Advanced Level Technical Dr�wing

Third edition, metricated

E. JACKSON, M.COLL.H.

..........
.. JJII ...
,T Longman
 LONGMAN GROUP LIMITED
Longman House
Burnt Mill. Harlow. Essex. U.K.

© Longman Group Limited 19-5

All rights reserved. No part of this publication may be

reproduced, stored in a retrieval system, or
transmitted in any form or by any means, electronic,
mechanical, photocopying, recording or otherwise,
without the prior permission of the Copyright owner.

First published 1968

Third edition 1975
Fifth impression ! 9S:

ISBN O 582 355::S -

Set in Monophoto Times New Roman in

Great Britain by William Clowes & Sons, Limited
London, Beccles and Colchester
Printed in Hong Kong by
Sheck Wah Tong Printing Press Ltd

J J
I I .i I I I J .I
. I� f J !
 Contents

1v Foreword
Part I Geometrical Drawing Part II Engineering Drawing
1 to 23 Plane Geometry 281 to 293 Rivet and Screw
24 to 25 1st and 3rd Angle Fastenings
Projection 294 to 328 Shafts, Bearings and
26 to 40 Pictorial Projection Seals
41 to 76 Conics 329 to 351 Pipes, Joints and
77 to 88 Cycloids Valves
89 to 99 Cams · 352 to 373 Pumps and Engine
100 to 123 Helices and Gearing Parts
124 to 155 Forces and Frameworks �378 to 381 Standards, Limits and
156 to 172 Centroids Symbols
173 to 190 Mechanisms 382 to 407 Machine Drawing
191 to 232 Inclined and Oblique Problems
Planes 408 to 414 Assembly Solutions
233 to 263 Simple Developments 415 to 423 Isometric Drawings
and Interpenetrations of Assemblies
264 to 272 Further 424 to 430 Electrical Circuits
Interpenetrations 431 to 434 Freehand Sketching
273 to 280 Development by Appendix 1 Tables
Triangulation Appendix 2 Questions
Index

'
 iv

Foreword
This book has been fully metricated in accordance with Worked examples and exercises are given in all sections
the national intention of adopting SI units by 1975 and of the book, from fundamental principles to problems of
by the stated intention of Examining Bodies to set examination standard. The appendices have been revised,
papers in Technical Drawing using SI units only by the tables giving information regarding the recommended
same date onwards. Since the Examining Bodies also metric sizes for ISO screw fastenings, rivets, materials
state that the provisions of the current BS 308 Engineering and components are included. Information of SI units,
Drawing Practice of the year of examination should be prefixes, multiples and submultiples, together with useful
observed in the preparation of all orthographic drawings related formulae for use in calculating forces in mechan­
and sketches, this volume has followed the suggestions of isms and frameworks are listed.
BS 308, Fourth Revision, 1972, the latest to hand. The EDWARD JACKSON
British Standard embodies much welcome definitive de­
tail of layout, lines, projection, lettering, sectioning, and
conventional representation of common engineering
features with simplification and clarification of many
items including the introduction of graphic symbols in
tolerance frames and the diagrams in this edition have
been amended to comply with these suggestions. ISO
symbols are shown in the schematic electrical diagrams
using BS 3939 as reference.
Examination Syllabuses remain broadly the same and
require an intimate knowledge of plane and solid geo­
metry as a basis for projections in both first and third
angle. The general layout of this edition is unchanged,
the first part dealing with plane geometry, conics, in­
volutes and gearing, cams and mechanisms, centroids,
vector polygons and frameworks, loading in SI units.
The projection of point, line, laminae and geometrical
solids to the main planes, to the inclined plane, and to the
oblique plane is progressively treated. The interpenetra­
tions of various solids are shown from simple examples to
angled offset cone, prism, cylinder, both rightand oblique,
and sphere. Developments by simple projection and
triangulation are also covered.
The second part of the book deals with engineering
components and conventional drawing practice using ISO
recommendations. Following present examination
trends, the machine drawing examples are shown as
pans to be assembled, from which sectioned or full views
are required in either first or third angle projection or
conversion from one projection to the other. Isometric
cutaway drawings of the assemblies are shown to assist
in the explanations, examples in pictorial projection
using perspective, oblique and trimetric views are given.

I I I I .I J I J I I 1 I
----

Part I
GEOMETRICAL DRAWING

Plane Geometry
-
2

I. Proportionate Division by intersecting parallels. Draw

the two lines at a convenient angle, join ends, and by GEOMETRY
parallels obtain the proportionate division of the new
line at A 1 , B 1 , C 1 , D 1 .
2. Rectangles of Equal Area Extend base to new length
at A. Erect perpendicular to height of original rectangle.
PROPORTIONATE DIVISION

0 :90
'
SQUARE
0
ROOTS
0
Draw the diagonal, and through B draw a parallel to the
base line.-The construction may be reversed. \ '1

3. Triangles of Equal Area Enclose triangle in rect·

angle. Extend base to new length, and obtain enclosing
\
\

rectangle for new triangle. Draw new triangle in new

rectangle, several solutions, using same base and height.
4. Square Roots Draw a right-angled triangle, two sides A 0
indicated one unit each. Hypotenuse gives root 2. Con­ ------,
tinue the construction with further right-angled triangles
I

to obtain roots 3, 4, 5 etc. by scale.

5. Logarithmic Progression Given the ratio and angle RECTANGLES OF EQUAL AREA LOGARITHMIC PROGRESSION
between two lines, draw the lines enclosing the angle
and join the ends of the lines. Draw the arc to give posi­
tion I, and draw further arcs and parallels to obtain
further points and lengths. These are used later in the
construction of the Logarithmic Spiral, see 110.
0 GIVEN:
ANGLE & RATIO
�
,,

·-·-,
,,

B
..�.'
6. Circles Touching Lines and Circle Common Tan­
� "':,, '

g,.ncy. Draw the two given lines. Obtain centre of first

circle from intersection of lines parallel to given lines,
distance R 1. Centre of second circle is at intersection of I
arc radius R 1 + R 2 from first centre and line parallel to
base line, distance R 2 . The common tangent to the circles -------'-. - . __ JA b
is also shown. This construction is also useful in diagrams
showing sections of touching spheres.

0
TRIANGLES OF EQUAL AREA CIRCLES TOUCHING LINES

0,
AND CIRCLE

t---+---,,,-->...----�---ii' .I

GIVEN
LINES---'

I I I I I I I J I ' l .I
 7. Touching Circles Deduct radii for internal circle and
add for externally touching circle. Point of contact on GEOMETRY
line joining centres, tangent at right angles to this line. TOUCHING CIRCLES AREA BY OR DINA TES
8. Three Touching Circles R 1 +R 2 , R t +R 3 , and
R 2 +R3 give the three sides of a triangle. Apices are
centres for touching circles. Conversely, given the tri­
angle, three touching circles may be found, the radii
being the length of the perpendicular from each side
intersecting the bisectors of the angles of the triangle.
9. Straight Line Equal to Arc Draw the arc and sector.
Extend the chord by half. Draw the tangent at A, and
cut off by an arc from B. Length required is AC. : j
2
10. Area by Ordinates Simpson's Rule, as shown. The ��'--"--�/�3---=-,c_.;__..L__.r____;'�

simpler method is to add the length of mid-ordinates of

each strip and multiply by the width of a strip.

0
11. Regular Polygon Given· the length of a side, draw
the semicircle on it, and divide into as many parts as sides. 3 TOUCHING CIRCLES REGULAR POLYGON
Join always point 2 to centre. Bisect the sides formed, and
from the centre obtained draw the enclosing circle, step
off the remaining sides.
12. Regular Polygon inside a given circle. Draw dia­
meter and from its ends draw intersecting arcs, radius as
diameter. Divide the diameter into as many parts as sides
of the polygon, a line drawn through 2 from A cuts the
circle giving the length of the first side of the polygon.

Rr
0

STRAIGHT LINE EQUAL TO ARC REGULAR POLYGON

r1l--------
.\\7\!
B GIVEN:
ENCLOSING
CIRCLE
I
I

A
.. I
�i 0 0
 4

13. Arc Equal to Straight Line (approx.). Draw the circle GEOMETRY
given the radius of the arc. Draw the line tangential.
Divide the line into four equal parts and with centre I and CIRCLE TO TOUCH
ARC EQUAL TO LINE
radius 1-4, draw arc to cut circle.
14. Circle to Touch Lines and Point All circles tangen­
APPROX
-0 I
POINT & CIRCLE

tial to the lines have centres on the bisector of the angle.

The line passing through P gives similar points P on
other circles. The dotted lines joining centre and P for
each circle are parallel.
II 0
A construction circle (two shown) should be drawn I
first and a line parallel to P 1 , 0 1 , passing through P gives
the centre of the required circle, 0. I ____::::,,.._...,-::::._____,t.:¥--,4---==:,,-..�====---+-
15. Circle to Touch Point and Circle Join P to the centre
of the circle and extend to cut the circumference in Q.
radius given l ftb_
PQ is the diameter of the circle required. A second solu­
tion with A as centre may be drawn.
16. Circle to Touch Point and Circle Draw the con­ I
struction circle centre P and radius as the given circle.
Obtain Q by drawing the perpendicular. Complete the
CIRCLE TO TOUCH LINES & POINT

61
i
TANGENCY
0
I
COMM ON TANGENTS
C'\

construction as in the diagram.

17. Common Tangents for Touching Circles Given the
centres of two circles A and B, radius R 1 and R 2 , draw
the arcs of touching circles radius R 3 and R 4, as shown.
Obtain the centre X of the outer touching circle by sub­ I
traction of radii, and then drawing intersecting arcs from
A and B. Obtain the centre Y of the inner touching circle
by addition of radii, and drawing intersecting arcs from
A and B. The points of common tangency are found by
joining centres and extending the lines to cut the circum­
ference. The tangents are drawn normal (perpendicular)
to the radius lines at the tangent points.
CIRCLE TO
TOUCH POINT
& CIRCLE

J I I J I ) I J I . I J I I
18. Tangency A circle drawn tangential to two lines
which meet. The centre will lie on the bisector of the G E OMETRY
LOCI
angle, also on intersection of Jines which are radius­
distant and parallel to the lines. These lines are the locus TANGE NCY LOCI TANGENCY

0 0
of the centres of such circles which roll on the lines inside
the angle.
19. Tangency Method of obtaining the radius of an arc
tangential to a given arc and to pass through a point P,
application of No. 1 6.
The method uses an auxiliary construction circle,
radius equal to that of the given circle and drawn to pass
through P. Join the centres, and erect a perpendicular bi­
sector to cut the centre line. The point of intersection c
is the centre for the required arc, Pc is the radius.
20. Tangency Second case using a constructional circle

8
to find the radius of an arc which is tangential to two
given circles. The diagram shows how inner and outer
constr'n circle:
arcs may be drawn. TANGENCY APPROX COMPASS FOR
The above constructions are frequently used in draw­ CURVE
ing metallic shapes which contain connected circular INV O L U T E

@ I
areas which surround bolt and shaft holes. See No. 382,
parts I and 2 .
--
C:

I
\ I
2 1 . Tangency The locus of the centre of a circle rolling

..
.c
.I
on a given line is shown. Notice how the circle passes the
re-entrant and salient corners, and how this affects the
locus line.

-�
:s!
22. Approximate Compass Curve for an Involute Curve.
G LA N D
;I;

\ '
This construction may be used in the representation of
COVER
involute gear teeth, when an arc approximate method is I
allowed. Draw the involute by the usual tangential �o i
method where the tangent is made equal in length to its
respective arc. Draw a chord in the working length of the COVER PLATE
involute, and draw a perpendicular to the mid-point of TANGENCY§ E N LA RG E M E N T
the chord. The centre of the approximate arc is C. Points construction BY RAD IALS
C for the teeth will lie on a circle radius CO. c i rcle:s
R b
23. Enlargement by Radials Enlarge the given shape in
ratio 4 : 5. Draw t he given shape. Draw radial lines
through the critical points of the shape from pole 0.
Enlarge the base of the shape by intersecting parallels.
The centres of arcs can be found by radials passing
through the first centres intersected by lines from new
positions, as shown at C .

5
 6

Exercises 1 3. A straight line is 90 mm long. Cut off graphically, an 24. Draw the approximate compass curve for the in­
arc of a circle 1 50 mm in dia, and equal in length to volute tooth shape using a base circle of 1 20 mm dia.
I. Draw a line 95 mm in length. Divide in the ratio of
the straight line.
I : 2 : 3: 4 : 5. Divide another line 1 75 mm long from the 25. Reduce the shape shown in No. 1 0 to a rectangle by
first. the ordinate method, make the base 120 mm. Check by
14 Two lines contain an angle of 85.5 °. A point P lies
30 mm from the intersection and 1 2 mm from one of the Simpson's Rule. Check the area by graphical integration
2. Given a rectangle 100 mm x 73 mm, draw a rectangle
lines. Draw a circle to touch the two lines and pass shown in 1 62-1 64.
equal in area but having a base of 65 mm.
through P.
26. Draw any suitable quadrilateral, find the C of G by
3. Given a triangle of sides in the ratio of 3 : 4 : I · 5, and l 5. Repeat the construction shown in No. 1 5, 0Q to be the median method.
the length of the first side 50 mm, draw a triangle equal 25 mm.
in area but with a base of 1 1 2 mm. 27. Drawa triangle ofsides 85 mm, 70 mm, 50 mm. Draw
1 6 . Draw a circle 55 mm dia tangential to a straight the inscribed circle and two other circles which touch two
4. Prove graphically that the square root of 5 is 2.236. line. A point P lies on the line 70 mm from the point of sides of the triangle and the inscribed circle.
tangency. Draw a circle to touch the given circle and
5. Draw a logarithmic progression given an angle of 50 ° passing through P. 28. Draw the ellipse given V to F 1 , 1 0 mm ; V to F 2,
and two vectors in the ratio of 1 .25 : I . 7 5. 70 mm. Construct the largest square contained by the
17. A line A B is 50 mm long. A and B are centres for ellipse, reduce to a triangle height 35 mm.
6. Two circles, 70 mm and 40 mm dia, touch each other circles of 65 mm and 40 mm dia. A circle, of 45 mm
and the smaller touches two lines which are at 85 ° . Draw dia, rolls without slipping round the outside of the 29. Given the medians, 65 mm, 85 mm, l 00 mm, draw
the diagram, showing several positions of the larger first two circles. Plot the locus of a point on the rolling the triangle, reduce the triangle to a rectangle, base
circle as it moves from contact with one line to the other circle. 40 mm.
line. Plot the locus of a point on the large circle during
the movement. 1 8. A rolling circle of 50 mm dia rolls without slipping 30. Draw two circles 40 mm dia and 60 mm d1,i �-
along a straight line 50 mm long, then up a slope 60 mm 75 mm centres. Draw one common external and one ,r.­
7. A circle of 50 mm dia stands on a straight line. A long and gradient of l : 3, then down a slope 40 mm long ternal tangent to the circles. Draw a third circle to WU<ch
second circle, 60 mm dia, touches the line and the first and at 45 ° to the original line. Draw dominant positions both tangents and the larger of the two circles. Find :.he
circle, then rolls round the outside of the first circle of the rolling circle from which the locus of the centre area of the space bounded by the two tangents and :he
until it comes to rest at the straight line. Plot the locus point may be seen. Draw also the locus of a point on the arc of the larger circle by Simpson's Rule, and check ::-:,
of a point on the moving circle which begins in contact rolling circle beginning at the point of tangency with the graphical integration.
with the straight line. first line.

8. Three circles, 33 mm, 55 mm, 66 mm in dia lie touching 1 9 . Take the shape drawn in No. 1 0, and enlarge by
each other. Draw the construction. radials in the ratio of l 00 : 1 30.

9. Draw the arc shown by a radius of 80 mm and sub­ 20. Two circles, 30 mm and 45 mm dia are tangential.
tended by an angle of 75 ° . Draw a straight line equal in A third circle, 85 mm in dia, encloses the other two
length to the arc. circles and touches both. Draw the figure.

l 0. Draw an irregular shape similar to No. I 0, in a rect­ 2 1 . Draw the diagram shown in No. 6 three times the
angle 90 mm by 70 mm. Find the area by Simpson's Rule printed size.
and also by simple ordinates; compare the results.
22. Draw the figure shown in No. 4 with the first side
1 1 . Construct a regular nonagon inside a circle of 1 0 0 30 mm, and reduce it to a square of equal area.
mm dia.
23. Draw three touching circles, diameters 40 mm, 5 5
12. Construct a regular polygon of seven sides, each m m and 7 5 mm, a s i n No. 8 . Reduce the centre triangle
side 37 mm. to a rectangle of equal area, base 60 mm.

I I I J I I I J I I I J r - 1 ..
ORTHOGRAPHIC PROJECTION
First Angle

Third Angle

PICTORIAL PROJECTION :
Isometric

Oblique

Axonometric
Perspective
Tri metric
 8

24. First Angle Projection The object is visualised as P ROJEC T I O N

being placed in a box with a bottom and three sides. The
A N GLE
bottom represents the horizontal plane on to which is

0
projected the plan. The three sides represent the three
vertical planes on to which the front elevation (centre) ISOMETRIC
and two end elevations (ends) are projected. The iso­ EXPLANATIO N
metric explanatory view should help to make this clear.
A typical drawing layout is shown ; the plan lies below
II
the front elevation, the end elevations are projected
· across ' the front elevation and are views of the · oppo­

i
site ' end of the object. , �p

I
To begin the drawing, the views required should be
noted, the overall sizes calculated for each view, and the
position of the X Y line fixed.

I
. "-
The front elevation and plan are usually drawn in I
I

related projected positions, and the end elevations then

projected. Special details may require special treatment
in projection. A centre line in either plan or elevation, or
a shaft or a boss centre may form the dominant point
from which a start is made.
· Heights remain the same in all elevations' and are
easily projected by the tee square.
Points in the plan can be returned to front elevation
by simple projection. Points in the plan can be returned
to the end elevations by using either the 45 c setsquare or
compass.
The angle of projection must be shown either in words D RAWING LAYO U T
or by symbol.

--- -
END E L ELEVATION
VP

I I
1 - - -r
I I

/
X y

Symbol

1ST ANGLE HP PLAN i

I
BLOCK -Et·

I 1 I ) I I I .I I I .I J J J
 25. Third Angle Projection In this projection the object
is placed in the third angle shown in the small sketch. TH I RD A N GL E PROJ E C T I O N
The plan and elevations are projected on to the planes
which are visualised as being transparent, the object 3 R D ANGLE 1
HP PLAN
B LOCK
G
[/ Ii i i)�
being viewed through them. When the planes forming
the transparent box are opened out, the plan appears D RAWING
above the front elevation, whilst the end elevations are LAYOUT
diagrams of the ends nearest to the plane and not of the
opposite end as in first angle projection. The student is
required to have a knowledge of both projections, and
to be able to interpret drawings set in either. All the ob­ I I !
jects and exercises can be projected in first or third angle, � I
and the projection should be stated.
- - -�I
I !:ND E.L I E. LE\/AT I
The drawing procedure is similar to that of the first
VP

angle. Note the views to be drawn, calculate the overall

sizes of the object, and so the space required for each
view. From this information, the position of the X Y line
may be fixed and the front elevation and plan begun. A
centre line forms the usual starting point, or shaft or
pulley centre, from which measurements may be taken.
The end elevations can be projected from the plan and
front elevation by either 45° setsquare or compasses.
More complicated layouts with lists of parts are given
in later drawings.
The angle of projection must be stated either in words
or by symbol.

TRANS PARENT
BOX PLANES
I S OMETRIC
1 1

EXP LANAT I O N
S mbol

9
 ---
IO
26. Isometric Projection A method of pictorial repre­
sentation on three planes arranged at 1 20° to each other, ISOMETRIC
easily constructed by the 60° setsquare. Measurements
can only be made on the three axes or their parallels.
When strict Isometric Projection is asked for, natural
scale measurements are diminished as in the diagram
from the 45 ° line to the 30° line. These diminished
measurements are to be used on all axes to give a pictorial
foreshortening. In ordinary isometric drawing-which
is commonly used-the ordinary scale sizes are used on
the axes. Note that sizes can only be used on the axes or
their parallels, since the diagonals of the face are not
equal.
The isometric drawing is best begun by drawing the
enclosing ' box' sizes obtained from the orthographic
projections, plan and elevations. Simple rectilinear
shapes can be drawn by projecting points on the faces
of the box from measurements on the axes. Two ordi­
nates will intersect to find a point or series of points.
27. Shaped Block Simple shapes and slopes are shown
in the exercise shown, and full construction lines are
given.
......___.....__.________,__.___
=----
SHAP E D B LO C K

s im pliz
s l o p ,a

l50M£Tl?IC Pl?O/ECi.!:f
Dimensions
can only be
m a de on axczs

I
Dime:nsions
cannot be
mad� on
diagonals
�
I S OM E T R I C DRAW I N G

J
J J
I I .I I I
J I J
 - -----------l 1----------------------
28. Compound Slopes Slopes shown in the figure will
require projections from two ordinates before the point ISO
- - ---- - -
S LO
- - -
P ES CURVES
in depth can be found to which an internal diagonal line
can be joined. Such a point is shown at A in the front
face of the object. The square window opening in the face
is drawn by using a centre line and ordinates from the AN G L E D
face of the ' box '. BRACKET
29. Curves in Isometric Compass or freehand curves of
an object such as the bracket are plotted by ordinates to
the curve from the axes. These are drawn in the elevation
first and the measurements transferred to the isometric

�-
w
' box ', where the ordinate is marked to the corresponding �

,___..... a:
length. A fair curve is drawn through the points. r

0
0
The projection of a circle in elevation to isometric can

r
be effected by ordinates, four ordinates and four points I

---· -B-· J II I
on centre lines being the minimum. I
T I - - -

II
The Four Arc Compass Method enables ellipses to be I I I

1
drawn more rapidly but approximately, the centres being I

easily obtained by cutting the centrelines of the face by L_

--- -
I
PR OJECTJ O N OF
lines joining the ends of the centre lines to the opposite A I
corner as shown. Four arcs using two radii aJ!ow the
I
projected circle to be executed neatly, and is especially
useful in engineering diagrams which employ many
cylinders. compound
s l o p<i s c i r c l <Z by
4 arc

I S OM E T R I C c l r c l <i by
DRAW I NG o r d i n atu

l1

...
 1:
JO. Compound Forms in Isometric Drawing Difficult
forms are best considered as being built up of simple COM POU N D FORMS
geometric solids, an enclosing · box ' being drawn as part

0
of the basic construction.
In the sectioned bearing block shown, a complete en­ B EA R I N G
BLOCK
closing box should be first drawn, the flange in the fore­
ground being easily drawn by simple ordinates. The
cylinder can be drawn using either ordinates to give the

CONE
ellipse, or by using the quick four arc method previously
shown. Having drawn the curves on the front face, the
centres may be projected at J0 ° to the thickness of the
block and the remainder of the rear face curves drawn.
The diagram should make clear the construction when
part of the bearing is removed and a sectioned face
exposed.
JI. Cone and Pyramid The enclosing box is drawn, the

ffi
base shape of the circle or. polygon constructed. A centre
line, vertical in these cases, helps in drawing the top face, O C TAGONAL
PYRAM I D
a point only for the cone, a small polygon for the pyra­
mid. In the special case of the sphere, a circle drawn to
simple measurements would appear too small, and the
radius of the circle is usually increased in the reverse

I
order to the isometric scale shown earlier. The larger
radius is used from the original base centre. >

B '
32. Geometric Solids, and m oulded forms. In the eight
examples shown, construction lines show how the forms

-t-
have been built up. An enclosing • box ' may be used
in all cases and the simpler forms drawn using ordi­
nates. Alternatively, the drawing may be started from
the basic · ground ' shape, ellipse in the case of the cone
for example, and a centre line erected therefrom, en­
abling the rest of the construction to be made.

D r a w n on
l s o m<Ztric: radius o l d c e n t re
of sphczre is
/ '-...
increased to
g ive improved
appeara nce
/ " l"_ _ _ '-...
.......

Is o metric:
D r.aw i n g
s cz c: t i o n cz d

I ) I I J ) J l J J
 C O M P OU N D FORMS IN ISO ME TRIC

G
CON E & CY L I N D ER HEX P Y R AM I D & P R I S M S Q PRISM & H E X PY R A M I D

CY L I N DER &
T R I A N G U LA R PRISM

' I
I I I
�i !
S P H ERE & CON E S HAPED BLOCK G E A R B OX EN D � S HA F T B R A C K E T

I3
 14

33. Oblique Projection In this pictorial projection,

three axes are used, two at right angles to each other, the OBL IQUE P R OJ EC T I O N

,... 0
third at 45 ° or 3 0 ° . The elevation is usually taken as the
R�OUCED
front face, the depth of the object being shown on third ...,1
axis. A cube is shown in Fig. I , full size measurements on DEPTH
the depth make the object appear distorted (cavalier
projection), and half depths (cabinet projection), or even
¾ scale is used to give a more proportionate diagram. A
series of simple solids are shown in oblique projection,
the shapes bounded by straight lines.
Block A shaped block having sloping sides. Draw the
enclosing ' box ' ; measurements must be made only on
the axes. CUBE 2
Block. 2nd Position The enclosing · bo x ' may be
drawn with the third axis pointing downward either left BLOCK

---
or right, and the object presented as shown. This angle
of viewing may be used to advantage where details of the

bi
bottom face have to be shown.
Pentagonal Prism on base. Simple solids are easily
projected by employing enclosing ' boxes · ; a centre line
!
i l �
may help in drawing objects whose front face does not i I

lie wholly in the same plane.

Octagonal Pyramid Draw the base octagon in oblique
projection, erect a centre line to the correct height and
construct the upper octagon. Join the corners of the
octagons. L/
Square Prism and Pyramid Begin the interpenetra­ LINEAR FORM BLOCK 2ND P O S IT I O N
tion by constructing the pyramid. From the elevations,
obtain height positions of the prism and · intersection
points. Project the prism from centre lines as shown. Pro­
ject intersection points and join to complete the drawing.

O CT PYRA M I D SQUA R E P RISM & P Y R. A M I D

.... .,
 34. Oblique Projection Objects with curves are best
presented with the curves in the front face and drawn by OBLIQUE P ROJECT I O N
compass or ordinate method.
Bracket The shape shown is arranged with the com­
pass curves in the front face, depth lines at 45 c _ Compass
radii for the rear face are the same as for the front face.
B R ACKET C I RC LES ON FAC E S CY L I N D E R
0
Circles on Cube Faces Circle on the front face mav
be drawn by compass. If the third axis is at 30 ° , compas�
arcs may be used in the ellipse, cavalier or full depth
being used. Ellipses in the top face and end face, if fore­
shortening is used, are drawn by simple projection on
centre lines and ordinates, a s in the isometric diagrams.
Cylinder Where possible arrange the ends of the
cylinder on the front and rear faces where the circles can
be drawn by compass.
Bearing Block Simple arrangement of grouped solids
drawn i n the easiest position for quick mechanical rep­
resentation.
Bearing Bracket Solid composed of cylinder and B E AR I NG B E A R I NG
BLOCK
two prisms. Construct the ' boxes ' for the three parts ;
the intersection of the sloping lines o f the prism and B RACKET
cylinder is tangential.
Casting Draw the centre line and centres for the two
bosses and draw the circles of the front plane. Draw the
third axis depth lines, and construct the rear face circles.

/
Sphere on Base Draw the base, foreshorten if desired.
The sphere is usually drawn on an extended radius to
make the drawing more proportionate. The radius can ,(
be lengthened from R to R 1 by using 30 ° and 45 ° angles
as shown in the diagram.

SPHERE
EXTEND ED
O N BASE
RADIUS

15
 16

35. Axonometric Pictorial Projection A pictorial me­

P ICTORIAL
thod of drawing used by architects. A plan is drawn as a AXONO M E T R IC PROJEC TION
basis, turned through 45 ° , vertical lines are then drawn

0
to give the height. Circles and arcs from the plan can be
drawn conveniently by compasses as shown in the dia­
grams on the top face of the object, but circles appearing
in the elevations are projected as ellipses in the pictorial
view. Measurements for axonometric projections can be
taken F S or to scale from the orthographic drawings
and used di rect on the axes or for ordinates. No measure­
ments can be made on diagonals.
36. Perspective Drawing In this type of pictorial coni­
cal projection, projectors from points on the object
converge at the eye point and pass through the vertical
picture plane on which the image is drawn. The inter­
section of the picture plane and the horizontal ground
plane gives the ground line-the horizon, on which lie
the vanishing points, and is the same height from the
ground line as the eye point is above the ground plane.
The layout of the object, picture plane, eye point and eye
line with the vanishing points may be drawn to a suitable
scale as shown in the diagrams.
Join points I to 1 0 in the plan to the eye point, dropping
perpendiculars at the points l ' to IO' on the picture plane.
The heights of these lines on the picture plane may be D
obtained by measuring their true height on the line A B
from the elevation, and joining to the appropriate vanish­
ing point V P 1 or V P 2 . This will give intersection points
from which the perspective view can be drawn. Curves
can be drawn by projecting points d, e , f, as in the dia­
gram and drawing a fair curve through the points.
Note that the vanishing points V P 1 and V P 2 are ob­
tained by drawing a line from the eye point parallel to
the edges of the object and cutting the eye line.
 P E R S P E C T I VE D R AW I N G

G
I EXPLANATORY

I
V I EW

PLAN

?
V P2

z P E RSPECTIVE
UJ

DRAWING
:i
.....
-
:I:

L
\/')

l::2
: i

UJ
:I:
I

i,
I
I
Ie..
UJ f �
PICT U R E
:::>
0:
t-
PLANE
i E LEV
le.

d.
ie

3RD A NGLE

!
17
l
�

i
 18

37. Tall Objects Lines which are parallel to each other P E R S P E C T IVE S H A D OWS
in the object, are drawn van1shing through the same V P ;

II \ 0
the picture plane may be any height, but the eye line is
always at the same height above the base line as the eye

I· ·,"
point is above the ground plane.
ii ____

I
Distant Vanishing Points To avoid quick diminishing

--
of the planes of the object viewed, a large drawing board
or table top may be employed with straight edges pivoted
at the widely spaced V P points. This will result in less
distortion.
Shadows in Perspective
Shadows may be caused by obstructions in the path of
light from either the sun, i n which case the rays are taken
as being parallel, or from a.lamp giving radial rays.
Sun Shadows The angle of the rays may be decided,
\
45 ° or 60° can be conveniently drawn by setsquare, and
lines are projected from the upper corners of the object.
Horizontal lines are then drawn from the lower appro­

---
priate points to intersect the former angle lines, and the
shape o f the shadow obtained by joining the points. TALL O BJ EC T SUN PARALLEL S H A D OW

Lamp Shadows Lines are drawn from the point

source of the lamp to the upper corners o f the object, to
intersect Jines from the ground point of the source.

38A. Shadows caused by parallel sun rays A solid of

revolution with projected shadows caused by parallel
light rays is shown. Sections are taken, points of inter­
section joined to the centre of arc shape. The intersection
of the radial and the centre line is returned to the section

----
--
line giving a series of points for the curve of the shadow
on the solid. The shadow can now be projected to the HP.
38B. Shadow caused by lamp source Cone of radial
light rays strike the salient points of the solids. Auxiliary
elevations give true intersections and heights with the
VP enabling the outline of the shadow to be drawn.
39. Nine objects are shown to be used for drawing as
above. One size only is given, the rest of the form should
be estimated by proportion, giving practice in visualising

I
�
the object.

RS
L AMP RADIAL S H A D OW
D I STANT VA N I S H I NG PO I N T S
I S U N S O U RCE P R O J E C TION OF S HA DOWS L A M P SOURCE

Projection of shadow
r"j;\
0

by p a r a l l e l ra ys of s u n �
s VP

Pictoria l
View

•
7

+ra mp

E LEV
P roj ection of shadow
by con e of rays
from lamp sou rce

A UX I L I ARY
E LEVS PLAN
G IVING TRUE
L E N GTHS

19
 20

EX E R C IS ES FOR P ICTO R IA L D R AW I N G

1 -- I N
--"- lo l
L_ _I
·1
I

-+---'
_ _ .1 _ _

- - - -i - ----
I

R 40 '
I
!

¢ 30

Q R

L.___ 10__..,...J I I I
p ; __L ___�_,
I

..
 40. Isometric Drawing The diagram shows a cuta\vay
isometric drawing of a Rotary Vane Pump, No. 385 in I SOM ETRI C DRAW I N G
the Machine Drawing section, which has been used a s a
subject.

8
SEE 385
Other involved isometric drawings are given in Nos.
4 l 5 to 423.
Begin by drawing the base, working from a centre line.
Extend the centre line, and draw the enclosing square of
the front circular face. Construct the isometric ellipse by
the four arc method shown in No. 29. Construct the
ellipse for the rotor, which is offset. The blades' position
can be found by ordinates. The shaft and oval pressure
plate holding the seal in place are projected from a centre
line.
40A. Tri metric Projection In this type of pictorial pro­
jection the three axes are arranged at unequal intervals
greater than 90° in an endeavour to lessen · distortion
which is felt in isometric views. Where two of the angles -
are equal, the projection is said to be ' D i metric ' .
In the example, the given axes are set at 1 00 ° , 1 50 ° and
1 1 0° . Measurements along the three axes are not equal
and the three scales are found by the method shown.
Draw the three axes, extend each and erect perpendicu­
lars to intersect at a. Draw the arcs, and project to the XY
line to find a', b', c'. Draw the semicircle, find z. Join z
to a', b', c'. Draw the constructional arc and project the
irrtersections to give the scale divisions as shown. These
may now be used on the relative axes and the bracket
drawn, note h o w ellipses may be drawn using radial inter­
ceptors, see No. 48.

R GTA R Y VA N E PUMP

21

'
 22

P I C TORIAL P RO J EC T I O N BRACKET
T R I M E T R IC

G IV E N : 3 A X ES

D R AW : TRIMETRIC V I EW
OF BRACKET No. 2 0 6

C
B

THE
AX E S

A
�\
8
f /

�I
I 1/i /
1 /

M E T H O D OF
>
I It O BTAIN I N G
1 00 1 00
° °
SCALES

a
D I M ET R I C PROJECTION
( W H E N TWO A N G LE S A R E EQ U A L }
 CONICS
Ellipse
Parabola

Hyperbola
Evolutes
 ,------- - - - ----- ------------- -- -,I
- ----------- - - -
24

41. The Ellipse When a sectional plane cuts all the

generators of a cone on one side of the apex, the section C ON I C S E L L I P SE
is an ellipse.
42. The Ratio of Eccentricity and position of directrix ·
and focal points are obtained by bisecting the angle at
V and drawing the circle which represents the focal
0 ,...,_
__M
_A�
A,C: 1!5
J_
O_R
_____,.,� 8
sphere. Lines through the tangential points shown in the
diagram show the positions. The ratio is that between the
distance of the focal point from the vertex and the
distance of the vertex from the directrix.
43. Ellipse Construction using a rectangle whose sides are
in the ratio of eccentricity. The longer sides are distances
from the directrix and the shorter from the focal point.
44. Parts of the Ellipse The diagram shows the parts of
the ellipse and methods o f drawing the normals and tan­
gents. Given the major axis and focal points, the ellipse
may be drawn by dividing the operative length of the
major axis any number o f times into lengths A and B and
swinging arcs of A from one focal point and lengths o f
B from t h e other focal point. This satisfies th e rule that
an ellipse is the locus of a point which moves at a constant
distance from t\VO focal points.

Exercises
I . Draw a cone, height 1 50 mm and base circle 1 00 mm
dia, cut by a plane inclined at 45 c to the H P and 1 5 mm
from a generator at the H P. Draw the derivations
(i.
0 ()
of the focus, vertex and directrix from the focal sphere as z� - )(

shown in the diagram. t,<

2. A cone, 120 mm high with a base circle of50 mm radius,
is cut by an inclined plane which is at 30° to the HP. The
major axis o f the section is 80 mm in length. Construct as
in the diagrams, draw the focal sphere and obtain data to a RATIO OF
draw the section ellipse. v,

FOCAi..
�PH'E.RE

PARTS OF ELLIPSE
®

- ..
45. Auxiliary Circle Method of drawing an ellipse. Draw
the two circles equal to the major and minor axes. Radials E L LIPSE
and projectors will give points on the ellipse.
46. Trammel Method Mark the half major and half
minor axes on a strip of card and use this trammel to
generate the curve, as in the diagram.
47. Enclosing Parallelogram.
0
48. Radial Interceptors These methods arc self explana­
tory from the diagrams given.

Exercises
I . Draw an ellipse by the auxiliary circle method, major
axis 1 3 0 mm, minor axis 50 mm.

2. Construct an ellipse by the trammel method. Major

axis 1 5 0 mm, focal points 25 mm from the vertices.

3. Construct an ellipse by the radial method in a rect­

angle 120 mm by 70 mm . . AUX ILIARY C I RCLE MET H O D ENCLO SING PARALLELOGRAM

4. Construct by intersecting arcs from the foci, an ellipse

8 8
whose major axis is I 1 0 .mm long, and foci 70 mm apart.
TRAMMEi...

I ½_ MINO.::. I
5. Given a focus 40 mm from the directrix, construct an
1--�-----i
..
ellipse of eccentricity l
,'')(1 $ ·,

6. A right cone height 100 mm base circle 75 mm, is cut

by a plane at 45 ° to the H P. Project the largest ellipse this
plane can produce from the cone.
-+- ·
I

T RAMMEL METHOD RA DIA L INTERCEP T O R S

25
 26

49. Arc Methods of drawing approximate ellipses. First

I
method. Draw the enclosing rectangle. Bisect the shorter ELLIPSE
side and join to the end of the minor axis. Join the corner
of the rectangle to the other end of the minor axis. The
intersection gives one point on the ellipse. Draw and
bisect the two chords s hown. This will give two centres
for the four arcs of the ellipse.
18
50. The Second Method shows the enclosing rectangle
with a diagonal drawn. Draw a quadrant with half the
major axis as radius. Deduct the radius of the small con­
struction circle from the diagonal. Bisection of the re­
maining portion of the diagonal gives the centres for the
four arcs required.
51. The Isometric Method of drawing an approximate
ellipse. Draw first the isometric parallelogram. Draw the

\ \ \11
two diagonals. Draw the centre lines through the inter­
section. Join apexes to mid points of sides and draw arcs
from the centres shown in the diagram. \ I
52. Section of Cylinder used to obtain an ellipse. Make
diameter of cylinder equal to minor axis. Make length of ARC METHOD I S OMETRIC METHOD
sectional cut equal to major axis. Project as in the dia­
gram and draw a curve through the intersections.

Exercise
\j ------ ·- - - - · - - - - ·

Draw .ellipses by each of these methods, 1 50 mm x 100

mm for I and 2; in an isometric square of I 00 mm for the
isometric method; and using a cylinder 1 20 mm high and
60 mm dia cut to give an ellipse of 1 20 mm major axis,
in the last method.

SECT I O N OF
CYLIN DER

ARC ME THOD 2

-
53. Ellipse Problems given the major axis and foci.
Draw the major axis and mark foci. Bisect axis and draw E L L I PS E
perpendicular for minor axis. With radius half major

G
axis and foci as centres, describe arcs cutting perpendicu­ Any two
lar to give minor axis length. Draw the ellipse by one of para J lq I
the methods shown previously. c h ord$

+
54. Given Two Conjugate Diameters, draw the diameters
as given and from P draw a perpendicular to diameter A
and equal to half A. Join B to centre 0, and draw circle,
B O as diameter. Draw a line from P to pass through the
centre o f this circle to give x and y. The major axis lies
on a line passing through xO, the minor axis passing
through yO. Half the major axis is Py, half the minor
axis is Px. N otice that the tangent of one conjugate
diameter is parallel to that of the other conjugate
diameter.
55. Given the Ellipse find the Axes. Draw any two paral­
lel chords. Bisect these and draw a line to pass through
the midpoints. Draw two other parallel chords, bisect
and draw a similar line. Where these two diameters cross, GIV E N ELLIPSE F I N D AXES
draw a circle which will cut the ellipse in four places.
MAJOR AXIS & FOCI G IVEN
Join to the centre, bisect the angles and draw in the axes.
56. Tangent to a Point on the Ellipse. Join th.: point to
both foci. Bisect the outer angle and draw tangent
through P. If the inner angle is bisected, the line through
P is the normal, and the tangent can be drawn at right
0 G
\
angles through P.
Tangent to a Point Outside the Ellipse. With radius
i Ge,y
equal to the major axis draw an arc from F. With centre
,-
T�
",c

P 1 and radius P 1 -F 1 draw an arc to cut the previous arc. I T

Join these intersections to F. Draw. tangents to pass

through the points where these lines cut the ellipse. !

I
I
GIV E N TWO CONJUGATE DIA METERS : TANGENTS TO POINTS OUTSIDE
CONSTRUCT ELLIPSE OR ON ELLIPSE

r
_I
 28
57. Centre of Curvature on a point of the ellipse. Join P
to the focus F. Draw the normal from P and where this ELLIPSE EVOLUTE
crosses the axis erect a perpendicular to cut the line from
F. Now erect a perpendicular to line P F to cut the normal
at C ; this is the point of curvature and a circle drawn on
this centre will be tangential at point P.
58. Centre of Curvature at the vertex. Describe the arc
centre F 1 and radius V-F 1 . Extend the line by distance
V F, and return a parallel to give C on the major axis.
59. Evolute of an Ellipse ( 1 ) Proceed by the method
shown in the previous diagrams plotting the centres from
points P 1 , P 2 , P 3 . Join the centres of curvature in a fair
curve, this is the evolute.
60. Evolute of an Ellipse (2) Where half the minor axis
is less than the distance F 0, then the two points of the
evolute will fall outside the ellipse, otherwise the pro­
cedure for plotting the centres is the same.

