n g t h e f o c u sw i t h t h e p o i n t o l c o n t a c t . The normal to the curve at any point is perpendicularto the tangent at that point. Probfem 6-15. To draw a tangent at.anypoint p on the conics(fig. 6-2, f i g . 6 - ' 1 1a n d t i g . 6 - 1 7 ) . (i) (ii) (iii) (iv)
J o i nP w i t h F . From f, draw a line perpendicularto pF to meet AB at f. Draw a line through f and p. This line is the tansent to the curve. Through P, draw a line NM perpendicular to lp. Ntvt is the normal to the curve.
1 1 6 E n g i n e e r i n gD r a w i n g
(b) Other methods of drawing tangents to conics: Method l: Problem 6-16. (lig. 6-22): To draw a tangent to an ellipse at any point P on it. (i) With O, the centreof the ellipse as centre,and one halfthe major a x i s a s r a d i u s ,d r a l v a c i r c l e . ( i i ) F r o mP , d r a w a l i n e p a r a l l etl o the minor axis,cutting the circle at a point Q. ( i i i ) D r a w a t a n 8 e n tt o t h e c i r c l e Ftc. 6-22 at the point Q cutting the extended major axis at a point R. From R, draw a line R5 passingthrough P. RS is the required tangent to the elliose. Method ll: When an axis of parabola is given. Problem 6-17. To draw a tangent to a parabola at any point P on it. (i) From P, draw a perpendicularPA to the axis, intersecting it at a p o i n t A [ f i g . 6 - 2 3( i )] . Mark a point I on the axis such that 8V : VA. D:rew a line from B p a s s i n gt h r o u g hP . T h e n t h i s l i n e i s t h e r e q u i r e dt a n g e n t . ( i i ) T h r o u g hP , d r a w a l i n e P Q p a r a l l e lt o t h e a x i s t f i g . 6 - 2 3 ( i i ) J . D r a w t w o l i n e s A B a n d C D p a r a l l e lt o , e q u i d i s t a n tf r o m a n d o n opposite sides of PQ, and cutting the parabolaat points A and C. D r a w a l i n e j o i n i n gA w i t h C . Through P, draw a line RS parallelto AC. Rs is the required tangent. Method III: When the focus and the directrix are given t{ig. 6-23(iii)1.
(r) (i)
(ll) Ftc.6-23
From P draw a line PE perpendicularto the directrix A8, meeting it at a point f. ( i i ) D r a w a l i n e j o i n i n gP w i t h t h e f o c u sF . (iii) Draw the bisector RS of angle EPF. Then RS is the required tangent.
C u r v e sU s e d i n E n g i n e e r i n gp r a c t i c e 1 1 7
Probfem 6-1S. ttis. 6-'t})t,To.draw a tangentto a hypeftola at a point ^ P on it when the axis and the foci are givin. . _?ju*^lil"r .joining P with foci F and F1. Draw the bisector RS of Z f P F 1 .R S i s t h e r e q u i r e dt a n g e n tt o t h e h y p e r b o l a . Note: In tig. 6-19, the line ,48 is the tangent to the hyperbola . at the point A.
6-2. Cycloidalcurves: curves are generated by a fixed point on the circumference of .These a c i r c l e , w h i c h r o l l s w i t h o u t s l i p p i n ga l o n g a f i x e d s t r a i g h tl i n e or a circle. circle is catleC generatingcircle and the fi;d straight line or l,!:,]"f!"q clrcte. ls termed directlng line or directing circle. Cycloidal irru", ar" used In tooth prolile of gearsof a dial gauge.
6-2-1. Cvcloid: C y c l o i d i s a c u r v e g e n e r a t e db y a p o i n t o n t h e c i r c u m f e r e n c e _ o{ a c i r c l e w h i c h r o l l s a l o n g a s t r a i g h tl i n e . I t c a n b e d e s c r i b e d by an equatron, y = a 1 1- c o s 0 ) o r x = a ( 0 - s i n o ) . Problem 6-1.9..ifig. 6-24)': To constructa cycloid, given the diameterof ., tne generattng circle.
Cycloid Ftc.6-24 (i)
With centre C and given radius R, draw a circle. Let p be the generating point.
(ii) Draw. a. Iine PA tangential to and equal to the circumference of the circle. (iii) Divide the circle and the line pA into the same number of equal p a r t s /s a y 1 2 , a n d m a r k t h e d i v i s i o n - p o i n tas s s h o w n . (iv) Through C, draw a line CB parallel and equal to pA. (v) Draw perpendiculars at points 1, 2 etc. cutting Cg at points C1, C2 etc. Assumethat the circle starts rolling to the right. When point 1, coincides -position with 1, centre C will move to Ct. ln this oi the circje, the generating point p will have moved to position pj on the circle, at a distanceequal_top,,from point 1. lt is evidentthat e-1lies on the horizontai t r n e t h r o u g h 1 ' a n d a l a d i s t a n c eR f r o m C . .
1 1 8 E n g i n € e r i n gD r a w i n g
SimilarlyP , 2 will lie on the horizontal ine through2'and at the distanceR {rom C2. Construction: ( v i ) T h r o u g ht h e p o i n t s 1 ' , 2 ' e t c . d r a w l i n e s p a r a l l e tl o P A . (vii) With centres C1, C2 etc. and radius equal to R, draw arcs cutting t h e l i n e s t h r o u g h1 ' , 2 ' e t c . a t p o i n t sP 1 , P 2 e t c . r e s p e c t i v e l y . Draw a smooth curve through pointsP, \, t h e r e q u i r e dc y c l o i d .
P 2 . . . . AT. h i s c u r v e i s
Normal and tangent to a cycloid curve: The rule for drawing a n o r m a l t o a l l c y c l o i d a lc u r v e s : T h e n o r m a l a t a n y P o i n t o n a c y c l o i d a lc u r v e w i l l p a s s t h r o u g h t h e c o r r e s p o n d i n gp o i n t o f c o n t a c t b e t w e e n t h e g e n e r a t i n gc i r c l e a n d t h e d i r e c t i n gl i n e o r c i r c l e . The tangent at any point is perpendicularto the normal at that point. Problem 6-20. (fig. 6-24); To draw a normal and a tangentto a cycloid at a Bivenpoint N on it. (i)
With centre N and radius equal to R, draw an arc cutting CB at M.
( i i ) T h r o u g h M , d r a w a l i n e M O p e r p e n d i c u l a tro t h e d i r e c t i n g l i n e PA and cutting it at O. O is the point of contact and M is the position of the centre oi the generatinBcircle, when the Seneratingpoint P is at N. ( i i i ) D r a w a l i n e t h r o u g hN a n d O . T h i s l i n e i s t h e r e q u i r e dn o r m a l . (iv) Through N, draw a line Sf at right anglesto NO. Sf is the tangent to the cycloid.
6-2-2.Trochoid: Trochoid is a curve generated by a point fixed to a circle, within or o u t s i d ei t s c i r c u m f e r e n c ea,s t h e c i r c l e r o l l s a l o n g a s t r a i S h tl i n e . When the point is within the circle, the curve is called an inferior trochoid and when outside the circle, it is termed a superior trochoid. Probfem 6-21. (fig. 6-25, fig.6-26 and tig. 6-27\: To draw a trochoid, given the rolling circle and the generating point. h\
lnferiot trochoid: Let Q be the point within the circle and at a distanceR1 from the centre C. (i) Draw the circle and mark a point Q on the line CP and at a d i s t a n c eR r f r o m C . ( i i ) D r a w a t a n g e n t P A e q u a l t o t h e c i r c u m f e r e n c eo f t h e c i r c l e a n d a l i n e C B e q u a l a n d p a r a l l e tl o P A . ( i i i ) D i v i d e t h e c i r c l ea n d t h e l i n e C B i n t o 1 2 e q u a l p a r t s .
Curves Used in Engine€ring Practice 119
Method I: (fig. 6-25): Determine the positions p1, p2 etc. for the cycloid as shown in problem 6-19. Draw lines qP1, C2P2etc. With centres C1, C2 etc. and radius equal to R1, draw arcs cutting C.tp1, C2p2 etc, at points Qj, Q2 etc. respectively. D r a w a c u r v e t h r o u g ht h e s e p o i n t s .T h i s c u r v e i s t h e i n f e r i o rt r o c h o i d .
Trochoids F t c .6 _ 2 5 .. /f iwethod II: (fig. 6-26): With centre C and radius equal t o R t , d r a w a u gffcle and divide it into 12 equal parts. ' T h r o u g h t h e d i v i s i o n - p o i n t sd, r a w l i n e s p a r a l l e lt o p A . With centres C1, C2 etc. and radius equal to R1, draw arcs to cut the l i n e s t h r o u g h 1 ' , 2 ' e t c . a t p o i n t s Q r , Q z e t c . D r a w t h e t r o c h o i dt h r o u g ht h e s e p o i n t s .
Inferior trochoid Ftc. 6-26 (b\ Superior trochoid: Let S be the point outside the circle and at a distance R2 from the centre.
?
"\
.. .-
S u p e r i o rt r o c h o i d Ftc. 6-27
e.
1 2 0 E n g i n e e r i n gD r a w i n g
Method I: (fig. 6-25): Adopt the same method as method I used for i n f e r i o r t r o c h o i d . P o i n t S w i l l l i e o n t h e l i n e C P - p r o d u c e da/ t d i s t a n c eR 2 f r o m C . P o i n t s5 1 , 5 2 e t c . a r e o b t a i n e db y c u t t i n g t h e l i n e s C 1 P 1 - p r o d u c e d , C2P2-producedetc. with arcs drawn with centres C1, C2 etc. and radius e q u a l t o R 2 . S , 5 1 , 5 2 e t c . a r e t h e p o i n t s o n t h e s u p e r i o rt r o c h o i d . M e t h o d l l : l I i g . 6 - 2 7 ) : S a m e a s m e t h o d l l f o r i n f e r i o rt r o c h o i d . N o t e that the radiusof the circle is equal to R2.A loop is formed when the circle rolls for more than one revolution.
6-2-3. Epicycloidand hypocycloid: The curve generated by a point on the circumference of a circle, w h i c h r o l l s w i t h o u t s l i p p i n ga l o n g a n o t h e r c i r c l e o u t s i d e i t , i s c a l l e d a n epicycloid. The epicycloid can be represented mathematically by x - ( a + b ) c o s o- , . o r [ e - ] - b g 1 , \a)
y - (a Ib/ sino-r rin f' * b ol
\a) where a is the radius of rolling circle.
When the circle rolls along another circle inside it, the curve is called a hypocycloid.lt can be representedby mathematicallyx = a cos30, - a sin30. Y Probfem 6-22, To draw an epicycloid and a hypocycloid, given the generating and directing circles of radii r and R respectively. Epicycloidtfig. 6-2S(i)l; With centre O and radius R, draw the directing circle (only a part of it may be drawn). Draw a radius OP and produce il to C, so that CP : r.
Epicycloid Frc.6-28(i)
*,* point.
. as centre,drawthe ,"""ri:;"::::
;;:;:t;:"r",:;
l n o n e r e v o l u t i o no f t h e g e n e r a t i n gc i r c l e , t h e p o i n t p w i l l m o v e t o a p o i n t A , s o t h a t t h e a r c p A i s e q u a l t o t h e c i r c u m f e r e n c eo f t h e generating circle. The position oI A may be located by calculatingthe angle subtended , by the arc PA at centre O, by the formula, Z POA 360'
arc PC _ 2nr. _ L c i r c u m f e r e n c eo f d i r e c t i n gc i r c l e R 2nR '! . .'. z POA : 360' x ,( (i) Set-off this angle and obtain the position of A. (ii) With centre O and radius equal to OC, draw an arc intersecting OA-producedat B. This arc Cg is the locus of the centre C. ( i i i ) D i v i d e C B a n d t h e g e n e r a t i n gc i r c l e i n t o t w e l v e e q u a rp a r r s . ( i v ) W i t h c e n t r eO , d e s c r i b ea r c s t h r o u g h p o i n t s 1 ' , 2 ' , 3 , e t c . (v) With centres Cl, C2 etc. and radius equal to r, draw arcs cutrtng the arcs through 1,, 2, etc. at points C1, p2 etc. D r a w t h e r e q u i r e de p i c y c l o i dt h r o u g ht h e p o i n t sp , p 1 , p 2 . . . . . . . A . H y p o c y c l o i dt f i g . 6 - 2 8 ( i i ) J :T h e m e r h o d f o r d r a w i n g t h e h y p o c y c l o i di s 'circle u: for epicycloid. Note that the centre C of thJ generaiing is :uT9 i n s i d et h e d i r e c t i n gc i r c l e .
Hypocycloid Frc.6-28(ii) When the diameter of the ro ing circle is halt the diametet of the d,irecting circle, the hypocycloid is a itraight |ine and is a diameter of the o t r e c u n gc r r c l e I t i g . 6 - 2 8( i i i )] . Normal and tangent to an epicycloid and a hypocycloid: Problem 6-23. tfig. 6-2B(i) and {ig. 6-28(ii)l: To draw a normal and a tangent to an epicycloid and a hypocycloid at a point N in each of them.
1 2 2 t n B i n e e r i n gD r a w i n g
Hypocycloid Frc.6-28(iii) (i)
With centre N and radius equal to r, draw an arc cutting the locus of the centre C at a ooint D. ( i i ) D r a w a l i n e t h r o u g hO a n d D , c u t t i n gt h e d i r e c t i n gc i r c l e a t M . ( i i i ) D r a w a l i n e t h r o u g hN a n d M . T h i s l i n e i s t h e n o r m a l .D r a w a l i n e 5f through N and at right anglesto NM. Sf is the tangent.
6.2-4. Epitrochoid 1tig.6-2eandfig.6-30): E p i t r o c h o i di s a c u r v e generatedby a point fixed to a circle (within or outside its circumference, b u t i n t h e s a m e p l a n e ) r o l l i n g o n t h e o u t s i d e of another circle. SUPERIOR
Epitrochoids Ftc. 6-29
Curves Used in Engineering Practice 123
S u p e r i o re p i t r o c h o i d Frc.6-30
6-2-5. Hypotrochoid (fig.6-3i): When the circle rolls insideanothercirclg the curve is called a hypotrochoid. The curve is termed inferior or superior, according to the position of t h e p o i n t b e i n g i n s i d eo r o u t s i d et h e r o l l i n ec i r c l e .
Inferior epitrochoid and hypotrochoid Flc.6_3i Problem 6-24. To draw an epitrochoid and a hypotrochoid, given
the rolling and directing circles and the generaringpoint's.
124 EngineerinBDrawing
T h e s e c u r v e s a r e d r a w n b y a p p l y i n gt h e m e t h o d s u s e d f o r t r o c h o i d s . N o t e t h a t a r c s a r e d r a w n i n s t e a do f h o r i z o n t a l i n e s . Epitrochaids:
Hypotrochoids:
M e t h o d l : S u p e r i o ra n d i n f e r i o r - seelig. 6-29
Method l:Superior and inferior - s e ef i g . 6 - 3 2
M e t h o d l l : S u p e r i o -r s e e f i g . 6 - 3 0 I n f e r i o r - s e ef i g . 6 - 3 1
M e t h o d l l : S u p e r i o r - s e ef i g . 6 - 3 3 I n f e r i o r - s e ef i g . 6 - 3 1 N o t e : W h e n t h e d i a m e t e r o f t h e r o l l i n g c i r c l e i s h a l f t h e d i a m e t e ro f t h e d i r e c t i n gc i r c l e ,t h e h y p o t r o c h o i dw i lI b e a n e l l i p s e .
Hypotrochoids Frc.6-32
hypotrochoids Superior Frc.6-33
6-3. Involute: The involute is a curve traced out by an end of a piece of thread u n w o u n d f r o m a c i r c l e o r a p o l y g o n ,t h e t h r e a d b e i n g k e p t t i 8 h t . l t m a y a l s o b e d e f i n e d a s a c u r v e t r a c e d o u t b y a p o i n t i n a s t r a i g h tl i n e w h i c h r o l l s w i t h o u t s l i p p i n ga l o n g a c i r c l e o r a p o l y g o n . l n v o l u t e o f a c i r c l e i s used as teeth profile of gear wheel. Mathematicallyit can be described by x = r c o s O+ r 0 s i n e ,y = r s i n O- r O c o s Ow, h e r e " r ' i s t h e r a d i u so f a c i r c l e
Curves Used in Engineering Practice l25
Probfem 6-25. (lig. 6-34): To draw an involute of a given circle. With centre C, draw the given circle. Let p be the starting point, i.e. . the end of the thread. S u p p o s et h e t h r e a d t o b e p a r t l y u n w o u n d ,s a y u p t o a p o i n t 1 . p w i l l move to a position P1 such that 1p1 is tangent to the circle and is e q u a lt o t h e a r c 1 P . P 1 w i l l b e a p o i n t o n t h e i n v o l u t e . Conslruction: ( i ) D r a w a l i n e P Q , t a n g e n t o t h e c i r c l ea n d e q u a lt o t h e c i r c u m f e r e n c of the circle. ( i i ) D i v i d eP Q a n d t h e c i r c l ei n t o 1 2 e o u a lD a r t s . ( i i i ) D r a w t a n g e n t sa t p o i n t s j , 2 , 3 etc. and mark on them points P 1 , P 2 ,P 3 e t c . s u c h t h a t l P t : P 1 ' , 2 P 2 : P 2 , , 3 P 3- p r ' s 1 g . Draw the involute through the points p, p1, p2....etc. Normal and tangent to an involute: The normal to an involute of a circle is tangent to that circle. Probfem 6-26. (fig. 6-34)t To draw a normal and a tangent to the involute of a circle at a point N an it.
lnvolute Ftc.6-34 (i) Draw a line joiningC with N. (ii) With CN as diameter describea semi-circlecutting the circle at M. ( i i i ) D r a w a l i n e t h r o u g hN a n d M . T h i s l i n e i s t h e n o r m a l .D r a w a l i n e 5I, perpendicularto NM and passingthrough N. Sf is the tangent to the involute. Problem 6-27. (Iig.6-35): 70 draw an involute of a given square. Let ABCD be the given square. (i) With centre,4 and radiusAD, draw an arc to cut the line BA-oroduced at a point P1.
26
Engineering Drawing
(ii) With centre I and radius Bpr (i.e. BA + AD) draw an arc ro c u t t h e l i n e C B - p r o d u c e da t a p o i n t p 2 . S i m i ! a r l y w i t h centres C and D and radii Cp2 (i.e. CB + BA + AD) and Dp3 (i.e. DC + CB + BA + AD = perimeter) respectively,draw arcs t o c u t D c - p r o d u c e da t a p o i n t p j a n d , 4 D - p r o d u c e da i a p o i n t p a . T h e c u r v e t h u s o b t a i n e d i s t h e i n v o l u t eo f t h e s q u a r e .
!
Pl
Frc.6-35
Ftc.6-36
Problem 6-28. (fig. 6-36):.Tracethe paths of the endsof a straightline An 10O mm long, when it rolls, without slipping, on a semi-circle having its diametet A8,75 mm long. (Assumethe line Ap to be tangent to the semi-circle in the statting position.) (i)
D r a w t h e s e m i - c i r c l ea n d d i v i d e i t i n t o s i x e q u a l p a r t s .
(ii) Draw the line 4P and mark points 1, 2 etc. such that Al : arc A 1 ' , A 2 : a r c A 2 ' e t c . T h e l a s t d i v i s i o n5 p w i l l b e o f a s h o r t e r l e n g t h . O n t h e s e m i - c i r c l em , ark a point p, such that S,p, : Sp. ( i i i ) A t p o i n t s 1 ' , 2 ' e t c . d r a w t a n g e n t sa n d o n t h e m , m a r k p o i n t s P1, P2 elc. such that 1'P1 : 1p, 2'P, : 2p...and S,Pu = 5' Po = 5P. Similarly,mark points A1, 42 etc. such that A11, : A.1, A r 2 ' - A 2 . . . . a n dA ' p ' : A P . D r a w t h e r e q u i r e d c u r v e s t h r o u g h p o i n t s P , P 1 . . . . a n dP ' , a n d t h r o u g h p o i n t s A , 4 1 . . . . a n d A , . l f , 4 P i s a n i n e l a s t i cs t r i n g w i t h t h e e n d A a t t a c h e dt o t h e s e m i - c i r c l e , the end P will trace out the same curve pp, when the string is wound round the semi-circle. F i g . 6 - 3 7 s h o w s t h e c u r v e t r a c e do u t b y t h e e n d o f a t h r e a d w h i c h i s longer than the circumferenceof the circle on which it is wouno. N o t e t h a t t h e t a n B e n t1 ' P r : 1 p , 2 ' P 2 : 2 p 2'P':Z'Ptt-14P.
etc. and lastly
Curves Used in Engineering Practice 127
Ftc.6-37
6-4. Evolutes: APB is a given curve (fig. 6-38). O is the centre of a circle drawn through three points C, P and D on this curve. lf the points C and D are moved to converge towards P, until they are indefinitely close together, then in the limit, the circle becomes the circle of curvature of the curve APB at P. The centre O of the circle of curvature lies on the normal to the curve at P. This centre is called the centre of curvatureat P. The locus of the centre of curvature of a curve is called the eyolute of the curve. A curve has only one evolute.
Ftc.6-38
Frc.6-39
Problem 6-29. (fig. 6-39): Io determine the centre of cutvature at a given point on a conic. Let P be the given point on the conic and f, the focus. ( i ) J o i nP w i t h F . (ii) At P, draw a normal PN, cutting the axis at N. (iii) Draw a line NR perpendicular to PN and cuttingPf or PF-produced at R. (iv) Draw a line RO perpendicularto PR and cutting PN-producedat O. Then O is the centre of curvatureof the conic at the point P. The above construction does not hold good when the given point coincides with the vertex. As the point P approachesthe vertex, the points R, N and O move nearer to one another, so that when P is at the vertex, t h e t h r e e p o i n t s c o i n c i d eo n t h e a x i s .
1 2 8 E n g i n e e r i n gD r a w i n g
In a parabola, PF will be equal to FR. Hence, when p is at y the vertex, the centre of curvature O is on the axis so that OF - VF. ln an ellipse or a hyperbola, the ratio of the distancesof p from the foci is equal to the ratio of focal distancesof N, i.e. + = *O . PF, NF, Hence,when P coincideswith V the vertex,the ratio becomesY
v\
= +
oF.,
Similarly, when P coincides with the other vertex y1, the ratio v1F| olF1 , : Decomes vrr * Probfem 6.30. (fig. 6-40)t To determine the centre of curvature O, when the point P is at the vertex V of a conic. (il
Parabola[fig. 6-a0(i)J: Mark the centre of curvature O on the axis such that OF = VF.
(ii) E//rpse:tfig. 6-a0(ii)l: Method I: (i)
D r a w a l i n e F 1 G i n c l i n e dt o t h e a x i s a n d e q u a l t o V F 1 . (ii) ProduceF1G to H so that cH : VF.)oin H with F. ( i i i ) D r a w a l i n e C O p a r a l l e l t o H F a n d i n t e r s e c t i n gt h e a x i s a t O . Then O is the required centre of curvature.
(.i)
(ii)
(rx)
(jv)
Ftc. 6-40 Method II: [fig. 6-a0(iii)l; (i)
Draw a rectangleAOCI in which AO =
11 malot axrs ano LU : t
2
minor axis. ( i i ) J o i nA w i t h C . ( i i i ) T h r o u g hE , d r a w a l i n e p e r p e n d i c u l atro A C a n d c u t t i n g t h e m a j o r axis at 01 and the minor axis 02. Then 01 and 02 are the centres of curvature when the poini P is at A and C respectively. (iii) Hyperbola l{ig. 6-40(iv)l: (i)
D r a w a l i n e F l G i n c l i n e dt o t h e a x i s a n d e q u a l t o F V 1 . (ii) On F1C, mark a point H such that HG = VF.Join H with F. ( i i i ) D r a w a l i n e G O p a r a l l e lt o H F a n d c u t t i n gt h e a x i s a t O . T h e n O i s the centre of curvature at the vertex y.
Curves Used in Engin€cring practi.e
129
Probfem 6-31, (fig. 6-41)t To draw the evolute of an elliose. T h e e l l i p s ew i t h m a j o r a x i s , 4 Ba n d m i n o r a x i s C D i s g i v e n . ( i ) M a r k a n u m b e ro f p o i n t so n t h e e l l i p s e . (ii) Determine the centres of curvatures at these points (as shown at the point P) and draw a smooth curve through them. this is the evolute of the ellipse. The evolute ruy rornutrn", go outside the e l l t p s ea _ ss h o w n i n f i g . 6 _ 4 1 ( i i )T . h e c e n t r e so f c u r v i t u r e a t p o i n t s A and C are shown by method ll.
(i)
Frc.6-41 ftc.6-42 Problem 6-32. (fig. 6-42')zTo draw the evoluteof a parabola. (i) Mark a number of points on the parabola and determine the centres ot curvature at these points (as shown at the point /,). (ii) Draw the evolute through these centres. Note that pF = FR. Probfem 6-33, (fig. 6-43)2To draw the evoluteof a hyperbola. (i) Mark a number of points on the hyperbola and determine tht: centres of curvature at these po;nts (as shown at the point p). (ii) Draw the evolute through these centres. To obtain ihe centre of curvature at the vertex, the position of the other focus f1 rnust be fou nd.
Frc. 5_43 Flc. 6_44 I t i s d e t e r m i n e db y m a k i n gu s e o f t h e f o l l o w i n gr u l e : The tangent at any point on. the curve bisects the angle made by lines . . joining that point with the rwo foci, i.e. Z FleC = Z Fat.
1 3 0 E n g i n e e r i n gD t a w i n B
Problem 6-34. (fig. 6-44): To draw the evolute of a cycloid. ( i ) M a r k a p o i n t P o n the cycloid and draw the normal PN to it ( P r o b l e m6 - 2 0 ) . (ii) ProducePN to Op so that NOp = pN. Op is the centre of curvature at the point P. ( i i i ) S i m i l a r l y ,m a r k a n u m b e r o f p o i n t s o n t h e c y c l o i d a n d d e t e r m i n e centres of curvature at these points. The curve drawn through t h e s e c e n t r e si s t h e e v o l u t eo f t h e c y c l o i d . l t i s a n e q u a l c y c l o i d . Probtem 6-35. (fig. 6-45): Io draw the evolute of an epicycloid'
FlG.6-46 Frc. 6-45 ( i ) M a r k a p o i n t P o n t h e e p i c y c l o i da n d d r a w t h e n o r m a l P N t o i t ( P r o b l e m6 - 1 4 ) . ( i i ) D r a w t h e d i a m e t e rP Q o f t h e r o l l i n g c i r c l e . . l o i n Q w i t h O , t h e c e n t r eo f t h e d i r e c t i n gc i r c l e . (iii) Produce PN to cut QO at Oo, which is the centre of curvature at the pointP. ( i v ) M a r k a n u m b e r o f p o i n t s o n t h e e p i c y c l o i da n d s i m i l a r l y ,o b t a i n centres of curvature at these points. The curve drawn through t h e s e c e n t r e s i s t h e e v o l u t eo f t h e e p i c y c l o i d . the line T h r o u g hO o , d r a w a l i n e p e r p e n d i c u l atro P Q o a n d i n t e r s e c t i n g j o i n i n g C ( t h e t e n t r e o f t h e r o l l i n g c i r c l e )w i t h O a t a p o i n t R . T h e e v o l u t e i s t h e e p i c y c l o i do f t h e c i r c l e o f d i a m e t e rN R , r o l l i n g a l o n g t h e c i r c l e o f radius OR. Problem 6-36. (fig. 6-46\t To draw the evolute of a hypocycloid. (i) Mark a point P on the hypocycloid and draw the normal PN to it ( P r o b l e m6 - 23 ) . (i i ) Draw the diameter PQ of the rolling circle. Join Q with O, the c e n t r e o f t h e d i r e c t i n gc i r c l e . (iii) Produce PN to cut OQ-produced at Oo, which is the centre of curvature at the point P. (iv) Mark a number of points on the hypocycloid and similarly, obtain centres of curvature at these points. The curve drawn through these centres is the evolute of the hypocycloid.
Cuftes Used in Engineering practice 131
Through Oo, draw a line perpendiculaL r o pO" and intersectinp O C - p r o d u c e da t a p o i n t R . T h e e v o j u t e i s t h e h i , p o c y c l o i do t t h E c i r c l e o f d i a m e t e rN R r o l l i n g a l o n g t h e c i r c l e o f r a d i u s O R . Problem 6-J7. (fig. 6-34): To druw the evolute of an involute of a circle. I n t h e i n v o l u t eo f a c i r c l e , t h e n o r m a l N M a t a n y p o i n t N i s t a n g e n t to the circle at the point of contact M- M is the centre of curvature at the p o i n t N . H e n c e , t h e e v o l u t e o f t h e i n v o l u t ei s t h e c i r c l e i t s e l f .
6-5. Spirals: l f a l i n e r o t a t e si n a p l a n e a b o u t o n e o f i t s e n d s a n d i f a t t h e s a m e t i m e , a p o i n t m o v e s a l o n g t h e l i n e c o n t i n u o u s l yi n o n e d i r e c t i o n ,t h e c u r v e t r a c e d o u t b y t h e m o v i n g p o i n t i s c a l l e d a s p i r a l .T h e p o i n t a b o u t w h i c h the Iine rotates is called a pole. Iine joining any point on the curve with the pole is called the ..The rcdius vector and the angle betweenthis line and the line in its initial position is called the vectorialangle. Each complete revoiution of the curve is termed the convolution. A s p i r a l m a y m a k e a n y n u m b e r o f c o n v o l u t i o n sb e f o r er e a c h i n gt h e p o l e .
6-5-1. Archemedianspiral: It is a curve traced out by a point moving in such a way that its movementtowards or away from the pole is uniform with the increase o [ l h e v e c t o r i a la n g l e f r o m t h e s t a r t i n gl i n e . T h e u s e o f t h i s c u r v e i s m a d e i n t e e t h p r o f i l e so f h e l i c a lg e a r s ,p r o f i l e s of cams etc. Problem 6.38. (fig. 6-47); To constructan Archemedianspiral of 1l; convolutions,given the greatestand the shortestradii. L e t O b e t h e p o l e , O P t h e g r e a t e s tr a d i u sa n d O e t h e s h o r t e s tr a d i u s . (i) With centre O and radius equal to Op, draw a circle. Op revolves 1
a r o u n dO f o r 1 1 r e v o l u t i o n sD . u r i n g t h i s p e r i o d ,p m o v e st o w a r d s O , t h e d i s t a n c ee q u a lt o ( O p _ O e ) i . e . e p . ( i i ) D i v i d e t h e a n g u l a rm o v e m e n t so l O p , v i z 1 l r e v o l u t i o n si . e . 5 4 0 " , a n d t h e l i n e Q P i n t o t h e s a m e n u m b e r o f e l u a l p a r t s ,s a y 1 8 ( o n e r e v o l u t i o nd i v i d e d i n t o j 2 e q u a l p a r t s ) . W h e n t h e l i n e O P m o v e s t h r o u g h o n e d i v i s i o n ,i . e . 3 0 . , t h e p o i n t P will move towards O by a distance equal to one division of QP to a point p1. (iii) To obtain points systematicallydraw arcs with centre O and radii 0 1 , 0 2 , 0 3 e t c . i n t e r s e c t i n gl i n e s 0 1 , , 0 2 , , 0 3 , e t c . a t p o i n t s P1,P2, P3 etc. respectively. I n o n e r e v o l u t i o np, w i l l r e a c ht h e 1 2 t h d i v i s i o na l o n ge p a n d i n t h e n e x t h a l f r e v o l u t i o ni t w i l l b e a t t h e p o i n t e q o n t h e l i n e 1 8 , _ O . (iv) Draw a curve through points p, p1, p1, p6. This curve is the r e q u i r e dA r c h e m e d i a ns p i r a l .
132 E n g i n e e l i n gD r a w i n B
Normal and tangent to an Archemedian spiral: The normal to an Archemedian spiral at any point is the hypotenuse of the right-angled triangle having the other two sides equal in length to the radius vecfor at that point and the constant of the curve respectively. The constant of the curve is equal to the differencebetween the lengths of any two radii divided by the circular measureof the angle between them.
A r c h e m e d i a ns p i r a l Ftc.6-47 OP and OP3 (ig. 6-47) a r e t w o r a d i i m a k i n g9 0 ' a n g l e b e t w e e nt h e m . = l . s z ' T h e r e f o r e ,t h e c o n s t a n t o f t h e I n c i r c u l a r m e a s u r e ,9 0 ' = | -_oPz
c u r v e c, :
9P
.
1.57 Problem 6-39. (fig. 6-47): To draw a normal to the Archemedianspiral at a point N on it. (i) Draw the radius vector NO. (ii) Draw a line OM equal in length to the constant of the curve C and perpendicularto NO. ( i i i ) D r a w t h e l i n e N M w h i c h i s t h e n o r m a lt o t h e s p i r a l (iv) Through N, draw a line 5r perpendicularto NM' Sf is the tangent to the spiral.
Curves Used in Engineering ptactice 133
Problem 6-a0. (fig. 6-49\t A link 225 mm long, swingson a pivot O . '75" from its vertical position of rest to the right throigh an ingte of and returns to its initial position at uniform velocity. During that period, a point P moving at uniform speed along the centie tine ol the tink from a point at a distance of 25 mm from O, reachesthe end of the link. Draw the locus of the point P
Flc.6_48 (i)
D r a w a v e r t i c a li i n e O A , 2 2 5 m m l o n g .
(ii) With centre O and radius equal to O,4, draw an arc. (iii) Draw a line OB makingZ AOB equal to 25. and cutting the arc at B. (iv) Mark a point P along OA and at a distance of 25 mm from O. ( v ) D i v i d e t h e a n g u l a rm o v e m e n to f t h e l i n k a n d t h e l i n e p A i n t o the same number of equal parts,say 8. The end A o{ the link moves to g and returns to its original p o s i t i o n .H e n c e ,t h e a r c A B m u s t b e d i v i d e d i n t o f o u r e q u a l p i r t s . (vi) With centre O and radii O1, 02, 03 etc., draw arcs intersecting lines 01', 02', 03' etc. at points p1, p2, \ etc. respectivelyl Draw a curve through p, p1....p4....A T .h i s c u r v e i s t h e l o c u s o f the point P.
6-5-2.Logarithmicor equiangularspiral: ln a logarithmic spiral, the ratio of the lengths of consecutive radius vectors enclosingequal anglesis always constant.In other words the values of.vectorial angles are in arithmetical progression and the corresponding values of radius vectors are in geometrical progression. The logarithmic spiral is also known as equiangularspiral because of . its that the angle which the tangent at any point on the curve .property makes with the radius vector at that point is constant.
1 3 4 E n g i n e e r i n BD r a w i n g
Probfem 6-41, (fig. 6-49): To construct a logarithmic spiral of one convolution, given the length of the shottest radius and the ratio of the Iengths of radius vectors enclosing an angle 0.
'
('
(t)
Frc.6-4g
Let the shortest radius be 1 cm long, 0 equal to 30' and the ratio
f; The lengths of radius vectors are determined from the scale which is constructedas shown below: (i) Draw lines AB and AC [fig. 6-49(ii)] making an angle of 30' between them. (ii) On A8, mark AD, 1 cm long. On AC, mark a point E such that
AE=ry xlcm=f;cm. v ( i i i ) D r a w a l i n e j o i n i n gD w i t h E . (iv) With centre ^ and radius AE, draw an arc cutting AB at a point .1. T h r o u g h 1 , d r a w a l i n e 1 - 1 ' p a r a l l e lt o D E a n d c u t t i n g A C a t 1 ' . Again, with centre A and radius A1' draw an arc cutting AB at 2. Through 2, draw a line 2-2' parallel to DF and cutting AC at 2'. R e p e a tt h e c o n s t r u c t i o na n d o b t a i n p o i n t s 3 , 4 . . . . . ' 1 2A . 1, A2 etc. are the lengths of consecutiveradius vectors. (v) Draw a horizontalline OQ and on it, mark OP, 1 cm long [fig. 6-a9(i)J. Through O, draw radial lines making 30' angles between two c o n s e c u t i v el i n e s . T h e s e a r e t h e r a d i u s v e c t o r s . ( v i ) M a r k p o i n t s P 1 , P 2 . . . . . . . . . . P to2n c o n s e c u t i v er a d i u s v e c t o r s s u c h = A'12. that OP1 = A1, OPz : A2..........OP\2 ( v i i ) D r a w a s m o o t h c u r v e t h r o u g h P 1 , P 2 . . . . . . O P 1T 2h . is curve is the r e q u i r e d l o g a r i t h m i cs p i r a l . T h e e q u a t i o nt o t h e l o g a r i t h m i cs p i r a l i s r = a € w h e r e r i s t h e r a d i u s vector/ 0 is the vectorial angle and a is a constant. Hence, logr =0loga. lf
0=0,
When
0 - 30":
r.e
toga:
logr=o
...r:1.