C • CENTRE OF C U RVATURE E VOLUTE OF E L LIPSE

Exercises
1 . Draw an ellipse having axes of 1 50 mm and 75 mm. OF POI N T P
Construct the evolute of the ellipse.

2. Construct an ellipse whose major axis is 1 30 mm and

foci I 00 mm apart. Construct the evolute.

3. Construct an ellipse 0. 7 eccentricity, distance of focus

GI
from directrix 40 mm. Show the centre of curvature of a
point on the ellipse 60 m m from a vertex.

_J

I I
4. Draw an ellipse which is enclosed by a rectangle of V 'F' C
base 1 5 0 mm and equal in area to a square 1 2 0 m m x 1 2 0

-l :c l
y

mm.

. I
5. Draw an ellipse. Data : major axis 1 50 mm, eccentricity
¾, vertex to focus 1 5 mm.

6. In the same diagram, from a focus 40 mm from the C = C E NTRE OF CURVATURE

directrix, draw an ellipse of eccentricity ½. OF VERTEX
EVOLU T E OF ELLI P S E 2

'1

--=
 C O N I-C------- - ------------
- PA R A B - L A- ----
-O-- --
6 1 . The Parabola I f a right cone is cut by a section plane
parallel to a generator, the parabola results. Details of S
the method of projectio n including the true shape of the

0
section are shown. The ratio of eccentricity is unity, i.e. PARABOLIC
the distance of the vertex from the directrix is the same SECTION '- - ----
/ P,._ RALlE l

as the distance of the focus from the ,·ertex. / I BFA.M

ABSCISSA
62. The Relations of the Directrix, the \'CrlCX and the
'/
II
focal point and method of derivation are given in the
' I

diagram.
63. The Parabola may also be constructed by dra\\'ing a
ct
� 1---�!V
centre line and directrix and plotting points of the curve
PARABOLIC
�
by taking equal distances from the directrix and the focal
TERMS
li. 1----�-.......
point. A tangent at P may be drawn by joining P to F and I.U
drawing a perpendicular from the directrix to pass
through P. Bisect the angle at P, this is the line of the Zi
tangent. The normal is drawn at right angles to the
tangent. A parabolic reflector will emit a para!lel beam
of reflected light if the filament is on the focal point. The " ,----,--..,:.--...,;;--�---,,,.J
rad i a I i:,. i1---'---'-.J.-"----'-=-_..;.L,____.
parabola is also the path of a falling body or a jet of
water. i nt«r­
0
64. Approximate Method . Fig. 1 of drawing the para­ s«ctors
bolic curve by radial intersectors in the enclosing rect­
angle. Fig. 2 shows the method of finding the focus and ' 1-1-,½:�'--- -i----'--_;____,
directrix by two constructional squares. The long dia­ axis
-+---,�---�- -
---:.._-:._-_-
__
-_
�_
-_
--
_
_-�
_-
__-�-�' 0 X iS

gonal gives the point on the parabolic curve through

which the square of the latus rectum can be drawn giving
both requirements.
65. Tangent to a Point outside the curve is obtained by
first joining P to F and drawing a circle with F P as
diameter. Draw an ordinate through the vertex and the
tangents are drawn through the points of intersection
and P. The actual point of contact is found by extending RATIO OF
the tangent line until it i ntersects the directrix and joining ECCENTR ICI T Y OIRE.CT R. t X.
a = J_
I
this point to F . If a perpendicular to this line is drawn
from F to cut the curve, then this is the actual point o f b
tangency.
FOCAL
SPHERE

0 TANGENTS TO A POINT
OUTSIDE THE PARABOLA

29
 30

C ON ICS H Y PERBOLA

8
P ROJ EC T J ON O F
AS'1'MPTO,E.

H Y P E R B OLA

RATIO OF
z
.. z
Q w u
)(
'&
�
w
8 A A, 0 8
et
A
� :J
/I
l
al
/

ECC.E.NTR\C.11Y 'II

!
I

G.�EATER
- =
b
THAN
UNITY

PLANE.
--- 1
B,
I ,,I I
I

/
0
CUTTING
B
.0
I
Dou BLE. C.ONE.
GIVEN ASYM PTOT ES RECTANGU LAR
_l_
, a IF & POINT ' p ' HYPER BOLA
I

�
 66. The Hyperbola If a section plane cuts the double 71. The Parabolic Evolute The loci of the centres of
cone on one side of the axis, the section is a hyperbola. curvature of the parabola is drawn in the same way as EVOL U T E

0
The diagrams show the projection of the true shape of that of the ellipse already shown. Take suitable points
the cut. Details of the focal point, the vertex and the on the parabola 1 , 2, 3, 4 and draw normals from these
directrix are obtained from intersections of the focal points. Draw also a line from each point through the
sphere and the section plane and are shown in the lower focal point. From the point where each normal crosses 4
diagram. The ratio of eccentricity is greater than l (unity). the axis draw a perpendicular to cut the focal chord.
67. Method of Drawing the Hyperbola Draw the trans­ From this point draw a perpendicular to the focal chord
verse axis and the conjugate axis. Draw the auxiliary to cut the normal. This second point on each normal is
the centre of curvature for the original point on the
circle. Mark in the two focal points and the two vertices.
parabola. In the case of the parabola, the C of C for a
Draw the two directrix. Draw the two asymptotes
poin.t at the vertex is the focal point itself. The complete
through their intersection with the auxiliary circle. The
method is shown in the diagram.
curve is drawn through points obtained by intersecting
arcs drawn from the foci. The radii of the arcs is obtained 72. The Hyperbolic Evolute Points on the curvature
by first marking off suitable points A, A 1 • A 2 , A 3 , and are obtained by the same method as above by drawing
using radius V A from one focal point and radius Y 1 A the normal and the focal chord, then the perpendiculars
from the other focal point. The four intersections give to arrive at the final point on the normal. The evolute is
points on the hyperbolic curve. Other points may be drawn through the points. In the case of the hyperbola.
found by using radii obtained from A 1 etc. the centre of curvature for a point at the vertex is shown PARABOLIC
in the small diagram. and it should be noticed that the
68. Given the Ratio of Eccentricity the hyperbola may be EVOLU T E
distances V 1 -F 1 and F 1 -0 are in the same ratio as the
drawn by first drawing the directrix and the transverse
ratio of eccentricity of the hyperbola.
I
axis and stepping off the vertex and the focal point in the
ratio given ( 3 : 2 in this case). The rectangular diagram

I
drawn to the same proportion gives values of a and b to
be used in obtaining points for the curve, a lengths being
perpendicular from the directrix and b lengths as inter­

I
secting radii from the focal point F 1 , F 2 , F3 , etc.

I
69. Given the Asymptotes and a Point on the curve. Draw
the two asymptotes A O and OB and the point P. Draw

I
lines A 1 and 8 1 parallel to A O and O B and passing
through P. Draw a number of suitable radials from 0.
Where these cut A I and B 1 , construct parallelograms.

,�
Compare the points x y and point Q in the upper drav,·­
ing. The curve is drawn through the corners of the similar I
I I )(
parallelograms.
70. Rectangular Hyperbola The construction is similar
to that used in the previous diagram. I .I �
I,
I �
I 0
i
0
;

I
HYPER BOLIC
EVOLUTE C. OF Ai
Y E �T I:. ')(

3l

I.·
 32

C O N ST R U C T I O N 1ST A N G L E SO L I D
73. Hyperboloid Given the generator, its distance from
the axis and the height, the hyperboloidic solid may be H Y PER B O LO I D
drawn by the method shown. Begin by drawing the axis,
the generating circle and the first four primary generators
in plan. Project the elevation to give Height, two Asymp­
totes, Throat and diameter from the plan. The envelope
may now be drawn by taking further positions of the
generator in plan and projecting their elevations. This
will give the two hyperbolas of the centre section. Hori­
0
zontal sections are circles ; inclined plane sections are
ellipses as shown. The projection of a point P is also
shown both views. Hyperboloidic forms are used in E L E VATION

\
generating skew gears for transmitting drive in two
planes, such as the rear axle or oil pump in an auto­
mobile.
H O R I Z O N TA L
IS A C IRCLE
Exercises
I . Draw F S in first angle, the hyperboloid to suit the
following data: height 75 mm, generating circle 20 mm in
diameter, generator 80 mm long in plan. Draw the true
shape of the section when the solid is cut by an inclined
plane on the line of an asymptote.

ASYM PTOTE
2. Draw the rectangular hyperbola given by a plane
which cuts a cone of 1 30 mm height and I 00 mm dia
giving a chord line in the base circle of 90 mm.

3. Draw the hyperbola from the following data. Focal

G E N E RAT I N G
length 45 mm, eccentricity 1, base line 1 1 4 mm.

CIRCLE
4. Draw the evolute of the hyperbola of the previous
question.

5. Given two asymptotes at 60° and a point lying on the

bisector of the angle 35 mm from the intersection, con­
struct the hyperbolic curve.

6. Construct a hyperbolic curve when two asymptotes G E N E RAT I N G

are at right angles and a point on the curve is 25 mm from PLAN LINE
the junction.
74. Elliptical Section of Right Cone, given the cone and
the section plane. Ora w the given elevation and the plane-. C ONICS E X E RC I S E S
Project the vertices V Y, this gives the major axis of the OF RIG HT
ELLIP TICAL SECTI O N
ellipse. Ora w the focal sphere; the centre can be obtained
by drawing the axis line of the cone and cutting this line
with a bisector of the angle at A. Project the centre of
the focal sphere to the major axis giving the focal point
CO NE
C:d Y � N CON E
E. SE Ci ION P I. A N E
0
F. The second focus can be also fixed.
The minor axis found by intersecting arcs form the
focal points radius V O or.half the major axis. The ellipse
can now be completed by one of the methods given on : 90
an earlier page.
The directrix line is projected from the plane which ELLIPSES
passes through the two tangent points of the focal sphere I
and the generators of the cone, as shown in the diagram. 11
75. Parabolic Section of a Right Cone, given the cone
and the section plane. Draw the cone and plane. draw
the focal sphere, and project the centre to give F the focal
point on the centre line. Obtain the directrix line by pro­
0 F V

jecting the line which passes through the tangent points

shown of the focal sphere. The parabolic curve can now
be drawn by the method shown on an earlier page.

G I Y E N CONE
76. Exercises
Ellipses
PARABOLAS
E. S E CilON

I. Construct the ellipse given by the section plane in the

PLANE

first sketch shown. (Turn the cone until the cutting plane
is horizontal, the question is then similar to that shown.) I

2. Draw the ellipse required by the second diagram: the

major axis will be 90 mm.
PARABOLIC
Parabolas
SEC TI O N
3. Draw the parabola which would be given by the OF R I G H T
cutting plane to the requirements of the third sketch.
CONE HYPE R B OLA
4. The fourth sketch shows a cutting plane cutting a cone
parallel to a generator and passing through the centre
point of the centre line. Construct the parabola. Con­
struct the hyperbolic curves shown when the double cone
is cut as shown.

33
 34

RELAT IVE POINTS

76A. Conic Problems
I . !fa focal chord is drawn through the focal point o f an
CO N I C P RO B LE M S
elliptical or parabolic curve as at A B in the diagram,
tangents to the curve at A and B will be at right angles to
each other and intersect in the same point on the directrix.

II. Given the asymptotes and the curve of the rectangular

hyperbola. find the focal point. Refer to No. 67 for con­
struction, draw the asymptotes, auxiliary circle, trans­
8
verse axis, conjugate axis and the directrix. From the
intersection of circle, asymptote and directrix, draw a
line at right angles to the asymptote to cut the transverse
axis in the required focal point. (The hyperbolic curve p
used in this example is that used in No. 70.)

III. A similar example is shown using the hyperbolic

curve shown in No. 69. Draw the curve first and the
asymptotes, proceed then with the construction as above.

IV. Given the asymptotes and the auxiliary circle, find

the focal point and draw the hyperbola. Draw the trans­
verse axis, auxiliary circle, conjugate axis, asymptotes
and directrix. From the intersection o f a circle, asymptote
and directrix draw a line perpendicular to the asymptote
cutting the transverse axis in the focal point. Draw the

B
hyperbolic curve by the method of intersecting arcs
shown in No. 67.

GIVEN ASYMPTOTES 1
GIVEN ASYMPTOTES & CURVE:
& A UX ILIARY C I RCLE
FIND FOCUS

FIN D FOCUS &

D R AW HYPER BOLA

.,
CYCLOIDS
Trochoids
Evolutes
 36

77. Cycloid The locus of a point on the circumference

of a circle rolling without slipping on a straight base line. �---C
�Y�C�L=-= D�S______E V O L U T E=--=-
O�l� S_ _ _
Draw the generating circle and the line. Draw the twelve
generating points on the circle, mark off twelve parts
on the line to equal the circumference. Project lines from
the circle to i ntersect the perpendiculars shown. From
the twelve centres strike arcs, radius as circle, to give the /

points on the cycloid. The process may be likened to the

!

positions of a wheel spoke through one revolution.

78. Epicycloid and Hypocycloid If the circle rolls round
the outside of a base circle, the plot of the point yields an
epicycloid. When the circle rolls inside the base circle
the locus is the hypocycloid. CYCL O I D 1!
Draw the base circle and the generating circles. Divide
the generating circles by twelve and draw arcs from 0 .
Step off o n th e base circle twelve divisions to equal the
EP I C Y C L O I D
circumference of the generating circles, and draw radials EV O L U T E OF CYCL O I D
from 0. Obtain the centres of the rolling circles, and
strike arcs, radius R 1 and R 2 to cut the arcs from 0.
Draw the curves through these points. °
ANG L E • 360 1" I<�
The length of the arc o n the base circle is also given by R
using the formula to give the subtended angle at 0 . Two l w L
special cases are shown : (a) When the radius of the rolling ct !
I-
:J \
circle is half that of the base circle the hypocycloid is a
C U RV E
straight line. (b) The rolling circle may be larger than the
&\
-{ I

base circle and thus have internal contact. F O R G E AR

79. Centres of Curvature for the cycloids are shown. TOOTH J UI'
V �\
Given point P, find its centre and draw the generating u.. :J
0 O ..J \]
circle. Find the point of contact with the base line. The
line drawn through P and this point is the normal. Con­ W ;;,, 'I\
IX uJ
struct the tangent at P . Draw the enclosing rectangle, I-

LI-
join the circle centre to 0. The Centre of Curvature is at
\
the intersection C. The evolute is the locus of such centres, �
u�
and is a curve identical to the cycloid itself. HYP OCYCL O I D I
80. The Centres of Curvature and the evolutes for the \
epicycloid and the hypocycloid are shown. Proceed as
ii
before. From P, find the centre of, and draw the generat­ G E H E QAT1r«:::
CI R C L E
ing circle. Join P to the contact point at the base circle,
giving the normal. Radials from O to the centre of the
circle and two points on its circumference, indicate a
smaller generating circle which forms the basis for an­
'
other epicycloid which is the evolute of the original I
epicycloid. E V O L U T E OF I
The Centre of Curvature for the hypocycloid is ob­ EPICYC L O I D
tained by drawing the generating circle for P 1 ; drawing 0
the normal and tangent. The intersection of the normal
and the centre line from O indicate the distant centre.

ff"" �:, =fllll"'!!

(C!f" ..�.
-
....___ ii..._ -=�
8 1 . Inferior Trochoid The locus of a point which lies
inside the generating circle. The construction layout is T ROC H O I DS
similar to that of the cycloid ; notice that the ' spokes ·
will lie in similar positions, but the radius is that o f the
generating circle.
82. Superior Trochoid If the point lies outside the
generating circle (on the extension of the 'spoke '), the
locus is a superior trochoid.
83. Inferior Epitrochoid The locus of a point lyjng in­
side the generating circle which is rolling outside the
base circle. Half the cur•,1e is shown and the construction
is similar to that of the epicycloid.
84. Superior Epitrochoid The locus of a point lying out­
side the generating circle which is rolling outside the base
circle.

I N F E R IOR T ROCH OI D S UP E R I OR T R O C H OI D

I
/ /
'/

:I
0
'j_ _j_

I N F E R I OR E PI T R OCHOI D S UP ER I OR EP I TR OC H OI D

37
 38

85. Superior Hypotrochoid Locus of a point lying out­

side the generating circle which is rolling inside the base S------- --E
H Y P O T ROC H O I D- -VO L U T E S
circle. Project as in the previous examples.
86. Inferior Hypotrochoid Locus of a point lying inside
the generating circle which is rolling round the inside of
the base circle.
87. Centres of Curvature for the hypotrochoids are shown.
Given the point P find the centre of and draw the generat­
ing circle at that position. Join the centre to O which is
the centre of the base circle. Join P and the contact point
on the base circle, also join O to the contact point. The
tangent is drawn at right angles to the normal and through
P. Draw the right-angled triangle having P and contact
point as base. Join as shown to the centre of the base
circle. The centre of curvature is shown a, the indicated
intersection. Draw the evolute through several such points
obtained by projections of P along the given epitrochoid.
In the case of the inferior hypotrochoid the centre of
curvature lies distant and is indicated by · the diagram.

Exercises
Trochoids
I . Draw (a) an inferior trochoid, (b) a superior trochoid
using a base circle of 65 mm dia and a point 1 0 mm
inside or outside the radius as required. Draw only half SUPERIOR H Y P OTR OCH OID I N F E RI O R HYPOTROCHOID
of the complete curve.

2. Construct (a) an inferior epitrochoid, (b) a superior

epitrochoid using a circle of 50 mm dia as rolling CENTRES OF
circle, with a point 5 mm from the end of the diameter.
The circles should roll round the outside of a 1 20 mm
C U RVAT U R E
diameter base circle.

3. Construct the superior hypotrochoid shown in the

diagram. Use a rolling circle of 50 mm dia turning in­
side a base circle of 75 mm radius. The point should be
1 5 mm outside the moving circle.

4. Draw the inferior hypotrochoid shown, using a rolling

circle of 50 mm dia rolling inside a base circle of I 00 mm
radius.
Draw the Centres of Curvature for the two last cases.
F O R .......,
FOR INFERIOR EPITROCHOID INFERIOR HYPOTROC D
1

·- . � " -
88. Exercises
/B
Cycloids
EX ERCISES ON LOC I
I . Construct a cycloidal curve for a point on a wheel ;A

0
..____
60 mm dia which rolls without slipping along a horizontal ¢ so
plane.
-i-
2. A circle 50 mm dia rolls without slipping round the
perimeter of a 1 25 mm dia circle. Draw and name the oJ
s
iocus of a point P o n the circumference of the rolling
t
�
circle.
!

3. Another similar circle rolls on the inside of the base

so
circle of the previous question. Draw the locus of a point
P beginning from a common point with the previous
locus, the circle moving in the opposite direction. Show P1 .0 1"'4 0
how the curves are used in gearing. ! c Il

I'
I

F',
'D '

\I
4. Construct the evolute of the curve drawn in Ques­
_,
tion I .
¢ so
\,

5. Construct the Centre of Curvature for one position

f'

l
� t>P
of a point P on an epicycloidal curve given when a circle
of 40 mm dia rolls on a base circle of I 00 mm dia.

�1
I
A p· 25
(a) Draw the three loci of points P, P 1 , P2 of the circle
which rolls without slipping on the contour shown.

(b) Plot the loci of the point P on the circle which rolls
¢ 35
without slipping inside the semi-circle shown. Draw only �
I• iF
i l
half the plot. 70 1
i
l
iI

(c) Plot the loci of the three points of the circle shown in
t he diagram, which rolls without slq:, ping on th e given
contour.

(d) A square rolls without slipping inside the circle as

shown. Plot the loci of points P and P 1 for one circuit. IA B cO I
->--- N I•

! .[ ____-'----
(e) A circle rolls round an ellipse as shown. Plot the locus
of the three points of the circle as far as B. l 1 E LL I P S E

(/) A triangle moves round the inside of the given circle

t
I

without slipping as indicated in the diagram. Plot the loci

of points P and P 1 of the triangle. ..j
.'L
C
 40

88A. Cycloidal Problems.

A. A circle rolls round an ellipse without slipping. Plot
C YC LO I DA L PROBLEMS
the locus of point P, starting point as shown, for one
revolution.
B. A circle rolls without slipping round an ellipse. Plot
the locus of Q from its shown starting point.
SUPERIOR
B
C. A circle rolls round an ellipse without slipping. Plot
the locus of R from the position shown. E P I T R O C H OI D
I N FERIOR
D. A circle rolls inside the ellipse without slipping. Plot
the locus of T for one revolution. EPITROCHOI D

E. A circle rolls round the inside of an ellipse without C

slipping. Plot the locus of S for one revolution.

Draw the ellipse, major axis 1 3 0 mm, minor axis 90 mm;

base circle A, B, C, 50 mm dia ; base circle D, 35 mm
dia; base circle E, 30 mm dia ; other dimensions left to
discretion.

C YC L O I D
SUPERIOR
l N FERIOA HYPOTROC H O I D
HYPOTROCHOI D
D
E

.,,
 CAMS
Followers
Uniform Velocity
Simple Harmonic Motion
Uniform Acceleration
and Retardation
Radial Plate Cams
Cylindrical Cams
 42

89. Plate Cams If a disc is shaped to a suitable contour

and mounted on a shaft it can be used to lift a follower RADIAL PLATE CAMS
and so a lever or a valve when the shaft turns. Followers KNIFE ROLLER -t --- OFFSET
may be knife ended, roller, flat or angled foot. They
may be in line with the camshaft centre or offset. Knife
followers can follow most contours but wear rapidly.
Roller and flat followers cannot follow hollow contours
and the cam profile must therefore be such that no bridg­ CAMS
ing occurs. Examples of simple cam assemblies are given. AND
FO LLOWERS
foot
90. The Design of a radial plate cam is shown. First the
performance graph is drawn showing the lift. the rest or
dwell, the fall and the lower dwell. The graph may be in
terms of degrees of rotation or in time, which would be
calculated on the basis of one or more rpm. The points
of the graph are now projected to the centre line of the
disc which is marked out in degrees, and brought to the
corresponding radial. These points are joined to give the u n iform v q l o c ity I d w12 1 1 u n iform v12 loc i t y d WIZ I I
contour of the cam. o n 12 I r n o l u t i on 11'D
All but simple symmetrical cams can only turn either o n q R PM
in clockwise or anticlockwise rotation to give the desired
I• ..
c o n v q r s i o n to t i m 12 q .g.
fo I I
performance, and this must be watched in plotting.
Os ( Os 1 30 s �O s
91. Uniform Velocity This term is applied to lifts or
60 s

falls which follow a straight line on the graph; in practice R A DIAL P L A TE CAM a nt i c l oc k w i s 12
the apex is rounded to avoid bounce of the follower. A
cam is shown which would give uniform velocity rise and
. fall of its follower. U N ! FO R M
92. Simple Harmonic Motion This is the sine curve, VE LOC I TY
and is constructed by projecting points from the half­
circle to the graph lines. 360
°

93. Uniform Acceleration and Retardation This curve

is drawn by fixing apex, base and midpoints, and in the
four quarters thus made, drawing six radials and six
verticals. The curve is drawn through the intersections.
The shape of the finished cam is shown.

P E R F O R MANCE GRAPHS UN !FORM ·..,\ \

ACCELERATION \\
RADIAL P L ATE CAMS & RETARDATION

... ... " (. ... ..

94. Radial Plate Cam Design a cam to fulfil the con­
ditions in the given data. Draw the graph lines showing RA D IA L PLATE CAMS
a 24 mm lift and twentv-four 1 5 ° vertical divisions. Draw
the straight line U V lift 12 mm for 1 20 °. Plot the S H M.
curve on t h e next 90 °, note th e small semi-circle for the
plot. A straight line horizontal gives the dwell or rest to
270° , and a straight sloping line to 360° gives the final
U V fall back to the base line. Draw the circle for the
cam and project the points on the graph to the centre

U V R ISE � SHM R IS6 R EST i UV FALL

line. Swing each point to its own radial and the i n ter­
.
' oo 2 10 ° 270 °
sections give the outline of the cam. This cam is suitable
1 20 °
°
only for -a knife follower as a roller or a flat could not
, 270
RADIAL PLATE C A M
follow the indentation.
/
95. Radial Plate Cam When a roller ended follower is
used, the lift i s measured from its centre, and the cam i \' \\ SHAFT (/) 10 BASE CIRCLE (fJ 25
I 120\i 00-120° U V L IFT 1 2
line is drawn tangential to the roller circles after plotting.

I
Draw the graph to satisfy the given data, 22 mm li ft and

I 1 2 0° - 2 100 SHM L I FT 1 2
24 positions. Construct a Simple Harmonic Motion
curve to lift 22 mm in 1 80 °, high dwell for 30° shown by
2 1 0 ° -270° DWE LL
I SHM
the straight line, Uniform Acceleration and Retardation I

270 ° 360 ° U V FALL 24

curve, to fall 22 mm to the base line. The heights of the

I S U ITA B LE FOR KN IFE FOLLOWER

positions on the graph are projected to the vertical centre

-----
line of the circle and each swung to its own radial. Roller

1 0°
circles are now drawn o n each centre, and the 1.:am profile
is obtained by drawing it tangential to the circles. This
�, /---ANTICLOCKWISE ROTATION -----<
r''\ 8
profile is also suitable for a flat ended follower as there --...___,
are no concave curves.
Quick models may be made of plate cams by cutting

-.S H M
in balsa- or plywood and mounting the shape on a pin,
testing the performance against a plywood follower.

I
I
I

RADIAL PLATE CAM

j

+
1 .
I

I SHAFT ¢ 10 BASE Cl RCLE ¢ 2 5

FO LLOW E R (/) 1 0
o0 - 1ao0 SHM LIFT 22
100 ° - 2 10 ° DWELL
2 1 0 ° 360 ° U ACC / RTD FALL 22
 44

96. Radial Plate Carn, Offset In this example the line

of the follower is offset to the right by 16 mm. Draw the CAMS
cam circle, draw in the line of the offset and draw the first
•ioHSU 16

---------'--_;.,�----lI,___ ______
low position of the roller with centre on this line and tan­ C LO C K W I S E -
gential to the centre circle. Project the centre of the roller
° 0
lc:,6 120 , 1so ·�:g.S
O
3�0

to give the base line of the graph. Draw the graph layout
I - ·- �
to give 25 mm lift and 24 divisions in 360° . Draw the x)i
Uniform Acceleration and Retard curve for' 120° and -}-

25 mm lift; straight line dwell for 30 ° , and a sloping

straight line to 225° for the Uniform Velocity section,
fall 6 mm; and a Simple Harmonic curve to 360° falling

_____ ____
.......
L,
1 9 mm to the base line. Before being able to project the l u ACC & RTD IR�I UV SH M
points to the circle, it is necessary to draw the tangent _.:.__
c.. oc.1C.w 1s£ !
��_.:.__

lines which are generated at the centre circle. The points

from the graph are now projected as in the diagram and RADIAL PLATE CAM OffS-�T TO ��MT 16
the roller circles drawn. The cam profile is drawn tan­
gential to these.
o0
SHAFT 9f 1 0 C / L TO ROLLER C 1 8
97. Carn and Radial Arm The diagram shows a cam - 1 :20
° u ACC & R ETARD L I F T 25
which lifts a radial arm instead of a vertical follower. 120 ° - 150° DWELL
Notice that the roller on the end of the arm moves for­
ward as i t passes through the centre of its travel. 1 50 ° - 225 ° U V E LO C ITY FALL 6
225 ° - 3 6 0° S H M FA L L 1 9
98. Cylindrical Carn If a groove is machined in a cylin­
der, a suitably shaped radial arm can be arranged to
follow the groove. A development is made of the cylinder CAM AND RADIAL ARM DEV E LOPM ENT SHM
and the motion plotted centre line first.

II

CYLIN D R I C A L
CAM

r- �
 EX E R C I S ES ----------- - --
;------- ---------------,.---------
Exercises
C AM
99. Cams
I A. Design a plate cam to the following data: 1 2 mm
R A D I A L P L ATE CAM (;;\. A I RA D IA L C A M
B

SHAFT (/) 12 M I N (/) 38 U FT 25 � S H A F T q) 1 5 M I N (/) 2 5

dia shaft, 38 mm minimum dia, 25 'mm lift. Clock­
ROLLER (/) 1 5 L I FT 25
wise rotation, knife edge follower. Performance: Uni­

I �jl
fonn Velocity 1 8 0 °, 3 0° dwell, Uniform Acceleration and
Retardation fail 1 5 0 °.
B. The sketch shows the requirements for a radial cam.
Drav.: the earn to give a rise of S H M, short dwell, and
fall of U V to the required time factor. (Five rpm gives

_JN I
1 2 s/rev, i.e. the cam rotates through 360 ° in 1 2 s =
30 ° per second.) j !
I It)

5
C. From the cam given, draw its performance graph.
o__/ ��. U ICC & RTD • M ___,: AL L
,-
! __ U_ s __S_
4_ H
iI

Y_ r e_
__ / 1�-
... __ ,s_
�! ;; ...,__ _
u __
v F__ -.-J' I s ....
I

D. Draw the double lift cam shown, and draw its dis­
0

placement graph.
E. The sketch shows the details of a cylindrical cam. 5 R PM
Draw out the development of the cam surface, and pro­
ject the required groove thereon, S H M.
DRAW
------.- \C l
D .
F. A simple quick return cam operating a plunger is
shown. Draw out the performance graph. State the form.
�1
_L
I
/I
/
, ! I
2. Draw a radial cam to give the following performance. I
Clockwise, 0 °- 1 20 ° S H M 45 mm lift, 1 20 °- 1 50" dwell.
___,__!! 0 u, I
1 50 °-300 ° 45 mm fall U V ; 300 °-360 c dwell. Shaft dia
10 mm. Min. dia 32 mm.
Ni
J:(__._-'--
/ -e.. , '&
-

3. Radial cam turns at I O rpm. Show its performance

graph to give a lift of 25 mm U V, a I second dwell, and a 1 \
l
D R AW
fall of 25 mm S H M. Flat ended follower, 6 mm offset.
GRAPH
Rise and fall equal duration, clockwise.
II
! E l/
4. Design a radial cam to work a radial arm, 50 mm long,
ir I

'
external diameter of cam 75 mm, minimum dia 40 mm.
S H M rise and fall with I 0 ° dwell and rest for a 1 2 mm 7 D R AW �l ROTATION '?
G RAPH
/ 50 lattZ r a l movGimGint i
dia roller ended follower. Anticlockwise rotation.
7
OEE..P
"" --.
j
I

°'!,
5. Design a radial cam to give a 30 mm rise and fall to
the following data. Shaft 1 5 mm dia. minimum dia 25
R . H . R O TAT I O N
st . II
mm. S H M for 1 00 ° ; dwell 20° ; fall 20 mm 1 00 ° U Y ;
dwell 20° ; fall 1 20 ° Uniform Acceleration and Retarda­
tion ; suitable for a IO mm dia roller ended follower. DRAW GRAPH �:!-·-
offset 1 0 mm. Rotation clockwise. ---+ O i
-e.. !
:

I DRAW
(/\

CY L I N D R ICAL C A M

.___________ J_
I ! 12 ! G RAPH
'--=
. --'¢
\...... 25 ,I L I FT �
=--...i
 46

99A. Cam Problems

A. A radial plate cam profile is shown. Draw the per­ CAM P R O B LE M S
formance graph of the cam.
B. The shape shown is to be used as a radial plate cam. A
A roller ended follower, 1 5 mm dia is used upon the
cam. Draw the modified profile of the cam to suit this B C
condition. Draw the performance graph.
C. Draw the cam profile shown, suitable for a 1 2 mm
dia roller follower. Draw the performance graph. \
D. The performance graph of a plate cam is shown.
Draw the cam profile which would give such an effect.

Describe the performance of the four cases shown above, % -

· I ¢' 1 2

I
(a) in terms of degrees, (b) in terms of time per revolution.
I
_..j lO

1. 14
5 .I s
I

I� 10 SECS. P E R. R EV.
�I

,. --
HELICES AND GEARING
Glissette

Roulette

lnYolute

Spirals
Spur Gearing
Bevels

Worms
 48

H E LI C E S SQUARE THRE AD SPRING -,

.__;,_____,_______....______..___..� ---r I
I
I __..- I

__, __
I
,
I __..-,
�--+--------�-____.
!.-..l-------'-----'----4---1 10

I
I
1----l--...J...------�--+----+-I 9
II
lll
---- :
e

I
0 I
I ---=-- �·
__.,.,..._,1:."{�
l
�-...J...-------------..J<--J 5 I
8
i I
I
i
""'°

I I
" 'v '
I
1.---'""'i
II

I __..- ____.
1 __...-:: II I
i
j
l;,�v.---.
I I
I
f-+--__.._--+--�-� 4
I i I

I
l
,-!-----+--,.-�---'----+-l 3
I
Y'""'
! i
__. I 1
I
1-,--�M---+---+- --+---l Q I

0
I

I i

I
I
0 3 €, "' 7 8 9 10 1 1
i---------- -- - - 'ii'D

II

0
g
JO

II
e

.
10
7 0
II

Jl
10
9

"
b
I
�
4
3
2
0

0
4

3
2 0 �

RH 0

B 8
RIG H T HAND
SQUARE TH RE.AD
VARIOUS HELICES S I NG L E START CIRC U LAR SECTION I
c_____ _____ ___________ _
_ ___________________ _J
 100. The Helix The locus of a point which moves
around the surface o f a cylinder so that its axial and radial S Q UARE T H R E A DS
speeds are in a constant ratio. The axial advance is the
lead, the pitch is the distance from a point on the helix to
a similar point for one revolution-centre to centre of
the screw thread for example. In a one start thread the
lead and pitch are the same, for a two start. the pitch is
half the lead. The projection of the helix is a straight line.
The helix may be right-hand or left-hand. To draw the
J:
a:
helix, project the twelve generators to intersect corre­
sponding twelve divisions of the lead. The curve drawn

....
is a simple harmonic motion curve. see cams. Various 0
helices are shown.
101. Right-Hand Square Thread single start. Pitch equal
to lead. Thread section a square P/2 side. Generate two
V)

a:
UJ
helices, one for the root of the thread, and one for the Vl

crest.
102. Left-Hand Spring, circular section. The helix to be
plotted is the centre of the circular section. The lines of
the spring are then drawn tangential to the construction
circles.
103. Two Start R H square section thread. Mark off the 2 START R H S QU A R E THREAD
lead on the cylinder blank, make the pitch half the lead,
the thread section is a square P/2 side. Draw the helices,
four for the crest, four for the root.
3 STAR T L H SQUARE THREAD
1 04. Three Start LH square thread, pitch is one-third
lead. The helices rise to the left-hand. six for the crests.
six for the roots. See also Worms in Spur Gearing.
Exercises
I. Draw a single start square section screw thread of 7 5
mm outside dia, lead 25 mm. Complete three turns,

------+--t- 0
L H.
2. Draw one and a half turns of a spring. 80 mm mean ....
diameter and 50 mm lead, R H, made from 1 5 mm dia
wire.

3. Draw a two start square section thread, L H 25 mm

pitch, outside dia 1 00 mm. Complete two turns.

�:
PITCH
"l LEAD 8
49
 50

105. Glissette When a line slides between two other

lines which are fixed, the locus of a point on the sliding G LISS ETTE ROULE TTE INVOLUTE
line is called a glissette. The lines may be at any angle and
may be straight or curved. The example shows a straight A
line moving through six positions between two straight
lines which are at right angles. Locus of P is the glissette.
106. Roulette When one curve rolls upon another with­
8
out sliding, the locus of a point P on the rolling curve
describes a roulette. The points can be plotted by tracings
or cutouts of the two curves in progressive positions.
107. Involute of a Circle If a straight generating line
rolls round a circle the locus of a point on the line is
the involute of the circle. The Evolute is the circle itself.
The involute curve is used in obtaining the form of spur
gear teeth. To construct the involute, divide the generat­
ing circle into twelve paris and draw the straight line
equal to the twelve divisions and thus the circumference. � �
Draw the first tangent and make it the length of one GLI S S E T T E R OULE TTE
I :t I> "I

division. Draw the tangent at 2 and make it two divisions

in length. Proceed in this manner until the twelve tangents
have been drawn. Draw a fair curve through the points,
this is the involute. If a cord is unwound from a cylinder,
a pencil placed in the end loop will trace the involute
curve.
I NVOLUTE OF A CIRCLE
§ 8
108. The Involute of a Square Draw the square, draw
the quadrants as sh·o wn from the corners. The radius
increases by the length of a side at each quadrant.
\
\
\
\

/I
Exercises
l. Draw the glissette in l 05 using a line 70 mm in length.

2. Draw a parabolic curve by the method of 64, using a

rectangle 70 mm x 40 mm. Roll the curve around a
II
quadrant of 70 mm radius. ,>�<:,,
..
3. Draw the involute curve when the evolute is a circle
of 35 mm radius. INVOLUTE
OF A
SQUARE
s
I I I J
3 � (, 7 8 g <O II I�
Spirals
The locus of a point mo�ing continuously in one direc­ S P IRA LS
tion along a line which is rotating about one of its ends. 8

109. Archimedean Spiral Given a pole P and limiting

vectors, A length and B length. Draw A and B in line from
the pole, divide A into twelve parts, and draw twelve
radial vector lines. Draw arcs to cut the radials pro­
8
gressively as in the diagram and construct a fair curve to
pass through the points of the spiral.
The normal and tangent to any point on the spiral can
be drawn if the value of C (which is a constant) be found
by the formula
= ___ r-
_ a
C
e (in radians)
where r is the radius length of the point
a is the initial radius vector length
0 is the vectorial angle in radians
( I ° = . 0 1 7453 rad).
Draw C at right angles to the line joining P and the point
on the spiral. Draw the normal from the end of C to the
point; the tangent is at right angles to the normal.
..... _ - ... 1 2
\
1 10. Logarithmic Spiral Given the angle between the I

vectors, the length of the first vector, and the ratio of the
vector lengths, 5 : 4. From the pole C draw the radial lines I�
of the vectors at the angle given. Obtain the lengths of
the vectors by drawing x = vector, y = ¾ x vector, angle
45 ° , and by arcs and parallels obtain the progressive vec­
tor lengths to be used on the numbered radials. A fair
curve through the points gives the spiral as shown in the
diagram.
1 1 1 . Conical Spiral Divide the vertical height of the
cone into the same number of parts as the base circle.
Join the plotted points with a fair curve. LOGARITHMIC
Exercises SPIRAL
I . Draw the involute of a 50 mm dia circle.
2. Draw an Archimedean spiral whose vectors are 50 mm
and 1 5 mm.
3. Draw the logarithmic spiral given two vectors 1 8 mm
and 25 mm at an angle of 40c .

4 . Draw the spiral of a cone of 50 mm height and 75 mm

dia base circle.
5. Draw the spiral of a hemisphere of 90 mm dia.
51
 52

1 12. Involute Spur Gears The involute curve o f a circle

is used in the profile shape of the involute gear tooth. The INVOLUTE SPUR GEARS
base circle i s the generating circle and is generated tan­

8
gentially to the pressure line.
1 13. Terms used in Spur Gearing. INVO LUTE OF C I R C LE
Pitch circle diameter PCD
Pitch point
Line o f action
Pressure angle 20 ° or 14½'
Addendum a
Dedendum d
Clearance c
Circular pitch p
Circular tooth thickness
Number o f teeth
p/2
T
Add� n d u m
Module m (metric, in mm)
Base circle diameter BCD
· The circular pitch is the distance centre to centre of

8
the teeth measured on the pitch circle. lnvo l u t (2
Module m = ---y-
PCD c irc I 12

WHEEL
It is usual to give the pressure angle; the number of
teeth, and the module of the gear to be drawn. Other
essential proportions and sizes can be found by the use
of the formulae shown.
1 14. First draw the touching pitch circles.
Draw the pressure line through the pitch point.
Draw the base circles tangential to the pressure line.

..
Draw the involute curves on the base circles.
Step off the pitch points on the pitch circles.

·- - I
Draw the addendum and dedendum circles.
a:
0
Draw the involute curves through the pitch points.
Complete the shape of the teeth by drawing the root radii. >- • .
N

I . -
Approximate compass curves may be used in drawing
0 "
>· (
V
the teeth, using one of the methods shown later. 0

Q.
u
-t.
1 H E EL ...;

.t:

I
'II I
ti,
i
I
� 3:
 46 '"'> _;:.
I
, (J /-
-f(i1

1 15. A Worked Example of Spur Gearing.