I radians, o
-6 ,rog 1 0
t
t^0 v
"
1=
1^0 1^0 = T lo*., ..lon "J " \i
o
Curves Used in Engineering Practice 135
Probfem 6-42. lfig. 6-49(i))':To draw a normal and a tangentat a gven point N on the loga thmic spiralin problem 6-41. The value of the tangent to the constantangle is obtained by the _ formulasiven below:
:logu, loga e = 2.71a
Ian 0
where
1C,92.718
. Ian C = =:_^
I .e .
9loe! n'9 s=78"-38'.
(i) Draw a line joiningN with O. ( i i ) D r a w a l i n e N f m a k i n ga n a n g l e o f 7 9 " - 3 8 , w i t h N O . N f i s t h e required tangent. NR drawn perpendicularto Nf is the normal to the spiral.
6-6. Helix: Helix is defined as a curve, generated by a point, moving around the s u r f a c eo f a . r i g h t c i r c u l a rc y l i n d e ro r a r i g h t c i r c u l a rc o n e i n s u c h a w a y t h a t , i t s a x i a l a d v a n c e ,i . e . i t s m o v e m e n ti n t h e d i r e c t i o no f t h e l e n e t h o f t h e c y l i n d e ro r t h e c o n e i s u n i f o r m w i t h i t s m o v e m e n ta r o u n d t h e s - u r f a c e o f t h e c y l i n d e ro r t h e c o n e . T h e a x i a l a d v a n c eo f t h e p o i n t d u r i n g o n e c o m p l e t er e v o l u t i o ni s c a l l e d the pitch of the helix. tf the pitch is say 20 mm and the point starts upwards from the base of the cylinder, in one-fourth of a revolution, the point will move up a distanceof 5 mm from the base. We shall now learn i n t h e a r t i c l e st o f o l l o w : 1.
A m e t h o d o f d r a w i n ga h e l i c a lc u r v e
2.
H e l i c a ls p r i n g s
3.
S c r e wt hr e a d s
4.
Helix upon a cone.
6-6-1. A method of drawing a helical curve: Probfem 6-43. (fig. 6-50): Io draw a helix of one convolution,given the pitch and the diameter of the cylinder. Also to develop the surfaie of the cylinder showing the helix in it. (i)
D r a w t h e p r o j e c t i o n so f t h e c y l i n d e rt f i g . 6 - 5 0 ( i ) J D . ividethe circle (in the top view) into any number o{ equal parts, say .12.
(ii) Mark a length P-12' equal to rhe given pitch, along a vertical side of the rectanglein the front view and divide it inio the same n u m b e r o f e q u a l p a r t s ,v i z . 1 2 .
136 tngine€ring Drawing
(i i i ) A s s u m e t h e p o i n t P t o m o v e u p w a r d s a n d i n a n t i - c l o c k w i s e d i r e c t i o n .W h e n i t h a s m o v e dt h r o u g h3 0 ' a r o u n d t h e c i r c l e ,i t w i l l have moved up by one division.To locatethis position,draw a vertical l i n e t h r o u g ht h e p o i n t 1 a n d a h o r i z o n t a l i n e t h r o u g ht h e p o i n t l ' , b o t h i n t e r s e c t i n eg a c ho t h e ra t a p o i n t P 1 ,w h i c h w i l l b e o n t h e h e l i x . ( i v ) Obtain other points in the same manner and draw the helix through them. The portion of the curve from P5,to P12will be on the back s i d e o f t h e c y l i n d e ra n d h e n c ei t w i l l n o t b e v i s i b l e .
(t Frc.6-50
(r)
Fig. 6-50(ii) shows the developmentof the surface of the cylinder. The helix is seen as a straight line and is the hypotenuse of a right-angled triangle having base equal to the circumference of the circle and the vertical side equal to the pitch of the helix. The angle 0 which it makes with the base, is called the hellx angle. The helix angle can be expressedas
tano:
pitch circumference of the circle
6-6-2. Helical springs: In a spring having a wire of square cross-section,the outer two corners of the section may be assumed to be moving around the axis, on the surface of a cylinder having a diameter equal to the outside diameter of the spring. The inner two corners of the section will move on the sur{ace o{ a cylinder having a diameter,equal to the inside diameter of the sprinS. The pitch in case of each corner will be the same. (a) Helical spring of a wire of square cross-section. ( b ) H e l i c a ls p r i n go f a w i r e o f c i r c u l a rc r o s s - s e c t i o n . (a) Helical spring of a wire of square cross-section. Probfem 6-44, (fig- 6-51\t Draw the projectionsof two complete turns of a spring of a wire of square section of 20 mm side. Outside diametel of the sPring : 110 mm; Pitch = 60 mm.
Curves Used in Ingineering practice 137
I n s t e a do f f u l l c i r c l e s ,s e m i - c i r c l e so f I l 0 m m a n d 7 0 m m d i a m € t e r s for the outside and the inside diameters of the spring may be drawn in the too view. 12
( i ) D i v i d e e a c h s e m i - c i r c l ei n t o , i . e . 6 e q u a lp a r t s . f ( i i ) Project up the division-points-aslines in the {ront view. Let peSR b e t h e s q u a r es e c L i o no I t h e w i r e . (i i i ) In one revolution,the point P will move to p,.so that pp, = 60 mm. S i m i l a r l yR , R ' , Q Q ' a n d S S ,w i l l e a c h b e e q u a l t o 6 0 m m . ( i v ) D i v i d e t h e s e d i s t a n c e si n t o ' 1 2 e q u a l p a i t s a n d p l o t t h e h e l i c e s traced out by P and R on the outer surfaceand by e and S on the i n n e r s u r f a c e ,a s s h o w n . T h e o u t e r h e l i c e sw i l l b e p a r a l l e lt o e a c n other. Note carefully the visible and the hidden parts of the curves. (b) Helical spring of a wire of circular cross-section. When the wire is of circular cross-section,a helical curve for the centre of the cross-section is first traced out. A number of circles of diameter equal to that of the cross-sectionare then drawn with a number of points on this curve as centres. Curves/ tangent to these circles, will give the front view of the spring.
Frc.6-5'l
atu. o-iz
Problem 6-a5. (fig. 6-52l: Draw two complete turns of a helical sp:ing . of circular cross-sectionof 20 mm diametei, Springindex and pitch ai 6 and 50 mm respectivelv.
1 3 8 E n B i n e e r i n gD r a w i n g
Mean diameterof a coil Diameterof wrre ( a ) M e a n d i a m e t e ro f a c o i l = S p r i n gi n d e x x Diameter of wire =6x20:12O mm. Spring index
(b) Outsidediameterof a coil = Mean diameter + D i a m e t e ro f w i r e = 120mm + 20 mm : 140 mm.
(c)Insidediameter of a coil : [il*"TT|. (i)
- Diameterof wire m m = 1 0 0m m .
D r a w t h e t h r e e c o n c e n t r i cc i r c l e so f d i a m e t e r s1 4 0 m m , 1 2 0 m m a n d 1 0 0 m m r e p r e s e n t i n tgh e o u t s i d e ,m e a n a n d t h e i n s i d ec i r c l e s o f c o i l o f t h e s p r i n g r e s p e c t i v e l iyn t h e t o p v i e w .
( i i ) D i v i d e t h e c i r c l e si n t o l 2 d i v i s i o n s M . a r k t h e d i v i s i o n so n t h e c i r c l e o f t h e m e a n d i a m e t e ra s 1 , 2 , 3 e t c . ( i i i ) D r a w t h e p r o j e c t o r sp a s s i n gt h r o u g h t h e s e d i v i s i o n si n t h e f r o n t view. Mark the pitch length along one of the projectors which is p a r a l l e lt o t h e a x i s o f t h e s p r i n g . ( i v ) D i v i d e t h e p i t c h l e n g t h i n t o 1 2 d i v i s i o n sa n d n a m e t h e s e d i v i s i o n p o i n t sa s 1 ' , 2 ' , 3 ' e t c . D r a w t h e h o r i z o n t alli n e s i n t e r s e c t i nt gh e previously drawn projectorsfrom 1, 2, 3 etc. ( v ) W i t h t h e i n t e r s e c t i o np o i n t a s c e n t r e a n d r a d i u s e q u a l t o 1 0 m m , draw circles. Draw the curve touching the top and bottom of c i r c l e o f t h e w i r e a s s h o w n i n f i g . 6 - 5 2 . R e p e a t h e p r o c e s sf o r o n e more pitch length.
6-6-3.Screwthreads: T h e s ea r e a l s o c o n s t r u c t e do n t h e p r i n c i p l eo f t h e h e l i x . In a screw thread, the pltch is defined as the distancefrom a point on a thread to a correspondingpoint on the adjacentthread, measuredparallel to the axis. The axial advance of a point on a thread, per revolution/ is called the /ead of the screw. l n t h e s i n g l e - t h r e a d e ds c r e w s , w h i c h a r e m o s t c o m m o n l y u s e d i n practice/ the pitch is equal to lead. Therefore, the pitch of the screw is equal to the pitch of the helix. Unless stated otherwise, suews arc always assumedto be single-threaded. Probfem 6-46. lfig. 6-53\t Project two complete tums of a square thread having outside diameter 120 mm and pitch 45 mm. I n a s q u a r e t h r e a d , t h e t h i c k n e s so f t h e t h r e a d = t h e d e p t h o f t h e t h r e a d = 0 . 5 x p i t c h . H e n c e ,t h e s e c t i o no f t h e t h r e a dw i l l b e a s q u a r eo f 2 2 . 5 m m s i d e a n d t h e d i a m e t e ra t t h e b o t t o m o f t h e t h r e a dw i l l b e 7 5 m m . P r o j e c tt h e t h r e a d si n t h e s a m e m a n n e ra s t h e s p r i n g i n p r o b l e m6 - 4 4 . T h e s c r e w d i f f e r s f r o m t h e s p r i n g i n h a v i n ga s o l i d c y l i n d e r i n s i d e ,w h i c h c o m p l e t e l yh i d e s t h e b a c k p o r t i o n so f t h e c u r v e s .
Curves Used in tngineering Practice 139
Frc.6-53
Ftc.6-54
I n d o u b l e - t h r e a d e sdc r e w s ,t w o t h r e a d so f t h e s a m e s i z e r u n p a r a l l e l to each other. The axial advance per revolution, viz. the lead, is made twice the lead of the single-threadedscrew, the pitch of the thread being kept the same in both cases.
H e n c e , i n d o u b l e - t h r e a d esdc r e w s , pitch of the helix = lead = twice the pitch of the screw. Fig. 6-54 shows a double-threadedscrew, having the same cross_section a n d t h e s a m e p i t c h o f t h e s c r e w a s i n p r o b l e m6 - 4 6 .
6 - 6 - 4 .H e l i x u p o n a c o n e : This curve is traced out by a point which, while moving around the axis and on the surface of the cone, approachesthe apex. The movement around the axis is uniform with its movement towards the apex, measured p a r a l l e lt o t h e a x i s . T h e p i t c h o f t h e h e l i x i s m e a s u r e dp a r a l l e lt o t h e a x i s of the cone. A s t h e w h o l e s u r f a c eo f t h e c o n e i s v i s i b l e i n i t s t o p v i e w , t h e h e l i x w i l l b e f u l l y s e e ni n i t . Problem 6-47. (fig. 6-55): Draw a helix oi one convolution upon a cone, diametet of base 75 mm, height 100 mm and pitch 75 mm. Also develop the surface of the cone, showing the helix on ii. (i) Draw the projections of the cone as shown. Divide the circle into t w e l v e e q u a l p a r t s a n d j o i n p o i n t s 1 , 2 e i c . w i t h o . p r o i e c tt h e s e p o i n t s t o t h e b a s e - l i n ei n t h e f r o n t v i e w a n d i o i n t h e m w i t h o , .
1 4 0 E n g i n e e r i n gD r a w i n g
( i i ) M a r k a p o i n t A o n t h e a x i s a t a d i s t a n c eo f 7 5 m m { r o m t h e b a s e . Draw a horizontai line through A to cut the generatorso' P at A,. D i v i d e P A ' i n t o t w e l v ee q u a l p a r t s . (i i i ) Let P be the starting point. When it has moved around through 3 0 ' , i t w i l l h a v em o v e du p t h r o u g ho n e d i v i s i o nt o a p o i n tp j , o n t h e g e n e r a t o ro ' 1 ' , o b t a i n e db y d r a w i n ga h o r i z o n t a l i n e t h r o u g h 1 , , . p 1 ' w i l l b e a p o i n t o n t h e h e l i x i n t h e f r o n t v i e w . I t s p r o j e c t i o nt o p 1 o n 0 1, w i l l b e t h e p o s i t i o no f t h e p o i n t i n t h e t o p v i e w . (iv) Obtain all the points in this manner and draw smooth curves t h r o u g ht h e m i n b o t h t h e v i e w s . ( v ) D r a w t h e d e v e l o p m e n to f t h e s u r f a c e o f t h e c o n e a n d l o c a t e P o r n t sp 1 l p 2 e t c .
t-tL.
b-J5
Ftc.6-56
Problem 6-48. (fig. 6-56): A conical spring of a bicycle's seaf has following specifications. Drcw the front. view and the top view of the spring. Only one-turn is sufficient. ( i ) O u t e r d i a m e t e ro f t h e c o i l a t t h e b o t t o m . . . ( i i ) O u t e r d i a m e t e or f t h e c o i l a t t h e t o p . , . . . . (iii) Wire diameter .. (iv) Pitch ..
72 mm 40 mm 10 mm 60 mm.
Curves Used in Engineering Practice 141
( i ) Draw a circleof meandiameterol 72 mm in the top view. ( i i ) Dividethe circleinto 12 divisions. Mark the divisions1, 2. 3
etc.as shownin fig.6-56. (i i i ) Draw the front view of the conewith the baselengthZ2 mm. ( i v ) Draw pitchlength60 mm alongthe line parallel to the axiso{ t h e c o n e . D i v i d et h i s p i t c h l e n g t hi n t o 1 2 d i v i s i o n sa n d numberthem as 1', 2', 3' etc. (v) Draw the verticalprojectors from thesepoints1,2,3 etc_to intersectthe horizontallinesfrom the points 1',2',3, eIc. With the intersection pointas centreand radiusequalto 5 mm, draw the circlesand the smoothcurvetouchingto the top and bottomof the circlesas shownin fig. 6-56.
6-7 Cam:
A cam is a machine-part which, whilb rotatingat uniformvelocity impartsreciprocating linearmotionto anothermachine-part calleda follower. The motionimpartedmay be eitheruniformor variable,dependinguponthe shapeof lhe cam profile.
The shapeof the cam to -transmituniformlinearmotion is determined by the applicationof the principleo{ Archemedianspiral, as shown in problem 6-38. Probfem 6-49. (fig. 6-57)zDraw the shapeof a cam to give a uniform rise and fall of 50 mm to a point, during each revolution of the cam. Diametetof the shaft : 50 mmi least radiui of the cam = S0 mm.
Cam Ftc. 6-57
142 [ngineering Drawing
.
(i)
Draw a circle (for the shaft) lvith centre O. With the same centre and radiusOB equal to 50 mm (the least r a d i u s o f t h e c a m ) , d r a w a n o t h e rc i r c l e .
( i i ) P r o d u c eO B t o C m a k i n g 8 C e q u a l t o 5 0 m m ( t h e r i s e o f t h e p o i n t ) . T h e p o i n t i s t o b e u n i f o r m l yr a i s e dd u r i n g I the shaft.
r e v o l u t i o no f
( i i l ) T h e r e f o r e ,d i v i d e 8 C a n d a n g l e 1 8 0 ' i n t o t h e s a m e n u m b e r o f equal parts,say 6. O b t a i n p o i n t s P 1 , P 2 e t c . a s i n t h e A r c h e m e d i a ns p i r a l a n d d r a w t h e curve through them. T h e p o i n t i s t o f a l l t h e s a m e d i s t a n c ed u r i n g t h e s a m e p e r i o d . H e n c e , t h e c u r v e w i l l b e e x a c t l yo f t h e s a m e s h a p e a s f o r t h e r i s e . The followers are generally provided with rollers to give smooth w o r k i n g . I n s u c h c a s e s ,t h e p r o f i l e o f t h e c a m i s d e s i g n e di n i t i a l l y ,t o transmitmotion to a point (the centreof the roller) and then, a parallel c u r v e i s d r a w n i n s i d e i t a t a d i s t a n c ee q u a l t o t h e r a d i u so f t h e r o l l e r .T h i s is done by first drawinga number of arcs with points on the curve as centres and radius equal to radius of the roller, and then drawing a s m o o t h c u r v e t o u c h i n gt h e s e a r c s , a s s h o w n i n f i g . 6 - 5 7 .
Vl Exercises 1 . D r a w a s t r a i g h t l i n e A B o f a n y l e n g t h .M a r k a p o i n t t 6 5 m m f r o m AB. Trace the paths of a point P moving in such a way, that the ratio o f i t s d i s t a n c ef r o m t h e p o i n t F , t o i t s d i s t a n c ef r o m A B i s ( i ) 1 ; (ii) 3:2; (iii) 2:3. Plot at least B points. Name each curve. Draw a normal and a tangent to each curve at a point on it, 50 mm from F. 2 . A f i x e d p o i n t i s 7 5 m m f r o m a f i x e d s t r a i g h tl i n e . D r a w t h e l o c u s o f a point P moving such a way that its distance from the fixed straight I i n e i s ( i ) t w i c e i t s d i s t a n c ef r o m t h e f i x e d p o i n t ; ( i i ) e q u a l t o i t s d i s t a n c ef r o m t h e f i x e d p o i n t . N a m e t h e c u r v e s . 3. The vertex of a hyperbola is 65 mm from its focus. Draw the curve ?
i f t h e e c c e n t r i c i t yi s D r a w a n o r m a l a n d a t a n g e n la t a p o i n t o n l. the curve, 75 mm from the directrix. 4. The major axis of an ellipseis 150 mm long and the minor axis is 1 0 0 m m l o n g . F i n d t h e f o c i a n d d r a w t h e e l l i p s eb y ' a r c s o f c i r c l e s ' m e t h o d . D r a w a t a n g e n tt o t h e e l l i p s ea t a p o i n t o n i t 2 5 m m a b o v e the major axis. 5 . T h e f o c i o f a n e l l i p s ea r e 9 0 m m a p a r t a n d t h e m i n o r a x i s i s 6 5 m m l o n g . D e t e r m i n et h e l e n g t h o f t h e m a j o r a x i s a n d d r a w h a l f t h e e l l i p s e s ethod and the other half by oblong method. b y c o n c e n t r i c - c i r c l em D r a w a c u r v e p a r a l l e lt o t h e e l l i p s e a n d 2 5 m m a w a y f r o m i t . 6. The major axis of an ellipseis 100 mm long and the foci are at a d i s t a n c e o f 1 5 m m f r o m i t s e n d s . F i n d t h e m i n o r a x i s . P r e p a r ea t r a m m e l a n d d r a w t h e e l l i p s eu s i n g t h e s a m e .
Cufves Used in Engineering Practice 143
7 . Two fixed points A and B are 100 mm apart. Tracethe complete path of a point P moving (in the sameplaneas that of A and B) in such a w a y t h a t , t h e s u m o f i t s d i s t a n c e sf r o m A a n d I i s a l w a y st h e s a m e a n d e q u a l t o 1 2 5 m m . N a m e t h e c u r v e . D r a w a n o t h e rc u r v e p a r a l l e l to and 25 mm away from this curve. B . I n s c r i b ea n e l l i p s ei n a p a r a l l e l o g r a h ma v i n gs i d e s1 S 0 m m a n d 1 0 0 m m l o n g a n d a n i n c l u d e da n g l e o f 1 2 0 " . 9 . T w o p o i n t s A a n d I a r e 1 0 0 m m a p a r t .A p o i n t C i s 2 5 m m f r o m A a n d 6 0 m m f r o m 8 . D r a w a n e l l i p s e p a s s i n gt h r o u g h A , B a n d C . 1 0 . D r a w a r e c t a n g l eh a v i n gi t s s i d e s i 2 5 m m a n d 7 5 m m l o n g . I n s c r i b e t w o p a r a b o l a si n i t w i t h t h e i r a x i s b i s e c t i n ge a c h o t h e r . 1 1 . A b a l l t h r o w n u p i n t h e a i r r e a c h e sa m a x i m u m h e i g h t o f 4 s m e t r e s and travels a horizontal distance of 75 metres. TracJ the path of the b a l l , a s s u m i n gi t t o b e p a r a b o l i c . 1 2 . A p o i n t P i s 3 0 m m a n d 5 0 m m r e s p e c t i v e l fyr o m t w o s t r a i g h tl i n e s w h i c h a r e a t r i g h t a n g l e st o e a c h o t h e r D r a w a r e c t a n g u l ahr y p e r b o l a f r o m P w i t h i n l 0 m m d i s t a n c ef r o m e a c h l i n e . 1 3 . Two straight lines OA and OB make an angle of ZS. between them. P is a point 40 mm from OA and 50 mm from OB. Draw a hvperbola through P, with OA and OB as asymptotes,marking at least ten points. Two points ,4 and B are 50 mm apart. Draw the curve traced out by a point P moving in such a way that the difference between its d i s t a n c e sf r o m A a n d B i s a l w a y s c o n s t a n ta n d e q u a l t o 2 0 m m . 1 5 . A c i r c l e o f 5 0 m m d i a m e t e rr o l l s a l o n g a s t r a i g h tl i n e w i t h o u t s l i p p i n g . Draw the curve traced out by a point p on the circumference,for one complete revolution of the circle. Name the curve. Draw a tangent to the curve at a point on it 40 mm from the line. T w o p o i n t s Q a n d 5 l i e o n a s t r a i g h ll i n e t h r o u g h l h e c e n l r e C o [ a c i r c l e o f 5 0 m m d i a m e t e r ,r o l l i n g a l o n g a f i x e d s t r a i g h t I i n e . D r a w a n d n a m e t h e c u r v e st r a c e do u t b y t h e p o i n t s q a n d S d u r i n g o n e r e v o l u t i o no f t h e c i r c l e . C Q = 2 0 m m ; C S : 3 5 m m . 1 7 . A c i r c l e o f 5 0 m m d i a m e t e r r o l l s o n t h e c i r c u m f e r e n c eo f a n o t h e r c i r c l e o f 1 7 5 m m d i a m e t e ra n d o u t s i d e i t . T r a c et h e l o c u s o f a p o i n t o n t h e c i r c u m f e r e n co e f t h e r o l l i n g c i r c l e f o r o n e c o m p l e t er e v o l u t i o n . Name the curve. Draw a tangent and a normal to the curve at a p o i n t 1 2 5 m m f r o r n t h e c e n t r e o f t h e d i r e c t i n gc i r c l e . 1 8 . C o n s t r u c ta h y p o c y c l o i d r, o l l i n g c i r c l e 5 0 m m d i a m e t e ra n d d i r e c t i n g -l circle 75 mm diameter.Draw a tangent to it at a point 50 mm from t h e c e n t r e o f t h e d i r e c t i n gc i r c l e . 1 9 . A circle of 115 mm diameterrolls on anothercircle of ZS mm diameter with internal contact. Draw the locus of a point on the circumference o f t h e r o l l i n g c i r c l e f o r i t s o n e c o m p l e t ei e v o l u t i o n . 20. A c i r c l e o f 5 0 m m d i a m e t e r r o l l s o n a n o t h e r c i r c l e o f 1 2 5 m m diameter. Draw and name the curves traced out by two points e and S l y i n g o n a s t r a i g h tl i n e t h r o u g h t h e c e n t r e C o f t h e r o l l i n g c i r c l e a n d r e s p e c t i v e l2 y0 mm and 35 mm from it, when it rolls (i) outside and (ii) inside the other circre.
1,14 tngineering Drawing
2 1 . S h o w b y m e a n so f a d r a w i n gt h a t w h e n t h e d i a m e t e ro f t h e d i r e c t i n g c i r c l ei s t w i c e t h a t o f t h e g e n e r a t i n g c i r c l e ,t h e h y p o c y c l o i di s a s t r a i g h t l i n c . l a k e t h e d i a m e t e ro f t h e g e n e r a t i n gc i r c l e e q u a l t o 5 0 m m . 2 2 . A c i r c l eo f 5 0 m m d i a m e t e r o l l s o n a h o r i z o n t alli n e f o r a h a l {r e v o l u t i o n a n d t h e n o n a v e r t i c a ll i n e f o r a n o t h e rh a l f r e v o l u t i o n .D r a w t h e c u r v e traced out by a point P on the circumferenceof the circle. 2 3 . D r a w a n i n v o l u t eo f a c i r c l e o f 4 0 m m d i a m e t e r A . lso,draw a normal a n d a t a n g e n tt o i t a t a p o i n t 1 0 0 m m f r o m t h e c e n t r eo f t h e c i r c l e . 2 4 . A n i n e l a s t i c s t r i n g 1 4 5 m m l o n g , h a s i t s o n e e n d a t t a c h e dt o t h e c i r c u m f e r e n c eo f a c i r c u l a rd i s c o f 4 0 m m d i a m e t e r .D r a w t h e c u r v e t r a c e do u t b y t h e o t h e r e n d o f t h e s t r i n g ,w h e n i t i s c o m p l e t e l yw o u n d a r o u n d t h e d i s c , k e e p i n gt h e s t r i n ga l w a y st i g h t . 2 5 . D r a w a c i r c l ew i t h d i a m e t e A r B e q u a lt o 6 5 m m . D r a w a l i n eA C 1 5 0 m m long and tangent to the circle. Trace the path of A, when the line AC r o l l s o n t h e c i r c l e ,w i t h o u r s l i p p i n g . 2 6 , .48 is a rope 'l .6 metres long, tied to a pegat I (fig.6-58).Keepingit always tight, the rope is wound round the pole O. Draw the curve traced out by the end A. Scale
1
full size. * Frc.6-58 2 7 . D ' a w a n A r c h e m e d i a ns p i r a lo f t w o c o n v o l u t i o n st,h e g r e a t e sat n d t h e l e a s tr a d i i b e i n g1 1 5 m m a n d 1 5 m m r e s p e c t i v e l yD . r a w a t a n g e n t a n da n o r m a l t o t h e s p i r a l a t a p o i n t , 65 mm from the pole. 2 8 . A p o i n t P m o v e st o w a r d sa n o t h e rp o i n t O , 7 5 m m f r o m i t a n d r e a c h e s i t w h i l e m o v i n ga r o u n di t o n c e ,i t s m o v e m e n t o w a r d sO b e i n g u n i f o r m with its movementaround it. Draw the curve traced out by the point p. 2 9 . A l i n k O A , 1 0 0 m m l o n g r o t a t e sa b o u t O i n a n t i - c l o c k w i s ed i r e c t i o n . A point P on the link, 15 mm away from O, moves and reachesthe e n d A , w h i l e t h e l i n k h a s r o t a t e dt h r o u g h o f a r e v o l u t i o nA . ssuming I t h e m o v e m e n t so f t h e l i n k a n d t h e p o i n t t o b e u n i f o r m ,t r a c e t h e p a t h of the DointP.
3 0 . Draw a cam to give the following uniform motions to a point: Rise 4 0 m m i n 9 0 " . I n u p p e r p o s i t i o n f o r 7 5 " . F a l l t h e s a m e d i s t a n c ei n 1 2 0 ' . I n l o w e r p o s i t i o n f o r t h e r e m a i n i n gp e r i o d o f t h e r e v o l u t i o n . D i a m e t e r o f s h a f t = 5 0 m m . L e a s tm e t a l : 2 0 mm.
3 1 . A c i r c l e o f 4 0 m m d i a m e t e rr o l l s o n t h e i n s i d e o f a c i r c l e o f 9 0 m m d i a m e t e rA . p o i n t P l i e s w i t h i n t h e r o l l i n gc i r c l e a t a d i s t a n c eo f l 5 m m from its centre. Trace the path of the point p {or one revolution of the circle. 3 2 . T h e d i s t a n c eb e t w e e n t h e f o c i o f a n e l l i p s e i s i 0 0 m m a n d t h e m i n o r a x i s i s 6 5 m m l o n g . D r a w t h e e v o l u t eo f t h e e l l i p s e . 3 3 . Draw the evolutesof the two curves,the data of which is given in Ex. 2. 3 4 . C o n s t r u c tt h e e v o l u t eo f t h e h y p e r b o l aw h o s e d a t a i s g i v e n i n E x . 3 .
C u r y e sU s e d i n E n g i n e e r i n gP r a c t i c e 1 4 5
3 5 . D r a w t h e e v o l u t eo f t h e c y c l o i d o b t a i n e di n E x . 1 5 . 3 6 . D r a w t h e e p i c y c l o i da n d h y p o c y c l o i dw , h e n t h e g e n e r a t i n gc i r c l e a n d the directing circle are of 50 mm and .1Z5 mm diameters respectively. Construct the evolutes of the two curves. 3 7 . l n a l o g a r i t h m i cs p i r a l , t h e s h o r t e s tr a d i u s i s 4 0 m m . T h e l e n g t h o f a d j a c e n rt a d i u s . v e c t o resn c l o s i n g3 0 . a r e i n t h e r a t i o 9 . 8 . C o n s t r u c o t ne revolution of the spiral. Draw a tangent to the spiral at any point on it. 38. Draw a triangleABC with ,48 : 30 mm, AC = 40 mm and / BAC = 45". I a n d C a r e t h e p o i n t s o n a n A r c h e m e d i a ns p i r a l o f o n e c o n v o l u t i o n of which ,4 is the pole. tind the initialline and draw the spiral. 39. ABC is an equilateraltriangle of side equal to Z0 mm. Tracethe loci of t h e v e r t i c e s A , B a n d C w h e n t h e c i r c l e c i r c u m s c r i b i n eA B C r o j l s w i t h o u t s l i p p i n ga l o n ga f i x e d s t r a i g h tl i n e f o r o n e c o m p l e t er e v o l u t i o n . 40. Draw the shape of the cam to give the same motions as in Ex. 30 to a roller of 25 mm diameter. 4 1 . A p o i n t i s r a i s e d b y a c a m u n i f o r m l y2 5 m m i n I o f a r e v o l u t i o n ; kept at the same height for j of the revolurion; then allowed to d r o p t h r o u g h1 0 m m h e i g h ta n d r e m a i nt h e r e f o r
B
of the revolution;
I o w e r e du n i f o r m l yt o i t s o r i g i n a lp o s i t i o ni n j o f t h e r e v o l u t i o na n d kept there for the rest of the revolution. Draw the shape of the cam. Diameter of shaft : 40 mm. Least metal = 25 mm. 42. fwo pegs A and B are fixed on the ground 6 metres apart. A string, I m e t r e s l o n g h a s i t s o n e e n d t i e d t o t h e p e g , 4 , w h i l e t h e o t l . t e re r l d i s p a s s e dt h r o u g h a s m a l l r i n g C a n d t i e d t o t h e p e g t s . A l i n k C D of 1 metre lengthis fixed to the ring C. The link with the ring slidcs a l o n g t h e s t r i n g .D r a w t h e b o u n d a r yo f t h e a r e ao f t h e g r o u n d b e y o n d w h i c h D c a n n o t r e a c h i a s C s l i d e sa l o n g t h e s t r i n g A B . 4 3 . F i l l i n t h e b l a n k s i n t h e f o l l o w i n g . w i r h a p p r o p r i a t ew o r d s s e l e c r e d from the list givenbelow: (a) When a cone is cut by planes at different angles, the curves of intersectionare called ( b ) W h e n t h e p l a n e m a k e st h e s a m e a n g l e w i t h t h e a x i s a s d o t h e g e n e r a t o r st,h e c u r v e i s a . (c) When the planeis perpendicularto the axis,the curve is a ( d ) W h e n t h e p l a n e i s p a r a l l e lt o t h e a x i s , t h e c u r v e i s a (e) When the plane makesan angle with the axis greaterthan what do the generators,the curve is a (f) A conic is a locus of a point moving in such a way that the ratio of its distancefrom the and its distancefrom the , _, is alwaysconstant.This ratio is calledthe . It is in case of parabola, in case of hyperbolaano - i n c a s eo f e l l i p s e . ( g ) I n a c o n i c t h e l i n e p a s s i n gt h r o u g h the fixed point and p e r p e n d i c u l atro t h e f i x e d l i n e i s c a l l e dt h e
1 4 6 E n g i n e e r i n gD r a w i n g
(h) Thevertexis the pointat whichthe -cutsthe (i) The sum of the distances from its of any pointon the two foci is alwaysthe sameand equalto the (j) The distanceof the endsof the ol an ellipsefrom the is equal to half the ----. (k) tn a ---the product of the distancesof any point on it from two fixed lines at right angles to each other is always constant. The fixed lines are called ----. (l) Curves generatedby a fixed point on the circumferenceo f a c i r c l e r o l l i n g a l o n ga f i x e d I i n e o r c i r c l e a r e c a l l e d- . . _ (m) The curve generatedby a point on the circumferenceof a circle, r o l l i n g a l o n g a n o t h e rc i r c l e i n s i d ei t , i s c a l l e da (n) The curve generatedby a point on the circumferenceof a circle r o l l i n g a l o n g a s t r a i g h tl i n e i s c a l l e da - . - . (o) The curve generatedby a point on the circumferenceof a circle r o l l i n g a l o n g a n o t h e rc i r c l e o u t s i d ei t , i s c a l l e d- . (p) The curvegenerated by a pointfixedto a circleoutsideits circumference, a s i t r o l l s a l o n g a s t r a i g h tl i n e i s c a l l e d (q) The curvegeneratedby a point fixedto a circleinsideits circumference, a s i t r o l l s a l o n g a c i r c l e i n s i d ei t , i s c a l l e d (r) The curvegeneratedby a pointfixedto a circleoutsideits circumference a s i t r o l l s a l o n ga c i r c l eo u t s i d ei t , i s c a l l e d (s) T h e c u r v e t r a c e d o u t b y a p o i n t o n a s t r a i g h t l i n e w h i c h r o l l s , w i t h o u t s l i p p i n g ,a l o n g a c i r c l e o r a p o l y g o ni s c a l l e d- - . (t) T h e c u r v e t r a c e do u t b y a p o i n t m o v i n gi n a p l a n ei n o n e d i r e c t i o n , t o w a r d sa f i x e d p o i n t w h i l e m o v i n ga r o u n di t i s c a l l e da ( u , The line joining any point on the spiralwith the pole is called (v) In the ratio of the lengthsof consecutiveradius v e c t o r se n c l o s i n ge q u a l angles is always constant. List of words for Ex. 43: 'L Asymptotes 9 . E q u a lt o 1 10. Epitrochoid 2. Axis 3. Cycloidal 1 1 . E l l i p s e 12. Eccentricity 4. Conic 5. Llrcte 13. Focus 14. Greater than I 6. Cycloid 7. Directrix 15. Hyperbola 8. Epicycloid 16. Hypocycloid
lar 1 7 . H y p o t r o c h o i d25. Rectangu '1 26. Smallerthan 18. Involute r/. )upenor 19. Inferior 20. Logarithmic 28. Spiral 29. Trochoid 21. Minoraxis 22. Major axes 30. Conics 31. Curves 23. Parabola 24. Radiusvector
Answercto Ex. 43: (a) - 30 (b) -23 (c)-s (d) -15 (e) -11 (f)-13,7, 12, e, 14 a n d 2 6 ( g ) - 2 ( h ) - 4 a n d 2 ( i ) - 1 1 a n d2 2 ( j ) - 2 1 , 1 3 a n d2 2 ( k ) - 2 s , 1 s a n d 1 ( l ) - 3 a n d 3 1 ( m ) - 1 6 ( n ) - 6 ( o ) - B ( p ) - 2 7 a n d2 9 ( q ) - 1 9 a n d 1 Z ( r ) - 2 7 a n d1 0 ( s )- 1 8 ( 0 - 2 8 ( u ) - 2 4 ( v ) - 2 0 a n d 2 8 .