M ETRIC I N VO L U T E G E AR AND R AC K NOT"ATION
Data given : module 10, teeth 25, pressure angle· 20'.
Draw five meshing teeth of two equal gears.
GIVE N : D RAW FIVE TEETH·!.11
/

8
Using the formulae shown, with the given data, obtain
MESH ING
the P C D, the addendum, the dedendum, the circular i TEETH = 25
pitch, and the tooth thickness.
M ETRIC MODULE = 1 0

I
EXAMPLE
Draw the touching circles, pitch diameters. Draw only
enough arc to enable the five teeth to be shown. I N VO LU T E P i tch c ! r c l � <ll 2 50 1

i 2 cf
! PRESSURE
20 i
0 I G E A RS
'i or 14.5
''
Draw the common tangent to give the pitch point. '
ANGLE
'
Draw the pressure line at 20° to the common tangent
:
II
and through the pitch point.
I

MODULE PC D
I
m
I
;
Construct the base circles tangential to the pressure
i T
10
line.
i CIRCULAR i j PC D ic 11'
i
Draw the addendum circle and the dedendum (root)
.circle. PITCH
! C.P.
T i 3 1.43
I P C D fj
or P !
T ort --
Construct the involute tooth curve on the base circle.
Step off the half-pitch distances along the pitch circles. :
TEETH I 25
i I
Draw the involute tooth curve through these points to

I
I

give the full tooth form. Use a tracing for reproduction

PITCH CI RCLE ! II 2 5 0
and reversal of the involute, or an approximate compass PCD Tx m
-f o
D IAMETER : i il
__

curve of the involute. See No. 22, page 5. I !I '

--- i
Complete the tooth shape by root curve radius W/7.
A D DENDUM
C

As a further problem, use the above data to draw the

wheel as above, then proceed to draw a meshing pinion CLEARANCE point
to give a reduction ratio of 5 : 3. Tabulate the necessary
information of both gears before attempting the drawing.
Since the wheel teeth are 25, the reduction ratio gives 1 5 DEDEN DUM
teeth in the pinion, T x m gives 1 50 mm as the pitch
II
:i
diameter of the pinion. The pinion tooth form must be
OUTSIDE 9f 0D (T+ 2) m 270
I
drawn using its own base circl_e and involute. j
CENTRE

!
I 16. The Form of the Involute Rack is shown. l½(T+t) m I
DISTANCE
c o 2so /

The involute curve taken to infinity becomes a straight

11 ,,,

fi R
/

line, the proportions of the rack are as the meshing gear,

the slope of the flanks of the rack teeth are the same ·ROOT ¢ l(T-2.3 1 4)m /J 2 2 6.861 I
angle as the pressure angle given, 20 ° . /

!;-...;
2Cf/ ',-I·--p-
/

31.4 -.. 3
TANGENTIAL
-+
--;
TO PRESSURE
! 01 I ADDM LINE
-- -, - 1-

-L___;- PITCH LINE - l n v o l ut�

DEDDM of. ban
cl rel�
t/')

I NVOLUTE RACK 8
53

0 -\'1/\. L
l
 54

1 17. Pinion and Rack The data given, module, pinion METRIC P I N ION AND RACK N O TATI. O N
teeth, pressure angle. Draw three teeth of the pinion
meshing with four teeth of a straight rack. G IVEN: DRAW:
Using the formulae and the given data, complete the m 5 3 P I N I O N TEETH
table before beginning the drawing.
Draw the part of the pitch circle required.
Draw the pitch line of the rack passing through the
T 25
ANG 20°
M ES H I NG RACK
SCA L E 2 : l
8
pitch point. I NVOLUTE S PU R GEAR
Draw the pressure line through the pitch point.
Draw the base circle tangential to the pressure line. ANG 20°
Draw the addendum circle and the dedendum circle.
Construct the involute curve on the base circle. , m PCD 5
T-
Step off half-pitch distances along the pitch circle.
-
Draw the involute curve through these points to give C P � 15.7
the.tooth form of the pinion, A tracing or an approximate T
compass curve may be used. PCD
Complete the tooth form by drawing the root curve. I T m 25
The rack may now be drawn. The flanks of the teeth
are straight lines at 20 ° to the rack pitch line, the same PCD T x m 125
angle as the.pressure angle. The root curve radius is again
W/7. Ad m 5
Compare this drawing inverted with No. 1 22 and
No. 396, noting how the rack form is used for the section Cl 0.157m 0.8
of the helical worm.
Dd U57 m 5.53
Exercises
¢ D PCD-2m 135
I . Two spur gears mesh to give a velocity ratio of I : 5, ¢ Root
the pressure angle is 14.5 °, the pinion has PC D 96 mm, ¢R PCD-2D 1 1 3.94
the module 6. Draw two teeth of the pinion meshing with
three teeth of the wheel.

2. The centre distance of two spur gears is 240 mm, the

pinion P C D is 1 20 mm, pinion teeth 20. Draw three
teeth of the wheel meshing with rwo teeth of the pinion.

3. A pinion has 20 teeth and meshes with a rack. Module

1 0. Draw the pinion meshing with five teeth of the rack. Pitch circf
Draw one tooth of a spur gear indicated by the following
data: pressure angle 20 °, teeth 22, module 8. Scale 2 : l . ¢ O utside
I
I
 -----
1 1 8. Mitre BcYel Gears The diagram shows the con­
ventional method of representing two equal bevel gears B EV E L GE AR WO R M GE A R
whose shafts are arranged at 45 c . and for this reason are

8
called mitre gears.
The axis li�es of the shafts are drawn first. followed by M IT R E BEVEL
the pitch and back cone lines. The teeth section can now GEARS REDUCTION B E V EL
GEARS
be inserted.
1 19. Reduction Be\•el Gears The ratio of the reduction
between pinion and wheel is in the same relationship as

,z
the pitch circle diameters.
120. Worm and Wormwheel A reduction gearing be­
tween two shafts whose . axes are at right angles. The .Q
toothed wheel is driven by the cylindrical worm whose z::

screw surface is an involute helicoid of single or multi­

�

G
start thread form. The worm tooth section is that of the

"'
involute rack (see 1 1 6). and the tooth section of the worm­
wheel is a suitable meshing involute spur gear, shovm in
rJ

detail in No. I 22. The reduction ratio is as the number

Q..

of teeth in the wheel as to the starts in the worm, i.e. a

30-tooth wheel driven by a single start worm gives a re­
duction of 30 : I, whereas a 30-tooth wheel driven by a
three-start worm gives a l O : l ratio. The worm is usually
made from case hardened steel or normalised high carbon P C D '-"' H 6 E L p C. 0
steel, and the wheel is made from phosphor bronze or
cast iron.
Proportions

D = Pitch dia of wheel

d = Pitch dia of worm

8
WORM

Pn = Axial pitch of worm

Pn = Normal pitch of worm GEARING

b = Dedendum of worm
a = Addendum of worm 0.5 p.,
C <ZOrCJnC<Z

t = Number of threads in worm

�-+-- · - -- · �1___

T = Number of teeth in wormwhcel

), = Lead angle of worm (tan i. = 1p0/ rrd) "TH ROAT
DIA
Useful guides: make d= 0.2D ; also a + b not greater than
d/2. Further reference to BS No. 72 1 .
D I ,..._

R.OOi /
W O R M W H E EL
OIA - D

'
I
1
�
55

\
\
 56
121. Bevel Gears The diagram shows details of a pair
of mitre bevel gears, 15 teeth, module 6 mm P C D 90 mm. MITRE BEVEL GEARS
Draw the pitch cones and the back cone lines. Draw
the generating tooth form, projecting the pitch circle
from the pitch angle line. The spur tooth form is drawn
conforming to the proportions given on previous pages.

G E N E RAT I N G TO O T H
The addendum and dedendum lines can now be swung

FORM P C D 1 2 7." 2 6
back to the back cone line and joined to the vertex point.
The shape of the teeth can be projected as in the given
partial elevation by projection from the section and by
S ECTI O N OF
B EVEL G EA R S
lines radial from the elevation centre to the generating
MITRE
tooth form as shown.

Exercises

I
1 . Draw the pinion of a pair of bevel gears which give a

A N G L E 4 5°
reduction of 2 : 1 , pinion has 20 teeth, P C D 200 mm.
C omplete two teeth.

2. Draw the sectional view of a pair of bevel gears pitch

angle 60 °, reduction I : 1.25. Pinion has 1 2 teeth, P C D
72 mm. Generate one tooth of pinion.

3. A pair of bevel gears work at right angles, the pinion

has 1 5 teeth, gear ratio 5 : 6, pitch circle diameter 1 50 mm.
Draw the sectional views of the gears, show the generat­
ing tooth form, and project three teeth of the pinion.

4. Draw one tooth of a spur gear, 20 teeth P C D 160

mm. Use this form to give the shape of the teeth of a
mitre bevel gear.

5. Draw the pinion of a pair of bevel gears the axes of

which are at 85°, P C D 60 mm, teeth 1 2. Complete the
two full elevations 2 : I Scale. P C D 90

GEAR D A TA 15 T E ETH
MO DULE 6
P C. D. 9 1,S

- .,,
 METRIC
122. ·worm and Wheel The diagram shows projections
of a 25. 1 5 mm lead, single start worm meshing with WO R M G EARS TRA I N & RACK
a 30 tooth, 25. 1 4 mm circular pitch involute gear worm­ 25•14 1)-.EAD
r--,r-;-;-:-,,:-77"1-=-.,-,!--�
LEFT H A N O STA R T
wheel whose P C D is 240 mm. --- ,.
1 -----

Draw the mid-section of the wormwheel first by the

methods previously shown-pitch circle, tangent, pres­
sure line, base circle, tooth spacing, tooth shape by
approximate compass form.
Draw the mid-section of the worm teeth to involute
rack shape, the tangent line forming the base line. The
rack form has been shown on a previous page. Make the
centre core of the worm not Jess than twice the depth of
the teeth, this enables the cylinder of the worm to be
found. Draw the construction for the single start helices,
25. 1 4 mm lead. The worm is shown completed in the dia­
grams.

123. Train of Gears A simple train of gears is shown.

the ratio of reduction or increase is equal to the number
of teeth in the final gears. the intermediate gears being
I
-
idlers which only aff;ct the direction of rotation.
In the compound train. the reduction or increase is INVOLUTE RACK S E CTION
obtained by multiplying the teeth of the drivers together
and dividing into· the ;umber obtained by multiplying 3 0 TEETH PC D 2 4 0 CIRCULAR PITCH .25 , 1 4
the driven teeth together. Rotation may also have to be INVOLUTE SPUR GEAR SECTION M O D U LE :=: 8
considered. usually effected by the introduction of an
S I M PLE TRAI N
equal idler at the beginning.
Two involute gears and rack are shown in oblique pro­ I NVOLUTE RACK I NVOLUTE SPUR
jection. GEARS S P U R GEARS

l
'I.
Further specialised information on gears may be ob­
tained from BS 436, BS 545, B S 72 1 , BS 2 5 1 9, I S O
R53.

8
�
- ----- - /
)

- ·
COMPOU N D
�---------- - - - -- - ...il.------------------ -
 58

123A Gear Angles and Forms: Worm. The diagram

shows a pitch helix given a worm lead angle of 30 ° . Draw G EA R A N G LES & FO R M S
D = Pitch circle diameter. Divide the semicircle into six
the centre line (axis) of the worm, draw the semicircle

pans, step off chordal distances O', 1 ', 2', 3' along the
base line as shown. Perpendiculars cut the slope line
giving the heights to be projected to the generators from
0, l , 2, 3 . The helix can be drawn through the intersection
points. Note the spiral angle, also the lead length.
Spur teeth simple involute form, can be shown sche­
matically by projection lines which are parallel to the
shaft axis of the gear.
Helical gears single helix, can be indicated by slope
lines at the same angle as the helix angle.
End thrust is engendered when single helical gears
work under load, and this is usually absorbed by ball or
roller races. S P IR AL __.,.r
Double helical gears are indicated in the third diagram. ANGLE
Gears having this form cancel out endthrust. I
Note that when drawing meshing gears of the last two I
cases, the slopes indicating the teeth will lie to the oppo­
site hand in engagement. �

Exercises
I . Draw the pitch helix o f a worm, lead angle o f 45° . The
diameter of the pitch circle is 60 mm. GEAR FORMS

2. Draw helix diagrams where the lead angle is (a) 60° ,

(b) 22½0 ; these angles are frequently used. Use the same
I

I
pitch circle diameter, 60 mm.

3. Draw the three gear forms shown in schematic form,

showing in each case a pinion of 50 mm meshing with a
I
70 mm diameter gearwheel. !
I'
See also BS 308 Part 1 for conventional representation of
gears.

S PU R H E LICAL H ELICAL

S C H E M ATIC

..
FORCES AND FRAMEWORKS
Triangle of Forces
Vector and
Stress Diagrams

Beam Moments
Funicular and
Link Polygons
Frameworks
 60

124. Resolution of Fore es

A. If two concurrent co-planar forces act upon a point, R ESOLU T I O N OF FOR CES
they must be equal in magnitude (amount) and direction
(line of action) if equilibrium (balance) is to be main­
tained.
B. An unstable couple is set up if the forces are equal but
not opposite. Rotation ensues.
STABLE
5
B
U N STA BLE COUPLE
C
5
STA B L E
5 5
D
8 5

'
C. Two parallel forces may be opposed by a single force
equal in magnitude and parallel in direction. R OTAT I O N

\
D. Two and two equal parallel forces in a state of equi­
librium.
E. Simple example of a common beam centrally loaded
by ten units balanced by left and right reactions R L and
R R of 5 units each. Mass of beam ignored. 5 10 5 5
F. Offset loading of simple beam by 1 2 units. The reac­
tions R L and R R are found by dividing the load in the
ratio 2: 6 shown in the diagram as the division of the
length of the beam between the supports. Notice that the E 10 F 12
greater reaction is at R L in this case, owing to the load
being nearer that end. The mass of the beam, force units
per metre = Nm would be divided equally between the
two reactions.
6m
125. G.. If two concurrent co-planar forces act at an
angle on a point P, their res4ltan1 (a single force equal to
the two forces) may be found graphically by drawing a
scale triangle (or a parallelogram) of forces. Represent
the two forces by vector Jines, length to scale, showing
magnitude, direction (angle) as given, to form two sides
of the triangle. The third side, closing the triangle, repre­
sents the resultant force in magnitude and direction R E S O LU T I O N OF E Q.U I L I B R A N T BY
GIVEN :
(angle).
TWO F O RCES VECTOR DIAG RAM
The equilibrant-a force equal to the resultant in mag­
nitude and direction (line of action)-acts on the point AT A POINT
in the opposite direction (pull or push) to the resultant. SCALE D I A G AAM
In the vector diagram, the arrows must · flow round ' the 6 VECTOR I
diagram; note how vectors I and 2 must be related to
obtain the ' flow '.

! p FORCE I

S C ALE OF U N IT S

-- - - ' . --==·--·- --
-- ---._:;_ ___ -
- ... "'
� - --
,_
 126.- Several Concurrent Forces at a point. A. Four forces
are shown acting at a point P. A fifth unknown force, the R ESOLUT ION OF FORCES
eq uilibrant, is needed to maintain balance. This is found
by drawing a vector diagram showing the known forces A
to a suitable scale·in both magnitude and direction to the
given data. The fifth, closing side of the polygon, shows
the equilibrant as a vector quantity, and its magnitude
and direction to scale. If the forces are arranged so that
CONCURRENT
FORCES AT A P O I NT
F3
8
the arrows ' flow' round the polygon, the push or pull
action of the equilibrant is indicated.
127. B. A further example of forces acting at a point.
When the vector polygon is being drawn, the closing line
showing the equilibrant, crosses the vector F2, but F 1 ,
F 2 and F3 must be drawn as shown in order that the
arrows can ' flow' round the polygon.
POLY G O N
128. C. Non-concurrent Co-planar Forces Given the U N ITS
direction and magnitude of four forces which are not
acting on the same point, find the force necessary to I
establish equilibrium.
Label the spaces between the forces as shown using
capital letters-this is known as ' Bow's Notation '. Con­
struct a vector polygon as above of the known forces to
scale and close the polygon so obtaining the magnitude
and direction of the equilibrant F5 •
B
G I V E N : THREE
FORCES AT A
VECTOR
POLYGON 8
Letter the points of the vector diagram by small letters POINT
a, b, c, d, e. Thus the vector ab represents the force F 1
lying between A B of four units magnitude and direction.
Join abcde to any pole o in the vector diagram. �'"'�
..
��
�
u.

I n space B, in the space diagram, draw a line parallel

to ob; in space C, a line parallel to oc, completing the
/
��
diagram with lines od, oe, oa. Intersection of oe and
oa will give a point through which F 5 , the cquilibrant,
may be drawn parallel to ae.

61
 62

129. Beam Moments The resolution of forces acting

FUNICULAR ( LINK) POLYGON BEAM POLAR POLYGO N
upon a beam may be found graphically using Bow's
Notation in a space diagram ; a load line and a polar
polygon ; a funicular or link polygon, all to suitable I OkN 2 o kN 3 0kN SPACE DIAGRAM
scales. A B I
C D
A beam is shown with loads of IO kN, 20 kN and

8
30 kN arranged at intervals shown. Find the reactions at Sm 6m 4m
R L and R R.
E
Graphic Solution Draw the beam as shown to a scale
of I O mm = l m. Label the spaces A, B, C, D, to conform
with Bow's notation. Draw the load line scale 20 mm =
1 0 kN, showing 1 0 kN a to b ; 20 kN b to c ; 30 kN c to d. ':I
Fix any pole o. Join abed to pole o, forming the polar F U N I C U LA R
RR
00
polygon. O R L INK
Begin the link polygon by drawing ao parallel to ao in POLYG O N 3 9. 1 66
the polar polygon. Draw bo, co, do as shown to give y in
the link polygon. Join xy with ea. Draw ea in the polar do
polygon parallel to ea in the link polygon, cutting the
GRAP H I C
bo
load line to give the magnitude of the reactions R L and
R R. SOLUTION
BEAM REACTIONS RL R
Beam Reactions by Calculation A ' moment' is ob­
BY CALCULAT I O N :-
G
tained by multiplying the force by its distance from the e. z
....: z
i�
point being considered.
Moments about R L are three, 1 0 kN x 6 ; 20 kN x 1 4 ;
3 0 k N x 20, total 940 kN. Moments about R R are, I Ok N 2 okN 3 0k N
A
�C dN
.,
_, z
II
R R x 24, that is, 940 kN. Therefore R R is 39. 1 66 kN, 8 C D
and R L is 20.834 kN.
Notice that the equilibrant, or single force required to E
"'
::,
"' a:
maintain the beam in equilibrium, is shown acting at a 6m � e
O -i---
point vertically above z in the link polygon, balancing 14m
the total load of 60 kN (beam mass has been ignored in 0
this example). 20 m
e
24 m

L:
MOMENTS A i R L -
10 X 6 - 60

"
20 X 14 = 280 POLAR
30 X 20 - 600 C P O LY G O N
MOMENTS AB �24XRR: 940 TOTAL
T ::,

z a:
940
RR = --r.i" = 39. 1 66 k N
l RL • 6 0 - 3 9, 1 66 = 20.834k N d
 130. Shear Force If a loaded beam were cut at any
SHEAR FORCE BEAM M O M ENTS SENDING MOMENT
point between the supports, the beam would collapse as
the forces acting upon it would be unbalanced.
SkN 1 2kN IOk N

8
The Shear Force is obtained by calculating the upward
forces (as moments), and the downward forces (as mo­ S PACE DIAGRAM
ments) at the section, and subtracting. Positive Shear,
A B C D
w

---l-,-·---
( + ve), is indicated when the force tends to turn the beam z
in a clockwise direction; Negative Shear', ( -ve), when 3m Sm 7 m 5 m
I :J
the beam tends to turn in an anti-clockwise direction.
The Bending Moment The force required to maintain
a cut beam in a state of equilibrium is known as the
o a
- -+- ---
Bending Moment, and is found by calculating the force i :
at the point of section.
POLAR
i
Graphical Solution Draw the beam and loads, to a
convenient scale. Label as in Bow's Notation. Draw the POLYGON
load line to scale, fix pole o and draw the polar polygon.
1
b -------.b -r
Draw the. shear force diagram. Construct the link or
funicular diagram. Find the reactions of the beam
supports, R L and RR .
Scale of the shear force diagram is the scale used for
the forces in the load line, of the polar diagram. e
Scale of the bending moment diagram is given by : e
Bending Moments in Nm
= space diagram scale in m c---------c C
x polar distance in mm
x load line scale in N SHEAR FdRcE DIAGRAM
'.

Moments of Beams The reactions RL and RR of

beams in the examples are resolved (a) by the graphic
method, using a load line ; a polar polygon; a link or
funicular polygon ; (b) by calculation, force x distance
= moment = Nm.
eo p

y
L l �K rLYGON
SHEARING ACTION ON
BEAM
co
BENDING - SHEAR FORCE :
MOMENTS DIAGRAM DIFFERENCE BETWEEN UP
& DOWN FORCES AT ANY
SECTION POINT ON BEAM.
BENDING M O M E NT : FORCE REQUIRED
:z TO MAINTAIN EQUILI BRIUM OF BEAM
AT A SECTIO POINT.
63

\
 64

131. Simple Central Load The topload on the beam of

40 units requires 20 units at each of the reactions R L MOMENT LINK FU N ICULAR
and RR to maintain equilibrium. kN a
132. Offset Load A load of 6 units rests on the beam p Q
2kN
A E
I m from R L and 5 m from R R . The diagram shows how the 3m
moments are obtained.
133. Multiple Loading A beam has loads of 20 kN, 30
kN and 1 0 kN force at the intervals shown. By the graphic
= f:::\
MOMENT = FORCE x DISTANCE y
method, notate the spaces A, B, C, D, and draw the load
momqnt about R L = 40 x 3 120
line to scale, abed. Join abed to any pole o. From x draw POLAR
the funicular or link polygon. Join x and y in eo. Draw eo moment about R R RRx 6 = 120 �
in the polar polygon parallel to eo, and read off the RR = 20kN
magnitudes of R R and R L to scale, on the load line. The
diagram shows the calculation of the moments also. 6 kN
p Q
Im
134. Overhung Beam The diagram shows details of the mom12nt about RL= 3x 2 + 5 x 5 + 2 x l 2 = 5 5
loading of an overhung beam. Notate the spaces A, B, Sm
C, D, E, and draw the load line to scale abcde. Join abcde S kN lkN mom12nt about RR= R R x B+ 2 x 3 = 55
to a pole o. Draw the link polygon beginning at x, obtain RL RR
.fo. Draw .fo in the polar polygon. Read off the magni­ 8 R R = 49

8
tude of RL and R R to scale on the load line. Calculation momt1nt about RL= 6x I =6
of the moments by arithmetic is also shown.
m o m t1 nt about RR = RRx6 = 6
Q R R = 6,125 &
kN
R L = S . 875
kN
135. Cantilever Beam Draw the beam shown to a suit­
able scale. Notate the spaces A, B, C, D, E. Draw the RR = l kN V
load line abed to scale. Join abed to a pole o. Draw the

!l
link polygon beginning at x. Notice how line co is taken 30kN IO k N a
lo the vertical line C D, do returning to give y on the C
Cl

reaction line R R . Line eo must lie between the reaction

D
b
lines. Draw eo in the polar polygon parallel to eo. Read 4m
off the magnitude of R L and RR to scale on the load line. R L RR ti - 0
Loading shown in kilonewtons (kN). X X

e.,. er-:---PO L A R
y
POLAR
do !�
d
D I AG
Ill R L

FUN IC�LAR DIAGRAM

momt1nt about RL= 20x3+30x 9 + !Ox 13 mom12nt about R L = 3x3+5x8+2 x 1 S = 8 5

= 460
mom12nt about R R = RRxl4 = 85
momoznt a bout R R = R R x 1 7 = 460
R R. = 6.07 k N

8
RR = 27.06
R L = 60 - 27,06 RL = 3 .9 3 k N
= 3 2 , 94 k N
 BOW S N OTAT I O N
136. Simple Framework Draw the framework to a suit­
able scale. Letter the spaces clockwise to comply with S I M P LE F R AM E WO R KS
Bow's Notation. Draw the force line ab to show 4 k N
load to a suitable scale. Construct the link o r funicular SPACE SPACE
diagram by drawing ad parallel to A D and bd parallel
DIAGRAM DIAGRAM
to B D. Draw de parallel to C D . Magnitude of force
in members A D and B D can be read off to scale. A
The direction of stress is indicated in the three smaller D
diagrams and show that the two members are in com­
pression and are struts. The member C D is in tension
C
RR
and is a tie; its magnitude can be scaled off from the link
R L
C
diagram also. Notice how the arrows follow round in 2kN � 2kN R R
one direction in the vector diagrams, to indicate direc­ SkN
tion of stress. If the framework has many members,
LINK
letters may be used on the outside of the frame, and F U N I CULAR LINK
numbers on the inner spaces. � FUNICULAR
137. Framework U niformly loaded beam. Reactions 8
are equal and half the total load. Draw the force line to
V)

a suitable scale, abed. Draw the link or funicular, start

at af, dj, parallel to A F, D J . Other vectors are parallel to POLAR
corresponding members. Notice how fe cuts the force
line to give the reactions to scale of 5 kN. Scale off the
magnitudes of members. Draw in arrows to represent
direction o f stress in members-strut or tie. Struts can
VECTOR
a
also be shown by heavy lines, and ties by thin lines, o r <f
arrowed.
b POLAR
138. Framework A more complicated example with
unequal loading. Notate the spaces. Draw the force line
abed to scale. Join abed to pole o. Draw the beam funicu­

0
lar to find the magnitudes of R L and R R ; starting from x �
draw ao parallel to ao of the polar polygon, bo parallel f .::-----+-----�-'
to bo, and so progress to y. Join xy with eo. Draw eo j
parallel to eo, this gives the magnitudes of R L and RR to
scale units. Draw the vector diagram, af parallel to A F,
etc. The completed force diagram enables the stress of 3.46
t
members to be read on the scale. The detailed vector V E C TO R
diagrams show the direction of stress in the members, STRESS 2
and is the key to later problems. d
Note : ' I ' is omitted in Bow's Notation, avoiding con­
fusion with numeral one.

65
 66

F RA M E W OR K WOR K E D E XA M P L E

\
2okN
V ECTOR

8
@
J
D I A G RA M S

-i - -n - - L
S PACE DIAGRAM b
kN 1
0 AF 87 C )�
0
BG I
Ej
78 C
- --- -
I
CJ 85 C 9

®
DK 1 00 C
8 y
LINK ! PO LYGO
�
i
FG
do
18 C
I

ao � a

I
.!! FE 77 T
GH
0 b
V
V)
29 T
HE 52 T

I
42 T
I
HJ
k 26 C

0
a

z
VECTOR
0
z
DIAGRAM
0
PO LAR
.J P)

P O LYG O N

®
2okN d

I
e

STRESS DIAGRAM f�

R
I g

37kN
 139. Beam with Varied Loads A beam, weight I 00 kN/m
VAR I E D BEAM LOA DS ANGLED
is varyingly top loaded at 500 kN/m as shown, plus a
bottom load of 1 00 kN. The three top loads will act at BEAM z z�
their respective centres of gravity (median points) , as in­ IOOkN/m LOADING SOOkN/m I SPACE

\9
'
DIAGRAM
dicated. The load for the left section will be 2 x 250 kN
and for the right section 4 x 250 kN. The centre load is
I lkN
\
1 x 500 kN, and the beam itself 7 x 100 kN. 2kN 1-S k N

I -....------
The final loading of the beam is shown in the notated
space diagram. The load line and polar diagram should I A B C D \
now be drawn, followed by the link polygon giving go X
1 0 0 kN
as the closing line. The line go may now be transferred to --
R L Im R R
eo ,E

'
the polar diagram and the reactions R L and RR obtained. 2m
The shear force and bending moment diagrams may be
completed as indicated. Refer also to No. 1 30. I
100
140. Beam with Angled Loads The beam, loads and load
angles given should be drawn say twice the printed size
and notated as shown. The load line abed should now be
t F
drawn to a convenient scale, and the points joined to a
pole o. Reaction R L is given as vertical, therefore the link
diagram should begin at the point of the unknown re­
action R R , draw do, co, bo and ao in that order, com­
plete the link polygon by drawing e(I. Transfer eo to the
polar diagram, ae is a vertical line, and de is the reaction
I "'
R R , magnitude and direction to scale. Complete the funi­ 18
cular diagram with ao and do extended, the Res/Equi ' C
line may be drawn through the intersection. Note that if
R R , R L and Res are extended, they pass through the 1 ....8
I
��-+--­
same point.
i
See also Nos. 1 55A and B. w
z
i§ AND l
I I
::i
"'
0
I SHEAR!

. ! 1 � .......-1---4--- �----. I <(

l -- __.i--
--
I Qf -+---1------- ,

+- � i :�
I I 0

I
1 I

I
I
FUN ICULAR
d
LINK POLYGON

67
 68

1 4 1 . Box Type Framework D raw the given frame.

Notate the spaces as in Bow's Notation, A, B. C, D, E, BOX F R A M EWO R KS WA R R E N
F, G, H, J. Draw the load line to a suitable scale show­

l
ing the loads ab 50 k N ; be 1 00 k N ; cd 100 kN. Join abed
SkN IOkN
to the pole o, forming the polar polygon. Draw the link D D
polygon beginning at x, making bo parallel to bo ; co A f B C D
parallel to co. Join xy giving eo. Draw eo in the polar
polygon parallel to eo, giving the magnitude of RR and S PACE
RL, the reactions. F
Draw the vector diagram, begin with ch, jh, hg, gf,
parallel to their respective n:embers in the space dia­

"-A
X
gram.
Notice that B F and E J have no. vector length and y
zero stress. R L 20 IOOkN
The resultant and equilibrant can be found, if required, R R
by drawing ao from x parallel to ao ; and do from y C 00
parallel to do from the polar polygon. The line of the

8
equilibrant is drawn through z and is parallel to the load
forces. LINK
The magnitude and type of stress in members of the
framework may either be shown in a stress diagram, or
in a table showing name of member, load, and type of
stress. A specimen table is shown.
z
a_�
11'1
N
8
UJ R R
Draw the frame to a suitable
..J
142. Warren Framework �
scale, letter the spaces to Bow's Notation. Draw the foad
line abc to a suitable scale and parallel to the direction of
I
I
I b_
the two load forces. Join abc to pole o. Construct the

'I
link polygon, join x and y to give do. Draw do in the 0
polar diagram giving d and the reactions RL and RR. f
I

Draw the vector diagram by vector lines parallel to the

members in the space diagram, ae from a and parallel to 0
-i
A E; de from d and parallel to D E ; bffrom b and parallel
to B F ; dg from d and parallel to D G ; cg from c and
parallel to CG. The length of the vector Jines in the vector
I POLAR

I
I

diagram are measured to the same scale used in the load

d
�
line and give the magnitude of the stress. The direction C
R L
of the stress is given by the arrows and indicates whether
the member is a tie or strut.
I
I
AF SkN
STRESS TABLE
C CH SkNJ C
STRESS
ci C:

�
8L -\
-sz
BF i 0 0 HJ 7kNI T
I
FG 7kN C JD I SkN C
GE SkN T JE 0 !o
GH SkN C AE IOk N! C
t i200 k N f 100 k N
'
 r
143. Wind Loads on a Framework On the previous
WIND FRAM E WORK LOADS DEAD & W I N D
examples, loads with a vertical line of action have been
shown. Roof trusses have to stand wind pressures, and l k o/ I kN
this example shows angled loading which is often found.
Draw the frame and letter the spaces. Draw the load
line to a suitable scale showing the three forces and
SPACE
D I AGRAM
8 SPACE
D 1 AGRAM
2
I
3 kN8
parallel to the action line of the forces. Draw the polar
polygon abed to the pole o. Begin the link polygon at
point C D drawing co, then bo, then ao. Complete the
I
link polygon by drawing eo. Draw eo in the polar diagram
to intersect a vertical line from a, ae gives the vertical
reaction R L . The vector de gives the magnitude and line
of R R .
Draw the vector diagram to give the scale loads in
RL
the members, begin by drawing af, ef, note that fgh lie
in the same point. Show the forces in the stress diagram.
144. Framework with Dead and Wind Loads Draw the
frame and letter the spaces. Draw the load line to a suit­
able scale showing the five loads. Draw the polar polygon
abcdef, join to pole o. Join af to give the res/equi line.
Begin the link polygon from the RR point x since RL is
given as vertical, drawing eo, do, co, bo, ao i n that order.
The funicular diagram is made by drawing ao and Jo,
giving the point through which the res-equi line af will
pass. Draw go to complete the link polygon and transfer
this line to the polar polygon to obtain the reactions RL z
and R R as shown. The lines of the reactions can now be
.Jit

drawn in the link diagram and will converge in point x.

Ill

1/
0
The vector diagram may now be drawn to give the loads V

to scale in the various members and the stress diagram ti) a

drawn as an exercise. POLAR

'I
C C 1/

'1
t
C 0 0 C
C j
t . t t

/
VECTOR
t
� POLAR
f

69

,,
 70

145. Canted Truss A canted frame with suspended

SUSPENDED CANTED TO P - W I N D - BOT TO M LOAD
loads is shown. Draw the frame and notate the spaces.

e
Ora w the load line bcde to scale and direction. Join bcde

8
to pole o forming the polar polygon. Draw the link SPACE DIAGRAM
A LINK
polygon. Draw ao, and find ao in the polar polygon.
'·
Construct the vector diagram, read off the magnitude of
stress in the vectors using the same scale as the load line.
Draw the stress diagram as an exercise. SPACE
DIAGRAM
146. Framework with top, wind and suspended loads.
Draw the frame and notate the spaces. Draw the load
line bcdd' e' and join the points to pole o. Join b and
e' to give resultant. Draw the upper link polygon, draw
the resultant from this polygon to meet RL. Join this
intersection to give true RR. Draw lower link polygon
e'o, d'o, do, co, bo, then closing link f' o. Draw f' o (a) in
polar diagram, measure up load line I kN to scale from
a to give f The vector diagram may now be drawn.
Tabulate the stresses in the members as an exercise. "o
FUNICULA R
147. Roof Truss with wind and dead loads. Draw the
frame and notate. Draw the load line bcdef and join
RL
points to pole o, to construct the polar polygon. (Note
that the wind and dead load at B C, C D, D E, must be
resolved into one force each by a local triangle or
parallelogram of forces; this gives the line of direction
or load line for the force.)
Draw the link or funicular polygon, note that the line
of RL is vertical. Draw R L line vertically downward from
b to meet ao. Draw afto give RR. Measure RL and RR on
the scale of the load line. Construct the stress diagram
and indicate the direction and magnitude of force in each
member. Tabulate the result.
148. Angled Roof Truss with wind and dead loads. Draw
�;.·.

the frame to suitable scale, and notate the space diagram.

Draw the load line to scale, join bcdefg to pole o. Draw
the link polygon. RL line of action is vertical, from b to
a, ag gives R R. Draw the vector diagram, measure the
vectors on the load line scale to obtain the magnitude in
the member. Draw the stress diagram. Tabulate the
:
results.
I I

/
VECTOR VECTO R /
0: r I
L--- - - - -;;_ WI N D AN D LOA DS

e
LOADS
IN kN

A N G LE D ROOF TRUSS

LOAD S I N K I L O N EWTONS

T RUSS 0

VECTO R
D IAGRAM

9'.,::::.._____---':;ct-

PO LAR
STR ESS
DIAGRAM g

VECTOR

71
 \ 72

B
C A N T I L EVER ANGLED TO P S US PE N D E D LOA DS
SPACE DIAGRAM
%

C).. SPACE \ \
f

11 � J\
DIAGRAM 'Qo
\ @e

i
l kN

C FUNICULAR SPACE
D I AG R AM
2
B

\
j
eo

E D
R L R L
LINK cd
00 d'o

LINK
FUN ICULAR
b

DIAGRAM Load ing i n

kilonewtons

D IAGRAM
VECTOR
VECTOR
g h
z
"' C ANTI L EVER TRUSS
C VECTOR
 149. Cantilever Truss A wall framework having top
; angled loads and a vertical bottom load is given. Draw C A N T I LEVER LI F T I NG C RANES
the frame and notate the spaces. Draw the load line
bcdefg to a suitable scale and parallel to the forces shown.
Draw the vector diagram, begin with di and el; then/k
and lk; cj and kj; gh andjh ; ba and ha. Join ag to find the
magnitude and direction of the reaction of A G . Draw the
stress diagram as an exercise.
150A. Angled Top Loads with R L given as vertical. Draw
the frame and notate. Draw the load line to scale, join SPACE
abc to pole o. Join ac to find resultant. Draw the link D I AGRAM
polygon, ao, bo, co, note that the force lines converge
in x. Draw the vector diagram ae, be; cf, ef; fd, ad; ed.
R L and R R are as shown, R R can now be drawn in the LOAD
space diagram. The stresses in members may be tabulated. 500 kN
l50B. Angled Top and Suspended Loads Draw the A
frame and notate. Resolve force AB by a parallelogram.
Draw the load line, abed', join to pole o. Join ad' to give 2m
resultant. Draw the link polygon, beginning at x, since
RL is given vertical, d'o, co, bo, ao, eo in that order. Trans­ LOAD
fer eo to the polar polygon, ea gives RL . Measure ed I to 500 k N
scale, cd gives R R in direction and magnitude to scale. LINK
The vector diagram may now be drawn and the stress
diagram drawn as an exercise.
151. Lifting Cranes In the first example A, draw the
frame and notate. Draw the polar diagram, load line ab, a
join a and b to o. Draw the link diagram to findjo. Draw
Jo in the polar diagram. Construct the vector diagram a
ac, be; ae, ce; ed, bd; df, bf; fj, bj. The magnitude and P O LAR
direction of R L and RR may be obtained as shown in
the diagram.
0

In the second example B, the direction of RL is given

and is extended to intersect the load line as shown in x.
a
The angle of RR can now be drawn to pass through the
same point. Begin the polar diagram by drawing the
load line to scale, join a and b to o. Construct the link
diagram to find co; draw co in the polar diagram. The
reactions R L and RR can now be found by drawing ac,
be. Draw the vector diagram, aj, bj; ad, cd; ae, de; cf, ef;
Jg, gj; gh, ch.

VECTOR f 9 hj
9 f VECTOR

73
 74

I 52. Bracket Loading Three cases are shown. In the

first case the load and one reaction direction are given. 3 CASES BRACKET LOADING SHEAR LEGS T R I PqD
Draw the frame and notate the spaces. The direction of
the second reaction must pass through point x, draw in.
Begin the vector diagram by drawing the load line to 50 S PACE DIAGRAM
scale kN. Draw vectors ae, be ; bd, ed ; de, ac. A line
parallel to the equilibrant gives be.
In the second case, the equilibrant must pass through x
which is the common intersection of the load line and the
C e ELE VAT I O N
� IC
t@ /;11I \
given reaction. The Frame should be drawn and notated
I

as shown after which the vector diagram may be drawn.

In the third case, extend the load line to intersect the

�� · r
----- I� \\
line of the given reaction to find x. The unknown re­

4
action must pass through this point and can now be VECTORS
drawn. The vector diagram is drawn as shown.
'-'
153. Shear Legs or lifting tripod.
The plan and elevation of a set of shear legs is given. X
Draw the given views and project the auxiliary view
/ -11 ii'
/ I I
B

71.
shown. Draw the load line 50 kN to a suitable scale.

l(tru� sho .1
Draw ac parallel to AC and cb parallel to the single force ---
i- - ,S
4
= -'m
-'--__,..
t± l,=
l m.l A UX � !
Ii ne B ' C ' . Oraw cd' parallel to C' D ' to complete the
.@ ,.. c.i

p
RL RR
12,2
vector diagram. AC is in tension, and the two front legs
� . '1
in compression. d 1
2. :

I
Calculate the reactions RL and R R by moments about
the two points.

/
I I

I
Angled forces and load lines Notice that the load lines
are always drawn following the load force lines on the C
fr ame. PLAN

VECTORS

a
@c
0
II)

VECTOR
C d'
D I AGRAM
Q
<
d g
b
z
,:¥
Col

1/)
154A. Overhung Frame with Top Loads Draw the
frame and notate the spaces. Draw the load line to scale,
I OVE R H U N G T & B LOA D S

8
since the loading is simple, the reactions R L and RR are B
equal and shown by point g. Begin the vector diagram SPACE
with ah, bh, hj, cj, jk, gk in that order. Draw the stress
diagram, tabulate the details of the members.
1 548. Framework with Top and Suspended Loads Draw
the frame and notate the spaces. Draw the load line
abcdef to scale. Since the loading is symmetrical with a
total of 1 0 kN, R L will be 5 kN, and RR will be 5 kN.
These can be drawn on the load line. The vector diagram
may now be drawn, and the stresses in the members cal­
culated by measuring to the same scale as the load line.
Tabulate the findings. Show the stress and magnitude o f
L M and N M.
155A. Beams with Irregular Loads Such loads can be
treated by dividing into smaller units, each unit being
considered as a concentrated load. If the limits of the
irregular load are drawn to a convenient scale and joined VECTOR
by a head line, the magnitude of the inner units is then
shown as a graph ready for transference to the load line
in the polar diagram. Draw the polar diagram, the shear
force diagram, the bending mom,tnt diagram, in that
order. Integrate the B M diagram to the new base line for
VECTOR
clarity.
1558. Hinged Beam with Irregular Loads and Three
Supports Draw the beam layout to a convenient size and
scale, notate the spaces. Draw the polar diagram. Draw
the shear force diagram except for k, I. Draw the bending
moment diagram, a-ko vectors first, then lo to pass
through the hinge point to intersect R 2 when ko can be
drawn. Draw ko and lo in the polar diagram and com­
plete the shear force diagram. Integrate the B M diagram
to new base line. See also No. l 30 for a simpler treatment
by medians of irregular loads.
S T R ESS

75
 76

B E N D I NG M O M E N T
SHEAR FORCE I R R EG U LAR BEAM LOADS
soN unl'form de:cre:a s e: 140N
µ • 130 N
ISON 70N
"'
I
B D C 8 A
8A J E

,
K J F E D C i G F
I
A.
f I I
H G H

I I 80
�

K K
R I R2 R 3a �----
b

RL Ill
RR RR
+

h
R L R I

+
R PO LAR
DIAGRAM
! i
j
·= "'
"'
z POLAR
D I AG RAM
"'
I
-0 -
k 0 0
O u
.!:. Z
...J V)

BrrN [!) I N
�
-0 �

MO EN
C C

1:
0 U 11

D ' AG RA�
...J V)

!1 1I I !
I!
!i

li
I,
i

:i �'
to ba e; Ii in tq·gratqd to bas<1 I i n«
I 11
e;

1,d
OVER H U N G B EAM
I JO I NTE D B E AM
 CENTROIDS
Simple
Calculation and
Graphic Methods
Integration of Areas
First and Second Moments
Formulae
 78

CENTROIDS
C E N TR O I O S
156. Centroids The centroid o f a n area i s the centre or

8
mean position of all the elements of which the area con­
QUADRI LAT E RAL

8
sists.
Square. Circle. Rectangle. Parallelogram Centroid is
the geometric centre.
Triangles The intersection of the medians is the
centroid.
Trapezium The method is shown in the diagram. . SQUARE CIRCLE
Quadrilateral First method. Divide into two triangles

l�I
by drawing one diagonal. Draw the medians in the two
triangles formed to find C 1 and C 1 ; join these. Draw the
second diagonal, and by medians find C 2 and C 2 ; join
these. The intersection gives C, the centroid.
R ECTANGLE
157. Second method. Divide the sides of the quadrilateral
into three equal parts. Join the appropriate points to
form the second quadrilateral. The diagonals of this
quadrilateral intersect in the centroid point.
TRIANGLES
I 58. The centroid of a re-entrant quadrilateral can be
found by the same method ofjoining third points to form 2ND METHOD
a rectangle the diagonals of which intersect in the
centroid.