Curves Used in Engineering Practice 147
M i s c e l l a n e o u sp r o b l e m s : 44. ABC is an equilateraltriangle of side equal to Z0 mm. Trace the loci of the vertices A, B and C, when the circle circumscribinqz4BCrolls w i t h o u t s l i p p i n ga l o n g a f i x e d s t r a i g h tl i n e f o r o n e h a l f f u v o l u t i o n . 4 5 . T h e m a j o r - a x i sA B o f a n e l l i p s e i s 1 4 0 m m l o n g w i t h p a s i t s m i d _ point. The foci f1 and F2 of the ellipse are 4g mm away {rom the m i d - p o i n t P . D r a w t h e e l l i p s ea n d f i n d t h e l e n g t h o f t h e m i n o r a x e s . 46. Draw an ellipse by 'concentric circles method, and find the lensth of t h e m i n o r a x i s w i t h t h e h e l p o f t h e f o l l o w i n gd a t a : 1 i ) M a ; o r a x e s = 1 0 0 m m . ( i i ) D i s t a n c eb e t w e e nf o c i 8 0 m m . 4 7 . P Q i s a d i a m e t e ro f a c i r c l e a n d i s 7 5 m m l o n g . A p i e c e o f s t r i n g i s tied tightly round the circumferenceof the semi-circle startine from P a n d f i n i s h i n ga t Q . T h e e n d e i s t h e n u n t i e d a n d t h e s t r i n g ," a l w a y s kept taut/ is gradually unwound from the circle, until it lies ilong the tangent at P. Draw the curve traced by the moving extremity oi the s r nn 8 , 4 8 . A h a l f - c o n ei s s t a n d i n go n i t s h a l f b a s e o n t h e g r o u n d w i t h t h e triangular face parallel to the Vp. An inextensibleitring is passed round the half-conefrom a point on the periphery and biought back to the same point. Find the shortest length of the string. Take the base-circle diameter of the half-cone as 60 mm and height Zs mrn. 4 9 . O n e e n d o f a n i n e l a s t i ct h r e a d o f i 2 0 m m l e n g t h i s a t t a c h e dt o o n e corner of a regular hexagonaldisc having a side of 25 mm. Draw the curve traced out by the other end of the thread when it is completelv wound along the peripheryof the disc, keepingthe thread always tighi. 50. A circus man rides a motor-bike inside a globe of 6 metres diameter. The motor-bike has the wheel of -l metre diameter. Draw the locus of the point on the circumferenceof the motor-bike wheel for one complete revolution. Adopt suitable scale, 5 1 . A _f o u n t a i n j e t d i s c h a r g e sw a t e r f r o m g r o u n d l e v e l a t a n i n c l i n a t i o n of 45' to the ground. The jet travels a horizontal distance of 2.5 metre from_ the point of discharge and falls on the ground. Trace the path of the jet. Name the curve. 5 2 . A c o i n o f 3 5 m m d i a m e t e rr o l l s o v e r d i n n i n g t a b l e w i t h o u t s l i p p i n g . A point on the circumferenceof the coin is in contact with the iable surface in the beginning and after one complete revolution. Draw the curve traced by the point. Draw a tangent and a normal at any point on the curve. 53. Draw a rectangleof 130 mm x g5 mm. Draw two parabolasin it with their axes bisecting each other. 54. Two concentric discs of 40 mm and 50 mm diameters roll on the horizontal line ,,48 150 mm long. Both discs start at the same point and roll in the same direction. plot the curves for the movement of the points lying on their circumferences. 55. Draw a path of the end of string when it is wound on a circle o{ 4 0 c m d i a m e t e rw i t h o u t s l i p p i n g .T h e l e n g t h o f t h e s t r i n g i s 1 5 0 c m long. Name the curve and write practicaltpplications.
148
EngineerinB Dlawing
56. A circleof 60 mm diameterrollson a horizontallinefor a hal{revolution and then on a verticalline for anotherhalf revolution.Draw the of the circle. curvetracedout by a point P on the circumference of circle path point a circumference Draw the traced out by a on 57. when it rolls,without slip, on verticalsurface,for the distanceequalto the Derimeterof the circleof diameterof 40 mm. 58. Draw a helix o{ pitch equalto 50 mm upon a cylinderof 75 mm diameterand developthe surfaceof the cylinder.Assumethe startinS point to be on the verticalcentre line in the top view. 59. Draw the projectionsof a helix having a helix-angleof 30", on a cylinderof 75 mm diameter. 25 mm x 20 mm. cross-section 60, A spiralis madeof a wire of rectangular 100 mm, diameter of the Outside Draw two completeturns spring. insidediameter50 mm and oitch50 mm. 61. Projectone completeturn of a helical spring o{ outside diameter of wire being a circle 75 mm and pitch 50 mm, the cross-section of 20 mm diameter. of threecoilsof a helicalspringof steelwire 20 mm 62. Drawthe projections diameter.Outsidediameterof the spring100 mm and pitch 50 mm. 63. Projecttwo completeturns of a triangularthread,outside diameter 125 mm, pitch25 mm and angle60". 64. A screw having triple-startsquarethread has outside diameter 150 mm, lead 120 mm and pitch 40 mm. Draw its projections. 65. Draw a helix of one convolutionupon a cone, diameterof the base 75 mm, axis 100 mm long and pitch 75 mm. Takeapex as the startingpoint for the curve. 66. A point P, startingfrom the base-circleof a cone, reachesthe apex, while movingaroundthe axisthroughtwo completeturns.Assumingthe parallelto the axis)to be movementof P towardsthe apex(measured uniformwith its movementaroundthe axis,draw the proiectionsand the developmentof the surfaceof the cone showingthe path of P in each,Diameterof the baseof the cone 75 mm; axis 100 mm long. 67. A propellerscrewof 20 mm diameterhasa helicalbladeweldedon its surface.The diameterof the helix is 80 mm and the pitch is 50 mm. Projecttwo completeturns of the screwwith the blade. 68. Two objectsA and 8, 10 m aboveand 7 m below the ground level are observedfrom the top of a tower 35 m high from respectively, of 45' with the ground.Boththe objectsmakean angleof depression the horizon.The horizontaldistancebetweenA and I is 20 m. Draw to scale1:250, the projectionsof the objectsand the tower and find (a) the true distancebetweenA and B, and (b) the angleof depression of anotherobjectC situatedon the groundmidwaybetweenA and 8. 69. An ant is travellingon a cylindricalsurfacein a circumferential directionat a uniformangularspeedin the clockwisedirectionas at a uniformratein advances observedfrom the top and simultaneously the axialdirection.l{ the diameterof the cylinderis 60 mm and the ant moves80 mm in the axialdirectionduringone turn, tracethe path of the ant and namethe curve.
PROJECTIONS
@
OF POINTS
9-0. lntroduction:
A p o i n t m a y b e s i t u a t e d ,i n s p a c e ,i n a n y o n e o f t h e f o u r q u a d r a n t , formed by the two principal planes of projection or may lie in any one or both of them. lts projections are obtained by extending projector p e r p e n d i c u l atro t h e p l a n e s .
One of the planes is then rotated so that the first and third quadrant are opened out. The projections are shown on a flat surface in their respective positions either above or below or in xy.
9-1. A point is situated in the first quadrant:
T h e p i c t o r i a lv i e w [ f i g . 9 - 1( i ) ] s h o w s a p o i n t A s i t u a r e da b o v e t h e H . p a n d i n f r o n t o f t h e V . P . ,i . e . i n t h e f i r s t q u i d r a n t . a , i s i t s f r o n t v i e w a n d a the top view. After rotation of the plane, these projections will be seen a s . s h o w n i n f i g . 9 - 1 ( i i ) .T h e f r o n t v i e w a , i s a b o v e i y a n d t h e t o p v i e w a below it. The line joining a, and a (which also is called a projector intersects xy at right angles at a point o. It is quite evident from the pictorial view that ab : ,Aa, i.e. the distance o{ the front view from !r' = the distance of A from the H.p. viz. h. Similarly,ao = Aa,, i.e. the distance of the top view trom xy : the distanceof A from the Vp viz. d.
-! t'rr t !tr. L\
Frc.9-1
Ftc.9-2
Projections of Points 171
9.2. A point is situated in the secondquadrant: A point I (fig. 9-21 is above the H.P. and behind the V.P.,i.e. in the s e c o n d q u a d r a n t .b ' i s t h e f r o n t v i e w a n d b t h e t o p v i e w . When the planes are rotated, both the views are seen above xy. Note that b'o = B' and bo = 8b'.
9.3. A point is situated in the third quadrant: A p o i n t C ( f i g . 9 - 3 ) i s b e l o w t h e H . P .a n d b e h i n d t h e V P . , i . e . i n t h e t h i r d q u a d r a n t .l t s f r o n t v i e w c ' i s b e l o w x y a n d t h e t o p v i e w c a b o v e xy. Also c'o = Cc and co = Cc'.
h
ll l----l-
Ftc.9-4
Frc.9-5
9-4. A point is situated in the fourth quadrant: A point E (fig. 9-a) is below the H.P. and in front of the VP, i.e. Ee i n t h e f o u r t h q u a d r a n t .B o t h i t s p r o j e c t i o n sa r e b e l o w x y , a n d e ' o : and eo: Ee'. Referringto fig. 9-5, we see that, (i) A point M is in the H.P and in front of the VP. lts front view m' is in xy and the top view m below it. (ii) A point N is in the VP. and abovethe H.P.lts top view n is in xy and the front view n'above it. ( i i i ) A p o i n t O i s i n b o t h t h e H . P .a n d t h e V . P l t s p r o j e c t i o no a n d o ' c o i n c i d ew i t h e a c h o t h e r i n x y .
172
Engineering Dtawing
9-5. General conclusions:
(i) The line joining the top view and the front view of a point is always perpendicularto xy. lt is called a projector. (ii) When a point is above the H.P, its front view is above xy; when i t i s b e l o w t h e H . P , t h e f r o n t v i e w i s b e l o w x y . T h e d i s t a n c eo f a point from the H.P. is shown by the length of the projector from its front view to xy, e,g. a'o, b'o eic. (iii) When a point is in front of the VP., its top view is below xy; when it is behind the VP., the top view is above xy. The distance of a point from the V.P. is shown by the length of the projector from its top view to xy, e.g. ao, bo etc. (iv) When a point is in a referenceplane, its projection on the other reference plane is in xy. Probfem 9-1. (fig. 9-1\t A point A is 2.5 cm above the H.P and 3 cm in front of the V.P.Draw its projections. ( i ) D r a w t h e r e f e r e n c el i n e x y [ f i g . 9 - 1 ( i i ) ] . (ii) Through any point o in it, draw a perpendicular.
A s t h e o o i n t i s a b o v e t h e H . P . a n d i n f r o n t o { t h e V .P . i t s front view will be above xy and the top view below xy. ( i i i ) O n t h e p e r p e n d i c u l am r , a r k a p o i n t a ' a b o v e x y , s u c ht h a t a ' o = 2 , 5 S i m i l a r l y m , a r k p cm. a o i n t a b e l o wx y , s o t h a t a o = 3 c m . a ' a n d a are the required projections. Problem 9-2. (lig. 9-6)t A point A is 2 cm below the H,P and 3 cm behind the V.P Draw its projections. A s t h e p o i n t i s b e l o w t h e H . P .a n d behind the V.P, its front view will be below xy and the top view above xy. Draw the projectionsas explainedin p r o b l e m9 - l a n d a s s h o w n i n f i g . 9 - 6 .
Frc.9-6
ExerciseslX
1 . D r a w t h e p r o j e c t i o n so f t h e f o l l o w i n g p o i n t s o n t h e s a m e g r o u n d line, keeping the projectors 25 mm apart. A , i n t h e H . P .a n d 2 0 m m b e h i n dt h e V . P . B, 40 mm above the H.P. and 25 mm in front of the V.P C, in the V.P.and 40 mm above the H.P. D, 25 mm below the H.P.and 25 mm behind the V.P. E , 1 5 m m a b o v et h e H . P .a n d 5 0 m m b e h i n dt h e V .P . 4 40 mm below the H.P.and 25 mm in front of the V.P. C, in both the H.P and the V.P.
Protections of Points 17
2 . A point P is 50 mm from both the referenceplanes.Draw its projection i n a l l p o s s i b l ep o s i t i o ns .
t rs 'hen :of ctor xy; ince ctor ther cm
3 . S t a t e t h e q u a d r a n t si n w h i c h t h e f o l l o w i n g p o i n t s a r e s i t u a t e d : (a) A point P; its top view is 40 mm above xy; the front view, 20 mn below the top view. (b) A point Q, its projections coincide with each other 40 mrr . oelow xy. A
A point P is 15 mm abovethe H.P and 20 mm in front of the V,p A n o t h e rp o i n t Q i s 2 5 m m b e h i n dt h e V . p a n d 4 0 m m b e l o w t h e H . p Draw projections of P and Q keeping the distance between their p r o j e c t o r se q u a l t o 9 0 m m . D r a w s t r a i g h t l i n e s j o i n i n g ( i ) t h e i r t o p views and (ii) their front views.
Projectionsof various points are given in fig. 9-2. State the position 01 each point with respect to the planes of projection, giving th€ d i s t a n c e si n c e n t i m e t r e s .
its 2.5 rd a Ftc.9-7
-9
6 . T w o p o i n t sA a n d B a r e i n t h e H . p T h e p o i n t A i s 3 0 m m i n f r o n t o l the V.P.,while I is behind the V.p The distance between their p r o j e c t o r si s 7 5 m m a n d t h e l i n e j o i n i n g t h e i r t o p v i e w s m a k e s a n angle of 45' with xy. Find the distance of the point B from the V.p
7 . A p o i n t P i s 2 0 m m b e l o w H . p . a n d l i e s i n t h e t h i r d q u a d r a n t .l t s shortest distance from xy is 40 mm. Draw its proiections.
und
B . A point A is situated in the first quadrant. lts shortest distanc f r o m t h e i n t e r s e c t i o np o i n t o f H . R , V i a n d a u x i l i a r yp l a n e i s 6 0 m m a n d i t i s e q u i d i s t a n tf r o m t h e p r i n c i p a lp l a n e s .D r a w t h e p r o j e c t i o n o f t h e p o i n t a n d d e t e r m i n ei t s d i s t a n c ef r o m t h e p r i n c i p a lp l a n e s .
9 . A point 30 mm above xy line is the plan-view of two points p and e. The elevation of P is 45 mm above the H.p. while that of the point e is 35 mm below the H.P. Draw the pro.iectionsof the points and state their position with reference to the principal planes and the q u a d r a n ti n w h i c h t h e y l i e .
PROJECTIONS
.f,@
OF STRAIGHT
LINES
10-0. lntroduction:
A s t r a i g h t l i n e i s t h e s h o r l e s t d i s t a n c eb e t w e e n t w o p o i n t s . H e n c e , t h e p r o j e c t i o n so f a s t r a i g h tl i n e m a y b e d r a w n b y j o i n i n g t h e r e s p e c t i v p r o j e c t i o n so f i t s e n d s w h i c h a r e p o i n t s . T h e p o s i t i o n o f a s t r a i g h tl i n e m a y a l s o b e d e s c r i b e dw i t h r e s p e c tt o the two reference planes. lt may be: 1. 2. 3. 4. 5. 6. 7. 8.
P a r a l l e lt o o n e o r b o t h t h e p l a n e s . C o n t a i n e db v o n e o r b o t h t h e o l a n e s . P e r p e n d i c u l atro o n e o f t h e p l a n e s . I n c l i n e dt o o n e p l a n e a n d p a r a l l e lt o t h e o t h e r . I n c l i n e dt o b o t h t h e p l a n e s . P r o j e c t i o n so f l i n e s i n c l i n e dt o b o t h t h e p l a n e s . Line containedby a plane perpendicularto both the referenceplanes. T r u e l e n g t ho f a s t r a i g h tl i n e a n d i t s i n c l i n a t i o n sw i t h t h e r e f e r e n c Dl a n e s . 9 . T r a c e so f a l i n e . 1 0 . M e t h o d s o f d e t e r m i n i n gt r a c e s o f a l i n e . 1 1 . T r a c e so f a l i n e , t h e p r o j e c t i o n so f w h i c h a r e p e r p e n d i c u l atro x y . '12. P o s i t i o n so f t r a c e s o f a l i n e .
10-1. Line parallel to one or both the planes (fig.10-1):
'., "';1. i
Projections of StraiSht Lines 175
S F S
( a ) L i n eA B i s p a r a l l e tl o t h e H . P a and b are the top views of the ends A and B respectively. The line joining a and b is the top view of ,48. lt can be clearly seen that the figure ABba is a rectangle.Hence, the top view ab is equal to A8. a' b' is the front view o! AB and is parallel to xy. (b) Line CD is parallelto the V.P. T h e l i n e c ' d ' i s t h e f r o n t v i e w a n d i s e q u a lt o C D ; t h e t o p view cd is parallelto xy. ( c ) L i n e f F i s p a r a l l e lt o t h e H . P .a n d t h e V . P ef is the top view and e' f is the front view; both are equal t o E F a n d p a r a l l e lt o x r . H e n c e ,w h e n a l i n e i s p a r a l l e tl o a p l a n e ,i t s p r o j e c t i o no n t h a t p l a n e i s e q u a l t o i t s t r u e l e n g t h ;w h i l e i t s p r o j e c t i o no n t h e o t h e r p l a n e i s p a r a l l e l t o t h e r e f e r e n c el i n e .
10-2. Line contained by one or both the planes (fig.10-2): re5. nce
xy.
F r c .1 0 - 2 L i n e A B i s in the H.P. lts top view ab is equal to ,48; its front view a' b' is in xy, Line CD is in the VP lts front view c' d' is equal to CD; its top view cd is in xy. Line FF is in both the planes. lts front v i e w e ' f ' a n d ef coincide with each other in xy.
the top view
H e n c e , w h e n a l i n e i s c o n t a i n e db y a plane, its projection on that p l a n e i s e q u a l t o i t s t r u e l e n g t h ; w h i l e i t s p r o j e c t i o no n t h e o t h e r p l a n e i s i n t h e r e f e r e n c el i n e .
'l76
EngineerinB Drawing
10-3. Line perpendicularto one of the planes (fig.10-3): Whena line is perpendicular plane,it will be paral to one reference to the other. o
b
(Third-angleprojection) Ftc.10-3
(a) Line AB is perpendicularto the H.P. The top views of its end c o i n c i d ei n t h e p o i n t a . H e n c e ,t h e t o p v i e w o f t h e l i n e A B i s t h point a. Its front view a' b' is equal to AB and perpendicularto xy (b) Line CD is perpendicular to the VP. The point d' is its fron view and the line cd is the top view. cd is equal to CD an perpendicularto xy.
H e n c e , w h e n a l i n e i s p e r p e n d i c u l atro a p l a n e i t s p r o j e c t i o no n t h a p l a n e i s a p o i n t ; w h i l e i t s p r o j e c t i o no n t h e o t h e r p l a n e i s a l i n e e q u a l t t its true length and perpendicularto the reference line.
ln firct-angle projection method, when top views of two or mor points coincide, the point which is comparativelyfarther away from xy ir t h e f r o n t v i e w w i l l b e v i s i b l e ; a n d w h e n t h e i r f r o n t v i e w s c o i n c i d e ,t h a which is farther away from xy in the top view will be visible.
ln third-angle projection method, it is just the reverse. When to1 views of two or more points coincide the point which is comparative nearer xy in the front view will be visible; and when their front view c o i n c i d e ,t h e p o i n t w h i c h i s n e a r e tx y i n t h e t o p v i e w w i l l b e v i s i b l e .
10-4. Line inclined to one plane and parallel to the other:
The inclination of a line to a plane is the angle which the line make with its prcjection on that plane,
( a ) L i n e P Q 1 t f i g . 1 0 - 4 ( i ) l i s i n c l i n e da t a n a n g l e 0 t o t h e H . P a n is parallel to the V.P. The inclination is shown by the angle { which PQ1 makes with its own projection on the H.P., viz. thr top view pgt .
The projections [fig. 10-4(ii)l may be drawn by first assumingthe lirx to be parallel to both the H.P. and the V.P, tts front view p, q, and th top view pq will both be parallel to xy and equal to the true length
Projections of StraiSht Lines 1
arallel
:s enos f is the r to xY. s front lD and on thal r q u a lt o rr mofe mxyin de, thal
W h e n t h e l i n e i s t u r n e d a b o u t t h e e n d P t o t h e p o s i t i o nP Q i s o t h a t m a k e st h e a n g l e 0 w i t h t h e H . P w h i l e r e m a i n i n gp a r a l l e lt o t h e V P , i n t l front view the point g'will move along an arc drawn with p'as cent a n d p ' q ' a s r a d i u st o a p o i n t g 1 ' s o t h a t p ' q 1 ' m a k e st h e a n g l e0 w i x y . I n t h e t o p v i e w q w i l l m o v e t o w a r d sp a l o n g p g t o a p o i n t g t o n t l projectorthrough q1'. p' q1' and pgl are the front view and the t( v i e w r e s p e c t i v e l yo f t h e l i n e P Q 1 .
(j)
Ql)
F t c .1 0 - 4
( b ) L i n e R S 1[ f i g . 1 0 - 5 ( i ) ] i s i n c l i n e da t a n a n g l e o t o r n e V P . a r , d parallel to the H.P. The inclinationis shown by the angle which R5.1makes with its projection on the VP., viz. the fro v i e w r ' s 1 ' . A s s u m i n gt h e l i n e t o b e p a r a l l e lt o b o t h t h e H . P .a t h e V . P , i t s p r o j e c t i o n sr ' s ' a n d r s a r e d r a w n p a r a l l e lt o x y a equal to its true length lfig. 10-5(ii)1.
hen top arative\ 'It view! ' i s i bl e .
H . P .a n Ftc. 10-5 , the li and t e lengt
W h e n t h e l i n e i s t u r n e d a b o u t i t s e n d R t o t h e o o s i t i o nR so that it makes the angle o with the V.P. while remaining para to the H.P, in the top view the point s will move along an z drawn with r as centre and rs as radius to a point s, so that /
1 7 8 E n g i n e e r i n gD r a w i n g
makes the angle o with xy. ln the front view, the point s, will move towards r' along the line r, s, to a point s1, on the projecto t h r o u g hs 1 . r s 1 a n d r ' s 1 ' a r e t h e p r o j e c t i o n os f t h e l i n e R S 1 .
T h e r e f o r e ,w h e n t h e I i n e i s i n c l i n e d t o t h e H . p a n d p a r a l l e lt o t h e V.P, its top view is shorter than its true length. but parallel to xy; its front view is equal to its true length and is inclined to x}/ at its tru€ i n c l i n a t i o nw i t h t h e H . P . A n d w h e n t h e l i n e i s i n c l i n e d t o t h e V . p a n d parallel to the H.P., its front view is shorter than its true length but parallel to xy; its top view is equal to its true length and is inclined to xy at its true inclination with the Vp. Hence, when a line is inclined to one plane and parallel to the other its ptojection on the plane to which it is inclined, is a line shorter than its true length but parallel to the reference line. lts projection on the plane to which it is patallel, is a line equal to its true length and inclined to the referenceline at its true inclination.
ln other words, the inclination of a line with the H.p. is seen in the front view and that with the V.p is seen in the top wew. Probfem 10-1. (fig. 10-6): A line PQ, 9 cm long, is in the H.p. and makes an angle of 30'with the V.p lts end p is 2.i cm in front of the V.P Draw its projections. As the line is in the H.P, its top view will s h o w t h e t r u e l e n g t ha n d t h e t r u e i n c l i n a t i o nw i t h the V.P lts front view will be in xy. (i) Mark a point p, the top view 2.5 cm below xy. Draw a line pq equalto 9 cm a n d i n c l i n e da t 3 0 ' t o x y . (ii) Projectp to p' and q lo q' on xy.
Frc.10-6 pq andp'q'are the requiredtop view andfront view respective
Probfem 10-2. (fig. 1O-7): The length of the top view of a line parallel to the VP and inclined at 45" to the H.P is 5 cm. One end of the line is 1.2 cm above the H.P and 2.5 cm in frcnt of the V.P. Draw the projections of the line and determine its true length. As the line is parallel to the Vp., its top view will be parallel to xy and the front view will show its true lengthand the true inclination with the H.P.
Ftc. 10-7
Projections oI Straight Lines 179
ivill :tor the its : r ue and DUI
ned her, han iane the tin and the ---l-{r
tI l
I I
Il
\q
vely. t;
b
,1.2 awa ;the also
Probfem 10-3. (fig. 10-B): Ihe front view of a 7.5 cm long line measures5.5 cm. The line is parallel to the H.P and one of its ends is in the V.p and 2.5 cm above the H.P. Draw the projections of the line and detetmine its inclination with the V.P. A s t h e l i n e i s p a r a l l e lt o t h e H . P . ,i t s f r o n t v i e w w i l l b e p a r a l l e lt o x y . (i)
M a r k a , t h e t o p v i e w o f o n e e n d i n x y , a n da ' , i t s f r o n t v i e w 2 . 5 c m a b o v ex y .
Ftc. 10-B ( i i ) D r a w t h e f r o n t v i e w a ' b ' , 5 . 5 c m l o n g a n d p a r a l l e tl o x y . W i t h a a s c e n t r ea n d r a d i u se q u a lt o 7 . 5 c m , d r a w a n a r c c u t t i n gt h e p r o j e c t o r throughb' at b. Join a with b. ab is the top view of the line. lts i n c l i n a t i o nw i t h x y , v i z . o i s t h e i n c l i n a t i o no f t h e l j n e w i t h t h e V . P
X(a) Exercises
1 . D r a w t h e p r o j e c t i o n so f a 7 5 m m l o n g s t r a i g h tl i n e , i n t h e f o l l o w i n g p o s i t i o ns : ( a ) ( i ) P a r a l l etlo b o t h t h e H . P .a n d t h e V .P .a n d 2 5 m m f r o m e a c h . ( i i ) P a r a l l etlo a n d 3 0 m m a b o v et h e H . P a n d i n t h e V . P ( i i i ) P a r a l l etl o a n d 4 0 m m i n f r o n t o f t h e V .p . a n d i n t h e H . p ( b ) ( i ) P e r p e n d i c u l a tro t h e H . P . , 2 0 m m i n f r o n t o f t h e V p a n d i t s o n e e n d 1 5 m m a b o v et h e H . R ( i i ) P e r p e n d i c u l atro t h e V . P . ,2 5 m m a b o v e t h e H . P . a n d i t s one end in the VP ( i i i ) P e r p e n d i c u l at or t h e H . P , i n t h e V .P .a n d i t s o n e e n d i n t h e H . p . ( c ) ( i ) I n c l i n e da t 4 5 ' t o t h e V . P ,i n t h e H . p .a n d i t s o n e e n d i n t h e V p . ( i i ) I n c l i n e da t 3 0 ' t o t h e H . C a n d i t s o n e e n d 2 0 m m a b o v e i t ; p a r a l l e lt o a n d 3 0 m m i n f r o n t o { t h e V . p ( i i i ) I n c l i n e da t 6 0 " t o t h e V P . a n d i t s o n e e n d 1 5 m m i n f r o n t o f i t ; p a r a l l e lt o a n d 2 5 m m a b o v et h e H . p 2 . A 1 0 0 m m l o n g l i n e i s p a r a l l e lt o a n d 4 0 m m a b o v e t h e H . p . l t s two ends are 25 mm and 50 mm in front of the V.p resoectivelv D r a w i t s p r o j e c t i o n sa n d f i n d i t s i n c l i n a t i o nw i t h t h e V p 3 . A 9 0 m m l o n g l i n e i s p a r a l l e lt o a n d 2 5 m m i n f r o n t o f t h e V p . Its one end is in the H.P while the other is 50 mm above the H.p. D r a w i t s p r o j e c t i o n sa n d f i n d i t s i n c l i n a t i o nw i t h t h e H . p . 4 . T h e t o p v i e w o f a 7 5 m m l o n g l i n e m e a s u r e s5 5 m m . T h e l i n e i s i n t h e V P . , i t s o n e e n d b e i n g 2 5 m m a b o v et h e H . p . D r a w i t s p r o j e c t i o n s 5 . T h e f r o n t v i e w o f a l i n e , i n c l i n e da t 3 0 " t o t h e V p i s 6 5 m m l o n e . D r a w t h e p r o j e c t i o n so f t h e l i n e , w h e n i t i s p a r a l l e lt o a n d 4 0 m m a b o v e t h e H . P . ,i t s o n e e n d b e i n g 3 0 m m i n f r o n t o f t h e V p 6 . A v e r t i c a ll i n e A 8 , 7 5 m m l o n g , h a s i t s e n d A i n t h e H . p a n d 2 5 m m in front of the V.RA line AC, i00 mm long, is in the H.p and parallel t o t h e V . P D r a w t h e p r o j e c t i o n so f t h e l i n e . i o i n i n gI a n c l C , a n d d e t e r m i n ei t s i n c l i n a t i o nw i t h t h e H . p
1 8 0 E n g i n e e j i n gD t a w i n g
7. Two pegs fixed on a wall are 4.5 metres apart. The distance betwee t h e p e g s m e a s u r e dp a r a l l e tl o t h e f l o o r i s 3 . 6 m e t r e s .l f o n e p e g i s 1 . metres above the floor, find the height of the second peg and thr inclination of the line joining the two pegs/ with the floor. L
Draw the projections of the lines in Exercises1 to 6, assuming then t o b e i n t h e t h i r d q u a d r a n t ,t a k i n g t h e g i v e n p o s i t i o n st o b e b e l o t h e H . P . i n s t e a do f a b o v e t h e H . P . ,a n d b e h i n d t h e V . P , i n s t e a do f i front of the VP.
10-5. Line inclined to both the planes:
( a ) A l i n e A B ( f i g . 1 0 - 9 ) i s i n c l i n e da t 0 t o t h e H . R a n d i s p a r a l l e lt r t h e V P . T h e e n d A i s i n t h e H . P .A B i s s h o w n a s t h e h y p o t e n u o f a r i g h t - a n g l e dt r i a n g l e ,m a k i n g t h e a n g l e 0 w i t h t h e b a s e . ,[',)o
,lB;
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Ftc.10-9
The top view ab is shorter than AB and parallel to xy. The fron view a' b' is equal to AB and makes the angle 0 with xy
Keeping the end A fixed and the angle 0 with the H.P. constan i f t h e e n d I i s m o v e d t o a n y p o s i t i o n ,s a y 8 1 , t h e l i n e b e c o m i n c l i n e dt o t h e V P . a l s o .
ln the top view, b will move along an arc, drawn with a as centn a n d a b a s r a d i u s ,t o a p o s i t i o nb 1 . T h e n e w t o p v i e w a 6 1 i s e q u to ab but shorter than 248.
In the front view, b' will move to a point b1' keeping its distan( f r o m x y c o n s t a n ta n d e q u a l t o b ' o ; i . e . i t w i l l m o v e a l o n g t h e l i r p q , d r a w n t h r o u g h b ' a n d p a r a l l e lt o x y . T h i s l i n e p q i s t h e l o c { or path of the end I in the front view.b1'will lie on the project t h r o u g hb 1 . T h e n e w f r o n t v i e w a ' b f i s s h o r t e rt h a n a ' b ' ( i . e .A { and makes an angle cc with xy. o< is greater than e.
(i) (ii)
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( b ) T h e s a m e l i n e A B ( f i g . 1 0 - 1 0 ) i s i n c l i n e da t o t o t h e V P a n d i s p a r a l l e l t o t h e H . P . l t s e n d A i s i n t h e V .P . A B i s s h o w r r a s t h e h y p o t e n u s eo f a r i g h t - a n g l e dt r i a n g l e m a k i n g t h e a n g l e O with the base.
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The front view a' b2' is shorter than ,48 and parallel to xy. The top view ab2 is equal to AB and makes an angle o with xy. Keeping the end A fixed and the angle o with the Vp. c o n s t a n t ,i f I i s m o v e d t o a n y p o s i t i o n , s a y 8 3 , t h e l i n e w i l l b e c o m e i n c l i n e dt o t h e H . P a l s o . fron' stant :omel tenrr( e qu a ;tanc{ e lint locur rJeCto e. AB
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ln the front view,b2', will move alongthe arc, drawn with a, a s c e n t r e a n d a ' b 2 ' a s r a d i u s ,t o a p o s i t i o nb 3 , . T h e n e w f r o n t view a' b3' is equal to a' br' but is shorter than AB.
ln the top view,b2 will move to a point b, alongthe line rs, d r a w n t h r o u g h b 2 a n d p a r a l l e lt o x f , t h u s k e e p i n g i t s d i s t a n c e from the path of a, viz. b2 o constant. rs is the locus or path of the end I in the top view. The point b3 lies on the projector through b3'. The new top view ab3 is shorterthan ab2 (i.e. AB) and makes an angle 0 with xy. B is greater than o. Here also we find that, as long as the inclinationof AB w i t h t h e V .P . d o e s n o t c h a n g e ,e v e n w h e n i t b e c o m e si n c l i n e dt o the H.P. ( i ) i t s l e n g t h i n t h e f r o n t v i e w , v i z . a ' b 2 ' r e m a i n sc o n s t a n t ; and ( i i ) t h e d i s t a n c eb e t w e e n t h e p a t h s o f i t s e n d s i n t h e t o p v i e w v i z . b 2 o r e m a i n sc o n s t a n t .
Hence, when a Iine is inclined to both the planes,its projections a r e s h o r t e r t h a n t h e t r u e l e n g t h a n d i n c l i n e dt o x y a t a n g l e sg r e a t e rt h a n t h e t r u e i n c l i n a t i o n s .T h e s e a n g l e s v i z . c r a n d p a r e c a l l e d a p p a r e n t a n s l e so f i n c l i n a t i o n .
1 8 2 E n g i n e e r i n gD r a w i n g
10-6. Projectionsof lines inclined to both the planes:
F r o m A r t . 1 0 - 5 ( a )a b o v e ,w e f i n d t h a t a s l o n g a s t h e i n c l i n a t i o no f 4 1 w i t h t h e H . P . i s c o n s t a n t( i ) i t s l e n B t hi n t h e t o p v i e w , v i z . a b r e m a i n c o n s t a n t /a n d ( i i ) i n t h e f r o n t v i e q t h e d i s t a n c eb e t w e e nt h e l o c i o f i t e n d s ,v i z . b ' o r e m a i n sc o n s t a n l . In other words if (i) its length in the top view is equal to ab, an ( i i ) t h e d i s t a n c eb e t w e e nt h e p a t h s o f i t s e n d s i n t h e f r o n t v i e w i s e q u a t o b ' o , t h e i n c l i n a t i o no f A B w i t h t h e H . P . w i l l b e e q u a l t o o . Similarly, from Art. 10-5(b) above, we find that as long as th i n c l i n a t i o no f A B w i t h t h e V . P .i s c o n s t a n t( i ) i t s l e n g t h i n t h e f r o n t v i e w viz. a' b2' remains constant, and (ii) in the top view, the distanc b e t w e e n t h e l o c i o f i t s e n d s , v i z . b 2 o r e m a i n sc o n s t a n t .T h e r e v e r s eo this is also true, viz. (i) if its lengthin the front view is equalLo a' b2 a n d ( i i ) t h e d i s t a n c eb e t w e e n t h e p a t h s o f i t s e n d s i n t h e t o p v i e w i e q u a l t o b 2 o , t h e i n c l i n a t i o no i 4 8 w i t h t h e V . p w i l l b e e q u a l t o o . C o m b i n i n gt h e a b o v et w o f i n d i n g s w , e c o n c l u d et h a t w h e n A B i s i n c l i n e at 0 to the H.P. and at o to the V.P. (i) its lengths in the top view an t h e f r o n t v i e w w i l l b e e q u a l t o a b a n d a ' b 2 ' r e s p e c t i v e l ya, n d ( i i ) t h distancesbetween the paths of its ends in the front view and the top viev will be equal to b' o and b2 o respectively.The two lengthswhen arrange with their ends in their respective paths and in projections with eac o t h e r w i l l b e t h e p r o j e c t i o n so f t h e l i n e A 8 , a s i l l u s t r a t e di n p r o b l e m 1 0 - 4 Probfem 1O-4, Civen the line AB, its inclinationsg with the H.P an g with the VP and the position of one end A. To draw its projections Mark the front view a' and the top view a according to the give position of A (fig. 10-12\. L e t u s f i r s t d e t e r m i n et h e l e n g t h so f A B i n t h e t o p v i e w a n d t h e f r o n view and the paths of its ends in the front view and the top view. ( i ) A s s u m e A B t o b e p a r a l l e lt o t h e V P . a n d i n c l i n e d a t 0 t o t h H.P /48 is shown in the pictorial view as a side of the trapezo A B b a l f i g . 1 0 - 1 1 ( i ) 1 .D r a w t h e f r o n t v i e w a ' b ' equal to AI t f i g . 1 0 - 1 2 ( i ) l a n d i n c l i n e d a t 0 t o x y . P r o j e c tt h e t o p v i e w a l parallelto xy. Througha' and b', draw lines cd and pq respective parallel to xy. ab is the length of AB in the top view and, c{ and pg are the paths of A and I respectivelyin the front viev ( i i ) A g a i n , a s s u m eA B 1 ( e q u a l t o A B ) t o b e p a r a l l e lt o t h e H . p . a n i n c l i n e da t 6 t o t h e V P . I n t h e p i c t o r i a lv i e w t f i g . i 0 - 1 1 ( i i ) J ,A B l i s h o w n a s a s i d e o f t h e t r a p e z o i dA B 1b 1 , a , . D r a w t h e t o p v i e w a b e q u a l t o A B t f i g . 1 0 - 1 2 ( i i ) Ja n d i n c l i n e da t s t o x y . P r o j e c t h e f r o r v i e w a ' b 1 ' p a r a l l e lt o x y . T h r o u g ha a n d b 1 , d r a w l i n e s e f a n d r respectivelyparallelto xy. a' bf is the lengih ofAB in the front vie{ and, ef and rs are the paths of A and I respectivelyin the top viev We may now arrange (i) ab (the length in the top viev b e t w e e ni t s p a t h s e f a n d r s , a n d ( i i ) a ' b 1 , ( t h e l e n g t h i n t h e f r o r view) between the paths cd and pq, keeping theri in projectid with each other, in one of the following fwo ways: I
Proiections of Strai8hl tines
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b c o m e s o n t h e p a t h r s , t h e l i n e A B w i l l b e c o m e i n c l i n e da r g r o t h e V . P T h e r e f o r ew , i t h a a s c e n t r e t f i g . 1 o - 1 2 ( i ) la n d r a d i u se q u a l to ab, draw an arc cutting rs at a point b2. project b2 to b2, on the. path pg. Draw lines joining a with b2, and a, with b2,. ab2 a n d a ' b 2 ' a r e t h e r e q u i r e dp r o j e c t i o n sC . h e c kt h a t a , b 2 ' = a , b { .