�
�.. a J. b �
TRAPEZIUM

1ST RE- ENTRANT QUADRILATERAL

METHOD
QUADRILATERAL
/

159. Centroid of L shaped figure. Divide the figure into

two rectangles. Centroid of each is given by diagonals. CT D BY M O M E N TS C T D B Y F U N I C U LA R
Join M and L. The centroid lies on this line which should /"

A = I 5, Area ofB = 20. Divide ML in this ratio. Note that

8
be divided in the ratio of the areas of A and B. Area of

the shorter distance is nearer the larger area.

160. Centroid of a figure with three rectangular units.
Divide the figure into t hree rectangular units, A, B and
C. Proceed as above to find N. Join N to o. Divide No
in the ratio of the areas of A B to C, i.e. 7 : 4 to give P 2ND
which is the centroid of all the elements in the figure. F U N I CU LAR
D I AG RAM
161. Centroids Found by Funicular or Link Polygons
Divide the figure again into the three composing reel·
angles and find the three centroids by drawing diagonals.

.1
A force diagram represented by the line abed drawn �
to any convenient scale, shows the area value in units of _J_
the three rectangles. Join abed to any pole o to obtain the 10
funicular polygon. Draw the three parallel lines from the �
centroids. Note these are parallel to the force line abed.
AREA OF :
From x draw Bo parallel to bo in the funicular. From its A 5X3
RATIO : .!_
intersection with B C, draw Co parallel to co in the funi­ AREA OF B = IO X 2 4
cular, to give y. From x draw Ao parallel to ao in the
funicular. From y draw Do parallel to do. The inter­
section gives z, which is the resultant of the three forces.
Draw the line of the resultant parallel to the force lines

8
from the centroids.
A second position for a second force diagram and funi­
cular must now be made, the force line ghij showing the

I
value of areas B C A, but labelled for clarity G H, H I o
and IJ, in the diagram. Proceed as before to obtain the 1 ST
second resultant line, and it should be clear that the FUN ICULAR(LINK)
centroid lies on the intersection of the two resultant lines. POLYGON

I
Further use of funicular polygons and force diagrams
will be found in Frameworks and Beams Diagrams in
another section of this book.

II
I

I
I I· 10
·I
I
I AREA OF A + B = 1 5 +20 = 3 5 7 CENTR OID BY
I

RATIO 4
AREA OF C � 5 X4 = 20 POLAR & F U N I C U LAR

L._. POLYGONS

79
 80

C ENTR O I D S 1 ST & 2ND M O M E N TS OF .A REA FO R M U LAE

K L H H

Area of
1ST 2ND 0
DER IVED FIG D E R I V E D FIG A = 49 x 2 x 30 (")

A1 A2 = 2 940 mm!J.
_.____ ___-31 pole

_
H . 49 X 2 H 47
C)i.
L �
Exa
A = 98mm x 30mm 2 n d m o m e n t of area a bout X X
p
: �• = 2 9 40 m m 2
= =
In
!xx 9 0 0 mm2 x 57mm x 57 m m
L
N
y u Ar = 47 m m x 3 0 mm - 2 9 2 4 0 0 0 m m4 or 2.924 x 1 0 6 m m 4
' "O
= I 4 1 0 m m2
0 45 · ../
2 n d m o m e nt of area a bout CG
A2 = 3 0 mm x 3 0 mm
= 9 0 0 m m2
X X TO C E NTROI D = Y - �x d = l x x - A (y) 2
A
d = 57 m m = 2 924 000 m m 4 - 2 9 4 0 mm 2 x (2 7 . 3 4 m m ) 2
2 N D MOM ENT O F
A R E A ABOUT XX
= Ixx = A2 x d 2
I 4 1 0 mm2 x . = 2 92 4 000 m m 4 - 2 1 9 8 0 0 0 m m 4
y = 57mm
2 9 4 0 mm 2
2 N D MOMENT OF = 72 6 000 m m 4 or 7.2 6 x 1 0 5 m m 4
2 == 27.34 m m
leg = !xx -Ax(y}
=
ARE A ABOUT CEN T RO I D
 162. Centroid of an Irregular Figure The centroid of 168. Centroids First and second moments of area.
the shape shown in the example may be found by using A = total area of figure. E X AM PLES
the first derived area (A 1 ) and the second derived area A 1 = first derived figure.

bA = area of element distance x from O Y and y distance

y
8
(A 2) called the first and second moments, in the formulae A 2 = second derived figure.
shown in the diagrams. This gives Ji which is the height
of the centroid from the axis X X . from O X.
Draw the shape standing on the axis X X and draw i = d istance of centroid from OY.
K L parallel to X X at d height. Divide d height into con­ _ii= distance of centroid from OX.
venient parts. Where each of these horizontal divisions Moment of area O A about O Y = xbA.
cuts the shape draw a vertical line up to K L and join to Moment of area O A about O X = ybA.
a pole O on- X X . Where this line cuts each horizontal a Moment of area (total) about O Y at centroid= Ax.
second point is obtained and a fair curve through these Moment of area (total) about O X at centroid = Ay.
gives. the first derived area or first moment. Repeat the Sum of moments about O Y = L xbA.
process to find the second area using the points of inter­ Sum of moments about O X = L ybA.
section of the first area. 0 X
x= L xM L yM .
B
The areas of the shape, the first derived area and the
fJ BO .--�
and y =
second derived area can be found (a) by squares, (b) by A A A
use of ordinates and Simpson's rule, (c) by graphical First moment of area about O Y = xbA.
integration, below: First moment of area about O X =ybA.
163. The Shapes of A 1 and A 2 are shown first brought Second moment of area about OY = L x 2 oA, shown Iyy.
down to an axis OX for convenience. The shape is now Second moment of area about O X = L y 2 bA, shown Ixx.
divided into equal sections and ordinates drawn in the Second moment of area about axis= A (Radius of
centre of each section. Each ordinate height is now Gyration)2 .
brought in turn to axis O Y and joined to a convenient
pole P on the extension ofO X. The.first two are numbered Distance of centroid from
on the diagram.
164. Diagram O P Y is a funicular and the sloping lines D
may now be used in the next diagram. Project the section Second moment of area = Ixx = A 2 d 2 .
lines down to axis O 1 and X 1 . From O I draw a sloping Second moment of area about axis through centroid =
line parallel to P 1 in the funicular. The line 1 -2 to section Ixx -Ay2 .
two is drawn parallel to P2 in the funicular. Continue
until the line is completed. The area of the shape can Graphical Method Summary
now be obtained by multiplying H by PO. By graphical I . Draw the figure. By integration obtain its area = A. ·
integration the shape has been converted to a rectangle 2. Draw the first derived figure. Integrate its area = A 1 .
of equal size and area. 3. Draw the second derived figure. Integrate its area F
165. The Area of the Second derived figure A 2 is shown Az.
4. Calculate distance of centroid from X X, i.e.
in a combined diagram.
_ A1d
166. The Area A of the original figure is also shown in a. =

5 . Calculate second moment o f area, A2 d 2 = Ixx.

y T·
combined diagram.
167. Formulae The values of A, A 1 and A 2 are used in
6. Calculate second moment of area about axis through,
the formulae shown, to find the value of _j,, i.e. the height
centroid,
o f the centroid above the axis XX.
Ixx - Ay 2 i.e. A2 d2 - Ay2 •
{Continued on page 83.)
EXERCISES

81

'------- - - - - -
- - -· ·
 82

C E NTRO I D WORK E D EXAMPLE

of s m e l l circltZ r<Zl'IIOV<Zd
�----���i
/�
�\ �
· �
_]
•· ·· ·· \· · ----

0
in

- "1
y

x --+------'-+-------"':�
i

· ..J12.s
_x__-+- - �� -_ _
i I

:�1
l !
G I V E N F I G U RE : 2nd dflrlvqd

C,
f i gurq
F I N D C E N-T R O I D
& M O M E NTS
A
= 1 a 4o m m 2
:�,_1_
l5t d°flrivqd
__ _ I =eoo\

..,
figurq ,__ _ _ __ _ __, 1---'-m:.:..:m:.:....2_. _L
20 �
� !

@
A = I 840 mm2 2 nd mom<Znt of area a bout· X X
= Ixx -
FO R M U L A E
800 m m 2 x 7 5 m m x 75 mm
A 1 = 1 040 mm 2 4 501 000 mm 4
°
or 4 . 5 0 1 x 10 6 mm 4
X X t o centroid = y = .£L x d
A A2 800mm 2
= 2n d moment of area about CG
2 nd momqnt
d = 75 mm = Ixx - A ( y) 2
of area about X X = Ixx = A2 x d 2
-= 4 501 OOOmm4 - · 1 8 4 0 m m 2 x (4 2 . 4 mm ) 2
= I 040 mm 2
y x 75mm - 4 50,1 OOO mm 4 -- 3 309 000mm 4
2nd moment of I 840 mm2
a r e a about CG =leg =Ixx -Ax(y)2 1 9 2 000 m m 4 or I . 1 9 2 x 106 m m 4
42.4 mm
 and x. This would mean all . the construction to find
The centroid could also be found by the intersection of ji
C ENTROID EXAM PL ES
to find x from YY.
distance y from X X , followed by a similar construction

169. Centroid Worked Example The diagrams shqw a

pierced example fully worked_ out to obtain the height of
ji (bar y).
8
Draw the shape standing on X X , and place the pole 0
on the intersection of X X and the centre line ; since the
figure is symmetrical the centroid will lie on this centre
line.
Draw the first derived figur·e by the method shown on
the previous page, notice that a derived figure must also
be drawn for the inner circle.
Draw the second derived figure, this also includes a
second figure for the inner circle.
Project the widths of the original figure A ; the widths
of the first figure, A 1 ; the widths of the second derived
figure A 2 to the area diagrams. Bring the widths to a base
line so that ordinates can be easily drawn. Notice that
the inner circle area is a negative area, and is deducted DERIVED FIGURES A R EAS OF F IGURES
from the area being drawn, as shown. Full projections
of the areas A, A 1 , and A are shown. By graphical inte­
gration, condense the are� to rectangles of equal area,
a common pole P distance (in this case 40 mm), helps in

-�.,.._ -
the comparison. K ____-.-____....,... L______
Apply the formula to obtain distance ji as shown. - -
170. Calculations and application of formula.

Two Centroid Examples

l 7 l. A further pierced example is shown, the first
and second derived areas are shown, and the area dia­
gram drawn. The graphical integration of the areas
should be drawn as in the previous examples as an
exercise.
172. A girder type figure is shown, and the area diagram
is also shown. The graphical integration diagrams
should now be drawn as an exercise.

X X

DERIVED A R EAS OF F IG U R E S

83
 84

172A. Centroid Problems Nine shapes are shown suit­

able for graphical integration, as described in the previ­ C E NTRO I D PROBLEMS
ous pages.
Draw the shapes larger, scaling up from the printed A B C

e
size; an indication is given in each. Some shapes will be
better turned to a more convenient position before
integrating to find first and second areas.

J.. 75
., J ... 60
./ 85

D E F

,.
0 90
� ,
.. 80
·I
!
�
R 65
@
..,
I
 MECHANISMS
Relative Velocity

Instantaneous Centre

Slider-crank

Four Bar Chain

Parallel Motions
Compound Motions
 86
173. Relative Velocities The motion of the ends of a
link may be shown by a vector diagram. The direction R E LAT I V E V E LO C I T Y
and magnitude of the given forces are indicated by vec­
tors to scale and a triangle of forces completed. The
8 8
/

magnitude and direction of the force at a second arm C

may be found by drawing the link polygon. Notice that B
the line ac in the polygon is at right angles to A C, and - --- G I V E N
similarly cb to C B. The line oc indicates the direction
and magnitude of the force at C.
174. Instantaneous Centre In this second method con­ ,p

..."'
sider the link A B in a circle centre I. If the dotted lines
are drawn at right angles to the forces at A and B they A SLIDER -CRANK CHAIN
intersect in centre I of the circle. The direction and magni­
tude of a force at C may be found by joining C to I and
drawing the force line at right angles. The triangle of
C:

force is shown.
:,

V(lc:tor
C Vector
Polygon
175. Slider and Crank Two methods of obtaining the
CJ
Polygon 0u
direction and magnitude of the forces acting at points on

I.
the mechanism are shown. In the first, the forces are
V) / C'lntre
shown in the polygon. Draw the crank and slider to a b
suitable scale and to the given data. The force V c is
Vp
0
metr'ls p e r second
always at right angles to the link C O and is indicated V E L OC ITY OF POINT C IN LINK
by rpm or by radians per sec. The velocity of P the �
\
slider is indicated in m/sec. To construct the polygon,
draw op parallel to O P and to a suitable scale. Draw oc
parallel to Vc, draw pc parallel to a perpendicular to C P. a
8 \ \� 8
The magnitude of V c is given to scale co. The force at D
is indicated by Vd, the position of d being obtained by
transverse parallels from C D arranged on a line from p
Va
IA = lB =
Vb Vab
AB 0
\ \
..�.....
Ve Vp
le = I p
=
Vpc
CP
!! t'!.
0
giving the same proportions as the division of the
connecting rod.
176. Instantaneous Centre In this method, the centre I ..
.,.
0 Angle Velocity
is found by drawing a perpendicular to PO, and extend­ of C
ing the line of the link C O to intersect. The magnitudes b =
ms-I
of points D on the line P C may be found by joining the
m
C e ntre .,
point to I, and its direction by drawing a perpendicular I B
ti
e
ct
Va "'-.-:- - , rad
as shown in the diagram.
The velocity of A relative to O = the vector sum of
�
,- - .,f C 90

Velocity of A relative to O = Vel. A + Yel. B, i.e.

velocity B. A f Vb
T

Ve
oa = ob + ba. p
Tht: angular velocity of A B is IN STANTANEOUS CENTRE INSTANTANEOUS
m s- 1 M ETHOD CENTRE M ETHOD
rad s - 1 link m
ab
AB
 I rev = 2n radians= 360 ° (radian = 57 ° 1 8 ').
M EC H A N IS M S
N rpm = N ra d s - 1

8 8
60
N
rad s - 1 FOUR BAR CHAIN
d
= 2n x
60

= n
3O
N rad s - 1

also w rad s - 1 = n
30
rpm

Relationship-linear velocity to angular velocity : length

angular 1v = v/r
of arc to angle in radians.

rx. = /Jr
Velocity linear v = wr
EQUAL L I N KS
Acceleration / = rx.r
Angular Velocity:

= 57.296 ° s- l
One radian per second = 9 . 5493 rpm

= 0.0 1 7453 rad s - 1

One degree per second = 0. 1 6667 rpm WAT TS STRAIGHT LINE
One rpm = 6 ° per s B M OT I O N
= 0 . 1 0472 rad s - 1
177. Four Bar Chain In this mechanism, given the
velocity of D, the relative velocity of C may be found by U NE Q U A L
the instantaneous ce centre method, by extending the
lines of links A D and B C until they intersect. A point E
may be fixed at any position on D C and the force indi­ 8
8
cated by joining to I, the centre, and drawing a ·perpen­
dicular to this line. The triangle of forces is also shown.
178, 179. Watts' Straight Line Motion Two diagrams
are given. In the first, equal radial links are joined by a
similar connecting link. The locus of a midpoint P on the
connecting link gives a straight line movement for much
of its travel. In the second case, unequal links are joined
by a connecting link, and the point P divides the con­
necting link inversely as the lengths of the radial links.
The locus of P is a straight line approximately for much
o f its travel.
180. Watts' Quick Return crank and slotted plunger.
This mechanism is used to operate slides requiring a
quick return on the recovery stroke. The link is adjust­ WATTS' QUICK RETURN MOTION
OA BO,
C R ANK & S L OT T E D LINK
able. The mechanism should be plotted through one
revolution. -=--
BP AP
a,
87
 88
1 8 1 . Crank and Slotted Link The locus of an end point
P is shown. The crank is drawn at twelve positions, and M EC H A N I S M S

8
the position of the link positioned at these twelve stations.

8
The curve can be drawn through the twelve plotted
positions of P.
182. Linked Cranks The cranks and links should be
plotted in twelve positions. Draw a fair curve through
the positions of P and P 1 . Notice that crank arms are p
rigid and can only describe arcs from their centres. 0
Notice that links are rigid and pinned to the cranks;
lengths measured along the length of the link are con­
stant, in any position.
183. Slotted Link and Crank To enable the crank pin
to work directly in the lin k , a suitable slot must be pro­ CRANK & SLOTTED LINK TOGGLE ACT I O N
vided. Other links and slides can be attached. Plot the
locus of a point P on the second link.
184. Toggle Action In the mechanism shown, the crank
gives a quick S H M clearing action for much of its
stroke, exerting compression as required clamping action
at the end of the stroke.
8 8
185. Cam and Plunger A simple circular cam actuates
a spring loaded follower. This mechanism can be used
to open a valve or a switch, or to operate a lever or link,
one or more times per revolution of the cam. See cams.
L I N KE D
186. Crank and Offset Linked Plunger The crank oper­
ates a pendulum link which is also linked to a pump ram. CAM & P L U N GER
The line of the crank centre is offset from the point of
attachment to the pendulum link, giving a quicker return
action to the ram. Plot the mechanism in twelve positions
8
8
and calculate the movement of the ram. Plot che locus
of a midpoint P in the link connecting the crank pin to
the pendulum link.

SLOTTE D L I NK
& C RA N K
CRANK & LINKED
P L U N G E R. OFFSET
187. Crank and Link A crank moves a link, the end of
which moves on the inside of an arc. Plot the locus of M E C HA N IS MS
any point P on the link. Substitute an elliptical curve for

8
the arc, and compare the locus. Other curves may be C RANK & L INK
substituted, parabolic. hyperbolic. B ELLCRANI<
& CAM
188. Radial Slotted Link The link swings from end A,
and a slide C moves from A to B and back whilst the link
swings through 90°. The diagram shows the locus of C.
A second locus of the slide is shown. Plot the complete
locus of C whilst the link rotates through 360° . Plot the
locus of C when the link passes through 360° and the
slide moves from A to B. Name the curve.
189. Bell Crank and Cam A cam giving S H M moves
a bell crank (two angled rigid arms) which has a swivel
shoe attached. Plot the locus of A. Design a similar sys­
tem to give 15 mm movement to the shoe, the ratio of the
arms being 2 : 3.
190. Two Slotted Links and a Cam A S H M cam

8
works two slotted links which have a common slide C.
Plot the locus of the centre of the slide. Vary the fulcrum
points of the links. Substitute other cam forms shown
on other pages.

RADIAL .
S LO T T E D LINK

89
 90

190A. Two Simple Harmonic Motions arranged at right

COM P OU N D M OTIONS AT 90
°
angles. Two rotating cranks work in slotted links, the
extensions of which combine to act upon a pin P. The
speeds of the cranks are equal, though the length of the
cranks may be unequal. To obtain the locus of P, divide
the sweep circle of each crank into the usual twelve divi­ TWO S IM PLE
sions and project the stations to give the points through HARMO N I C M OT I O N S
which the locus may be drawn. Note that the locus is an
AT 90
°
ellipse.
1908. Unequal Speeds ff two cranks turn at different
speeds, the locus may be plotted as shown in the example. EQUAL R PM
Two cranks of unequal length are arranged to work in
slotted links at right angles (only the schematic diagram
is shown). The speed ratio of the cranks is 2 : 1 . The locus
is plotted as shown by projecting from the twelve points
on the slower crank circle, and six points (used twice) on
the faster turning crank circle.

LOCUS
Exercises OF P
Use a crank radius of 50 mm in the following exercises.

I . Plot the locus of P when the cranks (in A) turn in the

opposite direction (anti-clockwise) .

2. Plot the locus when the two cranks start together at

top dead-centre.

3. Plot the locus (in B) when the top crank starts at No. 1
position, and the lower crank starts at top dead-ce.ntre.

4. Plot the locus when the speeds of the two cranks are
interchanged.

5. Plot the locus when the speeds of the two cranks are SPEED RAT I O 2:I
in the ratio of 3 : I (one revolution of the slower crank
will be sufficient).
INCLINED PLANE
True Lengths
Auxiliary Projection

OBLIQUE PLANE
Traces
Skew Lines
Dihedral Angle
Tangent Planes
 92
INCLINED PLANE
1ST ANGLE TRUE L E N GT H S
191. Projections of a Line on a plane inclined either to
the main horizontal or main vertical plane. The projec­

8
P RO J E C T I O N OF
tions of a line and traces of the inclined plane are give n ;
the isometric pictorial view should assist in en visualising A LINE ON A N
the position (a) of the line itself, (b) the position of the A U X I L IARY PLANE
plane. The true length of the line is obtained by rabat­
INCLINED TO H P
ment. The line is treated as being part of the hypotenuse
of a triangle which is rotated until it lies parallel to the
main plane, when its true shape can be seen, and the true Pictorial
length of the line measured off. This system is known as viczw
' triangulation ' and is much used in obtaining develop­
ments; it is dealt with in a later section. Where an inclined
plane cuts a solid, an auxiliary plan (for a plane inclined
to H P), or an auxiliary elevation (for a plane inclined to First anglcz
A u x i l iary
V P) would have to be projected at right angles to the pla ncz
inclined plane to obtain the true shape (length of lines) of

8
the section face. This is shown in the following diagrams.
192. Traces of a Line which intersect in the second angle.
The isometric pictorial view should help to make this

"
clear. x T

193. Traces of a Line which intersect in.the fourth aqgle.

Projections and an isometric view are given.
194. Section on an Inclined Plane A solid is shown cut

H
by an inclined plane, the traces of which are given. This
is an application of the plane to give a sectional view of H
the object. An auxiliary plan normal (at right angles) to
the plane will give the true shape of the sectioned face of
the object.

Auxiliary
p l a n cz 8
V

I I
"·
SQUARE P R ISM

D
y
CUT BY PLANE
First a n g l er I N C L I N E D TO
projcrc t i o n
PLAN
H
H p

----------
 195. Projections of a Line on a plane inclined to the main
V P. Traces of the inclined plane are given, VT H. The 1ST ANGLE I NC L I N E D P L A N ES
isometric pictorial view should help in understanding the
position of the line and the plane. The true length of the PROJECTIONS OF
line is obtained by rabating the line until it is parallel to
the main V P, projecting into the elevation, then joining A LINE ON AN
to the original height line. A U X I L IARY PLANE
196. Square Prism Cut by Inclined Plane. The plane is I N C LINED
inclined to the vertical plane and its traces are V T H .
Simple projection o f the points where the plane cuts the
V P
prism in the plan are projected to the elevation. The true
shape of the sectioned surface would be found by pro­ First a n g l er
jecting an auxiliary elevation on to the inclined plane.
projcrctions
The method is shown in the next diagrams.
197. Auxiliary Plan shows true shape of the sectioned
surface when a prism is cut by a plane inclined to the V
main horizontal plane. Widths remain the same when

8 8
measured from the X Y line.
I 98. Auxiliary Elevation shows true shape of the sec­
tioned face when a solid is cut by a plane inclined to the
main vertical plane. "-:=----a-+--.1---+-+---="-

P L A N E I N C L INED TO H P

8 V
PLANE
I N CLI NED TO
VP

X y
T

S Q UARE PR ISM
C U T BY A U X
VERTICAL PLANE 8
H
93
 - -
94

3RD ANGLE
199. An Hexagonal Pyramid tilted at 30 ° to the main
horizontal plane is cut by a plane inclined at 45 ° to the I N CLINE D PLA N E S
TO V P
H P. The true shape of the sectioned face is obtained by
projecting an auxiliary plan. PLANE INCLINED TO H P H PLANE INCLINED

200. A Cylinder and Prism are cut by a vertical plane

which is inclined at 30° to the VP. The true shape would
be shown by an auxiliary elevation.
8
201. A Roof and Stack are cut by a plane inclined at 30°
to the V P. An auxiliary elevation gives the true shape of
the sectioned face and length of lines thereon.
202. A Pierced Cylinder and Prism are cut by a vertical T y
plane which is at 30 ° to the main V P. An auxiliary eleva­
tion shows the true shape of the section.
y

8
PLANE INCLINED TO V p PLANE I NC L I N E D TO V p

8 8 H

V V I
 Exercises
I NC L I N E D PLANE E X ERC I S E S
G �'.

B
203. Inclined Plane Scale up the diagrams three times A 0

/
size shown. V ,V
A. Obtain the true length of the line shown.
B. Obtain the true length of the lines shown. Show the
traces of the planes. �
X y X y T y .A_

/
C. The elevation and trace of a line are shown, find its
true length.
D. Find the true length of the line, and the traces of the �
plane it lies in.
H� H
E. Obtain the true length of the line shown and indicate
the traces of the plane.
F. The plan of a line and the traces of the plane it lies in B E H Ki
are shown. Draw the elevation and find its true length. V
V
G. An inclined plane, traces V T H , cuts a triangular
pyramid. Draw the completed plan and elevation and

y y y X!
I
p
project the true shape of the cut face.
H. A triangular prism is cut by an inclined plane V T H X X X T y

<J
through point.P. Draw the plan and elevation, and the T
true shape of the cut face.
I. The cone shown is cut by an inclined plane, traces
V T H, which passes through point P. Draw the plan,
elevation and the true shape of the cut. H
H
J. A grooved rectangular prism is cut by an inclined
plane VT H shown. Draw the projections. Project an
auxiliary elevation to obtain the true shape oflhe cut face C I

I
F i v L

I
/v
1
v V
II
of the prism.
K. The tube shown is cut by a plane, traces V T H. Pro-
ject the true shape of the cut face. Name the auxiliary
I!
/ y y
j
/x
view required.
IT x/
! Ti
X
'\'"
X y I

I I

I
L. The two interpenetrating cylinders are cut by an in-

"
T
I
clined plane, traces V T H . Draw the true shape of the cut I
11
face.
'1
'
I.

J
�
I . Work out the exercises as shown in First Angle pro- ii,,
"" H
jection. H
11

2. Work out the exercises using Third Angle projection.

95
 96

204. Auxiliary Projection Auxiliary plans and eleva­

lions projected on planes inclined to the principal verti­ AU X I LI A R Y PROJ ECTION 1ST A N G L E
cal and horizontal planes of first and third angles are
DRAW : AU X I L I A RY E L EV ON X Y 1
1
required when special details are not shown on the prin­
cipal planes.
In the diagrams, a first auxiliary elevation is shown 2ND A U X I LIARY PLAN O N xa y� .
projected from the plan on a new x 1 y 1 line. Heights re­
main the same in all elevations, and the projector from

----y... 8
lhe plan is cut off to the same height as in the original
elevation, as H.
Second Auxiliary Plan A second plan of the object
may now be projected from the auxiliary elevation just
completed. 12
A datum line x 2y 2 is drawn at the angle required or
stated, and projectors from points in the first auxiliary
elevation drawn. These are cut off to the same width
measurements W below the datum line of corresponding
points. The marked points on the projectors are joined
ELEVATION
to give the shape of the second auxiliary plan.
Notice that the same W measurement is used for two
projector lines, since in the plan, the lower and upper H
points of a front edge are vertically above one another.
First Angle projection has been used in this diagram.
X

PL AN

y�

�,I
i
I

, ___....,__,,'r-1
_L +-

P2

2ND AUXIL IARY PLAN

205. Auxiliary Projection In the problem given, third
3RD ANGLE
,01
angle projection, a first auxiliary elevation is required, AUXILIARY PROJECTION
and a second auxiliary plan projected from the auxiliary
elevation. The procedure is similar to that of the preced­ GIVEN
PL,A.N
DRAW: 1ST AUX ELEV ON x' y'

�ccmT ..
ing page, first angle drawing.
Draw the new x 1y 1 line at the stated angle and in a 2ND AUX PLAN ON x::A. y<:l
"
convenient position. Project the salient points from the
plan, and cut off to corresponding heights H as shown. y -
Join the points to complete the first auxiliary elevation. 2ND AUXILIARY PLAN
Y�lr11
8
Second Auxiliary Plan Draw the datum line x 2 y2 and x•
the new x 2 y 2 line at the required angle and in position.
40 (20� �
Project from the auxiliary elevation, and mark off the
W measurements on the projectors. Qfl.
Join the points to complete the second auxiliary plan.

PLAN

ELEVATION

97
 98

206. Auxiliary Projection A shaped bracket is given;

draw in first angle, the original plan and elevation, fs, AUXILIARY PROJECTION 1ST ANGLE
project an auxiliary first plan. From this project a second
auxiliary elevation. A further auxiliary elevation is re­
R22 DRAW: AUXILIARY PLAN ON X Y'
1
quired to show the shape of the hole in the bracket in
�25 GIVEN:
that position.
2ND AUX ELEV ON · x�YJ.
I
Project the first auxiliary plan at the required angle
from the original elevation, on x 1y 1. x2. t

e
Draw the datum line and new x 2 y 2 line at the stated AUX ELEV ON x3ys
angle. Project the second auxiliary elevation fr om the

.I
first auxiliary plan. _y_ 11

!I
j1
The curve may be projected from centre line intersec­
tions, and four ordinate points on the compass arc shown

l
in the construction on the original elevation.
When the examples shown have been understood, any
of the objects shown for isometric exercises may be used +!./ 2ND AUXILIARY ELEV
Ii
for auxiliary projection. Common angles of the setsquare
are usually asked for 30° , 60 ° and 45 ° . First and Third
Angle projection should be used in successive drawings
to gain practice.

PLAN
207. The Oblique Plane The traces of an oblique plane !ST ANGLE OBLIQUE PLANE 3RD ANGLE
and one projection of two points P and Q are given, in
both first angle projection and third angle projection.

8 8
Notice that neither trace can be at right angles to the
xy line.
208. First Angle Projection The traces VTH of the
oblique plane and the projections of the two points P and
Q on the main HP are given. Before the projection of P
and Q can be made to the VP, their heights above the
xy line must be obtained by first projecting an auxiliary
elevation on x 1y 1 which is at right angles to the hori­
zontal trace TH, and converts the oblique plane into an
inclined plane. The height HVis the same for all eleva­
tions of point 0. Points P and Q are now projected to
the inclined plane and shown at P 1 and Q1 giving their
true heights above the xy line. These heights may now

1-
be measured on the projectors in the main vertical plane
giving P2 and Q2 completing the required projection of

I
the points to the main planes HP and VP.

i
209. Third Angle Projection The traces of an oblique
plane HTVand one projection on the HP of two points
P and Qare given. Complete the projections of the points

I
to the vertical plane.
Draw the xy line and traces as given; place the points
on the HP as given. Project an auxiliary elevation to
-<.��
II
convert the oblique plane into the inclined plane, height C,�
VHat point 0. Project lines from P and Qto cut the line i
of the inclined plane in P 1 and Q1 . Transfer these heights H W
to the main VP to obtain P2 and Q2. Auxiliary
The two methods are similar in procedure, though the Vertical

I
I
position of the main planes differ. Plane
H V0

I
H GI
Hp '(
)(,

I
Cil
V H

-f
Auxll lary
w I"\ Vcrtlc:al Plane

8 H

99
 100

210: A. Projections ofa Line on an Oblique Plane; plan 212. An Oblique Plane is inclined to both horizontal and
_
of lme given. The traces of the oblique plane VTH are vertical main planes, and its traces appear as indicated
OB L IQUE 1ST ANGLE I

given. By means of an auxiliary elevation or plan pro­ VTH. Auxiliary El�vation

jected in line of the appropriate trace, the oblique plane The traces are the lines of its intersection with the shows h�ights
is changed into an inclined plane. Lengths projected to main vertical and horizontal planes.
the inclined plane are then transferred to a second
213. The True Angles may also be found by projecting
auxiliary plan or elevation-to give true lengths.
A line AB is given in plan. Project an auxiliary ele­
auxiliary elevations and plans in line with the traces as
shown in the middle diagram. Notice that heights re­
vation in line with TH and on x 1 y 1• This gives the line
on an inclined plane, and heights H I and H 2 which can main the same in all the elevations, and widths below the
xy lines remain the same in all the planes, of associated
now be transferred to the original elevatiqn. Projectors
projections.
from the plan intersecting the height lines give the ele­
vation of the line. The true' length of the line can be 214. The True Angles of Inclination to the main planes
obtained by projecting an auxiliary view on x 2 y 2, this may be obtained by projecting the point of intersection
line being parallel to TH. of a perpendicular to VT and TH to the xy line and
joining this to V and H respectively. This is the rabatment X
The heights a3, b3 are L 1 and L2 • This view shows the of a triangle taken normal to the oblique plane and main
true shape of the inclined plane and thus the true length
of the line AB.
planes. The isometric diagram should help to make this
PROJECTIONS
clear.
OF A LINE ON AN
21 I. B. The elevation of a line lying on an oblique plane 215. Given a Triangle projected on an oblique plane, the
traces VTH is shown. Project an auxiliary plan. Width
traces of which are also given, find the true shape. OBLIQUE P LANE PLAN '\:
GIVEN
from the X Y line remains the same, and the line now Project an auxiliary elevation on the line of TH. The
appears lying on the inclined plane. W 1 and W 2 intersect VTH
po1�ts a 1 , bl , c 1 and L 1, L 2, L3 are true lengths on the
·
projectors from the elevation and the plan of the line mchned plane. The points are now brought to a position
established. Develop the surface of the inclined plane, where the widths may be projected lengths L 1, L 2, L 3•
V
ELEVATION OF LINE GIVEN
transfer the height lines to this plane, and the true length

81
This is shown in the true shape diagram.
of the line is found. �
216. Given the Position of Point Q and the traces of an
The isometric sketches in the following diagrams

I;
. Oblique Plane VT H, project the plan. Draw the traces
should help m understandmg how the oblique plane is
and the g!ve� poi�t Q in elevation. Project an auxiliary
I

transformed into an inclined plane by projecting the

elevation m !me with TH. H 2 remains the same height in
auxiliary elevations or plans.
all elevations, and enables the inclined plane (end view
i
of the oblique plane) to be drawn. Transfer H to the
au�iliary elevation, the intersection gives Q 1 . Project this 11 :,.
y
pomt back to the plan to intersect a vertical projector
from Q, resulting in Q 2, the required plan.
The dotted line in the auxiliary elevation is the true
distance of Q from the HP.
 TRACES OBLIQUE PLANE DETAILS TRACES 1ST ANGLE

TRACES GIVEN�

B Auxiliary
Plane 8 GIVEN:

POSITION OF
Q
V

8
E. iRACES
)(
"p

T
Angles of inclination
y
to H P and V P required H

ANGLES BY AUXILIARY
PLANES

8 EXPLANATORY
I SOMETRIC VIEW

�
)l '(

Angle of
inclination
to H P
GIVEN:

TRACES
l<..____..,..o-J..._,.........__�__'<_!..__

P L AN
PROJECTION OF Q
Project
H Elevation , TO OBLIQUE PLANE
ANGLES OBTAINED BY RABATMENT True Shape

101
 102

217. Dihedral Angle The dihedral angle between two

DIHEDRAL ANGLE 1ST ANGLE
planes is shown on a plane which cuts the two planes
when normal to both.
Square Pyramid I. Draw the plan and elevation of

8
the solid. Project an auxiliary elevation as shown on
x 'y l . Draw in a plane shown by the line ab, which is at
SQUARE
right angles to the projected angle line where two sur­

PYRAMID
faces of the pyramid meet.
Rabat point b to the plan, join to the two corners of
the base square. The dihedral angle is shown lined in, and
represents the true angle between the two adjacent sides
of the pyramid.
Triangular pyramid 2. In the projections of the tri­
angular pyramid shown, the cutting plane ab can be
drawn in immediately as angle between the two meeting
3
HEXAGONAL
sides is shown in true elevation. Point b can be rabatted

PYRAMID
to the plan and joined to the two corners of the base tri­
angle giving the dihedral angle as shown.
The diagram also shows the method applied in Fig. 1,
using an auxiliary plane enabling point c to be rabatted
to the plan.
Hexagonal Pyramid 3. Draw the projections of the
pyramid as shown. Draw the auxiliary plane, and the
cutting plane. Rabat to the plan and join to the two
corners as in the diagrams. Note that the cutting plane
affects only two adjacent surfaces shown, and is not taken
to the centre point of the base.
 1ST ANGLE 3RD ANGLE

"
218. Solids on an Oblique Plane First Angle Projection.
Given the traces of an oblique plane VTH, and a cone SOLIDS ON OBLIQUE PLANE
standing on it, point P of the cone base circle touching
the HP.
�!&CONE PYRAMID

�
Draw the traces of the oblique plane and the xy line,
in first angle projection.

,l
Project an auxiliary elevation converting the oblique
J.., 045

y,
plane into an inclined plane, on a new x 1 y 1 line. Draw
the cone resting on the inclined plane.
Fix P by measurement on TH, and draw the ellipse in
plan by projecting from the auxiliary elevation; the major

�
axis of the ellipse will be the undiminished diameter of
SOLUTION
I.LI
the original base circle, 45 mm. The apex of the cone is z
obtained by simple projection from the auxiliary ele­ ....J
vation.

8 8
The elevation of the cone on the main VP is con­
structed by projecting points on the plan ellipse to the
VP, and transferring the respective heights. H2, H 3 (two),
H 4 (three), H5 (two) and P (zero). The base circle of the

H
cone appears as an ellipse also in the vertical plane.

SOLUTION
219. Pentagonal Pyramid on an oblique plane. Third
Angle. Draw the traces. Project the inclined plane in line
with HT. Draw the pyramid resting on the inclined plane.
Project the base shape and apex back to the HP. Com­
plete this main plan view.
Project the main elevation from points in the plan,
transferring heights from the auxiliary elevation to their
respective projectors on the main vertical plane.
Note that the method is the same in both cases, the
object is drawn on the inclined plane which is the oblique HI
plane viewed' edge-on'. y

1ST ANGLE 3RD ANGLE

103
 1 04

O B L IQUE PLA NE S EC T I O N S 1ST ANG L E

O B L I QUE
PLANE

INCLINED
TRUE SHAPE PLANE TRUE SHAPE
OF SECTION OF SECTION

//
�,t
/

X y X y

..,,
,()

)'
�
S QUARE PYRAMID

8
T R IANGULAR P R I S M C'�
CUT BY CUT BY
OBLIQUE PLANE OBLIQUE PLANE
220. Oblique Section of a Triangular Prism, given the
traces and the position of the prism. Draw the traces and OBLI QUE PLANE SECTIONS
plan. Project an auxiliary elevation on x 1 y 1 , project abc
to the inclined plane giving a 1 , b 1 , c 1 . Obtain the lengths
L 1, L 2, L 3 • Project a new view on T 1 , H1 . Project points
abc from the plan to intersect L 1 , L 2 and L 3 , giving the
true s hape of the section.
8 ISOMETRIC
V I EW
ISOMETRIC
V I EW

221 . Square Pyramid cut by an Oblique Plane, given the

traces and position of the pyramid in plan. Draw the
traces and plan. Project an auxiliary elevation on x 1 y 1
and obtain heights H, H1 , H2 , H3 . The front elevation
may now be projected. Project a new view on T 1 H1 ,
project abed to intersect L, L 1, L 2, L3 ; thus giving the
true face of the section.
222. Oblique Section of a Cylinder given the traces and
position of the cylinder. Draw the traces and simple
orthographic views of the cylinder. Project an auxiliary
elevation on x 1 y 1 • The line of the inclined plane cuts the
cylinder giving heights which can be transferred via the
plane to the original elevation. A fair curve drawn
through the points completes the elevation. The true
shape of the section may be projected below x 2 y2,
widths being taken from the original plan.
223. Oblique Section of a Cone, given the traces and
position of the cone. Notice in this case the angle between
the traces is obtuse, and the isometric sketch should help C ONE CUT BY
LO show the direction of the plane which rises to the
observer from the xy line. Draw the traces and the OBLIQUE PLA NE
simple lines of the cone. Project an auxiliary elevation
on x 1 y 1 • The line of the inclined plane cuts the genera­
tors of the cone and thus yields heights. These can be
used in the original elevation being projected back
through the plan.

CY L I N D E R
C U T BY
OBL IQUE PL A NE

105
 1 06

Oblique Plane Problems

OBLIQUE PLANE PROBLEMS 1ST ANGLE
224. Given the Projections of a line A B and the traces of

GIVEN : PROJECTIONS O F LINE A B

an Oblique Plane VTH, find the point of intersection

& TRACES VTH OF O P

of the line with the plane. Draw the xy line, projections
of the line ab, and a 1 b 1 and traces of the plane VTHas

FIN O � P O INT O F INT ERSECT ION

shown. Project an auxiliary elevation on x ' y ' which is
at right angles to TH. Convert the oblique plane into an

, elevation from the plan to give a 2 b 2, their heights being

inclined plane. Project the given line also to the auxiliary

line a 2b 2 and the inclined plane is the point P , which may

· taken from the original elevation. The intersection of the

now be projected back to the plan P 1 and thence to the

elevation P 2 .
b
G IVEN: TRACES O F T WO O P
'(

Draw the traces as given. Project the points a and b which

225. Dihedral (True) Angle between two oblique planes.
VT H & P Q R
FINO: DIHEDRAL (TRUE)
are the projections of the line of intersection between the b
two planes. Project an auxiliary elevation on x 1 y ' which A NGLE BETWEEN
T HEM
is parallel to ac. Draw de normal to the inclined plane,
rabat to the plan as shown to give the true shape which
shows the true angle between the two oblique planes. See
also 227.
226. To Find the Traces of an Oblique Plane Draw the
given plan and elevation of the laminae, extend ab and
ad, a 1 b 1 and a 1 d 1 to give, by rabatment, intersection
points l and 2 through which THmay be drawn. Find 3
and 4 by projection from cd, parallel to TH to cut the
Q
y
xy line, thence to intersect horizontal projectors from c 1
and d 1 , through which VT may be drawn. The auxiliary

to verify heights a 2 , b 2 , c 2, d2 .
elevation shown may be drawn as in previous problems

This problem may now be extended as in Nos. 220 and

221 to find the true shape of the laminae when required.