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F t c . 10 - 1 2 ( b ) S i m i l a r l y ,i n c a s e ( i i ) i f i g . 1 0 - 1 1 ( i i ) 1 i,f t h e s i d e 8 1 6 f i s t u r n e d about Aa' till bf is on the path pq, the line A8.' will become i n c l i n e da t e r o t h e H . p H e n c e ,w i t h a ' a s c e n t r e [ f i g . 1 0 - 1 2 ( i i ) ]a nd r a d i u se q u a l t o a ' b 1 ' , d r a w a n a r c c u t t i n g p q a t a p o i n t b 2 , Project b2' to b2 in the top view on the path rs. D r a w l i n e s j o i n i n g a w i t h b 2 , a n d a , w i t h b2'. ab2 and a' b2' are t h e r e q u i r e dp r o i e c l i o n sC . h e c kt h a t a b , = a b .
| 84
t n g i n e e r i n g Dr a w i n g
F i g . 1 0 - ' 1 3s h o w s ( i n p i c t o r i a la n d o r t h o g r a p h i cv i e w s ) t h e p r o j e c t i o n obtained with both the above steps combined in one figure and as d e s c r i b e db e l o w . F i r s t , d e t e r m i n e( i ) t h e l e n g t h a b i n the top view and the path pg in t h e f r o n t v i e w a n d ( i i ) l h e l e n g t ha ' b l ' in the front view and the path rs in the top view. Then, wlth a as centre a n d r a d i u s e q u a l t o a b , d r a w a n a r c c u t t i n g r s a t a p o i n t b 2 . W i t h a ' a s c e n t r e a n d r a d i u se q u a l t o a ' b 1 ' d r a w a n a r c c u t t i n gp q a t a p o i n t b 2 ' .
F t c .1 0 - 13 j o i n i n g Draw lines a with b2 and a, with b2,. ab2 and a,b2,are the r e q u i r e dp r o j e c t i o n sC . h e c kt h a t b 2 a n d b 2 , l i e o n t h e s a m e p r o i e c t o r . I t i s q u i t e e v i d e n tf r o m t h e f i g u r et h a t t h e a p p a r e n ta n g l e so f i n c l i n a t i o a and b are greater than the true inclinations0 and o respectively.
1 0 - 7 .L i n e c o n t a i n e d b y a p l a n e p e r p e n d i c u l a r t o b o t h t h e reference planes: As the two referenceplanes are at right angles total of the inclinationo s f a linewith the two planes b e m o r e t h a n 9 0 ' . W h e n 0 + a = 9 O ' ,t h e l i n e w i l l p l a n e c a l l e dt h e p r o f i l e p l a n e ,p e r p e n d i c u l at ro b o t h
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Proiections of Straight Lines 185
AS
:n .'s an 1,
A l i n e f F ( f i g . 1 0 - 1 4 ) , i s i n c l i n e da t 0 t o t h e H . P . a n d a t o [ e q u a l t o ( 9 0 ' - 0 ) l t o t h e V P T h e l i n e i s t h u s c o n t a i n e db y t h e p r o f i l e p l a n e marked P.P. T h e f r o n t v i e w e ' f ' a n d t h e t o p v i e w e f a r e b o t h p e r p e n d i c u l a tr o xy and shorter than fF. Therefore, when a line is inclined to both the reference planes and c o n t a i n e db y a p l a n e p e r p e n d i c u l a tro t h e m , i . e . w h e n t h e s u m o f i t s i n c l i n a t i o n sw i t h t h e H . P a n d t h e V . P i s 9 0 ' , i t s p r o j e c t i o n sa r e p e r p e n d i c u l atro x y a n d s h o r t e r t h a n t h e t r u e l e n g t h .
'l0-8. True length of a straight line and its inclinationswith the referenceplanes:
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When projections of a line are given, its true length and inclinations w i t h t h e p l a n e sa r e d e t e r m i n e db y t h e a p p l i c a t i o no f t h e f o l l o w i n gr u l e : When a line is parallel to a plane, its projection on that plane will show its true length and the true inclination with the other plane. T h e l i n e m a y b e m a d e p a r a l l e lt o a p l a n e ,a n d i t s t r u e l e n g t ho b t a i n e d by any one of the following three methods: Method I: M a k i n g e a c h v i e w p a r a l l e l t o t h e r e f e r e n c el i n e a n d p r o j e c t i n g the other view from it. This is the exact reversal of the processes a d o p t e di n A r t s . 1 0 - 5 ( a )a n d 1 0 - 5 ( b )f o r o b t a i n i n gt h e p r o j e c t i o n s . Method ll: R o t a t i n gt h e l i n e a b o u t i t s p r o i e c t i o n st i l l i t l i e s i n t h e H . P o r i n thc VP Method lll: Projectingthe views on auxiliary planes parallelto each view. ( T h i s m e t h o d w i l l b e d e a l tw i t h i n c h a p t e rX t ) . The following problem shows the application of the first two methods a n d p r o b l e m 1 0 - 2 9 a n d p r o b l e m1 0 - 3 1 s h o w a p p l i c a t i o no f t h i r d m e t h o d . Problem 10.5. The top view ab and the front view a' b' of a line AB are given. To determine its true length and the inclinations with the H.P. and the V.P. Method I: F i g . 1 0 - 1 5 ( i )s h o w sA B t h e l i n e , a ' b ' i t s f r o n t v i e w a n d a b i t s t o p view. lf the trapezoidz48bais turned about /4a as axis, so that AB becomes p a r a l l e lt o t h e V . P , i n t h e t o p v i e q b w i l l m o v e a l o n g a n a r c d r a w n w i t h c e n t r ea a n d r a d i u se q u a l t o a b , t o b 1 , s o t h a t a b 1 i s p a r a l l e lt o x y . I n t h e f r o n t v i e w , b ' w i l l m o v e a l o n g i t s l o c u sp q , t o a p o i n t b l ' o n t h e p r o . i e c t o throughb1. ( i ) T h e r e f o r e ,w i t h c e n t r e a a n d r a d i u s e q u a l t o a b t f i g . 1 0 - 1 5 ( i i ) 1 draw an arc to cut ef at b.. ( i i ) D r a w a p r o j e c t o rt h r o u g hb l t o c u t p q ( t h e p a t h o f b , ) a t b 1 , . (iii) Draw the line a' 61' which is the true length of 48. The angle 0 , w h i c h i t m a k e s w i t h x y i s t h e i n c l i n a t i o no f A B w i t h t h e H . p .
1 8 6 E n g i n e e r i n gD r a w i n g
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A g a i n , i n f i g . 1 0 - 1 6 ( i )A B i s s h o w n a s a s i d e o f a l r a p e z o i dA B b ' a ' . lf the trapezoid is turned about Aa' as axis so that AB is parallel to t h e H . P . ,t h e n e w t o p v i e w w i l l s h o w i t s t r u e l e n g t h a n d t r u e i n c l i n a t i o with the V.P.
( i ) W i t h a ' a s c e n t r e a n d r a d i u s e q u a l t o a ' b ' l f i g . 1 0 - 1 6 ( i i ) 1d, r a w ^,,+ -) Lur LU
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( i i ) D r a w a p r o j e c t o rt h r o u g hb 2 ' t o c u t / s ( t h e p a t h o l b ) a t b 2 . (iii) Draw the line ab2, which is the true length of AB. The angle o which it makes with xy is the inclination of AB with the V.P.
(i)
(t) Frc.10-16
F i g . 1 0 - 17 ( i ) s h o w s t h e a b o v et w o s t e p sc o m b i n e di n o n e f i g u r e .
T h e s a m e r e s u l t s w i l l b e o b t a i n e d b y k e e p i n gt h e e n d I f i x e d a n d t u r n i n g t h e e n d A [ f i g . 1 0 - 1 7 ( i ) ] , a s e x p l a i n e db e l o w . (i)
With centre b and radius equal to ba, draw an arc cutting rs at a; (thus making ba parallel to xy).
I
I
I
,.1'
P.ojection. of straight Lines 187
(ii) Project a1 to a1' on cd (the path of a,) a1' b, is the true length a n d 0 i s t h e t r u e i n c l i n a t i o no f A B w i t h t h e H . p (i i i ) S i m i l a r l yw , i t h c e n t r eb ' a n d r a d i u se q u a l t o b , a ' , d r a w a n
arc cuttingpq at a2'. (iv) P r o j e c ta 2 ' t o a 2 o n e f ( t h e p a t h o f a ) . a 2 b i s
true length
a n d o i s t h e t r u e i n c l i n a t i o no f A B w i t h t h e V p .
a. to ion
F l c . 1 0 -l 7 Method II: Referringto the p i c t o r i a lv i e w i n f i g . 1 0 - 1 8 ( i )w e f i n d t h a t A B i s t h e line, ab its top view and a' b' its front view.
F t c .1 0 - 1 8 f n t h e t r a p e z o i dA B b ' a ' ( i ) a ' A and b' B are both perpendicularto a' b' and are respectivelyequal to ao1 and bo2 (the distancesof a and b from xy in the top view), and (ii) the angle between AB and a' b' is the angle of inclination O ol AB with the V.P
1 8 8 IngineerinBDrawing
A s s u m et h a t t h i s t r a p e z o i di s r o t a t e da b o u t a , b , , t i l l i t l i e s i n t h e V . p
I n t h e o r t h o g r a p h i cv i e w [ f i g . 1 0 - t S ( i i ) J ,t h i s t r a p e z o i d i s o b t a i n e d by drawing perpendicularsto a' b', viz. a, A1 (equal to aot) and b, B.l (equalto bo2) and then joining A1 with 81. The line 41 81 is the true I e n g t h o f A B a n d i t s i n c l i n a t i o na w i t h a , b , i s t h e i n c l i n a t i o no f , 4 8 with the VP. S i m i l a r l y ,i n t r a p e z o i dA B b a i n f i g . 1 0 - 1 9 ( i ) ,A B i s r h e l i n e a n d a b its top view. Aa and Bb are both perpendicularto ab and are respectivel e q u a l t o a ' o 1 a n d b ' o 2 ( t h e d i s t a n c e so f a , a n d b , f r o m x y i n t h e f r o n t v i e w ) . T h e a n g l e 0 b e t w e e n A B a n d a b i s t h e i n c l i n a t i o no f A B with the H.P.
(ri)
(.i)
F r c .1 0 - 1 9 This figure may now be assumed to be rotated about ab as axis, so that it lies in the H.P.
I n t h e o r t h o g r a p h i cv i e w [ f i g . l O - 1 9 ( i i ) 1 ,t h i s t r a p e z o i d i s o b t a i n e d b y e r e c t i n gp e r p e n d i c u l a rtso a b , v i z . a A 2 e q u a l t o a , o 1 a n d b B 2 e q u a l t o b ' o 2 a n d . j o i n i n g4 2 w i t h 8 2 . T h e l i n e A 2 8 2 i s t h e t r u e l e n g t h o f A B a n d i t s i n c l i n a t i o n0 w i t h a b i s t h e i n c l i n a t i o no f A B w i t h t h e H . p Note.' The perpendiculars on ab or a, b, can also be drawn on its other side assumingthe trapezoidto be rotated in the oppositedirection.
10-9. Tracesof a line:
W h e n a l i n e i s i n c l i n e dt o a p l a n e , i t w i l l m e e t t h a t p l a n e , p r o d u c e d i f n e c e s s a i yT. h e p o i n t i n w h i c h t h e l i n e o r l i n e - p r o d u c e dm e e t s t h e p l a n e is called its trace. T h e p o i n t o f i n t e r s e c t i o no f t h e l i n e w i t h t h e H . p . i s c a l l e d t h e horizontal trace, usually denoted as H.T. and that with the V.p. is called the vertical trace or V.F.
\t'
Proiections of Straight Lines 189
Referto fig. 10-20.
( i ) A l i n e A B i s p a r a l l e lt o t h e H . P a n d t h e V . p l t h a s n o t r a c e . ( i i ) A l i n e C D i s i n c l i n e d t o t h e H . p a n d p a r a l l e lt o t h e V .p . I t h a s o n l y t h e H . T .b u t n o V T . ( i i i ) A l i n e E F i s i n c l i n e d t o t h e V . p a n d p a r a l l e lt o t h e H . p t t has only the V.T.but no H.T. T h u s ,w h e n a l i n e i s p a r a l l e tl o a p l a n e i t h a s n o t r a c e u p o n t h a t p l a n e .
1ir,5 ::,
;;;";s z
(t
(.it) Frc.10-20
z
7z ."
(rrl) F t c . 10 - 2 1
Referto fig. l0-21.
ed to 4B
( i ) A I i n e P Q i s p e r p e n d i c u l atro t h e H . p . t t s H . T . c o i n c i d e sw i t h its top view which is a point. lt has no V.T. ( i i ) A l i n e R S i s p e r p e n d i c u l a tro t h e V .p . l t s V . T . c o i n c i d e sw i t h i t s front view which is a point. lt has no H.T. H e n c e , w h e n a l i n e i s p e r p e n d i c u l a rt o a p l a n e , i t s t r a c e o n t h a t . plane coincides with its projection on that plane. lt has no trace on the other olane. Refer to fie. 10-22.
:eo Ine
J^'1.
the t5
Ftc.1O-22
1 9 0 E n g i n e e r i n gD r a w i n g
( i ) A l i n e A B h a s i t s e n d A i n t h e H . P a n d t h e e n d B i n t h e V . P .l H . T . c o i n c i d e sw i t h a t h e t o p v i e w o f A a n d t h e V T . c o i n c i d with 6'the front view of 8.
(ii) A line CD has its end C in both the H.P and the VP. Its H. and V.T. coincide with c and c' (the projections ol C) in xy.
Hence, when a line has an end in a plane, its trace upon that plar coincides with the projection of that end on that plane.
10-10.Methods of determiningtracesof a line: Method I:
Fig. 10-23(i) shows a line ,48 inclined to both the referen p l a n e s .I t s e n d A i s i n t h e H . P a n d I i s i n t h e V .P . a' b' and ab are the front view and the top view respectiv Ifig. 10-23(ii)1.
T h e H . T . o f t h e l i n e i s o n t h e p r o j e c t o rt h r o u g h a ' a n d c o i n c i d with a. The V.T. is on the projector through b and coincides with b'.
(rl)
(j)
Ftc. 10-23
Let us now assume that AB is shortened from both its ends, i inclination with the planes remaining constant. The H.T. and V.T. of tl new line CD are still the same as can be seen clearlyin fig. 10-24(i
c' d' and cd are the projections of CD ffig. 10-24(ii)1.lts traces mi be determined as described below. (i)
Produce the front view c' d' to meet xy at a point h.
(ii) Through h, draw a projector to meet the top view cd-produce at the H.T. of the line. (iii) Similarly, produce the top view cd to meet xy at a point y.
(iv) Through 4 draw a projector to meet the front view c' d'-produce at the V.T. of the line.
Projections of Straight Lines 19.t
;1,/
1e
(1)
(ri; Ftc. 10-24
:ly Method ll:
c' d' and cd are the projections of the line C D l t i g . 1 0 - 2 5 ( i i ) 1 Determine the true length C1D1 from the front view c' d' by trapezoid method. The point of intersection between c, d'-produced and C 1 D 1 - p r o d u c e di s t h e V T . o f t h e l i n e . S i m i l a r l y , d e t e r m i n e t h e t r u e l e n g t h C r D 2 f r o m the top view cd. Produce them to intersect at the H.T. of the line.
T h e a b o v ei s q u i t e e v i d e n tf r o m t h e p i c t o r i a vl i e w s h o w n i n f i g . 1 O - 2 5 ( i ) .
r
J
"-lda
\\'cft,.4oiF+.
\ r/
its he
,/.
=- ,,F?-4,1
(r)
aY
F r c .1 0 - 2 5
10-11.Tracesof a Iine, the projectionsof which are perpendicular
to xy:
..1
When the projections of a line are perpendicular to xy, i.e. when t h e s u m o f i t s i n c l i n a t i o n sw i t h t h e t w o p r i n c i p a lp l a n e so f p r o j e c t i o ni s 9 0 ' , i t i s n o t p o s s i b l et o f i n d t h e t r a c e s by the first method. Method ll must, therefore, be adopted as shown in fiB. 10-26.
1 9 2 E n g i n e e r i n BD r a w i n g
,|, I
I I I
Fic.10-26
10-12.Positionsof tracesof a l i n e :
Although the line may be situated in the third quadrant, its both traces may be above or below xy, as shown in problem 10-6 and in fig. 10-27 a n d { i g . 1 0 - 2 8 .W h e n a l i n e i n t e r s e c t sa p l a n e ,i t s t r a c e so n t h a t p l a n ew i l l b e c o n t a i n e db y i t s p r o j e c t i o no n t h a t p l a n ea s s h o w n i n p r o b l e m l 0 - 7 .
1r"rr1l*
t'.ii
F t c . 10 - 2 8
Projections of Straight lines
193
P r o b l e m 1 0 - 6 . P r c j e c t i o n so f a l i n e p e a r e g i v e n . D e t e r m i n e t h e positions of its traces. L e t p g a n d p ' q ' b e t h e p r o j e c t i o n so f p e ( f i g . 1 O - 2 7a n d f i g . . 1 0 - 2 8 ) . (i) Produce the top view pq to meet xy at v. Draw a pro.;ector through v to meet the front view p, q'-producedat the V.T. ( i i ) T h r o u g hh , t h e p o i n t o f i n t e r s e c t i o nb e t w e e np , q , - p r o d u c e da n d xy, d.aw a projector to meet the top view pq-produced at the H.T. Note that in fig. 10-27, the tracesare below xy while in fig. 10-28 they are above it. Problem 1O-7.A point A is S0 mm below the H.p. and 12 mm behind the V.P A paint B is '10 mm above the H.p. and 25 mm in front of the V.P.The distance between the proiectors of A and B is 40 mm. Detetmine lhe Lraceso{ the line joining A and B. Draw the projectionsab and a,b, of the line AB. M e t h o d I : ( f i g .1 0 - 2 9 ) : ( i ) T h r o u g h v , t h e p o i n t o f i n t e r s e c t i o nb e t w e e na b a n d x y , d r a w a projector to meet a'b' at the VT. of the line. ( i i ) S i m i l a r l y ,t h r o u g h h , t h e p o i n t o f i n t e r s e c t i o n b e t w e e n a ' b ' and xy, draw a projector to cut ab at the H.T. of the l i ne .
:S t.7
i
|
\4; /'|-''t
)X
t.'/" Frc.10-29
F t c .1 0 - 3 0
Method ll: (fig. 10-30): A t t h e e n d s a ' a n d b ' , d r a w p e r p e n d i c u l a rtso a , b , , v i z . a , 4 1 e q u a l t o a o 1 a n d b ' 8 1 e q u a l t o b o 2 o n i t s o p p o s i t es i d e s ( a s a a n d b a r e o n o p p o s i t es i d e s o f x y ) . Draw the line 41 81 intersectinga, 6' at the V.T.of the line. Simifarly,at the ends a and b, draw perpendicularsto ab, viz. aA2 equal t o a ' o r a n d b 8 2 e q u a l t o b ' 0 2 , o n i t s o p p o s i t es i d e s ( a s a , a n d b , a r e o n o p p o s i t es i d e so f x y ) . J o i n 4 2 w i t h 8 2 c u t t i n ga b a t t h e H . T .o f t h e l i n e . Note that A1 81 = A2 82 = AB and that 0 and O are the incljnations of AB rvith the H.p. and the Vp respectivelv.
1 9 4 E n g i n e e l i n gD r a w i n g
ADDITIONATILLUSTRATIVE PROBLEMS:
I n t h e f o l l o w i n g p r o b l e m s ,t h e e n d s o f t h e l i n e s s h o u l d b e a s s u m e t o b e i n i h e f i r s t q u a d r a n t ,u n l e s so t h e r w i s es t a t e d . Problem 10-8. (fig. 10-31): A line AB, 50 mm long, has its end A in both the H.P and the V.P lt is inclined at 30' to the H.p. and at 45" to the V.P.Draw its ptojections.
(,
lii)
(iii)
F t c .1 0 - 3 1
As the end A is in both the planes, its top view and the front view w i l l c o i n c i d ei n x y . (i) AssumingAB to be parallelto the V.P and inclinedat 0 (equalto 30") to the H.P., draw its front view ab' (equal to AB) and proiect the top view ab. ( i i ) A g a i n a s s u m i n gA B t o b e p a r a l l e lt o t h e H . P . a n d i n c l i n e d a t s (equal to 45') to the VP, draw its top view abi (equal to AB) Project the front view ab1'. ab and ab1' are the lengths of AB in the top view and the front view respectively,and pq and rs are the loci of the end I in the front view and the top view respectively. (iii) With a as centre and radius equal to ab1', dtaw an arc cutting pq in b2'. With the same centre and radius equal to ab, draw an arc cutting rs in b2. Draw lines joining a with b2' and br. ab2, and ab2 arc the required projections.
Frc.10-32
Proiections of Straight Lines t9l
3ed lin 'lo
Fig..10-32shows in pictorialand orthographic views, the solutio . obtainedwith all the abovesteps combinedin one figure only. (fig. 10-33):A line pe 7s mm long, has its end p ir , llgbfuT 10-9. the V.Pand the end Q in the H.p. The tineis inclinei at 30" to the H.F and at 60" to the V-P Draw its proiections.
(t 30') the' al 6
(.t,
(ih)
F t c .1 0 - 3 3
The top view of P and the front view of e will be in xy. As shown i n t h e p r e v i o u sp r o b l e m , d e t e r m r n e
AB).
(i)
iront the
(ii) the p1 qf in the front view and the path cd of the end e .length in the top view.
t nS \'an rired
the. length of PQ in the top view, viz. q, p and rhe path ab of the end P in the front view;
(iii) Mark any point p2 (the top view of p) in xy and project its front view p2' in ab. (iv) With p2' as centre and radius equal to p1 ql, x y i n q 2 ' . t t c o i n c i d e sw i t h p 2 .
dtaw an arc cuttrng
(v) With p2 as centre and radius equal to g, p, draw an arc cuttrng in q2, p2q2 and q2' are the required projections. Thei ?.d .p2' l i e . i n a l i n e p e r p e n d i c u l a lro x y b e c a u s et h e s u m o f t h e t w o i n c l i n a t i o n si s e q u a l t o 9 0 . . "f ^^ e.rob]em 10.10. (fig. lA.3e: A line pe 100 mm long, is inclined at 30' to the H.p and at 45. to the V.p. Its mid-point is in the V.p. and 20 mm above the H.P Dtaw its projections,if its end p is in the third quadrant and Q in the first quadrant.
The front view and the top view of p will be below and above xy respectively,while those of e will be above and below xy respectively. (i) Mark m, the top view of the mid-point in xy and project its front view m', 20 mm above x/.
196 fngin€ering DrawinB
(ii) Through m', draw a line making an angle 0 (equal to 30") with xy a n d w i t h t h e s a m e p o i n t a s c e n t r ea n d r a d i u se q u a l t o PQ, cut 1 it at P1 below xy and at Q1 above xy. Project P1 Q1 to p1 91 on xy. pt gt is the length of PQ in the top view. ab and cd are the paths of P and Q respectivelyin the front view.
F r c .1 0 - 3 4
( i i i ) S i m i l a r l y ,t h r o u g h m , d r a w a l i n e m a k i n g a n g l e s ( e q u a l t o 4 5 ' ) with xy and cut it with the same radius at P2 above xy and at Q2 below it. (iv) Project P2 Q2 to p2' q2' on the horizontal line through m', pz' gz' is the length of PQ in the front view and ef and gh are the paths of P and Q respectivelyin the top view' (v) With m as centre and radius equal to mp1 or me1, draw arcs cutting ef at p3 and gh at 93. With m' as ceatre and radius equal Io m' p2' or m' q2', draw arcs cutting ab at p3' and cd at q3'. p3 q3 and p3' q3' are the required proiections. Probfem 10-11. (fig. 10-35): Ihe top view of a 75 mm long line AB measures 65 mm, while the length of its front view is 50 mm. lts one end A is in the H.P and 12 mm in front of the V.P.Draw the prcjections of AB and determine its inclinations with the H.P. and the V.P. (i)
Mark the {ront view a' and the top view a of the given end A. ( i i ) A s s u m i n gA B t o b e p a r a l l e lt o t h e V . P , d r a w a l i n e a b e q u a l t o 6 5 m m a n d p a r a l l e lt o x y . W i t h a ' a s c e n t r e a n d r a d i u s e q u a l t o 75 mm, draw an arc cutting the projector through b at b'. The line cd through b' and parallelto xy, is the locus of I in the front v i e w a n d 0 i s t h e i n c l i n a t i o no f A B w i t h t h e H . P . ( i i i ) S i m i l a r l yd , raw a line a' bf in xy and equalto 50 mm, With a as centre and radius equal to 48, draw an arc cutting the projector throughb1' aI 4. ef is the locus of B in the top view and o is the inclination of /48 with the V.P. ( i v ) W i t h a ' a s c e n t r e a n d r a d i u se q u a l t o a ' b 1 ' , d r a w a n a r c c u t t i n g cd in b2'. With a as centre and radius equal to ab, draw an arc cutting e/ in b2. a' b2' and ab2 are the required projections.
P r o j e c t i o n so f S t r a i g h tL i n e s 1 9 7
I
1 Frc.10-35
I
F t c . 10 - 3 6 Probfem 10-12. (fig. j 0-36): A tine AB, 65 mm long, has its end A ^^ 20 mm above the H.p. and 2s mm in frlont of the vii rne end B is 40 mm above the H.p and 65 mm in front of the V.p Draw the ptojections of AB and show its inclinationswith the H.p and the V.p (i) As per given positions,draw the locj cd and gh of the end A, and ef and lk of the end I in the front view and thi top view respectively. (ii) Mark any point a (the top view of A ) i n g h a n d p r o i e c ri t t o a , I o n c d . W i t h a , a s c e n t r ea n d r a d i u se q u a l i o 6 S r n . , iru* un-ur. j c u t t i n g e f i n b , . J o i n a ' w i t h b , . 0 , t h e i n c l i n a t i o n. i ,, f , *iif, ,". I
theHP'Project b't"'b o;'t ;D-fl iiJT#,.J Ti il :l'ifl"-,,;|
( i i i ) W i t h a a s c e n t r e a n d r a d i u se q u a l to 65 mm, draw an arc cutting J jk in b1. Join a with b1. o, the inclination of ab1 with xy, is th!. i n c l i n a t i o no f A B w i t h , t h e V . p p r o j e c tb . , t o b 1 ; ' o n cd.'a, b1, is rhe lengrhof AB in the front view. b e t w e e nt h e i r r e s p e c t i v ep a t h s a s s h o w n . a , b 2 , - -nr o Y "aqD: ^2ra, 0 , uired a r e" : t"h! :e' , trre, q p r o j e c t i o n so f A g . P T g b f " _ T1 0 - 1 3 . { f i g . 1 0 - 3 7 a n d f i g . t 0 _ 3 S ) : T h e p r o i e c r o r s of the ends . ot a ltne AB 5 .are cm aparl. the end A is 2 cm above the H.p. and 3 cm tn tront ot the U.P The end B is 1 cm below the H.p and 4 cm behind the VP. Determine the true length and t/aces of AB, and ,t, irrtiiitioi with the two planes. D.rawtwo projectors 5 cm apart. On one projector, mark the top view a and the front view a,of the end A. On the 6ther,'rnuit tn" top view b and the front view b, of the end B, as per given Jii"".", ab anda, b, a r e t h e p r o j e c t i o n so f 4 9 . D e t e r m i n et h e t r u e l e n g t h , t r a c e sa n d i n c l i n a t i o n s . .. by any one of the following two methods:
198
DBr a w i n g Engineerin Method I: B y m a k i n g t h e l i n e p a r a l l e lt o a p l a n e ( { i g . 1 0 - 3 7 ) : ( i ) K e e p i n ga f i x e d , t u r n a b t o a p o s i t i o na b 1 , t h u s m a k i n g i t p a r a l l e l to xy. Project b1 to b1' on the locus ol b'. a' bf is the true l e n g t h o f A B a n d 0 i s i t s t r u e i n c l i n a t i o nw i t h t h e H . P
.
( i i ) S i m i l a r l y t, l : r n a ' b ' t o t h e p o s i t i o na . ' b ' a n d p r o j e c ta l ' t o a 1 o n the path of a (becausethe end a has been moved). a1 b is the true length of AB and g is its inclination with the VP. Traces: (i)
Through v the point of intersection of the top view ab with xy, d r a w a p r o j e c t o rt o c u t a ' b ' a t t h e V . T .
(ii) Through h the point of intersectionof the front view a' b' with xy, draw a projector to cut ab at the H.T. of the line.
Frc.10-37
Flc, 10-38
Method II: By rotating the line about its projections till it lies in H.P. or V.P. (fig.10-38): (i)
At the ends a and b of the top view ab, draw perpendicularsto ab, viz. aA1 equal to a' o1 and b81 equal to b' o2, on opposite s i d e s o f i t ( b e c a u s ea ' a n d b ' a r e o n o p p o s i t es i d e s o f x y ) . 4 1 8 1 is the true length of 48. 0 (its inclination with ab) is the inclinationof AB with the H.P. and the point at which 4181 intersects ab is the H.T. of A8.
(i i ) Similarly, at the ends a' and b' of the front view a' b', draw p e r p e n d i c u l a rtso a ' b ' , v i z . a ' 4 2 e q u a lt o a o 1 a n d b ' 8 2 e q u a l t o bo2, on opposite sides of it. A2B2 is the true length of A8, o ( i t s i n c l i n a t i o nw i t h a ' b ' ) i s t h e i n c l i n a t i o no f A B w i t h t h e V P . and the point at which 4282 intersectsa' b' is the V.T. of A8. P r o b f e m 1 0 - l a , ( f i g . 1 0 - 3 9 ) :A l i n e A 8 , 9 0 m m l o n g , i s i n c l i n e da t 4 5 " to the H.P and its top view makes an angle of 60' with the V.P.The end A is in the H.P and 12 mm in front of the V.P Draw its front view and find its true inclination with the V.P.
Ploiections of St.aight LiDes 199
( i ) M a r k a a n d a ' , t h e p r o j e c t i o n so f t h e e n d A . ( i i ) A s s u m i n gA B t o b e p a r a l l e lt o t h e V . R a n d i n c l i n e da t 4 5 " t o t h e H.P, draw its front view a' b' equal to AB and makingan angle of 45' with xy. Project b' to b so that ab the top view is parallel to x/. Keepingthe end a fixed, turn the top view ab to a position a b l s o t h a t i t m a k e sa n a n g l e o f 6 0 ' w i t h x y . P r o j e c tb 1 t o b l ' o n the locus of b'. Join a' with b1'. a' bf is the front view of A8. ( i i i ) T o f i n d t h e t r u e i n c l i n a t i o nw i t h t h e V . P . ,d r a w a n a r c w i t h a a s c e n t r e a n d r a d i u se q u a l t o A 8 , c u t t i n g t h e l o c u s o f b 1 i n b 2 . J o i n a with b2. s is the true inclination of ,48 with the V.P.
F r c .1 0 - 3 9
Frc.10-40
Probfem 10-15. (fig. 1o-40)'tlncompleteprciectionsof a line PQ, inclined at 30" to the H.P are given in fi& 10-40(i). Complete the projections and determine the true length of PQ and its inclination \\'ith the UP. ( i ) T u r n t h e t o p v i e w p q t f i g . 1 0 - 4 0 ( i i ) lt o a p o s i t i o np g r , s o t h a t i t is parallel to xy. Through p', drayv a line making an angle 0 (equal to 30") with xy and cutting the projector through 91 at q1'. p' gt' is the front view of PQ. ( i i ) T h r o u g h q 1 ' , d r a w a l i n e p a r a l l e lt o x y a n d c u t t i n g t h e p r o i e c t o r throughg aI q'. p' q' is the front view of PQ. ( i i i ) W i t h p a s c e n t r e a n d r a d i u se q u a l t o p ' q l ' , d ' a w a n a r c c u t t i n g the locus ol q at q2, (iv) Join p wilh q2. o is the inclinationof PQ with the V.P Problem 10-16. (fig. 1O-41)tThe end A of a line AB is 25 mm behind the V.P.and is below the H.P The end B is 12 mm in frcnt of the V.P.and is above the H.P. The distance between the projectors is 65 mm. The line is inclined at 40" to the H.P and its H.T. is 20 mm behind the V.P Draw the prcjections of the line and determine its true length and the V.T.
2 0 0 E n g i n e e r i n gD r a w i n g
Draw the top view ab and mark the H.T. on it, 20 mm aDovexf. W e h a v e s e e n t h a t t h e l i n e r e p r e s e n t i n gt h e t r u e l e n g t h o b t a i n e d by t h e . t r a p e z o i dm e t h o d , i n t e r s e c t st h . e top viEw or the top"view-produced a t t h e H . T .a t a n a n g l ee q u a lt o t h e t r u e i n c l i n a t i o no f t h e l i n e w i t h the V.p ( i ) Hence, at the ends a and b, draw perpendiculars to ab on its o p p o s i t es i d e s ( a s o n e e n d i s b e l o w t h e H . p . a n d t h e o t h e r e n o above it). Throughthe H.T.,draw a line makingangle 0 (equalto 40") w i t h a b a n d c u t t i n g t h e p e r p e n d i c u l a ras t ; 1 ; d 8 1 , a s s h o w n . Ar81 is. the true length of AB. aAl and bB1 are the distancesof t h e e n d s A a n d B r e s p e c t i v e l yf r o m t h e H . p (i i ) Project a and b to a, and b', making a, o1 equal to aA.1 and b, o2 e q u a l t o b \ . a ' b ' i s t h e f r o n t v i e w o f A B . T h r o u g hv , t h e p o i n i of intersection between ab and xy, draw a prolectJr cutting a,b, at the VT. of the line. b: b,
Flc. 10-41
Ftc.'tQ-42 Problem 10.17. (fig. 10-42):A tine A8,90 mm long, is inctined at 30. to . the H.P lts end A is't2 mm above the H.p and Z0 im in tront of the V.p. Its front view measures65 mm. Draw the top view of AB and determine its inclination with the V.p. ( i ) M a r k a a n d a ' t h e p r o j e c t i o n so f t h e e n d , 4 . T h r o u g ha , , d r a w a l i n e a ' b ' 9 0 m m l o n g a n d m a k i n ga n a n g l eo f 3 0 . w i t h -x y . (ii) With a'as centre and radius equal to 65 mm, oraw an arc c u t t i n gt h e p a t h o f b , a t b 1 , .a , 6 1 , i s t h e f r o n t v i e w o f A g . (iii) Project b' to b, so that a6 is parallel to xy. ab is the length of A8 in the top view. (iv) With a as centre and radius equal to ab, draw an arc cuttrng the . p r o j e c t o r. t h r o u g h b l a t b 1 . J o i n a w i t h b . , . a b 1 i s t h ! required top view. D e t e r m i n eg a s d e s c r i b e di n p r o b l e m1 0 _ 1 4 .