G IVEN: T HE P ROJECTIO N S O F A LA M IN�

ABCO LYING O N O P
FIN O : T RACES OF O BLIQUE P LAN E
 227. Two Skew Lines, shortest distance between two
straight lines neither parallel nor intersecting, one com­ TWO SKEW LI NES 1ST ANGLE
mon perpendicular (shortest distance between), said to
be at right angles when their projections on any plane
normal to their common perpendicular are at right G I VEN : REQUIRED: S H O RTEST DISTA NCE
angles.
Draw the given lines and the xy line. B E TWE E N TWO SKEW LINES
Draw a construction line E F parallel to C D and E 1 F 1

8
parallel to C 1 0 1 and obtain the intersections.
Extend and project A B and A 1 B 1 to give further inter­
sections. Join the intersection points to give the traces
VT and T H. Project an auxiliary plane on x 1 y 1 which
is at right angles to T H. This converts the oblique plane

X
--
into an inclined plane, and the lines A B and C D can now
be projected from the plan to the inclined plane. The y
common perpendicular may now be drawn 0 3 P3 , and ------...:-,_

these points projected back to the plan, giving O and P.

Simple projection of 0, P to the main vertical plane give
0 1 and P 1 and completes the elevation.
0 3 P3 is the length of the common perpendicular and
e'
the shortest distance between the two skew lines.
The isometric sketch should help in visualising the
oblique plane and the skew lines.
Note that it is essential that points of intersection in
plan and elevation must lie in the same projection.

SOLUTION

107
 1 08

1ST A NGLE
228. Tangent Planes Given the plan of a sphere and a
point P, find the traces of the plane tangential to the TA NGENT PLANES
sphere at the point.
Draw the plan and point given, project the elevation. GIVEN:
ISOMETRIC
SPHERE & P OINT
P will lie on a section plane shown in the plan, and P I is
projected as shown. SKETCH
Project an auxiliary plan on x 1 y I which is parallel to

8
P 1 0 1 . P2 is obtained by projecting from P 1 , and the in­
clined plane is tangential to the circle at P2 . VT can now
be drawn parallel to P 1 P2 , and T His drawn from T and REQUIRE D:
TANGENT PLANE
at right angles to P 0.
The auxiliary elevation completes the projection, the
true angles of the plane to the HP and the VP are shown
in the auxiliary elevation and plan. An alternative method
is shown by the dotted lines. Draw the plan and elevation.
Draw P 1 a at right angles to P 1 0 1 to give a on xy. Find
b by simple intersection as shown. Pc is at right angles to
P O. Find d by simple projection. Draw VT through d
and parallel to P 1 a. Draw T H through b and parallel to
Pc.

A UX
ELEV

�J - >-,

J·
'(

• l(

planiz
SOLU TION
 11ST ANGLE
229. Tangent Plane to Cone Given point P on the plan
TANGENT PLANES
\
of a cone, find the traces of an oblique plane tangential
at the point P.
Draw the plan and point P. Project the elevation of the SOLUTION
cone. Draw a generator in plan passing through P; pro­
ject to the elevation to obtain P1 .
Draw TH tangential td the base circle and at right
angles to P 0. Draw Oa pirallel to TH. Find b by simple
projection. Join T to b to give trace VT.
Project an auxiliary elevation on x 1y 1 to give the
true angle of inclination of the oblique plane to the HP.
In the case shown, the oblique plane lies in contact on
the cone surface on the generator which includes P.
230. Tangent Plane to Cylinder Given the one projec­ '(

tion on the HP of a point P on a cylinder inclined to the

i G IVEN :
VP, find the traces of the oblique plane touching the j
cylinder at the point.

�
Draw the plan and point P. Project an auxiliary eleva­ CONE & POINT
I
I
tion on x 1 y 1 which is at right angles to the axis of the
ISOMETRIC
REQUIRED: TANGENT

.
cylinder. Draw a projector from P parallel to the axis to
SKETC H
I PLANE
give P 1 on the circle. Join P1 to 0, and draw the line of
:

I
�
' ==========:=:!
the inclined plane tangential at P1 . TH may now be
drawn by projection from x 1 y 1 , and is parallel to the
axis. The height of P1 is shown in the auxiliary elevation, 1tt::=
1 ====��
8
and can be transferred to the main VP to fix the position
ISOMETRIC
SOLUTION
of P2 . Vis found by drawing a perpendicular from c and I

I
is height HVtransferred from the auxiliary elevation.
Alternatively, simple projection will give points abc,
VT being drawn by joining Tb and extending to V.
In this case, it will be seen that the oblique plane rests ''
on the cylinder giving a line of contact including point P,
the line being parallel to the axis. 1�
• iI
i

!
'{
C
I i'
G IVEN: I
�- -·- - I
/
CYLINDER
& POINT

REQUIRED:
TANGENT PLANE

1 09
 1 10

1ST ANG LE
231. True Shape of a Triangle or of any plane area. The
true shape ofa plane figure may be found by the following T RUE SHAPE
method if the plan and elevation are given.
Draw the given elevation abc and project the plan
G IV EN � PROJECTION OF TRIANGLE
a 1 b 1 c 1 . Draw a constructional horizontal line cd in the
elevation, and project to the plan to give c 1 d1 .
FIN D : TRUE SHAPE ANGLE TO H p
Project an Auxiliary Elevation on x 1 y l normal (at
right angles to) line c 1 d 1 , to give a 2 , b 1 , c 2 . Heights of
these points are the same as in the original elevation from

8
xy. Project an Auxiliary Plan on line x 2 y 2 , which is
- TRUE SHAPE
parallel to a 2 b 2 c 2 , to give a 3b 3 c 3 • Width distances of
these points are taken from x 1 y 1 to the original plan as
shown in the diagrams.
The true shape of the triangle is given in the auxiliary
plan as indicated.
Refer also to the chapter on auxiliary plans and eleva­
a ELEVATION
tions first and second, given earlier.
The true shape of any figure may be found by the above Horizontal
method if the limit points of the horizontal construction
line are projected with sufficient boundary points of the Constru c t i o n l i n e
figure.
H
The true shape of the triangle could also be found by 'C
finding the true length of each side in turn by rabatment
-see triangulation-and thence drawing the triangle L :; '
from the true length of the sides.
r,

y
l

ANGLE TO H P -
 1ST ANGLE
232. Line of Intersection between two planes. Where two
planes intersect, given the projections, plan and eleva­ D I HEDRAL A N GLE
tion, the line of intersection may be found by the use of
the horizontal construction line (as in the preceding
GIVEN � P ROJECTIONS O F TWO TRIANG�ES

8
diagram), by auxiliary elevations and second plan (sec
204-206) which will give the · edge view' of the two planes
and consequently the true or dihedral angle. (See also FIN D : T RUE (DIHEDRAL) ANGL E BET W E EN PLANES
Dihedral Angle diagrams, No. 227.)
Draw the given elevation and plan with the horizontal
construction line ah as shown. Project an Auxiliary
Elevation on x 1 y 1 normal to a 1 h 1 , to give the points of C
EL EVATION
intersection of the two planes, one of which, a 2 b 2 ci is
seen in ' edge view'. Project the points of intersection of
the planes back to the original plan and thence to the
original elevation. The line of intersection between the
two planes may now be drawn as shown in the diagrams.
The True Angle or Dihedral Angle between the planes
may now be found by first projecting a second Auxiliary

y
Elevation on line x 2 y2 as shown, and projecting from
this view, a second Auxiliary Plan on line x 3 y 3 normal
to the line of intersection in the last view. This results in X.
both the planes being shown in ' edge view '. and the true
angles between the planes shown, as indicated in the

P L AN
diagram.
The true shape of each of the triangles can be found as
a further exercise using the same method as shown in the
previous page.

AUX
E LEV

c+

AUX P LAN

111
 1 12

232A. True Shape and Dihedral Angle Two views of a

piece of folded plate are given. To find the true shape of AUX PLANS TRUE SHAPE & DIHED RAL ANGLE AUX ELEVS
the triangle bed, first draw the construction line ed in the
elevation, project to the plan and project the auxiliary
elevation as Fig. 1. From this project a plan, Fig. 2, to G IVEN : PLAN & E L EV
obtain the true shape. Similarly, the construction line bf
in the elevation projected to the plan and a new elevation FIND: I TRUE SHAPE
and plan as Figs. 3 and 4 will give the true shape of 2 D IH EDRAL ANGLE
triangle abc. Figs. 2 and 4 complete the development.
To find the dihedral angle, project the auxiliary eleva­
tion as Fig. 5, then project a further elevation as Fig. 6,
obtain distances gk from the plan. Refer also to 23 I.

Exercises
I . Using No. 39L, draw the plan and front elevation
when the object is cut by an oblique plane VTH whose
traces are at 45° to the xy line inclined to the right, and
T being 5 mm from the lefthand point of the object. First
angle. E LEVATION

2. A cube of 50 nun side stands on the H P at an angle of

30° to the left and touching the VP. It is cut by an oblique
©
plane VT and T H both being at 30° to xy, inclined to the
left. T is IO mm to the right of the cube. Project a plan
and two elevations and the true cut section face. First
angle.

3. A pentagonal prism of side 30 mm and height 60 mm

stands on the HP one side parallel to and 5 mm from the
VP is cut by an oblique plane, VT is at 20° to the xy and
rising to the left, HT is at 30° to the xy and falling to the
right. T lies over the centre point of the solid and on the
xy. Draw the plan and elevation and the true shape of the
section. First angle.

d
PLAN 6
DEVELOPMENT

0
------ ----

SIMPLE DEVELOPMENTS
AND INTERPENETRATION
Rectilinear
Cylinders
Pyramids
Cones
Spheres
Palmate
 1 14

S I M P LE
.. D O DECA HEDRON
I

-- · -1---
I
--:r
' £/

PR ISM

DEVEL O P M ENT
i
i

12 P EN TAGONAL SIDES
DEVELOPMENT

I
Ii
I

1 32

D E V EL O P M EN T

CYLIN D E R
Simple Developments
233. Rec(angular Prism Simple projection of the faces
SIMPLE D E V E L O P M E NTS

'
DEV ELOPMENT OF TETRAHEDRON

8
in line. True lengths are shown either in plan or elevation
and require no rabatment. SPHERE

C") !
234. Cut Cylinder Intersection of the twelve generators o!
with the lines of the cuts give points which may be pro­
jected to the development. The top face would be an
ellipse. The lower face would be part circle.
235. Dodecahedron A solid having twelve regular pen:
tagonal faces. The projections and development are
shown. DEV ELOPMENT OF OCTAHEDRON
236. Tetrahedron A solid having four equilateral tri­
angular faces.
237. Octahedron A solid having eight equilateral tri­
angular faces.
238. Hood Rectangular section. The development of
the front face is shown, widths being taken from the CAN ISTER
elevation, true length of the line O to 9 being stepped off by
dividers. RECTANGUL AR SEC T I O N
239. Sphere The development of the surface of a sphere
may be approximated by drawing lines, whose length is
nR. The curved shape is obtained by ordinates from sec­

1-
RECTANGULAR SECTION H O O D
z T
tions of the hemisphere as shown.

w
240. Canister Rectangular section. Part development

-+
of an end is shown, widths being obtained from the plan,
0 to 1 3 being stepped off by dividers. � -+
Q.

I
0
..J 0
>
w C")

w
0

!1
ii

1 15
 116

2-H. Interpenetration of Cylinders Equal diameters.

Tee joint. Draw the plane and elevation. Divide the I N TE R P E N E T R ATION
(D TEE
circles into twelve parts and project generators. The
intersection of the appropriate generators give points on @ CORNER
the line of interpenetration, in this case a straight line.
The developments of the cylinders are best made in line
with the portion developed. Twelve chordal distances
have been taken to represent the unrolled cylinder, but
since the developments are symmetrical. only three points
need projecting.
242. Right-angled Junction Equal diameters. Only the
elevation need be drawn, a construction semi-circle gives
the generators. The line of interpenetration is a straight
line. The development is symmetrical, and only a half
development need be drawn, the other half being iden­
tical. DEVELOP M EN T
Greater accuracy may be obtained by drawing twice as
many generators and by calculating the circumference as
nD, but the twelve generators are usually accepted.

'--------�3-
DEVELOP MENT

I S OM ETRIC
8
VIEW

I NTERPENETRATION OF CYLI N DERS OF EQUAL DIAMETERS

243. Interpenetration of Cylinders Unequal diameters.
Right angled. Draw the outline plan and elevation. Draw C YLINDE R S
the generators of the smaller cylinder, obtain their points

8
of intersection with the larger cylinder in the plan. Pro­
ject these points from the plan to the elevation. The line A
of interpenetration is obtained by the intersection points
of the generators, through which a fair curve should be
drawn. The developments of the cylinders should be
drawn in line.
TEE
244. Two Unequal Diameter cylinders interpenetrating
at an angle of 60°.
Draw the outline of plan and elevation. Draw the
generators of the smaller cylinder and obtain their points

DEVELOPMENT
of intersection with the larger cylinder in plan. Project o I .1. .3
these points to the elevation to intersect the numbered

I NTERPENETRATION
generators there. The curve of interpenetration can be

OF CYLIN DERS
drawn through the points of intersection. The develop­
ments can now be made in line as before.

OF UNEQUAL DIAMETERS

CO -AXIAL

117
 1 18

245. Cylinders of Unequal Diameter, Offset Draw the

plan and elevation. Draw the generators of the smaller OF F S E T C YL I ND E R S
cylinder, obtain the points in the plan. Project these
points into the elevation to intersect generators which
give points on the curves of interpenetration. Note that

8
the front and rear curves are not the same. The develop­
ments may be drawn in line with the portion of the
cylinder being developed.
The isometric pictorial view shows a horizontal section
which gives a circle for the vertical cylinder and a rect­
angle for the horizontal cylinder. It should be seen that
if a number of sections are visualised with intersections
of the two shapes at the varying points the line of inter­
penetration becomes easier to understand. The line of the � a 4- 6
generators in the elevation show such sections.

IN T ER PENETRATION
OF T WO CYLINDERS OF
UNEQUAL DIAMET ERS
O FFSE T
I
$3 60 I
. I

I
i
I

.__________J
 I

246. Interpretation of Two Cylinders Unequal dia­

meters. Offset and angled. O F FSET AN D ANGLE D

e
Draw the plan and elevation. Draw the generators of
the smaller cylinder in the plan and obtain their points of
intersection with the circumference of the larger vertical
cylinder. Draw the generators of the smaller cylinder in

INTER P EN E TRATION
the elevation. Obtain the points of intersection of the
generators and thus the line of interpenetration. Note
that an additional generator must be drawn for points I OF TWO CYLINDERS OF
UNEQUAL DIAMETERS
and I I .
The developments may b e drawn in line with the

ANGLED & OFFSET

cylinder being developed.

U')

DEVELO P M E N T

1 19
 120

247. Two Projections of a Cone and a Pyramid are shown

interpenetrated by a cylinder (not shown). CONE D E VELOPM ENTS P Y R A M ID
The projections and development are shown. The

8
generators of the cone are drawn from the plan and their
points of intersection with the cylinder line projected a
back to the plan to give the line of interpenetration. The
sector of the circle for the development, arc as the twelve
divisions of the base circle is drawn, and the points on
the generators projected to the line of true length before
being swung to the corresponding line in the develop­
ment.
248. The Pyramid The projections are similar to that
of the cone, generators being drawn through convenient
points in the cylinder curve in the elevation, and projected
to the plan. This enables the line of interpenetration to
be drawn in plan. The true length of O C is found by
rabatment, this is a generator of the · enclosing cone ' and
enables the development to proceed. The base true
lengths A B and B C are stepped off along the develop­
ment curve, and generators drawn in. Points I to 7 must
be projected to the line of true length before being
fI
brought to the corresponding generator in the develop­
ment. o
0-
/
249. Cylinders and Cones Three cases of interpenetra­
tion of these solids are shown. Generators ofthe cylinders
are drawn first, sectional circles of the cone are drawn in
the plan, points on intersection obtained, and projected
back to the elevation. The line of interpenetration can be
drawn through these intersections. Horizontal sections
of the solids show the intersection of a circle for the cone
and a rectangle for the cylinder.
Refer to the pictorial isometric. No. 32.
The developments of the cone are made radially from
the generator, the development of the cylinder is best
made in line.

75
a: n c lo s i n g
concz

¢75
9
 C Y L I N D ER AND CONE

8
8

¢ 75

3 STANDA R D E X A M PL ES Horizontal

OF C O N E & CYL I N DER

Sections
cone g i v e s
IN T E RPENETRATION
cylinder circle
g i ves rec ta n g l¢
DEVELOP M EN T
O F CONE
DEVELOP M E N T O F CYLIN D E R

121
 122

250. Cylinder and Cone The interpenetration of a right

cone and a vertical cylinder is shown. Draw the traces of I N T E R PE N E T R ATI ONS
four horizontal sections in the elevation, each section

8 8
line will give a circle for the cone and a circle for the I SPHERE &
CYLI N D E R I
I
HEXAGONAL
\
cylinder. The intersection of the circles shown in plan \
& CONE
when projected back to the elevation section line will give PRISM
four points through which the line of intersection may
be drawn. Draw First Angle.
25 1 . Cylinder and Sphere Horizontal sections give a
constant circle for the cylinder, and varying diameter
circles for the sphere. Draw Third Angle.
252. Sphere and Prism A sphere and vertical hexagonal
prism are shown. Simple projection gives the intersection
of the sphere and the edges of the prism, but a midpoint
of the prism must be projected as shown to give midpoint
intersection through which the curve may be drawn.

,...0
Draw Third Angle. 0
'Q
Q
253. Cone and Sphere Horizontal sections give varying
circles for both cone and sphere. The projection of their 'Q
intersections from the plan to the elevation give the line
of interpenetration. Draw First Angle.
254. Cylinder and Sphere Horizontal sections give a
constant circle for the cylinder and varying circles for the
sphere. Projections from the plan will give the inter­

8 8
CYLINDER C O NE &
penetration curve points. Draw First Angle.
S P H ERE SPHERE
255. Two Cones If two cones interpenetrate so that the
focal sphere contained touches the four generators, the
lines of intersection will be straight. Draw Third Angle.
256. Cone and Hemisphere Two horizontal sections
give points on the intersection line, obtained by simple
projection. Draw in First Angle.
257. Two Solids of Revolution Four horizontal sections
are used to obtain points of intersection. Intersecting
circles in plan show the curve. Draw in First Angle.

I
258. Palmate Section If a turned rod is cut by a plane
as shown in the diagrams, the curves of intersection are
ol
0
obtained by projection from horizontal sections as
shown. Draw First Angle.
'Q
 VAR I OU S I N TERPE N E TRAT IO N S SEC T I O N
CONE & H E M I S PHERE PALMATE SECTION

9
CYLIN D E R
& S P HE R E

0
0
'S.

&

123
 1 24

259. lnterpenetrating Spheres The section is a circle,

and the· p rojected line of intersection in the given example S P HERICA L

8
is a straight line.
S PHERES
260. Touching Spheres Problems of this nature are S!lctfon Z /Z
shows true
based on common tangency between adjacent spheres. l <Z ngth of p i n
The distance from one centre to the centre of the touching
circle is the sum of the radii. The tangent is at right angles

8
to the line joining the centres. In the example shown of
five equal spheres arranged four and one , the four are
drawn first touching in plan, then projected to the eleva­
tion and the fifth sphere drawn tangentially.

TOUCHING
X
261. Sphere P ierced by Pin True positions on the sphere

SPHERES
are found by auxiliary sections in this case on the line of
the pin obtained by joining A and B in plan. The section
Z / Z enables true heights H1 , H 2 and H3 to be found and
used in the elevation where projectors from the plan
allow the projections of the pin to be completed.
262. Cylinder and Touching Spheres Given a cylinder

SPHERE
and two spheres draw them in mutual contact. Draw the

PI ERCED BY P I N
sphere l and cylinder in contact in both plan and eleva­
tion. Draw construction circles of the smaller sphere in
contact with the cylinder and larger sphere, and swing
the centres of the smaller sphere to intersect as arcs at C
in plan. The points of contact are then obtained by pro- R'
jection P, P 1 , P 2 . x __.:::_+-�--.::,,,.�---�-_.:::-...,:::::_-__:_
Y
:..:_ c y l i nd!lr

8
y

& SPHERES
 263. Spheres in Contact with a Pyramid and each other.
The large circle representing the sphere is drawn tangen­ SPHERES IN C O N TA C T
tial to the horizontal plane and the side of the pyramid.

S P H E R ES IN C O NTACT
Point of contact can then be projected to the plan. The
smaller circle is then drawn tangential to the HP and

WITH PYRAM I D
the pyramid. Before this can be projected to the plan, the
two circles must be drawn as in the right of the diagram
tangential to each other and to the HP obtaining the
projected length of R 1 and R 2 . This distance is used in
the plan to find the centre of the smaller circle. It must be
seen that the points of contact are only shown when a
mid section of both of the touching spheres is drawn, and
often an auxiliary elevation or plan has to be projected
first before these circular sections can be drawn.

X y

A U X I L I ARY
E LEVAT I O N
¢ so

1 25
 1 26

MATCHING CONE & CYLINDER

263A. Interpenetration of Cone and Cylinder
A. Given the cone and cylinder sizes, and the angle of
I NTERPE NETRATION
interpenetration, find the· line of intersection and draw
the developments.
Draw the cone elevation. Draw the circle shown equal
in diameter to that of the cylinder, and tangential to the
slope lines of the cone. The circle represents the ' en­
velope d ' sphere which is common to both cone and
cylinder. The generators of both cone and cylinder are
tangential to this sphere, and it should be seen that the
cylinder can be rotated to any position around the sphere.
The centre line of the cone and cylinder can be drawn at
the given angle. Draw the outer generators of the cylin­
der, these will cut the cone. Join these points to give the
intersection line which will, in these cases, be a straight
line.
The developments can be drawn by methods shown on
ENVELOP ED
previous pages.

,/"'1
B. Given Cone and Line of Intersection, find the angle SP HERE

@
and diameter of the cylinder to intersect the given cone.

,,� /_ .,
Draw the given cone and line, the given line cutting the

<
cone will yield an ellipse (see 52) which will also be a
section of the desired cylinder.
C. Bisect the line of intersection to give C. Cut the cone
horizontally through this point, draw the semi-circle on
this line. Drop a perpendicular from C to cut the semi­
circle, and draw the arc, centre C and radius C B as the
perpendicular. .
Draw the generators of the cylinder tangential to this
arc and to meet the line of intersection as shown.

CONE G IVEN : CONE & LIN E

SECTION OF INTERSEC TIO N
LINE
---- -

FURTHER
INTERPENETRATIO NS
Angled Prism and Pyramid
Angled Cylinder and Cone
Co-axial and Offset
 128

P RISM AND P Y RA M ID 1ST ANGLE JUNCTION

ELBOW & J U N C T ION

Part
D ll V ll lopment 8

'

�---
/

8 I

S Q UA R E P R ISM & T R IA N G ULA R PY RAMID

 _______ _
.....,.--

264. Square Prism and Triangular Pyramid, interpene­

tration. The projections and developments are shown. ANGLED PRIS M & PY RAM I D
Draw the plan and project the elevation. Points of inter­

8
section obtained from the elevations can be projected
back to the plan. The · enclosing cone' must be projected
from the plan before the development of the pyramid can
HEXAGONAL PYRAM I D & ANGLED TRIANGULAR PRISM
proceed.
265. Elbow and Cylinder Junction Projections of a
cylinder and a quadrant section of pipe interpenetration
are shown. Draw the twelve generators of the cylinder
and obtain points of intersection with corresponding
horizontal section generators of the quadrant pipe.
266. Hexagonal Pyramid and Angled Triangular Prism.
Draw the plan and elevation; all the points of interpene­
tration can be obtained from the projections, except one
at the lower right hand. A generator of the pyramid in
plan projected into the elevation to meet the prism line
at a completes the lines of interpenetration. Points of
intersection on the pyramid must be projected horizon­
tally to the true length line which in this case is the outer
generator, before being swung to the development. The
part development of the prism has been projected in line
from the elevation ; true lengths at B and C are found from
the true end shape of the prism by a part auxiliary view.

3RD ANGLE
1 29
 1 30

267. Cone and Inclined Cylinder Draw the plan and

elevation, project the inclined cylinder to the H P to A NGLED C YLI N DE R AN D C ON E 1ST A NGLE
obtain the ell(P._ s_�_(f\uxili.ary plan), Project the apex of the
cone to the H P, and draw several trace lines to cut both
circle and ellipse in the H P. Project these points into the I N TER PENETRATION OF CONE
elevation as generators of cone and cylinder��r_i9.ter­ & I N CL I N E D CYLI N DE R
section in the elevation gives.the line of i nterpenetration,
and the curve is drawn through them. The points should

8
be projected back to the plan, and the line of interpene­
tration drawn. The developments are projected in the
usual method, in line for the cylinder and sector for the
cone.
Take base dia of cone as I 00 mm.

poi n t
to

c y l l n d c r e l l i pse
on H P ..j
 J ST A N G L E INCLINED SOL IDS
268. Interpenetration of Cone and Inclined Prism. First
angle projections of the interpenetration are shown. INTERPENE TRAT I O N
Draw the plan and elevation. Project an auxiliary plan

8
at right angles to the axis of the prism. A true section of
the prism will be shown. Widths below xy and x 1 y 1
CONE &
lines are the same in both original and auxiliary plans. ELEVA TION
Draw the generators of the cone in the plan, project to P RISM

C O -A X JA L
the elevation and thence to the ellipse in the auxiliary
plan. J:'l.�_1?ber the points of intersection of the prism
section with the generators in the auxiliary plan. Project
. .these points to the elevation and to their respective
genera.tor. Join these points in the elevation to give the
_curve of interpenetration. Project the points to the
original plan to the respective generator. Join these points
to give the line of interpenetration in the plan.
Draw the development oft he cone. Draw in the genera­
tors. Mark off the true length of the point of intersection
on each generator, draw a fair curve through the points
to complete the development of the cone.
Project the development of the prism at right angles to
the axis of the prism.
Make the developments in thick paper and assemble
as a model to prove the method.
Take base dia of cone as I 00 mm.

P LA N

3
c.___________ __ _ _ ____ __ ____ ___________________________________ ___,
_

131
 1 32

3RD ANGLE
269. Interpenetration Oblique Octagonal Pyramid and
an Inclined Cylinder-offset. I N TERPENE TR AT I O N
Third Angle projections of the interpenetration are
given.
The problem may be set showing the incomplete plan
OBLIQUE OCTAG ONAL PYRA M I D
and elevation, or by description of the two solids and the
detail of the offset axes.

I I N CLINED CY L I N DE R ­
Draw the incomplete plan and elevation in simple pro­
jection, labe! the xy line.

O F FSET
Project an Auxiliary Plan as shown, using a new x 1 y 1
line which is normal (at right angles to) the axis of the
inclined cylinder. Widths remain the same in the new
plan, i.e. points 0, 1 , 2, 3, 4, 5, 6, 7 will be the same distance
from x 1 y 1 as from xy in their respective plans.
The auxiliary plan will show points of intersection of
the cylinder with the faces of the pyramid.
These points may now be projected back to the respec­
tive generator in the elevation, giving points through
which the line of interpenetration may be drawn.
The line of interpenetration in the original plan may
be obtained by projecting the points from the elevation
to this view and drawing a curve joining the points.
Note that the points will be a similar distance from
their respective xy line and this gives an additional y
check on accuracy in projection.
Developments of the cylinder and pyramid can now be
made to prove the method.
Take length 1-2 as 30 mm.

8
I.

I
I

I
 270. Interpenetration Oblique Cone and Inclined Cylin­
der, offset. F I RST ANGLE I NT ERPENETR ATION
First Angle projections of the interpenetration are
given. The problem may be set by incomplete views or
by a written description.
Draw the views as fully as possible from the given--0ata. O B L I QU E CONE
Project an Auxiliary Plan from the elevation on x 1 y l ,
normal (at right angles to) the axis of the cylinder. Widths & I N CL I N E D
remain the same as in the original plan, note width ' Z ' . CYLINDER
.Draw the generators of the cone, project these to the
. Auxiliary Plan. Where these cut the circle which repre­ - OFFSET •
·sents the section of the cylinder, points 1, 2, 3, 4, 5 , 6, 7, 8, : . :
·9; 10, 1 1 , 12 and a, b , c, are found. These are projected
back to their respective generators in the original eleva­
tion. The fair curve drawn through the points gives the
line of intersection.
The positions may now be projected to the original
plan and the line of intersection shown in this view.
Check that the points in both plans are the same
' widths ' from the respective xy lines.
Draw developments of the cylinder and cone on strong A U X I LIARY
paper or card with construction flaps and erect the model.
PLAN
Reference to the development of the oblique cone and
pyramid by the triangulation method shown in the next
section should be made.
Take the base circle dia of the cone as l 00 mm.

8
133
----- --------------------------------- - - - - -

134

271. Interpenetration Right and Oblique cones. First

angle projections are shown including an A uxiliary Eleva­ O BLIQUE & RIGHT CONES I NTERPE N E TRATION 1ST ANG LE
tion . Dr.aw.first the simple outlines in plan and elevation,
taking the diameter of the right cone as 100 mm, other
dimensions by proportion. Project the. simple outlines of

-,,t...
. th"- Auxiliary Elevation. Draw the lines of the five section

0
. planes in both elevations. Project the centres and limits
/
of the planes to the plan, and draw the circles in plan. The
intersection points of the two circles on each section
· plane indicate the line of intersection in the plan and EL E VATION
these points may now be projected back to their respec­
tive section plane in the elevations indicating the line of

may now be found giving a. Not.e that this lies. on a line

interpenetration in these two views . .A further section line

joining the two centres on a section line obtained from Section planes

5 __:a
�======�=====t���=::/."=/
the auxiliary elevation.

4 ---- --!.---�:.._++....,..:hf----,:'li....J[,L-
See also Nos. 245- 258 for uses of the horizontal section
planes to obtain intersection points. (Based on an A EB
exam question.) 3 -------+---,,&- - -+- -M,
2 ---��-+--+7'c:....,t +-::.a1-1�/..,+--I-+-�
- '---l-+--l-.lla'l.-.t-

0 "'---'--+���..;.i,i<'---H-�¥1-1-#-�'/A/
I

PL A N
 Exercises
272. lnterpenetrations and Developments
E X E R C I SE S

8
A G
A. Draw the developments of the two cylinders shown
in the elbow. fl)'4 0

/:

"-,
/
/ i
:
- -r-
B. Draw the developments of the two pieces of tubing
shown in the elbow. If)

�i
CD

C. Draw the projections of the two offset unequal dia­

meter cylinders which meet at right angles. Draw the
developments of the two cylinders.
65
D. A cylinder and cone interpenetrate at 90 , co-axial.
°

Draw the projections and show the line of interpene­

tration. Draw the developments.
B ¢ so H

E. A right cone and cylinder interpenetrate as shown,

co-axial. Draw the full projections, and developments.

F. A cone and square prism interpenetrate at right

angles as shown. Draw the full projections, and draw the
developments.

G. A square pyramid and a sphere interpenetrate as

shown in the diagram. Draw the projections, Draw the
development of the pyramid only.

r
C F
..,
H. A sphere and cylinder interpenetrate as shown. Draw
the projections ; project the development of the cylinder.
1 r ¢ so

I . A hemisphere and two cylinders interpenetrate co­

axial as shown in the diagram. Draw the plan and com­
pleted elevation. Draw the developments of the two
cylinders.

i....,.___.....0
,c._8,,.__
5 _ ___
.., �
13S
 1 36

D
Exercises
E X E RC I S E S
272A. Interpenctrations F
A. A cone and cylinder interpenetrate as shown. Draw
the full elevation and plane. Draw the developments, in
First Angle.

B. A cylinder and triangular prism interpenetrate, angled

and offset as shown. Draw the completed plan and eleva­
tion. Draw the developments, in Third Angle.

C. A cylinder and two cones interpenetrate as shown.

Draw projections which show the line of intersection of
surfaces, and draw the developments, in First Angle.

0 30 B G
D. A triangular pyramid and cylinder interpenetrate as
shown in the diagrams. Complete the line of interpene­

·91
tration, and draw the developments, in Third Angle.

�
E. From the figure shown, draw the projections and de­

'
/'t�
velopments of the four parts, in First Angle. '< I
, ,
E

�
F. An octagonal pyramid and a cylinder interpenetrate
as in the diagram. Complete the projections, and draw '

the developments, in Third Angle.

O FFSET
G. Two cones interpenetrate as shown. Draw the full
projections and developments, in First Angle. C H

H. A cone of I00 mm dia base circle and !00 mm

height is pierced by a triangular prism positioned as in
the given plan. Complete the projections, and draw the
developments, in Third Angle.
DEVELOPMENT BY
TRIANGULATION

Oblique Pyramid

Oblique Cone

Transition Pieces

Junction Piece
 138

273. Oblique Pyramid The true lengths of A 0, B 0, __

r-O __
BL ___ _P Y_ R
___---�-,
CO and DO are obtained by rabatment from the plan. I Q UE A M ID TRA N S I T I ON P IECE

8
Arcs are swung from O in the plan to a line parallel to
the xy line, from A, B, C and D, and rabatted to the
xy line. A l , B 1 , C l , D 1 , are then joined to O l giving the
true lengths of the corners of the pyramid.
The true lengths should be seen to be the hypotenuse o'
of a right-angled triangle which has been rabatted to the
vertical plane so that its true shape may be seen. This
system is shown in the chapter on Lines on the Inclined

II
Plane.

e.
The true lengths of A B, BC, C D and D A are given
directly from the plan. The development of the pyramid A O C.
is now drawn by constructing, by the arc method, the
four triangles shown, the first, for example, has sides
using true lengths, O l C 1 , O l D 1 , C 1 D 1 .
))
·"/
Where the pyramid is truncated, the true lengths of the
corners may be obtained by horizontal projectors from
the elevation cutting the true length lines as shown, and ,'
transferring these points to the development.
274. Transition Piece Square to square, angled. From
the plan and elevation given it will be seen that the faces
are triangles. The bases of the triangles are rabatted as
before to the elevation, where heights remain the same,
and the new hypotenuse gives the true length of the line
required. Lengths of the squares I , 2, 3, 4, and A, B, C, D, TRA NSI TION PIECE
are truly given in the plan, and the development may be SQUARE TO SQUARE
• A NG LED •
drawn as shown.

DtZve lopm4t nt by
Notice that in this example, there are two sets each of

Triangu lation
four similar triangles. The examples should be drawn to
a convenient size, say twice the printed size, cut out and C
pasted together (make pasting flap at 3A), as proof of the
method.

P ROJECTION OF A N D«v4t lopmtZnt

OBLIQUE PY RAMID A
 275. Transition Piece Rectangle to circle. Pipes and
ducts for conducting air, gases, powdered solids, etc . , TRIANGULATION 3RD ANGLE
often require two dissimilar sections to be smoothly
joined usually by a transition piece such as the one shown.
A pattern of the development has to be made before TRUE LENGTHS PLAN TRUE LENGTHS
cutting, folding and joining the sheet metal to form the
transformer piece. C. C.
The method used is that shown in the previous ex­
ample, triangulation, the circle being divided into twelve
parts, the points being joined to the nearest corners of
the rectangle to give connecting triangles. The true shape
of the triangles is then obtained by rabatment, and the
final development drawn by constructing the triangles in
their adjacent order.
The plan and elevation of the piece is given in Third
Angle projection. A B, B C, C D, DA are true lengths as
shown in the plan. 0 l , l 2, 2 4, etc. are chordal. B 0, B l ,
82, etc. are rabatted as shown, and a new hypotenuse
drawn to the height line from the elevation gives the true
length. The development is now drawn by constructing
the triangles in the correct order, using true lengths.
The piece should be drawn to a convenient size, take
AB as I 00 mm, cut out and pasted together as an example.
Chordal distances on the circle are usually allowed to

work would demand that the circle be taken as D =

be taken as the third side of the triangle, but accurate

3 . 1 4 1 6, and divided by 12 to give the true length of the

small arc.

ELEVATION

TRANSITION

DEVELOPMENT

1 39
 1 40

276. Transition Piece Square to circle. Draw the plan

and elevation. Divide the circle into twelve parts, join as DE VE LO P MENT BY T R IANG U L ATION
in the diagram to A B C D. Obtain true length of C 6 and
C 7 by rabatment to the elevation. Draw the development
by constructing the triangles. Lengths O I, 1 2, are
chordal.
277. Transition Piece Shaped hood. Draw the plan and
elevation. Divide the circle into twelve parts. Join to the
shown points A B C D E F G, to give triangles. Draw the
rabatted triangles in the elevation to obtain true lengths
of A 1, B t, B 2 , etc. In the diagram, in this instance, the
bases of the rabatted triangles have been stepped off by
dividers to the elevation instead of arcs from the plan.
The hypotenuse of the triangle is the required true length
to be used in drawing the triangles in the development.
Begin the development by drawing O A as a centre line
since the shape is symmetrical about this line.

D
p

TRANSITION
TRA N S I TI ON PIECE
P I ECE
S haped Hood

S q uare to C i r c l e

8 8
Deve l o p m Q n t
D A
 278. Oblique Cone Notice the circular plan of an
oblique cone. OBLIQUE C ONE
A
Divide the base circle into twelve parts. Draw genera­
tors. Drop a perpendicular from the cone apex A to cut
the xy line in A 1 .

8
Rabat points O to 6 to the xy line to give points 0 1 to
6 . Join these to A. These are the lines of true length of
1

the generators shown.

The full development of the oblique cone may now be
drawn by constructing triangles using true lengths ob­
tained above, and the chordal distances from the base
circle. (True length of the arc is given by 0 = 3. 14 16/12
for more accurate work if required.)
Three section lines of the cone are shown. True lengths
from the section lines can be obtained by projecting y
horizontally from the intersections of the section lines
and the respective generator to the correct line of true
length. These distances are now transferred to the respec­
tive line in the development and joined by a fair curve.
Notice the shape of the development of the oblique
cone as compared with that of the right cone, shown on
an earlier page.
Draw the example of the oblique cone shown on this
page, base circle 75 mm dia, height 70 mm, generator

A
A 3 at 45 ° to the xy line. Draw the development when the
cone is cut by a plane parallel to the HP and 40 mm
above it.

PROJECTION OF A N
OBLIQUE CONE

Deve lopment

141
 1 42

279. Junction Piece Three cone. Third Angle projec­

tions of a junction piece in which three cones meet at DEVELOPMENT 3RD ANGLE
1 20° .

8
Draw the plan and elevation as shown, using a base
circle of 75 mm dia, cone height 75 mm, upper circle of THREE CONE JUNCTION PIECE
cone 40 mm dia.
Draw the generators. Obtain the points and lines of
intersection.
Draw the part development shown (a) by simple coni­
cal development, using th·e apex of the cone, (b) by tri­
angulation.
PLAN
In (a) find the apex of the cone by extending the two
extreme generators. Obtain the true length of A 0 1 , and
draw the full development of one cone. Draw horizontal
projectors from the intersection points in the elevation
to the line of true length A O 1 , and cut the generator
lines in the development by arcs from A. Join these points
by a fair curve to give the final development.
In (b) set out the diagram of part of a cone as shown;
obtain first the true lengths of 0 0 1 , 1 1 1 , 2 2 1 and 3 3 1 by PART
triangulation as in the previous examples. Draw in the
diagonals of the three panels of the cone shown, and
obtain their lengths by triangulation. Check the lengths
on the first development.

ELEVATION
 e
Exercises
TRIA N G ULA T I O N E X ERCISE S
280. Development by Triangulation A D
A. Make the development of the oblique cone shown.

B. Make the development of the oblique octagonal

pyramid shown. y
y

,..., +-+-- -+---f--:::

C. A transition piece circle to square is shown. Make a
development of the surface of the piece. Joint to be o n Ill
centre line o f the front face.

D. A hood is shown; develop the surface by the method

E
of triangulation, joint to be left rear corner seen on the
plan. B

E. Develop the transition piece shown by triangulation,

joint to be on the rear centre line.
X
X y
F. Make a development of the surface of the truncated
oblique cone shown in the diagrams. Joint to be on the
shortest generator.

G. A transition piece, rectangle to circle is given.

Develop the surface by triangulation. Make the joint on
the centre line of the left face.
C F
H. Make a development of the hood shown in the dia­
grams.

X
I. Develop the fairing shown in the diagrams. Make the
joint in the middle of the front face. X

The projections given in the above problems should be

scaled up from the one dimcnsioq given. The develop­
ments could be made finally on drd cut-outs to prove
the answer.

143
 144

Exercises
DEVELOPMENT TRIANGULATION E X E RCISES

K
280A. Development by Triangulation
J. A transformer or transition piece is shown. Draw the
views as shown, draw the generator lines. Obtain the

"
lines of true length. Draw the development.

Ill
K. A hood is shown, circular at the base, transforming

°'0
to a rectangle at the top. Draw the given views, draw in
the generators. Obtain the lines o f true length, and pro­
ceed to construct the development.

.¢ 100 ' 25

\
L. A transformer piece is shown, base shape encloses

ft
three circles, changing to a circular top. Draw the given
views, draw the generators. Obtain the lines of true
length, construct the development.

M. A shaped hood is shown, circular base, changing to

an angled square. Draw the given elevation, project a
plan, draw in the generators. Obtain the lines of true
length, construct the development.

The developments should be cut out and pasted together L

to prove the solution.
 Part II
ENGINEERING DRAWING
RIVET AND SCREW
FASTENINGS
Rivets
Riveted Joints
Screw threads
Bolts and Screws
Conventions
Lockouts
Locking Devices
 1 46

¢ R I VET H EA DS BUTT JOIN TS

I
S NAP HEAD COUN T ER SUNK C OUNTERSUNI< FL AT H E A D U N IVE RSAL HEAD
L2
1.6 ':
20 2 D
FLAT H EAD
1.5 D
ROUNDED HEAD
2 0

0.0 2 D 1"
""'0
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5_
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2 I.I)
0.6D /"

3 �
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4
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I, D .
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" I
.&:

6 �
s !:; C:
0,43 0
D
...J
·D
B
10
12 D = 6 /thkknns of pl at�

36 SI NGL E LAP
•• DOU8 LE R O W LAP

l I
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$ $-
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+ + gz
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uJ

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$ I
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+ +
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 281. Rivets A rivet is a permanent method of fastening
two or more pieces of metal together. The head is pre­ R I VET E D J O I N TS J O I NT FA I L U R E S

I
formed by stamping, the second head closed by pneu­
matic hammer or hydraulically whilst red-hot. Small
rivets are closed by hammer and set. The shape and de­
81
tw/7� - I
tailed form of common rivet heads are shown. Proportions
of rivets shown in terms of D = 0 of Shank.
Rivet Spacing A simple spacing based on the dia­
meter of the rivet is shown for single row lap riveting, for
double row chain riveting, for zig-zag riveting, and for S I N G L E SH EA�
a butt joint with double cover straps, the outer row left
alternate.
282. Seven riveted joints are shown. Corner joints in DOUBLE SHEAR
tanks may be made by bending the plate or by the use of
angle. Slides and box constructions can be built up from
standard sheet, strip, angle, channel and tee sections.
283. Joint Failures A rivet may fail by single or double
shear stress. A riveted joint may fail by the plate tearing
on the line of the rivets, or by marginal tearing, or by
elongation of the rivet holes. CORNERS
Stresses in Rivets These may be calculated by using
the following formulae, where d= rivet dia, p = rivet
pitch, t = plate thickness,/, =safe tensile stress for plates,
f. = safe shear stress for rivets,!,= safe crushing stress for
rivets or plates. (Rivets, 386 MNm - 2 to 463 MNm - 2 ). F LU S H
Single shear= V, = rrd2f./4.
Double shear = Vd = rrd2f./2.
Crushing= V, = dtfc.
Tearing resistance= (p -d)tf, .
d may be found by equating f. and!,,

= 4rrf.
SLI D ES
d tfc

p may be found by equating tearing resistance and

shearing
1td2f. .
P = 4 t/, + d x number of rivets x rows.

147
 148

�/ V E T S R IVETED ST RUCTURES
I
8 ROOF

8
t = t h i c k n ess t h c o r e t ica I n e a r e s t std

i
of p l a t e 0 r i vet 0 r i ve t
I I TRUSS
+
'
1 .0

I
6,0 6 'I

I
! I
I ,2 6.0 6
I
1.6 6.0 6 ! + GUSSET
! 2 8.4 8
2.5 9.5 10
+
l
I
I
i 3 10,5 10 + + + + +
4 1 2. 0 12
5
STA NCHION
1 3. 5 1 2
6 1 4. 8 16
8 1 7. I 16
10 1 9. I 20 GUSSET
12 20.9 20
16 2 4.0 24
20 2 7. 0 24
24 29. 6 30
30 36

i
33. I
36 36 36
A method of obtaining a rivet of suitable diameter,
given the plate thickness.
284. Riveted Structures The structural engineer uses
standard mild steel hot rolled sections for the fr�mework
of buildings and roofs. The members are cut to length,

CHA N NEL
drilled or punched for the rivets or bolts and assembled.
The diagram shows a stanchion with its soleplate
riveted in position using plate gussets, and angle pieces.
The joists are fixed in position with brackets made from
angle sections, and the top plate is also bracketed by
angle sections.
285. Roof Trusses Designs for these are shown in an­
other chapter, and the frameworks are made in angle or
other section, riveted together by means of plate gussets,
as shewn in the sketches.

L_,_____ _ _ _ ____
Sections of angle, tee, channel and beam are shown,
these are made to standard sizes and sections. ..L__ _
_ ________ _ _ __J
_
286. Special Rivets for thin plates. Thin plates require
rivets which can be closed without distorting the sheet. THIN R I V E T I NG
Aluminium and duralumin, extensively used in the air­
craft industry, not easy to solder and weld, are riveted
using rivets shown in the diagrams.
Tubular Short lengths of tube set by a shaped
plunger.
M ushroom Large head, second end closed to the flat
shown.

�
Snap Solid or part hollow. A small explosive
charge detonated by a hot plunger enables

� ! ?§«<{
the rivet to be set in difficult positions.
Flush Countersunk rivets giving a flush or level
surface. Where the sheet is stamped or Jm� I
T U B ULAR MUSHR O O M FLUSH
' dimpled ' to form the countersinking a
stronger joint results.

I· nso.1
Mandrel Where the plate is inaccessible from one
Type side, the hollow rivet is expanded by means
of a mandrel which is then cut off level.
(Avdel type).
The Chobert hollow rivet allows the
mandrel to be withdrawn after expanding
the rivet, a plug then sealing the hole if
required.

� R:$ l�
Two forms of rivet spacing are shown in addition to those
previously given. The first shows alternate spacing in the
outer row of rivets, the plate being less weakened. The
A V D EL MA N D R EL
second shows a three, two, one arrangement for a strap
end which also tends to weaken less.
See B S 64 l and B S 4620 for further information on S NA P FLA T
rivets.

+ + $
-+ 4-- - -+- - -+-
---- - --
+ + + -+-
+ +
A L T E RNA T E S PA CING S T RA P END

1 49
 150

CONVENTIONS DES IGNAT ION

287. I S O Bolts Nuts and Washers The conventional
M E T R IC N U TS B O LTS
with proportions in terms of D = shank diameter. The
method of drawing the metric screw fastenings are shown &

------.
3
MET RIC BOL T EXT ERNAL Overall Length

__
SCREW Metal I Nominal Length
head may be drawn using a constructional circle 2 D in

]
_____..,--
.,____
diameter, enclosing the hexagon, or alternatively, the
THREADS

[
End /

r-: ,I
.-----,
f -
chamfer circle may be drawn first l . 75 D in diameter, and
-.---,-0:::--:::::--+-""""'==--

&
enclosing it with the hexagon.
Q

Qj_ .____,.____,__,._....,
The conventional method of rendering external screw­
threads on a stud is shown. Proportions are varied to

4
suit the design of the component which is to be fixed.
I nternal screwthreads require the hatching to extend
over the threaded portion. � j.e1a;, ./. N ot End
Screwthreads assembly shows the bolt clear and the

"
internal thread hatched. ,0 ! INT ERNA L SCREW T HREA D S
Thread inserts are shown in clear lines. +i C:
The designation of I S O bolts require: material, head al ...J

d:u
shape, thread diameter, pitch, length, type of fit. N I

i
Example : Steel, Hex Hd Bolts, I

M I O x l . 5 x 50 -6 g
Steel, Hex Nuts,
M 1 0 x l.5 - 6 H
Types of fi t for bolts are:
close fit, 4 h ; medium fit 6 g ;
free fit, 8 g.
Types of fit for nuts are :

SCREW
close fit, 5 H ; medium 6 H;
free 7 H.
See No. 293, page 156 for table of sizes and other THREADS
ASSEMBLY
relative information. Also tables in Appendix l.
B S 3 LS I972 part l .
B S 1936 I S O Metric screwthreads.
BS Handbook No. 18.

fil]_..:::::::_.J,.._:::::..�
LOCK T HREA D
NUT INSERTS
Designation:
Matcrrial, Head, ISOdetails
e.g.
ST EEL HEX HD BOLTS
M 2 0 x 2. S x 55
 APPLICATION FORM
288. Nut and Bolt Assembly The view shows two cast­
ings held together by a nut and bolt with a washer in­ S C R E W T H R EADS
serted under the nut. Note how the casting is sectioned, NUT &
whilst the bolt washer and nut are clear. BOLT SQUARE
Stud and Nut Assembly In certain cases, studs are ASSEM BLY
screwed tightly into the body casting and left in position
where they act as positioning devices enabling the second
component to be returned to its former place with
accuracy.
Note the method if showing the stud end and the
internal thread, with clearance at the bottom of the hole.
Bolts may, in some cases be used instead of studs, again,
the internal thread is longer to give clearance enabling the
ACME

..,
bolt to tighten.
p I
Screwthread Forms Sectional views of a number of
screwthread forms are shown. I·
STUD &
The thread follows the path of a helical groove cut or

NUT
pressure rolled on the cylinder of the bolt or stud, and
may be right-handed or left-handed, single or multi­
start (see helices). The pitch is the distance measured ASSEMBLY
from crest to crest of two adjacent thread forms. The
major diameter is the full diameter over the crests. The
BUTT RESS

Il
minor diameter is the core or root diameter.

p
i
Metric Thread A 60° angle thread similar to the
Sellers thread, sizes in millimetres. See No. 293.
Special purpose threads, Square. Used for threads
I

.\
taking pressure in both clockwise and an1 i-clockwise

ii
direction. Used extensively in machine parts for operat­

;1
ing slides.
Acme threads used for lead screws. BOLT

·I
METRIC p
I
Buttress threads. Use where pressure is exerted in one
I·
;

direction. :ffC,

o.�

Xlll

15l
 152

289. Machine Screws I S O machine screws take the

form shown in the diagrams. The proportions are given M AC H I NE SCREWS
in terms of the shank diameter.

I.
COUNTERSUNK RAISED CSK PAN C HEESE
2D 20
Countersunk head machine screws are used when a flush
finish is required on the outer surface. The head is either
20
slotted as shown, for use with a plain flat screwdriver, or 1.75 D t

1�.250� 1_o.25�1
recessed. A recessed head has a crossed closed slot re­

' _L
quiring the use of a Philips screwdriver.

Raised countersunk head screws give a small projection

of the head beyond the general surface.

Pan head screws require no countersinking and all the

head J:lrojects beyond the plate surface.
V o.to v J:

z
uJ
C�eese head screws also give ahead projection beyond D .J
the plate surface.

Square Head Set Screw is often ·used to fix collars,

GRUB SCREW HEX SOCKET SCREWS

gears and wheels to shafts, though the head can give

SLOTTED
imbalance.
CAP
-1t •! G RUB
I.SD
Grub screws, slotted for screwdriver give a non-projec­

��
tion finish, and are useful for fixing collars, wheels, gib
I I
strips. I I

,...,..
I I I
I I
1.....3:--. -;. l
Hexagon Socket screws have that shape recess in the
head, weakening the head less than the slot, and are
l
... ...
tightened by the use of a hexagon socket key. This is a
1.25 D
piece of hexagonal sectioned rod with rightangled bend
to give torque. D J: J:
\ I
z

.J I
Cap socket screws are the shape shown, and are used
z
C) C)
4 5° D
for securing covers and flanges which is counterbored to uJ

�i
take the caphead giving a flush finish.

Countersunk head screws also have hexagonal sockets

_t_

0 0
for use with a key.

SCREW
The ends of machine screws may be rolled, chamfered,

ENDS
D
radiused, coned, cup, w point, dog as suited to the special
purpose.
Tables in Appendix I give the size A/f of the hexagon

ROLLED
socket relative to the head size of various socket screws.

CHAMFER RADIUSED CONE

 ---- ------

290. Nut Locking Devices

A. Locknut A nut may be secured on a stud against N UT LO C K I N G DE V I C E S

8
vibration by the use of another nut as shown.
B. Slotted nut The nut may be secured by using a taper
pin or a split cotter pin which passes through a hole in
B ¢ as A/F ·j C
the stud and lies in a slot cut in the nut.

1
;...._.....u
_
C. Castle Nut Similar to the slotted nut, the turned re­
duction allows the folded ends of the split pin to lie with­
out projection.
Three I S O nuts are shown, the thin locknut, the
slotted nut, and the castle nut. The proportions are shown
in terms of the diameter of the shank of the bolt.
Ring Nut A set screw is tightened into the turned

LOCKNUT S LOTTED N U T CASTLE N U T

annulus after the nut has been tightened. A taper pin
may also be driven through the end of the bolt as an
additional safeguard.
Ring Nut Where a ring nut is used away from the
edge of a plate or casting, an additional collar has to be
used and located by a dowel pin as shown.
Wile's Nut This invention adopts a slit nut which can
be compressed by a set screw to give additional friction.
Spring Washers Split spring washers, with or without
serrations on the outer faces, may be employed to prevent
a nut from loosening.
Simmond's Nut This type of nut has a fibre or nylon
ring set in an annulus cut inside the nut. As the nut is

W I LE J NUT
tightened the ring is forced into the thread of the stud or
bolt and the compression and friction give the required
locking action. Frequently used on automobile con­
struction, but should be replaced on dismantling.
Taper Pin After tightening the nut, a taper pin may
be driven through a hole drilled and taper reamed to fit
the pin, the ends of which are opened after insertion to
prevent accidental withdrawal.
B S Hand book I 6. 1 50 Rec R 288.

S P RING WA SHER S I M MOND S NUT TA P E R PIN

153
 1 54

291. Locking Devices Nuts and bolts may be prevented

from loosening by adopting various anchoring devices. BOLT & NUT LO C K IN G DE V I CE S
Locking Plate This plate has a double hexagonal
aperture cut in one end which allows it to be dropped LOCKING PL A T E T WIN TA B WA S H E R
over the tightened nut or bolt head. The plate is secured
to the component by a set screw.
BENT TM
Twin Tab Washers The stamped sheet metal strip is
tightened under the bolt heads and the tabs of the washer
are then bent up closely to the side of the hexagonal face.
Single tab washers may be used at corners and edge
locations by turning the tab over the corner of the com­
ponent. When the tab washer is used in a central position,
a hole is drilled in the component to receive the tab.
Wired Bolts In this method, the bolt heads are drilled
to receive a wire which is passed through the holes after
the bolts are tightened, the ends of the wire are twisted
together to fasten off. It is essential that the direction of
the wiring prevents loosening of the bolts.

.i
I

CO R N E R

CEN T R A L

S P R I NG WA SHER T A B WA S H E R
TA B WA S H E R S
 292. Riveted Joints 3. Show the disposition and size of rivets in a double 14. Draw I S O Metric screwthread section, 25 mm dia
d= rivet diameter. cover riveted joint fastening two steel plates 150 mm twice full size.
p = rivet pitch. wide and 12 mm thick.
t = plate thickness.
I, = safe tensile stress for plates. 15. Draw I S O Metric screwthread section pitch 40 mm.
I, = safe tensile stress for rivets. 4. Show details of a riveted joint zig-zag fastening two
fc = safe crushing stress for rivets and plates. plates 200 mm wide and 1 8 mm thick, flush finish.
I 6. Draw the Acme thread and the Buttress thread sec­
Ultimate strength = safety factor.
tions pitch 50 mm; draw three crests in the above

=
Working stress
5. Sketch three methods in which a riveted joint may diagrams.
Single shear (p - d) x t xi, fail. Give reasons for the failure.

=
= ¼n: x d2 x .fs
I 7. Make a conventional drawing of a nut and bolt as in
d x t xfc.
6. Draw diagrams showing the riveted joint between a No. 287. Take shank diameter = 50 mm.

= ½n x d2 xf.
Double shear = (p - d)t x I xi, stanchion I section, 200 x 150 mm, and a beam of

=
similar dimensions. Scale I : 2
dx t xfc. 18. Draw any five of the various bolts or screws shown
in No. 289. Take shank diameter=25 mm.
U nwin's Formulae, 6Jr, if t is greater than 8 mm; 7. Give a sketch of a roof truss No. 137 made from
if less, make shear resistance = crushing resistance. 75 mm x 9 mm steel angle, span 10 m, joints having
gussets made from 10 mm plate. Show the centres only 19. Draw any three methods of nut locking devices
I = --
d2 of the rivets. shown in No. 290. Take shank diameter= 30 mm.
l .44
cl = 2.55t, single shear. 8. Sketch four methods of riveting thin sheet, two giving 20. Draw sectional views of three of the locking devices
d = l .2751, double shear. a flush surface, one for inaccessible positions. shown in No. 291. Take shank diameter = 30 mm.
Strap thickness,
Single cover, 1.25 thickness of plate.
Double cover, 0.75 thickness of plate. 9. Draw a sectional view of two 12 mm thick steel plates 293. ISO Metric Screw Thread Form
held together by a suitable Metric nut and bolt. Details of the form for internal (nut) and external
Steel Safety Stress Values, (bolt) metric screw threads are shown. The condensed
Plates,/1 85 M N m - 2 table gives a short range of sizes in general use.
Single rivets,/, = 65 MNm- 2 1 0. Draw a sectional view showing two 12 mm thick Metric threads are designated in drawings for example,
plates held by a nut and bolt giving a flush finish to one as follows: M6 x l.O meaning a metric thread six mm in
Double rivets, .fs = 60 M N m - 2
side. dia with a thread pitch of one mm.

=
Single,/, l 10 M N m - 2
Double,/, 1 44 M N m - 2 See also Appendix I.
11. Draw the sectional view showing a cylinder head
50 mm thick held down by a 12 mm dia stud.

Exercises 12. Draw a sectional view showing a 25 mm thick plate

fastened to a casting by suitable Metric set bolt.
1. Draw full size, two plates each 25 mm thick fastened
together with a snaphead rivet.
1 3 . A brass plate 6 mm thick is set screwed to a casting
2. Draw the five type rivets shown in No. 281 full size, to give a flush finish. The plate is 100 x 60 mm ; show
D = 30 mm. spacing and one setscrew. Scale 3 : I
1 55
 I SO METRIC SCREW THREAD FORM

BOLTS & SCREWS

ill
SCREW THRE A DS
p

- � C

�
..,
4: H F
R
� ••, o t:±1--1
t:
< 0

� a ltc,naHv< h�ads

ll.J--o.4mm

NUT BOLT

Diameter of
Nominal Pitch of unthreaded Width across Diameter of Height of Radius Thickness
size thread shank flats Washer Face Head underhead of Nut

D Coarse Fine B A E F R F1
M3 0.5 0.35 3.00 5.50 5.07 2.00 0.2 2.40
M4 0.7 0.50 4.00 7.00 6.53 2.80 0.3 3.20
M5 0.8 0.50 5.00 8.00 7.53 3.50 0.3 4.00
M6 1 .0 0.75 6.00 10.00 9.53 4.00 0.4 5.00
M8 1 .25 1 .0 8.00 1 3 .00 1 2.48 5.50 0.6 6.50
MIO 1.5 1 .25 10.00 17.00 1 6.48 7.00 0.6 8.00
Ml2 I . 75 1 .25 · 1 2.00 19.00 1 8.37 8.00 1 .0 1 0.00
M16 2.0 1 .5 1 6.00 22.00 23. 1 8 10.00 1 .0 1 3 .00
M20 2.5 1 .5 20.00 24.00 29. 1 0 1 3.00 1 .2 1 6.00
M24 3.0 2.0 24.00 36.00 35.00 1 5.00 1 .2 1 9.00

-
M30 3.5 2.0 30.00 46.00 44.95 1 9.00 1 .6 24.00
M36 4.0 36.00 55.00 53.83 23.00 1 .6 29.00

Ref BS Handbook 1 8 . B S 3643. ISO Rec R272

SHAFTS, BEARINGS AND
SEALS
Keys and Keyways
Cottered Joints
Shaft Couplings
Rigid and Flexible
Universal Couplings
Clutches
Bearings
Seals
 158

294. Keys and Keyways Keys are used to fix shafts and
wheels and collars together, either rigidly or with limited KEYS AND K E V WAY S

8
axial movement.
F EA T H E �
5imple
RECTANGU LAR G I B HEAD
proportion
Rectangular Key A rectangular recess is cut in both
wheel and shaft to the proportions shown, and the key
fitted.

1 • i w- + 6
Gib Head Key This tapered key. provides a rigid fixing W e ::;DIA
when driven into position. The gib head enables the key

�
lo be withdrawn for dismantling.
Feather Key The parallel-sided key is setscrewed into
an end milled slot in the shaft. This key allows axial
movement of the wheel on the shaft while still trans­
mitting rotary movement. It also fa_cilitates assembly of
parts under difficult conditions.
Woodruff Key A segmental shaped key which fits
into a corresponding recess cut by a standard size cutler.

•I
This key is used on tapers and other positions where self
aligning is required.
w
o o=o
� up to 75
Saddle Keys Used for temporary fixing or light loads.
Round Keys These may either be screwed or tapered.
The diagram shows a screwed pin, the head of which is
sawn off after insertion to give a Aush finish.
SADDLE FLAT
WOODR U F F
K EYS

""" "' + t>

I

,- - .Q.
12.

ROU N D TAPER & SCREW TAPER P I N

 295. Splined Shafts A shaft may be grooved to form
splines or keys enabling gears or similar components to S P L I NES COT T E R E D JOINTS
S P LI N E S S HA F T
slide axially whilst still transmitting rotary motion. This
method is adopted in gearboxes when the gear shift lever
engages or disengages a gear by means of the shift claw ,llde
working in an annulus as shown in the diagram.
296. Split Tapered Cotters are used to retain the collar
for the spring on poppet valves in the internal combus­
tion engine.
297. A Tapered Cotter tightened by nut and washer can
be used for locking a shaft and lever or crank together.
The cotter requires a flat machined on the shaft.
298. A Flat Tapered Cotter is used for locking a shaft and
base or wheel together. Simple proportions are given for
general use, a taper of I : 30 is usually adopted.
299. Cottered Joint for Rods Two rods may be joined
by a tapered spigot joint locked by a flat tapered cotter.

in terms of D = diameter of rod.

The proportions shown give a joint of uniform strength shift

300. Cotter and Gib for strap end of engine connecting TA PERED COTTERS
rod. The bearing brasses of the big end of steam engines
are connected to the connecting rod by means of a strap
which in turn is locked to the rod with a gib and cotter.
The jaws of the gib prevent the strap from opening as the

C O T TE R
cotter is driven home, and since its outer edge is tapered
as the cotter, a parallel wedging action takes place.

LEV E R l SHA FT

1 59

\_
.-

160

JO I . Muff Coupling for Shafts Two shafts of equal dia­

meter may be rigidly connected by means of a muff S H A FT COUPL INGS
MUFF ANO KEY
8
coupling. This consists of a cylinder machined in two
halves and secured on the shafts by nuts and bolts. The
shafts and muff are slotted for a long rectangular key.
The general proportions of the muff are shown in terms
of the diameter of the shaft.
302. Flanged Coupling Two shafts may be rigidly
coupled by using flanged coupling plates. The couplings
are machined with matching turned projection and
annulus for location and are press fits on the shafts. The D
shafts and couplings are slotted for taper keys driven in (0
from the meeting faces. The flanges and shafts are finally
bolted together the bolts being tight fits in the flange
holes.
The general proportions are given in terms of the

4
shaft diameter. See tables.
To � O

FLANGED

p C. 0
2.2 o+ 35
8
 V A R I O U S C O U P L INGS
VAR IOUS C O U PLINGS

8
303. Flexible Coupling Moulded Rubber Insert. Where

8 M O U LD E D RU B BE R INSERT
two shafts are to be connected and some slight mal­
alignment and _torsional strain is anticipated, a suitable S T E E L -RU B B E R B O NDED
- F L EX I B LE - - F LE X I B LE -
flexible coupling is o btained by the use of a moulded rub­
be·r insert between the coupling dogs. The metal flanges
are keyed to the shafts and have projections which fit
into the moulded recesses in the rubber insert, through
which the drive is transmitted and shocks absorbed.
304. Bonded Steel and Rubber Flexible Coupling. This
coupling employs a flexible rubber disc bonded to two
steel plates which are bolted to flanges keyed to the shafts.
The bolt holes are arranged alternately and irregularities
or transmission are absorbed by the disc. An Isometric
View is given later.
305. Rigid Compression Coupling The diagram shows
a compression coupling in which the coned flanges com­
press a mating double conical sleeve which is slit to allow
it to be reduced radially. This type of coupling grips by
friction and uses no keys; it is capable of being dismantled
without disturbing the shafts.
306. Chain Coupling Flexible. The flanges have chain
sprocket wheels cut at the meeting faces, and are coupled

C OM PR E SSI O N C O U PLING C HAI N C O U PL I N G - FLEXI BL E -

by a length of duplex chain by which the drive is trans­
mitted. The flanges are keyed to the shafts, and the coup­
ling is shrouded by a cover. The chain allows for some -RIG I D -
flexibility in mal-alignment of the shafts, and the drive
can be dismantled by the removal of the cover and chain.

8 8
161

- ·
 1 62
U N I V E R S A L J O I NT C O U P LI N G S
Where drive has to be transmitted through shafts having U N I V E R SA L J O I N TS
an angularity of more than 5° , a universal joint coupling
H O O K E'S TY PE J O IN T
8
is employed. This usually consists of two yokes and a
cross trunnion piece.
307. Hooke's Type Universal Joint The diagrams show
a typical Hooke's joint consisting of two yokes keyed to
the ends of the shafts, and connected to a spider or cross
in which screwed pins form bearings for the yokes. Pro­
portions are given in terms of the diameter of the shafts.
This joint will transmit drive through 20° of angularity. 0

r- ­
� -­
308. Automobile Type Universal Joint A universal
,
r--
coupling used in transmitting drive from a flexibly
mounted engine to the sprung back axle unit of an auto­
mobile. The coupling has to work under varying loads I
often at high speeds and through rapidly changing L__
angularity. Needle roller bearings are used on the trun­
nions, held in position by spring circlips. Splined shafts
are employed to allow extension of the shaft since a uni­
versal joint is used at both ends of the transmission shaft
and a double angularity is often experienced.

AUTOM O BIL E TY PE UN IV ER SA L

8 CtRC.LtP
E.l(TeNS!ON
Sf>LIHES.

NEEDLE
ROL.LERS

L
t- _-�_--:_--:_--
�-�­
L- - - - - - - ­
,- - - - - - - --1---��--.,,,_.
-
--'-...1..--1

- ...,._,,�c--"'C"-"tr
= =r==:::1
F...:--..=-_:- ..=--=
1-- - - - - - -�-....{
L _ _ _ _ _ __
- -- air � ............
___,----:,.,-
CLUTCHES
Where a drive needs to be completely disengaged and C L U T C HE S
re-engaged frequently, some form of clutch must be intro­

8
duced. One side of the coupliog is moved on a splined C LAW
PL ATE
shaft by means of a foot or hand operated lever often
returned by a spring. DRY
309. Claw or Dog Clutch One side of the coupling is
keyed permanently to one shaft. The other coupling
slides on a splined shaft or on a feather key and is

- -�
operated by a lever. Ma ting projections on the flanges
transmit the drive, and this simple type of clutch can
only be re-engaged at low speeds. - --j

310. Cone Clutch The clutch has two mating conical

surfaces, one surface may be coated with leather or
bonded lining. The cones allow easy disengagement and
re-engagement of the drive from rest position, the drive is
by friction between the two engaging surfaces. An Iso­
metric View is shown later.
3 I I. Dry Plate Clutch This type of clutch is used in

8
many automobiles, and consists of a light steel plate with
asbestos friction lining rings compressed between the fly­
wheel surface and the driven presser plate connected to
the gearbox. On depressing the clutch foot-pedal, the
plate is freed and the drive lost, being automatically re­
engaged as the foot pedal is released. A carbon pressure
CONE

i
ring or ball bearing takes the thrust of the foot-pedal, the

I
surface pressure on the plate linings is usually between
1 kg/cm 2 to 2 kg/cm 2 .
SPLll'IED _____.:,,

I
SHAFT

I
I,

I
I
8 I
,!

1 63
 ..
1 64

SHAFT BEARINGS
SHAFT BEARING S

8 ii
Shafts and axles need supports lined with anti-friction

i
I
surfaces.and provision for lubrication.
The support is usually an iron casting or forging either SIMPLE Q HALVED
i
in one piece or split with inset bearing surfaces of bronze, '1
�
gunmetal, white metal, lead bronze, sintered metals, and
plastics, depending upon the use and conditions.
312. Simple Bearing A simple bearing with a pressed
bronze bush, oil hole drilled in top surface.
313. Halved Bearing The support is made in two parts
I
bolted together, the ' brasses ' are also split and located
by flanges.
314. Footstep Bearing Used for the lower end ofa ver­
tical shaft. A hardened steel plate takes the weight of the
shaft. A ball thrust race may alternatively be used, and
is shown in a later diagram.
315. Ring Oiler Bearing A loose ring rolled by friction
from the shaft dips into an oil well and continually
lubricates the bearing.

Other methods of simple lubrication are by drip feed,

wick feed and felt oilsoaked pads.

I
R I NG OILER

8 8
FOOTSTEP
 BALL BEARINGS
316. Single Row Ball Bearing The race is composed of
BALL B EAR I N G S

8 8
an inner and outer hardened grooved ring between which

SINGLE R OW SELF-ALI G N I N G
run hardened steel balls carried and spaced in a cage. Ball
bearings are made in a wide range of sizes. the smallest
are found in watches whilst the largest are used in turrets
of heavy naval guns.
The simple type shown usually fits into a recess turned
in the housing and is retained by a collar or by shoulders
on the shaft.
317. Self-aligning Ball Bearing The outer shell is
ground to the same spherical line as the housing enabling
the bearing to align itself to the shaft line which may vary
as the load is applied. The bearing also adjusts itself to
small inexactitudes arising if the housing is incorrectly
laid.
318. Thrust Race Ball bearings may be used to take end
thrust as shown in the footstep bearing for a vertical
shaft. The axial thrust is taken by the balls carried in a
cage and moving in circular grooves ground in hardened
steel plates.
319. Spring-loaded Ball Bearing Where some axial
movement of the shaft may take place due to expansion

TH RUST RACE S P R I NG - LOA D E D

or load, the race may be held to the shaft by a screwed
collar and the bearing moves against a spring held in the
housing.
320. Ball Races require lubrication to avoid becoming
seized under load. The housing may be arranged to act
as an oil bath requiring sealing-shown later-or the
race may be packed with grease sealed in with spring fit

S EALED
plates as shown, the lubricant lasting the life of the bear­
ing.
Ball races may be single or double row, designed to

8
take radial and/or axial thrust. and since the balls are in
point contact with the shells much less friction is gener­
ated than with the simple journal bearing.

8
1 65

�-- -
 1 66
321. Roller Bearings In bearings which carry heavier
loads , roller bearings are used, the line contact of the ROLLER B EA R I NGS

8 :.
roller and shell giving better radial capacity. Rollers are
spaced in cages and may be arranged to run in a groove
cut in the inner shell whilst the outer shell being plain, SINGLE NEEDLE ROLLERS
allows of some axial movement. The rollers may be
ii
ROW

Ii
.,
arranged in single, dou ble or triple rows to carry greater II

'1
loads.

!1
322. Needle Roller Bearings Caged needle roller bear­

,,
ings can be used where space is restricted. The end of the
roller is reduced and rounded to prevent it acting as an

Ii
,I
end mill and cutting the cage. The rollers run on the shaft
which requires to be hardened.

j!
323. Sealed Roller Bearing The diagram shows a heavy
duty roller bearing. The inner shell is secured to the shaft
by a screwed collar and tab washer. The housing is

,,
sealed by synthetic rubber or plastic rings set in grooves.

i
The housing acts as an oil reservoir.

I
I
324. Tapered Roller Bearing Where radial and axial
thrust is present, tapered rollers can be used. The dia­
gram shows a bearing with double tapered rollers used I!
'.---=---=-=-=-================#=========================d
I
in automobile front wheels capable of absorbin g radial
and double axial thrust at high speeds. The bearing is
lithium grease packed and sealed with a garter spring
seal, the bearings are secured and adjusted by a slotted
SEALED ,,
I
TA PERE D R O LLERS
8
I
nut and split pin.
Shafts may employ both ball bearings and roller bear­

I
ings. The ball bearing at one end may be used to absorb
axial movement, and the roller bearing at the second end
to bear the heavier load.

I
Other types of seals are shown later.
Refer to B S 292.

I
1!

DOUBLE THRUST
 SEALS
Shafts and rams in hyd raulic systems often require high S EA LS FOR S HA F TS AND RAMS

8
pressure sealing against leakage. Thisjs accomplished by
a sealing ring forced against both shaft and housing. LABY R I N T H METALLIC SEAL SPLIT RING SEAL
Early types of seals, leather, braided hemp, cotton, flax,
have largely been superseded by synthetic rubber, butyl,
silicone, nylon and other plastics. These materials stand
up to modern solvents.
325. Labyrinth Seals The seal and gland shown, con­
sists of a number of rings of vee section, spring loaded.
The rings may be of metal and plastic, or metal and metal.
White metal and a bronze are one combination.
326. Split Ring Seal The seal rings are packed in sec­
tions inside cups, and a garter spring applies concentric
pressure closing the seals on the shaft and to the housing.
A number of units may be. used in line, held in place by
a keeper plate and bolts. The units may be replaced with­
out disturbing the shaft.
327. ' U ' Seal A moulded U section seal of leather or
synthetic material is held in position by the gland. The
pressure of the pumped liquid tends to force open the

I
U seal and improves the sealing.

I... .=-'-���----
328. Garter Spring A moulded leathe_r o r synthetic
material seal held in place by a garter spring. Used for
.

8
light pressures, automobile shafts and gearboxes. 1
LEAT H E R 'u' SEAL LEAT H E R · , G A R T E R
SPRING

1 67

l
 1 68

328A. End Thrust Bearing A shaft and its endplate

designed to take end thrust loads. The load is taken by PLAIN END TH R U ST B EA R I N GS B A L L BEAR I N G
the two hardened washers, one with a convex face to
lessen friction, which are housed in a casing. This com­
ponent may be called a footstep bearing.

328B. End Thrust Bearing In this second case, the shaft

rotates in a ball bearing, and the endthrust is taken by a
second ball race which considerably reduces friction.
PIPES, JOINTS AND VALVES
Pipes and Couplings
Expansion Joints
Non-return Valves
Safety Valves
Stop Valves
 1 70

329. Pipe fittings A selection of standard pipe fittings

are shown, sizes are given in the example for pipes of a
PIPE FITTI NGS

8-
nominal bore of 25 mm. The ends · of the pipes are
threaded and a socket piece is used to join straight
CAP

-
lengths. An end may be sealed by a cap fitting. Tee SO C K ET TEE

I
junctions enable pipes to be joined at right angles.

-�"i=-=i
Elbow pieces may be right angles or 45 °. Pipes may also

T t=t
be joined by flanged bends which are bolted together.

'° I
Standard sizes and proportions are listed in detail in the
BS Handbooks.

��
BS IO : 1 962. Pipe flanges. ()
'<:f
M ��--1-­
M

�
BS 27. Pipe threads.

- 1
BS 1 256. Copper pipe fittings.
BS l 43. CI and copper pipe fittings. i - L.. l=t=:::l-.--J-\---t:==:t=:1 rs..
BS 1 740. Wrot pipe fittings B S P thread. .._

I
_i_

30.5 ..1
76.2
I I I
$2f 1 14.3

TUBE

flJ 37.3 I
14 BOLTS
M 12 x 1 .75

\
Js.,\_
'°
\ N
\

0
u

FLA N G E D
a.

BEND
ELB OW
330. Flanged Pipe Couplings Pipes may be joined by
screwing flanges to the pipe ends and bolting the sections FLANGED P I PE C O UPLI N G S
together. A taper thread is cut on the pipe and flange. The

344 .74kNm-'l
diagram shows details of flanges suitable for liquid and S C R EW E D H EAVY DUTY S C R E W E D I. W E L D E D
gas pressures up to 344. 74 kNm - 2 , and on pipe of
nominal 25 mm bore. pressure ---l-r-r"7'""1

Heavy Duty Flange The diagram shows details of a

coupling capable of withstanding higher pressures up to
3. 103 MNm - 2 • The pipe and flange are taper screwed to­
]-

I: M 16X2
�OUR

gether, the flange is heavier, and the bolts greater in

'Q
diameter. (\j

Screwed and Welded Flanges In this type, the heavy

flange is screwed on a taper thread and a vee-cut is

;�
welded up. The joint is machined to fit, and the sections
bolted together.
331 . Hydraulic Couplings Extra high liquid pressure
systems using pressures of I 0.342 M Nm - 2 require special 'Q

3 , I M N m-2
couplings. The heavy flanges are screwed on the pipes,
one pipe penetrating into the opposite flange, the seat _j_
being made by compression on a soft copper ring. The prczssurcz
flange shape is shown, and has two bolts.
Pressure/stress is shown in newtons per
square metre = Nm - 2 M 12x 1.75
or in newtons per square millimetre = Nmm- 2

lI
See tables and definitions on page 260.
An interesting comparison with erstwhile imperial H Y D RA U L I C
measure may help in visualising equivalent quantities : C O UP L I N G
50 lbs per square in. = 344.738 kNm- 2

i
0.345 Nmm- 2
300 lbs per square in. = 2.068 MNm- 2
= 2.068 Nmm·- 2

-----1--= Ii
I
I

:
�1
I
i
I0.34MNm-1
pressure 8
1 7l
 1 72

332. Hydraulic Pipe Couplings A further coupling is

shown having square flanges. The sealing is effected as HYDRAULIC P I PE C O U PLINGS
before by the compression of a soft copper ring between
the ends of the meeting pipes. Sizes are given for a pipe SQUARE FLANGE CO UPLING

·i �-
of nominal 25 mm bore, pressures up to 1 5. 5 1 3 MNm - 2 .

•!
27 27 95.2
S M A L L B O R E H Y D R AULIC J O INTS
pc D
333. Coned Union A separate double coned sleeve is
placed on the tube and is compressed when the union 82 .6
nut is tightened. The coned sleeve may be soldered o r
brazed t o the tube.
334. Nut and Sleeve An in line joint between two tubes
@
can be effected by the use of an internal coned sleeve and
outer ring which compresses the end of the tube to a seal
when the union nut is tightened.
335. Brazed and Screwed Sleeves are hard soldered to
the tubes, the solder having being contained in the annu­
lar groove in the sleeve. The flanges are screwed to the
sleeves. The joint seal is effected by compressing a joint
ring when bolting up the flanges. This type of coupling is
used on thinner walled tubing.
1 5 . SM Nm-). pressure

CONED UN ION BRAZ E D & S C R EW E D

NUT & SLE E V E

8
336. Expansion Joints and Bends Expansion of pipes
carrying steam or hot liquids require special provision EXPA N S I O N JO I NT A N D B E N DS
and methods adopted are lyre bends and the U bends in

8
the pipe line. The bends may be made up of bends built
to the form shown. E X PANSION BEN OS
337. Expansion Joint The unit consists of two tubes
one of which slides inside the other, the gland sealing the
joint whilst allowing axial movement. The movement is
limited by the length of the slide bars. The seal ring is
asbestos fibre. The joint is drawn suitable for pipes of
25 mm nominal bore.

E X P AN SION JOINT

1 73
 1 74

N O N - R E T U R N V A L V ES
N O N - RE T U R N VALVES
POPPET I
To prevent the back flow of gases and liquids in systems

VALVE I RELEA SE VALVE

a number of types of non-return valves are employed.
338. The Poppet mushroom headed valve is adopted in
most internal combustion engines for inlet and exhaust
purposes. The valve is opened mechanically by a cam,
and returned by the spring to its seat after a measured
8 ii
lift, dwell and fall, generated by the cam shape.
339. Release Valves The diagram shows a spring­
loaded needle valve seating on a circular seating. The
flow lifts the valve on reaching a pressure exceeding that
of the pre-set loading of the valve. This type is used on
hydraulic systems.
340. Feed Check Valves This type of valve is used in
the flow of boiler feed water and uses a disc valve with
three or four aligning vanes to restore the non-return
valve to its seating, helped by a light spring.
341. Ball Check Feed Valve This type of non-return
valve employs a bronze or stainless steel ball as the valve.
Plastic balls may be used where the pressure is low as in
cooling liquid systems.

FEED CHECK VALVE BAL L VA LVE

8 8
342. Non-Return Box Flap Valve An in-line gravity fall
non-return valve for use in gas pipes. The flap valve is NON - R ET U RN RELEASE SAFETY VALV ES l
8
pivoted at the top edge and closes the pipe if an explosive
blow-back ignites the gas beyond the valve. It is· used as
BOX FLAP VALVE
a safety measure in gas systems. In normal use the flap
valve is opened by the flow of gas. The sealing face is of
asbestos lining.
343. Ball Release Valve In hydraulic and steam or gas
systems a ball valve may be spring loaded to lift off and
release excess pressure at a safe limit.
344. Release Valve Spherical ended plunger, spring
loaded. The release valve is similar to the previous ex­
ample, the valve being a plunger, hollow, containing the
spring. The cap is bolted down. The spherical ended
plunger requires relief holes drilled in it above the seal
line to avoid pressure build-up in the spring tube, which
could interfere with the normal lift-action of the valve.

8 8
R ELEASE VALV E R ELE AS E VA LVE

175
 1 76
345. Disc Non-Return Valve A spring-loaded face
valve, plastic face backed
' by sheet metal. The valve is DISC & LINE VA LV E S SAFETY VALVES
used in lift pumps.
346. Line Pressure Valve A spring-loaded valve set to
open at a fixed pressure. The union joints are made by a
seal ring which enters a groove in the pipe end.
347. Steam Safety Valves A gunmetal valve is held on
D I SC 8 LINE P R E SS U R E 8
its seating by a lever and weight. The valve is vaned to
direct the valve during re-seating.
The safety valve is bolted to the boiler lop and is set
to lift at a safe pressure.
348. Ramsbottom Safety Valve Two vaned valves are
used, held down by a common lever which is held down
by a spring and slotted link. Both valves can be manually
tested at set intervals by easing the lever up and down.
Used on steam locomotives.

SAFETY L I F T � OFF
VALVES R A M S B OT TOM

LEVER & W E IG H T

I
.

--*--
--- ---
. .

l
\ /
)
8 8

--
STOP VALVES
Screw down valves are used for cutting off a supply in
NEEDLE VALVE STOP VALVE
hydraulic steam and gas systems.

8
349. Needle Screw Down Valve A 45 ° pointed screw NEEDLE S C R E W - DOWN
plunger closes on a seating to shut down a supply. A
screw down gland and sealing ring make the plunger gas
tight. The inlet and outlet pipes are connected by coned
unions. Suitable for high pressure small bore systems up
to 1 0 mm dia, and low pressure to 20 mm bore.
350. Screw Down Valve The diagram shows a standard
type screw shut down valve flanged for bolting into pipe
systems.
The body is a casting. The spindle screws down through
a bridge carried by the top plate. A gland makes the
spindle gas tight and the sealing ring is compressed by the
two bolts o f the gland. The valve seats on the centre cast
web, and is located by a pin. The valve has an open-sided
slot enabling it to be fitted on the end of the spindle.

S C REW S H UT D OW N

1 77
 1 78

351. Needle Line Valve The diagram shows a needle

valve for an in-line joint. The conical ended plunger NEE DLE VALVE HYDRAU LIC H.P. VALVE
screws down to seat. A gland holds a sealing ring. The

8
pipes are attached by coned union joints.
351A. Heavy Duty Hydraulic Valve The screw down
stop valve has a heavy cast body and stout spindle which
works in a screwed collar. The collar and gland arc
secured by bolted flanges. The oval flanges carry two N EEDLE
heavy bolts. LINE VALV E

H p HYD R A ULIC
VALVE 8
 PUMPS
Ram. Oscillating. Centrifugal
Gear. Diaphragm. Disc Valve
Screw. Cam and Plunger

ENGINE PARTS
Piston. Crosshead
Connecting Rod
Crankshaft
Eccentrics

I.C. ENGINES
Two Stroke
Four Stroke

PARTS
Carburettors
Ignition
 1 80

PUMPS
352. Ram and Eccentric A cylindrical ram worked by
I
I
H Y D RAUL I C P UM P S

8
an eccentric crank draws in a charge lifting the lower

8
valve and closing the upper ball valve. The delivery stroke RAM & E C C E NT R I C
closes the lower valve and the upper valve opens to pass
the c.harge. The ram is sealed by a screwed gland and seal
ring.
353. Oscillating Cylinder The oscillating cylinder is
carried on a trunnion or shaft. As the cylinder oscillates,
the single hole port in the cylinder coincides with the port
in the standard. The diagram shows the plunger halfway
down on the delivery stroke.
354. Centrifugal Vane Pump The offset rotor carries
twelve sliding vanes which keep contact with the outer
casing by centrifugal action. This type of pump can be
used as a circulatory pump or a pressure pump.
355. Gear Pump Two equal spur gears working in a
housing can be used as a pump. Used in automobile
engines as oil circulatory pumps for lubricating bearings
and pistons, pressure 1 38 kNm - 2 to 276 kNm- 2 .

O S C I LL A T I N G
CYLINDER

G E AR

8 8
C E NTR I F U GA L VA N E

--
356. Automobile Petrol Type Pump The pump shown
employs a plastic diaphragm worked by a cam and lever H Y DRAU L I C P U M PS
from the engine camshaft. Metal or plastic disc valves,

8
kept against their seats by light springs, control the flow
o f fuel. A wire mesh filter is included in the top casing.
The pump has to work under light pressures to keep the
carburetter level constant against the float and needle
valve (shown later).
357. Disc Valve Pump A vertical spindle lifts a plunger
which has a face disc valve on its upper face. A lower face
valve located by a centre pin allows liquid to be drawn
in as the plunger is lifted , and on the downward delivery
stroke, the lower non-return valve closes and the upper
valve opens to allow delivery. This is the same principle
used in the century-old well pumps.

I
I
/

D I A P H R A GM PUMP
-- ;

- /
..____-·-"-

(!r�-
-· ----- -- DISC VALV E
PUMP

I81
 1 82
H y o R A u u c Pu M P s :-------�H--:--v:-::o=-R=--A---:-u-=-L:---r c-:--- --:p=--u-M
_P_s________---,
358. Screw Pumps Two spindles, geared together, have

SCREW PUMP
left- and right-hand square threads i ntermeshing. Liquid
is drawn in in the lower orifices, is passed along the upper
and lower helical grooves o f the square thread to the
upper delivery opening. A seal must be included in the
drive shaft and this is shown in a later diagram.
359. Plunger and Eccentric Roller bearing eccentrics
drive a plunger (spring returned) and spring loaded
poppet valves, non-return, control the flow of the liquid.
Since the stroke is limited and short, a number of plungers
are worked from the same camshaft, two, four or six
units arranged to give continuous flow at high pressures.

PL U N G E R & E C C E N TR I C 8
STEAM ENGINE PARTS
STEAM E NGINE PA R T S
I n steam reciprocating engines, the steam entry t o the
cylinder is controlled by either a slide valve or a piston
valve. The steam pressure on the piston slides the piston P I STON l ROD CROSSHEAD & S HO E S PIN
in the cylinder and the piston rod actuates the crosshead.
The connecting rod then turns the crankshaft on which
are shrunk the wheels, the reciprocating action of the
piston thus being turned into rotary movement.
360. Piston and Rod The cast-iron or steel piston has
cast-iron piston rings set in it, and is a coned fit on the
piston rod secured by a cottered nut. Cone taper I : 6.
Crosshcad and Shoes The crosshead is usually a steel
casting, with shoes on pins which allow for alignment.
The piston rod is coned and cottered to the crosshead by
a flat tapered cotter. The little end of the connecting rod
works on the crosshead pin which is taper socketed into
the cheeks of the crosshead, secured by a cottered nut.
361. Connecting Rod The connecting rod is a forging,

8
the little end bronze bushed. The big end has a strap
holding flanged bronze split bushes, bolted and cottered
with a compensating wedge taking up the slope of the
cotter. Two bolts fix the cotter. In some cases the little end
is strapped, bolted and cottered to hold the bearing
brasses.
C ON N E C T ING ROD

L I T TL E EN D BIG E N D

8
183
 1 84

C R A N K S H A FT S
Crankshafts consist o f crankpin, web(s) and mainshaft.
CRANKSHAF T S
362. An Overhung Crankshaft has one web, forged or
O V E R H UNG
cast, with crankpin and mainshaft shrunk and pinned in
position. It can also be forged from solid steel and ma­ C RANKSHAFT
chincd·fo finished form. CAST WEB
The web may be extended and shaped to balance the
revolving weights of the crankpin and big-end of the con­
necting rod and the reciprocating weights of the little-end
and the piston. A simple rule is ' balance all the revolving
weights and half the reciprocating weights '.
363. Locomotive Type Crankshaft A built-up, closed
crankshaft is shown, with integral balance weighting
formed in the webs. The crankshaft could also be forged
in steel, and machined to final form.
Crankshafts may have several · throws' equal in num­
ber to the connecting rods and cylinders.
Some internal combustion engines use a hollow cast
steel crankshaft, giving strength and rigidity with light­
!)ess.

8
LOCOMOTIVE TYPE
C RANKSHAFT

B A LA N C E
W E IG H T
364. Eccentrics for Steam Engines The slide or piston
valve of the steam engine is actuated either by an eccen­ ECC E N TR I C S FOR ST E A M ENGINE

8
tric crank on the mainshaft or by an overhung crank from
an outer crankpin. ECCENTR IC & S TRA P
Eccentric and Strap The eccentric may be a casting
VALVE ROD E C C E N TRIC ROD
or solid, and may be split to allow it to be erected on the
shaft. It may be keyed and cottered to the shaft. The
strap is made in two halves and is grooved for retention, l GLA N D
and bolted together.
Eccentric Rod The eccentric rod is a forging and a
palmate end is bolted to the strap. The little end is forged
into a yoke to take the eye of the valve rod. The pin has
a ring and taper cotter pin.
365. Stephenson's Link Reversing Motion. A schematic
layout of Robert Stephenson's link motion using two
eccentrics, one for forward motion and the other for
reverse motion. The valve rod ends in a die block which
slides in the expansion links. The expansion link is lifted
or lowered by the weighbar and lever. The neutral posi­
tion is shown when neither eccentric will work. Economy
of. steam can be obtained after starting in full gear, by
' notching up·, i.e. moving the lever nearer to neutral to
restrict the valve rod movement, lessening the valve
movement and opening.

8
S T EA M
S T E P H ENSON S

CHEST
L IN K R EV ERS I N G

WEIG H BA R
M O T I ON

EXPA N SION
L I N KS
DIE B L O C K
'------.

R EVERSE
..f--t-- - --

ECC E N TRIC

185
 186

C
366. Internal Combustion Engines The two-stroke pet­
rol engine draws in an explosive charge of petrol and TWO STRO KE E NG I NE
air through a carburettor into the crankcase where it is
compressed by the downward power stroke until the

8
piswn uncovers the transfer ports and allows the charge
lo travel to the cylinder where it is compressed by the
rising piston and ignited by the spark.
At the end of the power stroke the exhaust gases pass
out by the exhaust port. The cycle is completed in two
strokes of the piston-once every revolution.
The lightweight piston-aluminium alloy-has a cast­
iron ring to give a gas seal, the little end of the connecting
rod has either a plain bush or needle rollers, the big end
has a plain bearing, ball race or rollers. The crankshaft
is usually built up, press fits or coned fits-nutted. The
drive to the gear box is by sprocket and chain or spur
gears. The ignition is by battery and coil, or flywheel
magneto which generates the high tension spark to the
spark plug. The timing of the spark is effected by a con­
tact breaker with tu ngsten points opened by a cam on
the crankshaft. The cooling of the cylinder is by air
currents generated by the passage of the machine or by
a shrouded fan driven by the engine itself. Lubrication
is usually effected by adding oil to the petrol fuel, but
variable oil pumps are now being fitted.
Two-stroke engines are made in working sizes from INLET
50 cc to 350 cc, popular range, and used for cycles,
lawn-mowers, boat engines and industrial compressors
and similar purposes. Racing two-stroke engines of 1 00
cc twin in line can now develop 25 bhp at 7 500 rpm.

S C H E M AT I C . LAYOUT
367. · Internal Combustion Engine Four-stroke. The
four-stroke petrol internal combustion engine completes FOUR STROKE C ENGINE
its cycle of induction, compression, power, exhaust in

8
four strokes of the piston, i.e. two revolutions of the
crankshaft. The control of the charge of petrol and air MUSH R O OM POPP E T VA LVES
from the carburettor to the cylinder is by a cam-operated,
mushroom-headed, poppet valve, and the exhaust gases
leave on the opening of a similar valve. The valves may
be duplicated giving four valves per cylinder to facilitate
rapid induction and exhaust giving greater speed and
power. Aluminium alloy pistons are generally used,
sealed by cast-iron piston rings. The gudgeon pin is lo­
cated by circlips, and the little end is a plain bush or
needle rollers. The big end is a plain bearing or ball race
or rollers. The valves are operated by push rods and
tappets from a two to one reduction cam shaft from the
main shaft. The ignition is by coil and battery, and a cam
operated contact breaker gives the spark timing. The
schematic diagram shows an air cooled, motor-cycle type,
petrol engine.

S C H EMA T I C LAYOUT

1 87
 1 88

368. I C Pistons and Connecting Rods. Two-stroke

Type. Details of a piston used in a two-stroke engine are I · C· P I STON S AN D CON N RO D S
shown. The top of the piston is slightly domed, the skirt

8 8
is cut away to suit the transfer ports, the gudgeon pin is
located by a circlip. T WO STROKE FOUR ST R O KE
The connecting rod is a forging with a bushed little end,
and the big end is ready to receive the rollers. A connect­ OVER SQUARE
ing rod with a closed big end must be threaded on the
crankpin during assembly of the built up crankshaft.
369. Four-stroke Piston and Connecting Rod. In some

��"""!:: =_=_=_=_=-::i_.,i
.....
automobile engines ' over square' proportions have been

�,
adopted, i.e. the length of the stroke is less than the

�I
diameter of the pisron, resulting in less piston movement
for the same power. Larger valves are needed to facilitate
breathing. Very short connecting rods can be used, large
diameter plain bearings in the big ends, and since hollow

I
cast or forged four- or six-throw crankshafts are used,

t
split bearings are used.
V)

F OR ONE PIECE
CRANKSHAFTS CAST OR FORGE D CRAN KSHA F TS
 370. IC Carburettor The carburettor is a means cf
obtaining a measured flow of petrol and air to the induc­ CARBURETTOR CONTACT BREAKER COI L
tion chamber. A diaphragm pump provides a steady flow

8 C O N TA C T B R EA K E R·
8
of fuel to the -float chamber and a needle valve and float
ensure a steady level at the jets. A venturi tube with a
reduced diameter giving an accelerated air flow over the
jet picks up the fuel when the butterfly main valve is CA R B U R E T T O R
opened. For slow running and fast speeds extra jets meter
the fuel to the engine's requirements. A choke valve,
cutting off part of the air supply for starting purposes is
also fitted. Choke
371. Coil Ignition The ignition coil consists of a core valve
of soft iron wire round which a primary winding of heavy N e e d le
copper wire is made. The secondary winding consists of v a lve
many turns of very fine wire. When a low tension current
of 1 2 volts is passed through the primary winding and
broken, a high tension current is induced in the secondary
winding of often 7 000 volts. This process can be used to
produce a spark across the electrode of the plug in the
cylinder head to ignite the petrol and air charge.
372. Contact Breaker Single cam operated contact FOUR CYLINDERS
breaker opens tungsten points to cause the igniting spark
at the plug as the piston approaches T D C. A capacitor

C O IL I G N I T I ON
8
across the points intensifies the spark.
373. Contact Breaker for four-cylinder engine.

PR IMA RY SEC O N DA R Y

8 C ON TA C T B REA K E R

PLUG

/1/1/1-
feATTERY · CA M q:>N TA CTS
SIN G L E CYLIN DER

1 89
 190

H Y D R AULIC S C H EMAT IC
373A. Car Braking System The diagrams show the nor­
mal hydraulic method of drum and shoe used in many CAR B RA K ING SYSTE M
automobiles. A footpedal actuates the hydraulic ram via
the pushrod, the relief valve closes, the fluid is com­
pressed in the system and thus pressure is obtained at the
wheel cylinder. The plunger in the wheel cylinder is moved
outwards moving in turn the brake shoe whose asbestos
fabric lining is forced against the brake drum bolted to

CA P VENT PUSH ROD

the wheel causing the wheel to slow down and stop,
bringing the car to a halt. Pressure can be maintained &
whilst the pedal is depressed, on releasing the pedal, the
plunger returns to its normal position, fluid returns to the
reservoir, the shoes are returned clear of the drum by the

\
return springs and the system completes its cycle. Air
must not be allowed to enter the pipes, ' bleeding' of air
bubbles is effected by pumping on the pedal and drawing
off the air by unscrewing a bleed nipple usually near the

RELIEF VALVE
wheel cylinder.
Disc type hydraulic brakes follow a similar system to
that illustrated, but the pressure is applied to brake pads

R E T U R N SPRING
working on a thick disc., of steel integral with the wheel
axle. S EAL & RAM
The car handbrake is made to work manually usually
by a ratchet lever, cable and pulleys working an arm near
PENDULUM
FEED PIPE ______,
the wheel cylinder, opening the brake shoes. The ratchet
lever is returned to rest by disengaging the pawl freed by
a press button on the lever, releasing pressure on the
brake shoes.

L I N ING

S HOE
FULCRUM
BLOCK
W H E EL

R E T UR N SPRING

DRUM REMOVED

� WHEEL ROTAT ION

STAND ARDS, LIMITS AND
SYMBOLS

Standards
Conventions
Tolerances and Limits
Geometric Tolerances
Machining Symbols
Surface Texture
Symbols
 1 92

LETTER I NG
374. Standards and Conventions The British Standard
308 1 972 Parts I, 2 and 3 covers fully the conventional C O N V E N T I ONS L I NES DIMENSI O N I N G SY M B OLS
methods of representing engineering components in
SECTION
drawing practice. It is the recommended reference book A BC D E F G H I J K L M N O P Q R S
for most engineering drawing examinations. It may be
necessary to refer to the standard frequently for fuller TUVWXYZ 1 2 3 4 5 67 8 9 0
explanations on difficult points as many items are a b c d c fg h lj k l mn opq rstu v w x y z
1 2 34 5 6 7 8 9 0
discussed in more detail than is possible in this volume. R 3.5 �
A selection of items is shown. A simple sans-serif line
ABCDEF G H I J KL M N O P QRSTUVW XY Z
1---L-+-1---1
��
block lettering is suggested, straight or sloping, with lower
case letters for explanations. Underlining is discouraged,
larger lettering to be used for emphasis. Seven types of 79 5 6.43 0.02 98 4 5 3 .067 5 9
m kg s A K cd mol rad sr
line are suggested, thick lines 0.7 mm and thin lines

f/) 7 5
0.3 mm in width. The type of projection is to be stated,
RS «> 2 0 S'FACE B M AX
either by symbol or words. Unit of dimension should be

w
suited in m or mm, also the scale, title, name, No. of
drawing in boxes.

TY P ES OF L I NE
Diameter is indicated by the symbol 0, prefixed,
radius by R, prefixed. Scales are I : I ; I : 2, I: 5 ; I : 1 0 ;
outlines/edge
I : 20. Also 2 : I ; 5 : I ; IO : I ; 20 : I.
± 0,03 TOLERANCES
A simplified drawing layout is shown, less involved
A
than the detailed layout for commercial work, more
d imensions
DRAW I NG L AY OUT
suited to the time factor limitations of examinations.
Further conventions are shown in the later adjoining B hatching
pages and in the Machine. Drawing section.

C � partial view
Selection of abbreviations

sections
Across flats ;\/F Internal I NT
Assembly ASSY Left hand L I-I
Centres CRS Long LG h idden
detail
Chamfered CHA M Material MATL
D
Cheese head CHHD Maximum MAX
E centre Ii nes
___ _ ___ _ __ extreme pstn
Countersunk CSK Minimum MIN

8
Pitch circle dia PCD
moveable pt
Countersunk
head C S K HD Right hand RH

F __ _ __ - - - -- cutting
Counterbore C'BORE Round head R D HD

planes Dimensions in mm
Cylinder CYL Screwed SCR
Drawing DRG Spot face S ' FA C E
External EXT Undercut U'CUT
G -- - -- - --
4
Hexagon HE X True position TP
Hexagon head HE X HD 3
2
I ARM C l
B S 308 1972 Parts I, 2 and 3.

3 RD ANGLE
I

1ST ANGLE 1ST ANG ITEM PART N O, REMARKS

SYM BOL
I
� rl __
Et@- OF SCALE DRG OF 2
P ROJECTION � t:::::f TITLE N AME
 MAX MIN DEP TH TA PE.R
375. Conventions Counterbores and Countersinks.
A. The hole may be shown as a fully dimensioned dia­
!- C S K C BORE CON VENTIONS

-----
gram, or with a simplified diagram with printed instruc­
� 1 0 DRILL
SCREWTHREAD
DE PTH
tions.
COUNTER B ORE
COU NTERS I N K C BORE 3 DEEP M l2 x 2 6H
C S K 90° TO
B. The hole may be shown as a fully dimensioned dia­
¢ 20
gram or simplified. with printed instructions.
C. In castings, the rough surface at the face of a drilled
hole may require spot-facing with a suitable cutter to
provide a flat surface at right angles to the bore enabling
the nut or bolt head to seat without distortion.
3
D. Screwthread Conventional Form. A quick simplified
25
method of drawing an internal thread showing maximum
and minimum depths. Notice how the hatching crosses the
thread lines only in the unoccupied portion of the in­
M </J IO M IN
ternal thread.

\ I,.....
E. Four methods of indicating a taper are shown in the
TOLERANCED TAPER
standard, one only being shown. In this case, the rate of COUNTER B O RE

-£;,-
--o -.-s : 1 1
2-
taper on diameter is shown, with a tolerance.

\
i
(/) 1 2.0 --,

</) 25.0
F. Shaft and Hole. The hole is usually taken as the stan­ :!:Q .04
dard part, executed by drilling and reamering. A plus 9
tolerance on the hole is usually allowed. The shaft is
usually machined or ground to a minus tolerance to give 22

w
the required running fit.

c' BORE </) 22 X 9 DEEP

2 0-0 . I
0
I- .,
SPOT FACING
TOLERANCED SIZES
;f 6 DRILL I

S FACE ¢ 1 6 ¢ 2 9 .99
�
1

- 0.01
�
I I
� 30.0
/
/
+ 0 .01

NOMINAL SIZE '30.0

193
 1 94

376. Toleranced Dimensions Holes and Shafts. · A

toleranced dimension defines limits of size of a feature, L I M ITS & F ITS HOLES . & SHAFTS
and also has a bearing on the geometrical form of the

8
SE LECTE D FITS
feature'. B S 308.
M odern industry often requires that components are U N I TS = O.OOlmm CLEARANCE TRANS ITION INTERFER ENCE
machined within specified tolerance zones of the nominal HO� Ha HOLE H7 HOLE H7
size, to allow interchangeability for assembly. 2 s . o so_�--�� LE �
40
Holes and Shafts A shaft may be fitted into a hole to
+
-!------,£-,+-..,L......

give: 30 _.,_____
20 ____
(a) An interference or press fit in which the shaft is
larger than the hole and requires pressure and/or 10 --+- -�
heat for a permanent assembly. 0 --..---- 0 __4-____,
(b) Transition or location fits in which the shaft is a 10
light interference fit allowing positioning and with­
drawal under slight pressure. � E
e
(c) Clearance fits, sliding or running (rotating) fits, in 0
which the shaft is smaller than the hole, allowing the 0

"'
40
shaft to slide or revolve in the aperture. q
in 24.9 so,._�,__,__,_______,
The ' hole ' basis is usually adopted, in which the hole is P.6 s6
the nominal size carrying a plus tolerance, since holes are shaft shaft
easiest produced by drilling and reamering, and the

8
shaft being machined to a plus or minus tolerance de­
pending on the type of fit. AVE RAGE CLEARANCE D I M ENSIONING
An average running fit and a precision running fit are
shown in the lower diagram; the exaggerated hatched F IT
portion should help in explaining the method. The upper
diagram shows a range of the three fits indicated on a
graph to a greatly enlarged scale using 0.001 mm as the
unit. A fuller explanation and range is given in BS 1 9 1 6 :
L953 and also in B S Handbook No. 1 8. I S O Rec R286.
See also page 264 in Appendix I .
I n dimensioning a drawing, the limits may be shown by
z q °'
(a) showing both limits, (b) size and limit in one direction, SHAFT UJ 8:
(c) size with tolerance above and below. See tables in f7
Appendix. ...J in �

P R ECISION CLEARANCE �0-0

377. Positional Tolerances Lengths, such as those
shown in the diagram should be measured from a datum F IT � t0.05
=:, 30.00
line to avoid multiple errors occurring.
!::c '!Q.05
0 . 0
! 0.05
56.00

12 .00
't 0 .05
SHAFT
96 !0.10
 TYPE STR j FLAT I I TOL
378. Geometrical Tolerances Geometrical Tolerances

EJ
G EO M E T R I C A L TOLE R A N C ES RD

--
of straightness, flatness, parallelism, squareness, angu­
Type of tol Characteristic Symbol
I
larity, concentricity, symmetry and position, are required
to ensure that component parts are interchangeable for STRAIG HTNESS TOL

8 S TRA IG HTNESS
assembly and function.
B S 308 1 972 Part 3 has now adopted the I S O symbols

i-1 -1 0. 0 3

--
CJ
• to enable tolerances to be set out in a more succinct form
jrool
F LATNESS 0.03
and to overcome the language barrier '. A list of the

� .. t
symbols is shown and a selection of applications of sym­

F-
bols and tolerances is given in the following pages. As
stated in the Standard, the principles are the same as in
ROUNDNESS
YIT -$-
r 100
recent years, the method is more visual.
For
Straightness Tolerance The diagram shows the toler­ slngle
Form �
CY LI NDRICITY
ance zone applied to a cylindrical component. This indi­
cates a tolerance of 0.03 mm error in a 1 00 mm length.
Note how the symbol is shown in the first compartment
features
/:/ STR TO L 0 . 0 3 - 1 0 0

of the box, the tolerance next, followed by the testing

length.
PROF ILE ( line) n FLATNESS TOL [Q]
PROF I LE(surfacc
Flatness Tolerance Tolerance applied to a certain
0

II
surface. Note the symbol, followed by the tolerance.

N O T C ONCAVE
Length for testing not stated this time.
�OJ o.o 4
PARALLEL ISM 0,0 4

--,
i
Roundness Tolerance This tolerance can be required
on cylinders, conPs and spheres, and the amount of

_L
ovality must lie within the tolerance zone. Note the symbol
SQUARENESS
-
and the tolerance in the box. Attltuda

A NG U LAR ITY L F LA T TOL 0.04 N O T C ONCAVE

For

-$- @]
rcilatcid
fciatur�s P OSI T I O N
I
ROUNDNESS TOL

(Q) I
-10 10.02 1

--
CO NCEN TRICITY

I
Location 0.02

--
• 1

+ +-; t-
Y
--
_L

Cylinder
-..__ _
SYMMETRY
I
/
!
� �
Composite RUN OUT
i
I

Maximum matcirlal con dit ion MMC

T
@ olso:
I

R D TO L 0 . 02 cone. sphere
Boxed dtme n,lon (true pos i t i o n ) D
195
 1 96

CYL TP PA R SQ A NG
379. Geometrical Tolerances
GEOM E T R ICAL TO L E R ANCES
CY L I N DR I C I T Y TOL PARALLE L I S M TOL
CZZl
Cylindricity Tolerance The symbol and zone of al­

n
lowed error are indicated in the diagram and shown in the
box. Note how this tolerance lies in two directions.
[22Jo.1 jA 0. 1
Q 0.02
cbl_
rT 8
Profile Tolerance (line) The shape is set out by or­
dinates, the ' boxed· measurements indicate the true
measurement, and the error tolerance straddles this
position. Note the symbol and box. This symbol is used
on templates and laminae.
0.02 Datum
�-L� A
1
Profile Tolerance (surface) The tolerated error again

-t-- I
straddles the true line, and is tangential to spheres 0.02.

Parallelism Tolerance Tolerance of error in relation

to a stated datum axis. Note the symbol, followed by the
CYL TOL 0 . 02 PAR T O L 0. 1 [AJ DATUM A
· - ------l

P R O F I LE TOL SQUARENE S S TOL

tolerance, then the datum letter.
( 11 n c )
Squareness Tolerance This indicates the tolerated

o_os-j f-
error of one surface at right angles to another-the datum
surface.
n o.2
Angularity Tolerance Again a datum surface with the
...L 0.05
second surface to be within a tolerance zone of the stated
angle.
The symbol of Flatness could also be used in addition
to the one used, if a second limitation was required.
[[] 6 ordi nates
Datum
sped at 1 0 • 6 0 SQ TOL 0.05
TP TOL 0.2

PROF I LE T O L ( surfacc) le] A NG U L A R I T Y TOL

� o 10.02 1 ...:::::. 0 . 1 A l-
�_,___.__1

TP TO L 0.02 ANG TOL 0. 1

 380. Geometrical Tolerances
POSN CONC SYM GEOMETRICAL TOLERANCES RUN-OUT
Positional Tolerance Boxed true measurements with
a tolerated error circular zone ofO.O l on centre position of
=
POSITIONAL TOL SYMMETRY TOL
3 HOLES
oo -
each hole. Note the symbol and the diameter symbol �

8.--�r---1§7_/_rt>_ _. _,_I \
preceding the tolerance. o.o3 A-Bl
Positional Tolerance (2nd position) Positioning of
holes on a pitch circle with a stated tolerance from a shaft
0.0�
hole.

Concentricity Tolerance Stated tolerance from a da­

tum axis within a zone.
25
Symmetry Tolerance Tolerance zone affected by two
datum lines or surfaces. ,8
POSN TOL 0.01 3 HOLES SYM TOL 0.03 DATUM A-B
Run-out Tolerance The diagram shows a surface
rotating being tested -by indicator clock for eccentricity,
the shaft being datum.
POSITIONAL TOL
4HOLES (/> 0.2 1±1 RUN-OUT TOL
[ZJ
-� 00.2 A
/0.1
Application The example shows the use of several
tolerances on one unit. First, concentricity, diametrical D
tolerance of 0.2, maximum material condition, lower
shaft, datum, minimum material condition on head,
concentric with shaft.
�
0.1-J/-

POSN TOL 0. 2 DATUM A RUN-OUT TOL 0,1 DATUM A

C ONCENTRICITY TOL \@I APPLICATION

@ o.o3 A

kf -;
�
-· f t a
Datum axis Cyl o.o3

CONC TOL 0,03 DATUM A

197
 198
381. Machining and Surface Texture Symbols
A. Machining Symbol Where a surface is to be ma­
chined to certain limits a machining symbol is used to
USAGE I MACHINING & SURFACE TEXTURE I SYMBOLS

ness value is shown at position a and is stated in micro­

®
indicate such surfaces and the limits required. The rough­ APPLICATION
a Roughness value
�
metres which can be shown also by a roughness number,
listed in the lower column. Treatment is indicated at b,
® b Method: treatment:
c oating ---- LAP

£
N� 30
b para to
c Sampling lcngth
method or coating. Direction of lay of finish is shown at d.
plane
-
Machining allowance of metal is indicated at position e.

w
d Direction Of lay
B. Two examples of the symbol, first to show that the
e Machining allowance
-
�
instructions are obligatory and must be done, followed
by the symbol containing a circle, which indicates that ..__
pcrp to
machining is not to be done.

plane
�
C. Application on a component of the symbol. Limita­
tions stated in micrometres.

D. Application
obligatory prohibited - LAY
®
o/ �.� �
This example shows that the surface
indicated is to be lapped to 0.05 micrometre of the stated
or
�
diameter, circular lay, over a length of 30 mm.
crossed
E. Application Finish by honing in the direction parallel
to plane, to the figure stated. The lower example means
that the direction of the grinding lay is to be radial and

® �
on the surface indicated to the limit shown by the figure.
APPLICATION

-fj
.__
__

-, I·
r,1 ·-,r51
... R multi lay
N4

-• _,
I

8J!
I
I
m or mm = I micrometra Symbol IJ,lm
�
FINISHES:

lLJ
I
0.8 circular
MAC HINING GRINDING
t
g
PLATING LAPPING
CHEM ICAL HONIN G
3
�ALL OVER EXCEPT AS STATED
® radial

micrometre 50 25 12.S 6.3 3.2 I. 6 o.e 0.4 0.2 0,1 o.os 0.02S I

Roughnen numb, Nl2 N7 N2 NI

NII NIO N9 NB N6 NS N4 NJ I
 MACHINE DRAWING PROBLEMS
Pump Parts
Cover Plate
Vane Pump Parts
Bearing Bracket
Face Valve Parts
Gearbox End Plate
Camshaft Pump
Worm Reduction Gear Parts
Screw Pump Parts
Cylinder Head
Overhead Swivel Pulley
Gear Box
Swashplate Pump

l
 200
::: :::
T �R:0:o =�AN=G�L=E=========�:::::::::1:P� !I ��������=��=1:::;
li_:::::::::.i'f!'
i
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- Iu�M�EP:::::::::::::::::::::::::::Je:� B:JJ[::::::::::::::::::
PARTS :::::::::::::::::: :::::::::r:::.
DRAWING NO. I OF 2

---+------r-+--.---.
Qt r:n- ---
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 THIRD ANGLE
MACHINE DRAWING PROBLEMS
T ESTS
The following drawings are set as examination type ques­
tions. Some of the data will require a search to be made
to test ability both in reasoning and in visualising the
20
parts and assembly of the unit.
Setsquare angles may be inferred, details and sizes not 155
shown are left to discretion. Most of the problems show
pans from which assembly views are required, and refer­
ence to solutions and isometric drawings shown later ¢12
should not be made until some attempt at a solution has
been attempted, to simulate examination conditions. A
number of the objects set are complete working units
such as pumps, and would form suitable projects for con­

��-------------------.-------,
struction in the workshop if time permits, bringing the lx45°
work to an ideal conclusion.
382 and 383. Pump Parts The parts of a water pump N

with an inclined ram are given. The valve seat screws into
the lower end of the body casting, a plastic or metal ball
valve seating upon it. The upper ball valve is retained by
RI
the union. The ram is sealed by the hemp ring and gland
fastened by two bolts. The gland, part No. 2, uses 19
Tangency construction to obtain its outline.
8
Draw:
(a) In First Angle Projection, a sectional front elevation ¢ 20 SPt\ERE.
with the parts assembled, ram at lowest position, fs. 6g
(b) Project a full plan from the required elevation.
(c) Project a full end elevation to the left of the front
elevation.
Complete the drawing by adding a list of parts, title,

8
border, and six dominant dimensions. A part solution is
given later, see 408.

Dlm�n,ions in mm
4 UNION BRASS 8 BOLTS 2 32 X M8 X 1.25

3 VALVE SEAT BRONZE 7 RAM STNLS ST

2 GL AN D BRONZE 6 SEAL RING HEMP

BODY BRASS 5 BALL VALV E 2 PLASTIC

NO.
I

ITEM! NAM E REMARKS ITEM NAME Off REMARKS

PUMP PARTS DRAWING NO. 2 OF 2

201
 202

384. Cover Plate A cover plate with bearings for three

shafts is shown in Third Angle projection, sizes in milli­ COVER PLATE 3RD ANGLE
metres. Dimensions in mm
Draw:
(a) In First angle, fs, the given plan as the new front
elevation.
(b) Project an end elevation to the left of the new front
elevation.
(c) Project a plan in accordance with the two above

Jl/10
views.
(d) Project an Auxiliary Plan from the new front eleva­
tion normal to a line drawn through the two centres
of the larger shafts, sectional.
Dimension the front and end elevations, and the
plan. Add titles and name. Symbol machined sur­
faces. Add tolerances for average fit shaft holes.
(e) Make a full size isometric drawing, natural scale, of
the cover plate, with the smallest shaft hole nearest
the viewer.

'°.�! ____
 3RDANGLE
385 and 386. Vane Pump Parts The parts of a rotary
VANE PUMP PARTS DRG I OF 2
vane pump are given. The sliding blades of the rotor are
kept in contact with the housing by centrifugal force. The <1>7 RB
rotor turns in a clockwise direction, fluid entering the
top orifice and being pumped out of the lower opening.
A sealing ring and gland are held in place with two bolts. 6
A cover plate closes the front of the housing sealed by a
0.5 mm gasket.
Gland plate No. 5 uses the geometrical construction of
No. I 9 to obtain its shape. N
0
___,_..._._.N,1,0
Draw:
(a) FS using First Angle Projection, a full elevation of
I 'Q[�
the pump with front cover removed but otherwise 7. .� =..J
assembled, and looking at the end of the rotor.
(b) Project a full elevation to the right of the front eleva­
tion.
(c) Project a plan, sectioned on the centre line of the O.QY
rotor shaft. 'V mm
(d) Draw a union and cone joint to suit 10 mm outside
diam. tube the pump connection.
Complete the drawing by adding a list of parts, titles.
Show six overall sizes.
See 409 for an assembly drawing.

)-- -.-
(/J

t 2"__ _

0 I. 12 ••
,,._2_
.......
i-La 3
2____,.,I
____:: -=-
......1,.
_..
118
I •

203

,,,/
 204

VANE PUMP PARTS THIRD ANGLE

0
____ _ _ _
___:_:c.....:...;rr-==------'---=--.;:...:...:.._ __-'-'--'-'-_:....--'---------------t---------·-----·--·--

�To
RS 387. Bearing Bracket A front and end elevation
of a Bearing Bracket are shown, in First Angle
R
Projection, sizes in millimetres.
"'Ii 3
Draw:

I
I 6 HOLES (a) Using Third Angle Projection, fs the front

(b) Project a full plan.

_j_ elevation as shown.

\
118 PCD
1
r-
_30_.., o
�L \lao (c) Project a full end elevation in the direction of
arrow A, but in keeping with the required

../
\
l a:1gle of projection.
(d) Project an Auxiliary Plan in the direction
indicated by line B.
Dimension the plan and elevations. Add suitable
titles.
I M6X 1.0

I
-r s(
! Bg
I 4
I
L=--=-- =-==::dk================j ROTARY VANE PUMP
Design a small leaf spring to hold van� in
contact with casing during rotation

Dimensions in mm ----

-+-'I
7 BODY GUNMETAL
°

11
lx45 6 SEAL RING HEMP

5 GLAND GUNMETAL

�
4 BOLT S 8 i M6xl.Oxl5 Bg
VANES
I 3 12 II NYLON

i 2 COVER
--·-,----�---
I
BRAS S

ROT OR I M s
i

ITEM! NAME ; NO.: REMARKS

_J_ I;----
25.95
--..;------65 ---------i DRAWING NO. 2 OF 2
 BEAR ING B RACKET 1ST ANG LE

e
Oim�nslonr in "''"
E 0� 23

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mm

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30

205
 206
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PCD 8 8 t � RIB S EQU I
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388 and 389. Face ·V alve Parts Parts of the pump are
shown. The pin screws into the valve plate; the dia­ 3 RD ANGLE FACE VALVE PARTS DRG 2 OF 2
phragm, washers and spring fit on the pin and are locked

8
in place by the nut. The locknut secures the valve plate.
The valve assembly drops into the top of the cylinder,
and is held by the cylinder head. ¢ 58
The safety-release valve screws, as an assembly, into
the cylinder head. M l6xl.5
Draw:
(a) F S First Angle Projection, a front sectional eleva­
tion of the parts assembled taken on a vertical centre
line of the cylinder. ¢s
(b) Project a full plan from the elevation.
(c) Project an end elevation to the left of the sectional
elevation. 08
The cylinder height may be reduced by break lines to
save drawing space.
A part assembly diagram is given on another page,
Third Angle, see 4 1 0 .

Dimensions i n mm T H I RD A NGLE

B RELEASE VALVE BRO NZE ss

7 SPR I N G S T EEL
6 WASHERS A B 2 M S
5 DIAPHRAGM NYLON g M S

4 I PI N & NU T S M s
3 VALV E PLATE C

2 CY L H EAD C
Ill
N C Y LINDER C AS T I RO N

NAME NO. REMARKS

,. ¢ :2 2
•I PUMP PARTS
207
 208

GEAR BOX END PLATE RD ANGLE

3
Dimensions In mm
// o.os

8
A
9/ 1 90
\
25 .i- 1 3 ., PCD
I I 165
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\ 1'9---------.. L.
--- - - - - - - - -1
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-_ _-_-_ -
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- - - -- S - :x,
1

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6 S HO L E S (/> a H9

@
S FACE
1
MSx l.25 x l 5 MAX �-!�_PAC E D -;
6 HO L E S
E QU IS PA C ED
i-$-1 <1> 0 - 2 l@i
80 -$- </) 0.2 �
390. Gear Box End Plate Third Angle Projections of a
gearbox end plate are shown. C AMSHAFT PUMP DRG OF 3
Draw:

8
(a) F S First Angle Projection, a front elevation lookine-
into the recessed interior of the plate.
(b) Project a sectional end elevation to the right of tl�e
front elevation. (;)

-*-··
/
(c) Proj ect a plan. Dimension the diagrams fully. Add I
titles. Indicate surfaces requiring machining.
I ¢ B
391, 392 and 393. Camshaft Pump Three sheets of dia­
grams showing parts of the camshaft pump are given.
The body casting is bored to take ball races which receive
the camshaft. This is retained in place by the cover plate.
The two rams are held in contact with the cam faces by M l2 x 1.25 69 I
springs. The two inlet and outlet valves are stainless steel M 6 x l.0
ball�.
Draw :
20 A/F
(a) F S, First Angle Projection, a sectional front eleva­ II
tion taken parallel to the axis of the camshaft.
(b) Proj ect a full plan.
(c) Project a sectional end elevation to right of the front M 20 x 1.5 69

0�
elevation, and on the line through the centre line of
the first cam.
(d) Project a full end elevation to the left of the front <1> 24 - </) 20
elevation. Add eight maj or overall dimensions. Title.
(el Draw 3 x fs a diagram showing how the contour of
the cam is obtained. Uniform acceleration and re­
tardation, lift 10 mm.
Assembly diagram see 4 1 1.

D im q nsions i n m m 1ST ANGLE

II PLUGS 2 M 2 0 x I.Sx8 6g

5 RAM 2 S STEEL 10 BOLTS 6 M 6 x 1.0 X l6 6g

9 U N IO N 2 S T EEL
.__......
4 j BALL RACES 2 RADIAL

3 COVER PLATE C 8 VALV E SEAT 2 STEEL

'
I

2 CAMSHAF T C STEEL H T 7 BALL VALVE 4 s S TEEL

CASING C 6 SPRING 2 COMPRSN.

ITEM NAME NO. REMARKS ITEM NAME NO. REM ARKS

CAMSHAFT PUMP PARTS

209
 210

1------ - .------'--- -
1 S T ANGLE
- ----
C AMS H AFT PUMP PARTS DRG 2 OF 3

o 90
ffmm 25 14
8

-r .
I
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....�--------
I

VALVE SEA T S -
I
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8 6
S1 FA CE
- i----t---t-- ,- _ ..,

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I I

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// 0.02 I 1--
A I o.o 2 A

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.� I L - - -I
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0.1
EQUI SPACED
B

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1

M 6 x l.O 6H
' 'bl
I : 12 M I N LENG T H
FULL T HR D

--+-------,.,>---,
12 12 6
// 0 . 02
54

' ' ¢ 50
A

A 1-'2'_J

¢a

-MACHG PLUGS
M 20 x 1.5
6H

F ILLE T RA D I I 3
 1ST ANGLE CAMS HAFT PUMP PARTS ORG 3 OF 3

0�
120 -.-l
"'m i..t-----------------
1----= :.c;_J

U N IFORM
A ACC & R T O

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21 1
 212

394, 395, 396 and 397. Worm Reduction Gear Parts The
parts of a worm reduction gear are given. The worm wheel WOR M REDUCT ION GEAR

8
and shaft run in bushes set in the gear housing. The worm
runs in bushes set in the casing and bearing support. A C L
thrust race fits on the shorter end of the worm shaft. The
" ¢ 6 .6
a:) i
wormwheel has twelve teeth. The worm tooth section is
a rack section.
Draw:
N'

+1
(a) F S First Angle, a sectional elevation taken on the
centreline of the worm with all the parts assembled.
The worm may be drawn by simple representational
straight lines, full helical curves need not be shown.
�
0
Draw only three meshing teeth of the wormwheel,
compass curves allowed.
(b) Project a full plan, from the elevation.

+
(c) Project a full end elevation which shows the bearing
support.
Complete the drawing hy adding a list of parts and titles.
Add ten important <limc:nsions.
Assembly diagrams are given on another page, Third
Angle, see 4 I 3 140

'.'
11'1 0,Qij

38
V mm
.,
I • o

..I I.
8 I
Dimensions in mm R D ANGLE
3
DRG NO. I OF 4 1 BEARING S UP T STEEL
3
6 THRU ST RACE STEEL SET 12 BOLTS 22 M 6 x 1.0 x lS 6 9

5 WORM S TEEL 11 ENO PLATE S TE; E L

4 BUSH A & B 2 BRONZE 10 END PLATE S TE E L

S HA F T & S C REW M s 9 C OVER PLATE M S

3
2 WORM WHEEL C a BU SH BR ONZE
C A S ING C 7 B U SH B RONZE
I TEM NAME NO. REMARKS I TEM NAME NO. R E M ARKS

WORM REDUC T ION GEAR PARTS

 DRG 2 OF 4

j//j 0.02j B ,.._

,�
-1

r-
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===-=--=--=-1
FILLET RADII 3

28
..,
-
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WOR M

-=-----=---_-�--14-�
-_ l......,--/
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GEAR

_j____
PAR TS

38
1 -40

......,..+,,-----,----+--------�----.---..-�
3 RD ANGLE

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M 6 x 1.0 6H
EQU ISPACED
$ «> 0. 1 D
10
-1----1------1 II -$- o .02 A
M 6 x 1.0 6ti
12 MAX
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p__________ 20
,___ _ _ _ _ _ _ __,
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130
20
____________---=-=-..., 8

213
 214

RD ANGLE WORM GEAR PARTS ORG 3 OF 4

3
12

8
I

EXCEPT P C 0 72 --
FOR WORM

6 H O L ES ¢ 6 . 6
EQ U I S PAC ED
-$- ¢ 0. 1 8 R 3
tl_i_J 0.02 j A I
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10
C LOSED
165

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3RD ANGLE WORM GEAR PARTS DRG 4 OF 4

A DDENDUM

DEDENDUM 6.9
6 ------- P C D 72
8
....
M Sx O.S x 1 2
H exagon Soc k et

-
BLANK Set scriiw
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-- - - � - --+--�

12 TEETH
C P 18.86 4
M ODULE 6

1 25

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0 0 +
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0
� ang led 10 ° RING

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mm
215
 216

WORM SCREW PUMP DRG OF 3

¢ :25.0 1
2 5. 0
20
MBx l.O 6H
I,. 62
8
- -r -
-l---

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_ ¢42

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63
16 20
�
-$- (/> 0 . 1 E-E 43 0�
mm
j_ 0. 0 3 A EXCEPT W H E R E STAT E D
Dime;nsions in mm

3 D R I VE N W O R M STEEL 6 GLAND BRONZE i 9 B O LTS 6 M8 > d.0 x 28 6g

2 D R IVER W O R M STE E L 5 E N D PLATE C I 8 B O LTS 2 M8 x l,O x 30 6g

CA S I N G C 1 4 BUSH 4 BRONZE 7 SEAL UNIT 2 GAR T E R

I TEM NAME NO. REMARKS ' ITEM NAME NO. R E MARKS ITEM NAME NO. R E MARKS

1 ST ANGLE SC REW PA R T S
___
PUMP
_ _l
1ST ANGLE W OR M SCREW PUMP DRG 2 OF 3

6
262

D B
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- -- ----- ------

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1x 45°

57 12 I 57 ! < ,_-;� � 6 83
2
¢ 3

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,
0,05 A-B
-- - - - --- - - - - - - --
RH LH

! 11 !
� r- ¢25 ¢ 50.02

,
MACH I N E ALL SURFACES
OF WORMS LAP

07
,

217
 218

M B x l.O

8
2.00 B
3
Q
¢ 50 03
C
50.00
R28
,- --- -;
r - �1
10 I 15 138 27
- I
I I
L ,z'�r - - - - - - - - - - - - - - - - I TW I N
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32

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EXCEPT WHERE STATE D
..!.__

1ST ANGLE DRG 3 OF 3

FILLETS 3 R 42
S CR E W PUMP PARTS
 398, 399 and 400. Hydraulic Screw Pump The parts of (c) Project a full end elevation showing the gland end. 401. Cylinder Head Two views are shown of a cylinder
a screw pump are given. The two geared shafts have (d) Project a sectional end elevation at the other side of head, valve, and spring and cover plate. See 4 2 1 .
intermeshing left- and right-hand square threads which the front elevation taken on centre line of the top I . Draw :
carry the fluid from the lower orifices to the delivery port orifice. (a) The given left-hand elevation fs.
at the top of the casing. The shafts run in bushes pressed Add titles and list of parts. Show ten major dimensions. (b) Project a plan, from (a) in Third Angle, sectioned on
in the casing. The upper drive shaft is sealed by a double The worms may be shown by conventional straight the centre line of the valve.
unit using a sealring and garter spring. the units are held lines and not full helical curves; the gears by simple con­ (c) Project both end elevations, full views.
in place by a gland plate. vention. Add ten major dimensions, and a list of the parts.
Draw: Part assembly views are given on another page. see 2. Draw:
(a) A sectional elevation of the assembled pump taken 4 1 2, also 358. (a) The given elevation as the plan in Third Angle Pro­
on the centre line of the shafts, fs first angle. Use Geometrical Construction shown in 19 to obtain jection.
(b) Project a full plan. shape of gland plate. (b) Project an elevation from (a).
(c) Project both end elevations, one of which should be
sectional to show the valve, seating spri_ng and cover
plate details.
Add ten dimensions. Add titles and list of parts.
(d) Make an Isometric Drawing, natural scale, of the
assembly, cut away to show the parts.

219
 220
,----------------- - ----- - ----- ----- ------- ,---- --- - ---
C Y L I N DER HEAD 1 ST ANGLE

FILLETS R3

8
84
I LIO I 63
!
I I .,..
j I
(j 175

8 HO L ES
PC D
1 50 ct> B. 5
O N BO PCD
57
IN

r I .
, 8 H O LES ¢ I I
E Q U I SPACED 1
e _j__ r N H E A D -
B HOLES
3 22 • ,
�lr._l
$-
I
I </J Q , I @ M8 x 1 .25-SH LIFT

_l ¢ 0. 1 B ,_____
'---- A T H ROUGH
E Q U I SPACED
D i m e n s i o n s In m m -$ <1> 0.1
 402. Overhead Swivel Pulley A sectional elevation is
given, and a part full end elevation. 1ST ANGLE SWIVEL PULLEY
Draw: fs, using First Angle Projection, detail work­

r
8
ing drawings of the ten separate parts, unassembled.
The vee pulley is free to turn on its shaft, which is held
in the fork by two collars and taper pins. The fork is ¢ 130

·1
-1!
screwed to take a circular retaining collar which is PC D I 1 8 6 HOLES
tightened by a peg spanner fitting in the two holes shown. ¢ & EQUI-SPACED
The downward pressure is taken by a radial needle roller ¢ 65
caged unit revolving bet ween two hardened steel plates.
The whole assembly is fixed by six bolts 8 mm diameter
¢ so I
I .
�-
spaced equally on the pitch circle shown. f-- 3 8 C R S
Show a detailed list of the parts in a suitable position
on the drawing.

1
Show a method of locking the retaining collar.
Draw the assembly as an Isometric Drawing, natural
<>
scale, sectioned on the vertical centre line, front quarter
removed. See 422.
s6 32
10-¢
28caged Q
radial
ro I I ers -.----_,.--,,... .,.,... RS

42

I
-{ ,..._-i'----�--t'-�
21 12
PIH

I
80
m
Di cznsions i n m m

22 1
 222

I GEAR BOX 3 RD ANGLE

HOLES B & C
_L o.o4 A

- -� R54
RS / 1
- -... r _ _
)
�"'--­ ...._,_ _ ------
'l
- --,r- - '
" 1--"-·--+----+---i.:� l

""'�
't
-+ - -
- -,
t- - -4 - �
h L--J. -;-....-----.-'_...J
1
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-- r - - /5
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-- - 1 19.02
.. .
� 1 2 ,7 15.85 \
I
'
;:5 38
, -
,- 142 .62 I 48,8

I
+--
85.73

I I - -- -
. T
- - -, --- 2 HO LES D
¢ 19
1
1
,.-- J 0 .04 AII L __
'
I
I I
(

R 12
'- - - - - - -
- - - - _ _ _ _ _ _ _ __ ,,
I
\ R41

....____· ���!__
' s'-
FA
_CE t·-o=·o/=2==m:m'----
_ - - �
_l____�_:_-
_-_
-_
-_ _ @
�_ :_ ::::::_ -=:r����;_-;j Dim •nsloas In mm
 403. Gear Box Third Angle Projections of a Gear Box
Cover are shown. S WASHPLATE PUMP
(a) Take the right-hand view given as the plan for new
projections, first angle, fs, and project a front eleva­ INLET VALVE OUTLE T VALVE
tion and an end elevation to the left. The plan and
end elevation should be full, but section the front
VALVE � SPRING

8
elevation on the centre line of the 25.4 mm dia shaft.

0
BODY
(b) Project a new sectional elevation normal to the line
X X. Take the section passing through the centre of U"I
the 25.4 mm dia shaft.
Dimension the full diagrams (2), add titles.
(c) Draw an Isometric View of the box, from the most
convenient point. NUT
404, 405, 406 and 407. Swashplate Pump The parts of a
swashplate pump are shown in Third Angle Projection.
Four plungers are actuated by a face cam, the rams are
kept in contact with the face by springs. The shaft runs
on a radial loaded ball bearing and a caged needle roller
SPRING
bearing. Four mushroom-headed inlet valves are ar­
ranged round a central inlet opening. Four outlet valves ¢2 0
are arranged round the periphery of the valve body cast­
ing. All the eight valves are lightly spring loaded to help
in re-seating. ¢12
Draw: VALVE
(a) FS a sectional front elevation taken on the centre
line.
(b) Project in First Angle, a full end elevation to the left of CLIP WASHER
the elevation.
(c) Project a full plan.
(d) Project a sectional end elevation to the right of the
front elevation taken on the centre line of the cored
holes of the inlet valves.
Calculate the output of the pump when running at 2 000
rpm.

..
 224

----i---1 ,j-r 0
RD ANG L E S WASHPLATE PUMP PARTS DRG 2 OF 4
3

1111 °- 02 ,A I-.+
' �
-

66
8
11.
-

1 2 .i_ 40

I ! ·r - - f
25 32 ,
----...---- ·- ·i
· -----
..---.--�-........
.........--
A f---•
co
N
I
!
PC D 128

(\I I
C"I '

M8 x l.25 S H
SHOLES

.____...._u.... ! ...
, --+-----'-'-- ::r--.J 1-Ett1 </> o. o 2 1
...
- M 24 X 2.0 sH
4----
-

FILLETS R3 R6
:
1
 - -·
3fl0 ANGLE SWASHPLATE PUMP PARTS
-- - -- - · - .. - I_ DRG 3 OF 4

lo�
8
MACH I N E
1 10 BALL N E E DLE
ALL ROLLER

+'l!'J
RACE

0
mm S U R FACES CAGE
------- .l--
I
45
- �o _J:�L- j i
2_
2_ ..
J,
� l3l,
"--,,{ ,

:, !I / !
�
¢'so I
/
I
;u 70 1
2
t I
I �
0 )" 0.01 A B
I '
¢

j
10
semi
�irculor 0
groove
_ .l
�.

"'I
'Q. , ¢ ,o______
-- . � -

·-+-
y)
12
5
2

PC D
.0' 1 0

64 , 5 1,.-
1 I ¢ 25

----
,-+--11--f-+--t (YJ 'Sf" -'-- - .
Cl) Ill :

� '& ,
1,5 I :
f6 38/

..L o .o4 _ i . D
------- - --- - _ _ _ _ _J

225
 226
---
l D RG 4
1

I
,-- - -3
- R_
D_A_
N_
GL_E_�, . SWAS HPLATE P�MP PARTS ·· ··- ·- -···- ·--· .

e
�� -�-- �

I/� 1 I 0.0
4
A
1� r�- -,
t- ·
I
.... ·- - · · -· -· - -·----!
75
!f\
8 HOLES </) B H S
</J 0.02 /

I
/ I
·r
/ /

- T -
I ,, - �

---
,--------·-i--:;

- �
T - -
� -:
I T_ i
I _... ___.
I

P C D 1 28

l I - �--'= � �I I t
I
i= el 57

/ O O . iI :l
N j
M ·

!
_,
0 I 1
'& '& '& '&
Q cX> l.ll �
, : I
I I

- - --
l_ j
i. -J
I
I

l_l_ -1._
_L
I

--- --
0.
_ __L

J_ A MAC H I N E

L.__.?_ �_._ J1�

0.0 2
P C D 64 \ ¢ 146 I ALL
_ 2_

I
... �� 1_....... , S U R FACES!
I
6 HOLES M 6x I 25 S H
EQUISPACED
FILLETS R 3 -$-/ </J 0.02 I 8
0

efm m
ASSEMBLY SOLUTIONS
Ram Pump
Rotary Vane Pump
Face Valve Pump
Camshaft Pump
Hydraulic Screw Pump
Worm Reduction Gear
Swashplate Pump
Ratchet Feed Gear
 228

3RD ANGLE
�--·- · - - - - -- - --'���
P U M P-___:��=-=--=.....::_.=��
S O L U T I ON ----
A SS E M BLY S O L UT I O N S
Orthographic projections in either First or Third Angle -- - - -------,

8
are shown of some of the problems set in the previous
pages.
I I
I ' I

�08. Ram Pump Assembly Th! rd Angle. A full plan is

_
shown. A sectional front elevation on the centre line of
t h e ram is projected below the plan.

A S S E M B LY
409. Vane Pump Assembly Third Angle. A sectional 3 RD ANGLE
plan is shown taken on the centre line, the rotor is shown ROTARY VANE PUMP A S SE M B LY
in full. The union is shown in full section for ease of

8
drawing.
The front elevation is shown with the front cover re­
moved to display the rotor.

UNION SHOWN

FRONT C OV E R
R E M OV ED

229
 230

410. Face Valve Pump Cylinder Head Assembly Third

Angle. The plan shows the cylinder head removed to FACE VALVE PUMP THIRD ANGLE
show the face valve. The sectional elevation shows the
valve in section. The release valve is shown in section
also. A scrap plan shows the true shape of the pipe con­
nection face of the cylinder head.
4 1 1 . Camshaft Pump Assembly First Angle. A sec­
tional front elevation along the axial centre line of the
CYLI NDER
R EMOVED
HEAD
8
shaft is shown. The camshaft and plungers are shown in
the round as are the ba)l valves.
The sectional end elevation is taken on the centre line
of one o f the plungers. The cams and plunger are shown
in the round.

-----8
A SS E M B LY
A S S E M B LY C A M S HA F T PUMP 1ST ANGLE

r 8

S EC TI O N A L VIEWS

23 1
 232

H YDRAULIC S CREW PUMP

8
 ---- ------

THIRD ANGLE
ASSEMBLY SO LUTIONS
Orthographic Projections in either First or Third Angle
SCREW PUMP
are shown of some of the problems set in the previous
rages.
412. H ydraulic Screw Pump Third Angle. A sectional
front elevation taken on the centre line of the shafts is
shown. The shafts are shown in the round, with the right­
and left-hand meshing screwthreads in simple conven­
tional form as are the gears. Scrap details show the shape
of the pipe flange coupling and recess for the soft copper
seal ring. An end elevation is shown, half as full view, and
half sectionalised on the vertical centre line of the outlet.
Worms are shown in schematic form only.
4 1 3. Worm Reduction Gear Assembly First Angle. A
sectional front elevation is shown taken on the centre
line. The worm is shown simple conventionally in full;
three meshing involute teeth of the wormwheel are
shown.
The end elevation is sectionalised on the centre line of
the wheel and shaft.
414. Swashplate Pump Assembly Third Angle. A sec­
tional fron·t elevation is shown; the swashplate shaft, two
rams and two valves are shown in full round form.
A full end elevation is shown.
· II
\
\
\
\
''
'............
---

233
 234

1 ST A NGLE WORM REDUC T I ON G E AR A S SE M B LY

8
ASSEMBLY SWASH PLATE PU M P 3RD ANGLE

6 14
8

7 I I

235
 236

414A. Ratchet Feed Gear The system consists of a

e
lever, a spring loaded pawl, and a ratchet wheel. The
RATC H E T GEAR I S O M E T R ICS 3 RD ANGLE
wheel is fitted to the square section on the shaft, the lever
is free to swing on the shaft and its oscillatory motion
actuates the pawl which turns the wheel and shaft
through a set number of degrees with every stroke. The
shaft has spur teeth milled in it, to mesh with a spur gear 125
wheel giving a reduction, and may be used to lift the table
of a milling machine in an automatic feed.
Draw :
(a) A sectional front elevation of the assembly, supply­
ing any details which are missing, keeping the pro­
portions similar to the given diagrams.
(b) Project an end elevation, First Angle.
( c) Project a plan from the front elevation. Dimension
the drawings, add titles.
( d) Draw detail drawings fs. Third Angle Projection of
the se p arate parts of the assembly-shaft ( teeth need
not be shown ) , lever, ratchet-wheel, washer, nut,
pawl, screw and washer, stud and spring, spur gear.

Sec 423.
 ISOMETRIC DRAWINGS OF ASSEMBLIES
Roller Bearing

Cone Clutch

Flexible Coupling

Camshaft Pump
Cylinder Head
Swivel Pulley
Ratchet Feed Gear

--
 238

R O L L E P. B E AR I N G I SOMETR ICS N O N � R E TU R N VAL V E S

8 8
VANED DISC
VALVE

V AL V E

8
I
C O NE C L UTCH ISOMETR ICS F L EX IBLE COUPLING

239
 240

ISO METRIC O R A WINGS CAMSHAFT PUMP

ISOMETRICS
Isometric Drawings of several examples given on earlier
pages, sectionalised to show hidden details.

e
415. Roller Bearing The diagram shows an Isometric
View of the housing, rollers and races, cover plate and
shaft. Begin the drawing by drawing the shaft, then the
housing and plate as simple · boxes '. Draw the ellipses
using the compass four-arc method. Sectionalise the
housing. Draw in the rollers.
416. Disc Valve with Vanes. The vanes are shown with
a helical twist which causes the valve to rotate to a fresh
position on re-seating.
417. Ball Check Valve A sectionalised valve box, cap
and ball valve are shown.
418. Cone Clutch An orthographic section is shown on
an earlier page. Draw the shaft first, then the ' boxes ' en­
closing the two halves of the clutch. Draw the sectioning.
Draw the ellipses by the four-arc method.
419. Flexible Coupling Draw the shaft first, followed
by the ·boxes' enclosing the coupling discs. Draw the
sections. Draw the · ellipses by the compass four-arc
method.
420. Camshaft Pump Begin by drawing the camshaft.
The cam curves are obtained by drawing the enclosing
· box · and ordinates. Draw the section surfaces. Draw
· boxes ' for the housing and plate. Draw the ellipses for
the housing shapes.
I S O METRI C DRAWI N G S
I SOME TRICS CYLINDER HEAD
421. Cylinder Head and Spring Loaded Valve. A sec­
tionalised Isometric Drawing of the cylinder head show­
ing the vaned valve and spring.
422. Swivel Pulley A sectionalised Isometric Drawing
of the swivel pulley showing the parts assembled. Draw
the top plate first. Draw the section surfaces. Oraw
ellipses for the shal't and retaining collars.
423. Ratchet Gear A sectionalised Isometric Drawing
of the ratchet feed gear. Begin by drawing the shaft. Oraw
the section surfaces. Draw ' boxes' for bearings, lever,
ratchet wheel and nut. Obtain the final shapes by ordi­
nates and ellipses. Draw the pawl, screw, stud and spring.
Enclosing · boxes ' will assist in arriving at the final
shapes.

241
 242

SWIVEL PULLEY

8
423A. Isometric Drawing The diagram shows a cut­
away isometric drawing of the Endplate for the worm RATCHET GEAR I SOMET R I C S

e
screw pump shown in No. 398 in the machine drawing
section. Begin by drawing the enclosing • box ' with centre
line. The semicircular ends of the base portion may be
drawn by the four arc method, but the upper shapes are
best drawn by a curve passing through points obtained
by ordinates. The top gland plate can be projected from
the upper face shape of the base.
Axonometric Drawing The diagram shows an under­
side sectional axonometric view of part of the face valve
pump, orthographic details given in No. 388.

243
 244

E N D PLATE I SOMETRIC GLAND AXONOMETRIC FACE VALVE

\
SEE 3 9 8 8. 4 1 2

SEE 38 8 & 4 10
ELECTRICAL CIRCUITS
 246

ELECTR I C A L C f RC U IT S S C H E M AT I C

, G
L A MP SW I T C H POWER INDEPENDENT S W ITC H R E L AY
e;
�

1
S ol�noid

S IMPLE
+ 2

8 E H
TWO LAMPS I LAM P 2 SW I TC HE S T R AN S F O R M E R

IN S ERIES
1 S E C ONDARY PRIMARY

2 L A MPS IN PAR A L L E L VARIA B LE S W I TC H F I C APAC ITOR A C RO S S PO I NT S

I
I

1
II
ii
POTENTIOMETER
11

,1
ii
,,
q

11
 ·-
424.
A. Simple circuit of lamp, battery and control switch. S YM BO L S I E L ECTR I C A L C I RC U I TS I S ELECTION
8. Two lamps in series. The current is reduced by the
equal lamps which glow at half power.
C. Two lamps in parallel, the two lamps glow at fu ll
illumination, double current being taken from the
I _CC\_

no connczct io n czlectto magnczt

�
i n ductor
8
frczq u cznc:y mtr
@
ohm mczt!lr
0
watt mczter
battery.
D. One lamp controlled by either of two switches.
E. One lamp controlled by either switch.
! r -.J\/\1\N- -0- 0 0 8
F. Variable switch (rheostat). One lamp controlled con n 1Z c t ll d rczsistanccz microphoncz a m m et ,u volt m cz t ii r synchronous
motor
from zero to full illumination by a slider contact on

=O ®
a resistance coil.
G. Soft iron core magnetised by the field in the coil when
the switch is closed. The armature is attracted and
closes a switch to operate a fu rther circuit.
_/0-
s i ng l cz p o l cz
-Jl..flJL-
riisistanc:e iiarphon11
®
d C: m o t o r a c motor
0.
L.:::=::::.J
piirm magnczt

+
H. Two coils wound round a common armature made 240V (,v SOHz motor

� I-
+ -
-.J\/VVV'-
up of silicon iron laminations, insulated. The input
and output are i n a ratio of the number of t u rns in
each. No connection between the two coils; the � 00
+ -
electromotive force (emf) is induced from the input s i ng l ll c iz l l variabliz ru 1 oudspiia kiir

-+--1--
primary winding into the output secondary winding. gqnqrotor
�
I . Tungsten or silver points \vhich make and break are
prevented from pitting by sparking by placing a foil
and insulation capacitor across the points.
m u l t i c: cz l l
-0- =ill]
f i la ment lamp moving coi I sp
siiri cs motor
motor

p h ases : 3 rv
motor/9<2n
I '""'

I
J_
- � -I I-
425. Electrical Symbols

i'"
The diagram shows a selection of I S O symbols taken
from B S 3939 Graphical symbols for electrical power etc. � �

kY
Parts 1 - 4.
I,.

ll O r t h chassis c a p a c i tor ha nds<1t

¢t
These can be used in circuits similar to those in this
section and in answer to this type of question set by
certain examining bodies who include an electrical section I �
��
I
in their syllabus.
c..., Y -'( l \...
--0 0--
� a c r e p u lsion motor d C: s h u n t s t a r t motor
f u s cz s push button
i
!
contoc:tor

� J /6
I
I

i
--0 0---

/'"YY'Y'I
91
winding two pol« sw I r ii lay
i
I (®)
�
� i -/� '"\.,

I
� I ! 3 phasci motor sq uirrii I cagll motor
tra nsformer transformiir c ontact 415V 3 rv SOHz I

247
 248
426.
A. Diagram for a shunt wound motor. The field and SYMBOLS E LEC TRI CAL C I RC U I TS SCHEMATIC
A

8
armature are connected in parallel.
j
1

B. Diagram showing a motor wound in series. The cur­ SHUNT MOTOR SYNCHRONOUS MOTOR
rent passes through the armature and then through I
the field. I
C. Starter and field windings are connected as shown.
D. Electro-magnetic field attracts a simple polar arma­
0
FIELD
ture. The rotation of the rotor is dependent on the ..J
number of poles and the cycle frequency of the alter­ I.LI
nating current. The rotor must receive a speed u..
impulse on starting. S mbol
E. Direct current motors require a resistance starter to
prevent current becoming excessive when starting. A
i� @!

I
solenoid holds the switch in the running position once
the normal motor speed has been attained.
F. The diagram shows a circuit used in the electrolytic E
deposition of metals. The system is also used in the SERIES M OTOR D C R H EOSTAT IC STARTER
shaping of metal by wasting electrolytically. Metal is
wasted away from the anode and deposited on the

I·
cathode as the current flows from positive to negative.

S mbol

®
C F
C O MP O UN D W OU N D MOTOR ELEC T R O LY S I S

- - - - - - - - - --
-�-!-tcTROLY
S m bol n-
 S CHEM ATI C
427.
A. Capacitor Start, split phase electric motor. In many ELEC T R I CAL C I R CUITS
II A
smaller type domestic single phase motors, the stator is
CONTAC TOR MOTOR SWITCH
laminated and has a field winding in series for normal
running. A second field winding with a capacitor or large CAPAC I TOR START
condenser aids in starting. After normal running speed SPLIT PHASE MOTOR
is attained, a centrifugal switch cuts out the starter coil.
OVERLOAD
TRIP
B. Contactor Motor Switch A switch operated by a

I{ �}
�Cl�
solenoid or electro-magnet. When the ON button is
pressed, current passes through the solenoid coil and the

�
armature is attracted closing the supply line contacts to
pass current to the motor. If the system overheats due to
F I ELD
Q.
overloading, the bi-metal strips at D 1 . 2. 3 , arc heated by
::>
0..

L3
resistance wires, and expand to operate an overload trip
II)
switch to break the circuit, saving damage to the equip­
- ON
ment. The motor is stopped in the normal way by press­
ing the OFF button which breaks the solenoid circuit
releasing the armature and line contacts.
I�· ARMATURE
--+
Most contactors are of the airbreak type with silver
OFF I / SOLENOID
I I
contacts ; oil immersed contacts to avoid arcing, are
I I
used for large motors.
CUT OUT
u
C. Loud speaker Circuit (centri f u gal ) I I
D. Telephone Circuit The telephone consists ofa micro­
phone and a receiver connected by wiring and energised
by battery or low power mains.
The microphone or transmitter consists of a steel dia­
phragm backed by a capsule of carbon granules which
pass varying amounts of current dependent on the pres­
sure on the diaphragm by the sound waves. The receiver C D
consists of a stalloy diaphragm which is made to vibrate
and cause sound waves in tune with the microphone. The . L OU D S PE AKER Cl Rcu·1 T T E LE P H ONE C I RC U I T
varying current passes through the coils of the electro­
magnet in the receiver catising a varying magnetic field
and thus vibrating the diaphragm.

3
M i c ro­
phon«

Spciak«r H a n d set

249
 250

428. Electrical Layouts

A. Thermostats (a) Bi-metal Strip. The bi-metal strip is E LECTRICAL L AYOUTS S CHEMATIC
A I C
L--
made from two strips of metal having widely different
coefficients of expansion. As the strip heats up, bending THERMOSTATS LIF T I N G MAGNETS
takes place opening the contacts and breaking the circuit.
(b) Invar Rod. A brass tube holds an Invar rod. As the
tube heats up and expands at a greater rate than the rod
the contacts open.

--
(c) A fluid-filled sealed capsule expands on heating to
MAGNET
open the contacts. - .::::.._
B. Magnetic Separation M ixed metallic scrap passes - - -==-. �
--::i
down the chute, the ferrous scrap is drawn further round
the revolving drum to drop in the second hopper. A con­ B IN VAR ROD
tinuous magnetic field holds the iron scrap to the drum.
C. Lifting Magnets A motor driven D C generator
supplies current to a large electro-magnet held by a crane �
(1����������
and is used to carry ferrous scrap . . The current passes
through a voltage regulator V and a contactor switch C. C F LUID CAPS ULE
The current may be reversed to speed up the drop of the
scrap after transport. As an example a 0.75 m dia
magnet working on 220 volts D C 1 3.9 amps will lift
5 500 kg.
� t::======::: M OTOR D C GENERATOR

D. In metallic arc welding, a low voltage-25 to 50

volts-and high amperage D C current is passed through
a welding electrode and the object to be welded. The arc B
temperature melts the electrode and enables welded o r M A G N ETIC S EPARATION WELDING GENERATOR
fused joints to b e made.
The diagram shows a circuit for a cross field series CIRCUIT
generator with a variable resistance.

CON T I N U OU S
MAGNETIC FIELD
l
:1/\l
I

-....=J
.�

F E R ROUS
NON-FERROUS
 429. Generation of Electricity Electricity is generated
ELECT RICAL LAYOUTS S CHEMAT I C
by three phase, 50 cycle alternators driven by steam tur­
bines. In the latest nuclear power stations, the high pres­
sure (up to 200 kg/cm 3) steam obtained from the cooling GENERATION AND DISTRIBU TION OF ELEC T RIC POWE R BY G RID SYSTEM
of the uranium reactors, is used to produce AC current
at 27 5 000 volts and this is fed into the Super Grid system.

8
Transformer sub-stations convert the current to 132 000 POWER
NUCLEAR
volts which is then fed into the National Grid system
being conveyed by the familiar overhead wire and pylon SUPER G RID
method. Electricity is also generated by steam turbines,
the steam for which is obtained from coal fired boilers. 275 000 V 3 "' 50 Hz
These run on much lower pressures 40 to 60 kg/cm 2 and
the current is 1 1 000 volts. This is stepped up by trans­
former sub-stations to 132 000 volts before being ac­
cepted by the National Grid.
Electricity is also generated by water power. The water T
is piped from highland reservoirs to water turbines GENERATOR G E N E RATOR
(Pelham wheels) which drive A C alternators supplying COAL POWER 1----1 ------1 WATE R POWER
current at I I 000 volts requiring transforming before
feeding into the grid. II OOO V II OOO V
The diagram shows details of the conversion of the
power supply from high pressure voltage carried by the 3 ,--.; so H z 3 rv 50 H z
grid, to voltages suitable for traction ( D C ) : three phase 132 0 V 3 "-" 50 H z
4 1 5 volts for industrial uses; domestic and commercial,
240 volts single phase.

T T : T RANSFORMER

33 0 V 3 "' 50 Hz

T RAC TION 4 1 5 V A C 50 H z
ELECTRlr3---1r1
RAILWAYS
T 1-----llNDUSTR!AL

660 V DC I I 000 V I I OOO V 2 40 V I rv SO Hz

3 1'V 50 Hz 3 .- SOHz

DOMESTICr------l T T

240 V 6 6 00 V 6 60 0 V 240 V
l tV SO Hz 3 "' SO Hz 3 rv 50 Hz ll'V SO Hz

251
 252

430. Domestic and Commercial Wiring The street main

D O M ES T I C E LEC T R I C A L L AYOUTS SCH E M AT I C
is usually 415 volts A C, 50 cycles, three phase, carried
on three conductors and a fourth neutral wire.

8
Two conductors-one phase wire and the neutral-are
brought to the house meter through a Supply Board DOM ESTIC W I RING
sealed fuse box, giving 240 volts A C, 50 cycles, single
phase current.
From the meter the lights circuits have a common
4 15 V 3 t"V 50 Hz STRE E T MAIN
main switch and then individual fuses and circuits con­
I
trolled by single pole switches on the ' live' wire. I
Ring Main Power points are arranged on a ring 2
circuit which consists of the phase · Jive' wire and the 3 L I V E/ BROWN !
neutral wire on a main ring switch and fuse, as shown in N NEUTRAL/B LU E '
the diagram. Power points are then tapped across the
two conductors at the required places without the need SEALED 240V I l"V' 50 Hz
for long individual wiring back to the fuse box. FUSE
Fused plugs are used wi,th each piece of apparatus to
avoid overloading.
A local earth wire forms the third wire in the ring cir­
cuit, and short circuits the current to earth should acci­
dental breakdown of the insulation take place, providing
METER
0 L I GH T ING C I RC U I T
an additional safeguard to the operator.

HOUSE
RING
SWITCH

E ARTH
GREEN/ -;-
.
/YELLOW
HOUSE RING MAIN
NEU T R AL
POWER POINTS
BLUE

LIVE
BROWN
FREEHAND SKETCHING
 254

4 3 1 . Freehand Sketching Freehand sketching without

instruments should be practised constantly. The simple FREEHAND S KETC H I N G
technique of outline and shading of a sketch based on an
isometric or oblique drawing of simple geometric solids
should be attempted first. The object is usually illumi­
nated from the left, and shadow shading effected by
hatching lines in two directions on plane surfaces. The
8
brightest effect is obtained by placing the deepest shadow
adjacent to the highest light. Analyse the object into its
basic geometric solid forms, and sketch boldly to obtain
the outline. Pencil shading will give a greater graduation
than the pen and ink diagrams shown; H B to 4B pencils
can be used.
432. Orthographic pencil sketches are often required,
and squared paper allowed. Plans and elevations, first or
third angle, sectional to show hidden detail, based on
BS 308, dimensioned, are shown in the diagrams.
433. More complicated freehand sketches are shown on
the following pages, sectionalised orthographic and iso­
metric diagrams are given, car components, lathe, drill­
ing machine, power saw, bench shears, and hand tools of
the workshop provide examples for further subjects.
Electrical and laboratory gear should also be sketched.
List of suggestions is appended.
434. Given an isometric drawing or a photograph of a
casting or engineering component, with one dimension
supplied, make orthographic drawings full size or to
scale on squared paper.
Two castings are shown and one dimension given for
each. Draw two elevations and a plan of each of the
castings, the first in Third Angle, and the second in First
Angle. Dimension the drawings.
 -

-�I
!_ l___ 1·- :___- -R;o;c-
___ I- --, ---! -
I
____: --1 _. ; ---'. - ' ·· - · -r-· - ··- : -�
, - ,-r
i ' '
E -·- ·· .. r ·�-· ; -·· ..
·
; r : ; '. -�- ; I
JL. : I � t-r--i--i--,:---,_r------------.i

-tfttRt;1sT-·
'

I l l ·-···-·-····· ····· · - : · - ... . :· - . ·r·· ·-r .

'

.
. : __ ; _' __ ·___ __! __ _' -· : _ _
I 1 •
r

;· r
• I

-�[� . --�+-+--.�
r·-· ,

� .... - �I
,__ ·-- ; - - · .. -- I
- -

t·- -:-· - · ! -- · - ··'· -; : ,

i =

--· sorN,----- --·- - ·'· ·-�R6 --i\NGt-e:.- ··-- -

...,-,-'..
! -- - ,,-�,--- .

� r:: N E.TE.D
!

25 5
 256

FREEHAND SKETCH ING

8 TME.RMOSTAT fOR C.A-R MOYIHG COi\...

vm..1/AMMe.Tf.R

CAR 'RADI A.TOR.

I S O ME T R ICS FOR FREE HAND SKETC H I NG

257
 258

APPENDIX I AI Metric Drawing Paper Sizes

A5 A3 Angular Measure
mm
AO 841 x I 1 89 Radians Degrees
Radians
SUGG ESTIONS FOR SKETCHING
Degrees
A l 594 x 841
Electrical Rear spring, axle mounting.
A2 420 x 594
Front wheel bearing. - A4 I 0.0175 0.01 0.57
Light switch. Circuits 424- A3 297 x 420
Brake drums.
430. A4 2 1 0 x 297
Brake master cylinder. 2 0.0349 0.02 I. I5
1 5 amp fused plug. A2 A5 148 x 2 10
Door catch. A6 1 05 X 1 48
1 5 amp switch and socket. Bonnet catch. 3 0.0524 0.03 1 .72
Pendant lamp holder. Brake, hand ratchet
Batten lamp holder. system. 4 0.0698 0.04 2.29
Fuse box, fuses. Wiper motor, rotary to
Mains flow meter. oscillating movement. 5 0.0873 0.05 2.86
Venner-type time switch. Flash indicator mechanism.
Dampers and action. 10 0.1745 0.10 5. 7 3
· Motors Brake light switch.
Synchronous clock type. Silencer. 20 0.3491 0.20 1 1 .46
Drawing Sheet Height (Min)
Battery type impulse Engineering parts.
motor. Parts of drilling machine. 30 0.5236 0.50 28.65
Metric Sizes Letters Figs Notes
Fractional HP series Parts of lathe.
wound commutator 60 1 .047 0.75 42.97
Al A2 A3 7 7 2.5
type. Lathe processes:
Squirrel cage motor. Centring in chuck, 90 1 .571 1 .00 57.30
A4 5 5 2.5
Centrifugal switch. facing,
Rheostat and switch. drilling and reaming, 120 2.094 1 .50 85.94
Contactor, solenoid tapping,
switch. tailstock die thread- 150 2.6 1 8 n/2 90.00
ing, screwcutting,
Car and Motor-cycle parts spot-facing, Metric Scales 180 3. 1 42 7f. 1 80.00
milling,
Battery section. 360 2n 2n 360.00
end-milling, Full Size I:1
Starter motor. keyway cutting,
Dynamo. taper and steady Enlarged 2: 1 5: l 10: I 20: l
Ammeter. turning.
Dynamo solenoid cut-out. Component parts 294-357, Reduced 1 :2 I: 5 I : IO l : 20
Ignition coil and circuit. bearings,
Contact breaker and seals,
condenser. couplings,
Valve gear, push rods, flanges,
rocker and valves brackets and clamps,
with springs. riveted joints, 281 -293, Angular Velocity
Carn shaft.
= 9.5493 rpm =
bolted joints,
Oil pump, filter system. nut locking devices. One Radian per second 57.296 ° per second
Pistons and crankshaft.
Clutch, gear box One degree per second = 0. 1 6667 rpm = 0.01 7453 rad s
schematic.
Universal coupling. One rpm = 6 ° per second = 0. 1 0472 rad s
Rear axle, schematic.
 I S O Metric Screwthreads ISO Metric Machine Screws

Coarse Fine Washers Length SLOTTED HEADS RECESSED

Shank of
Nominal Tapping Tapping Clearance 0/Dia Thickness Pitch Thread CSK R CSK CHEESE PAN CSK R CSK PAN
- - - -
0 Shank Pitch Size Pitch Size H 13 Size 2x0 0.2 0
15 ./ ./ ./
-
M 1.6 0.35
-
- - -
M 1.6 0.35 l .25 1.8 3.2 0.3
.,I .,I ./
-
M2 0.4 16 -

M 2 0.4 l.6 2.4 4.0 0.4

./ ./ .,I ./ ./ ./
-

-
./
-
M 2.5 0.45 18
M 2.5 0.45 2.05 2.8 5.0 0.5
./ ./ ./ ./ ./ ./ ./
-
M3 0.5 19
M 3 0. 5 2. 5 - 3.4 6.0 0.6
M4 0.7 22 ./ ./ ./ .,I ./ .,I .,I
M 4 0.7 3.3 - - 4.5 8.0 0.8
0.8 25 ./ ./ .,I ./ .,I .,I ./
-
M 5
M S 0.8 4.2 - 5.5 10.0 1.0
.,I ./ ./ ./ .,I .,I ./
-
M6 1.0 28
-
./ ./
M 6 1.0 5.0 6.6 12.0 1.2
M8 1.25 34 ./ ./ ./ ./ ./

./
M 8 1.25 6.8 1.0 6.75 9.0 16.0 1.6
M IO 1.5 40 ./ ./ ./ ./ ./ ./

- -
M 10 1. 5 8.5 1.25 8.75 1 1.0 20.0 2. 0
M 12 1.75 46 ./ ./ ./ ./ ./

- - - -
M 12 1.75 10.2 1.25 l 0.75 14.0 24.0 2.4
M 16 2.0 58 ./ ./ ./

- - - -
M 16 2.0 14.0 l .5 14.75 18.0 32.0 3.2
BS M 20 2.5 70 ./ ./ ./
l 8.75 22.0 40.0 4.0 4183
M 20 2.5 17.5 l.5
I S O Metric Hexagon Socket Screws
M 24 3.0 21.0 2.0 22.0 26.0 48.0 4.8
HEXAGON SOCKET A/F
M 30 3.5 26.5 2.0 28.0 33.0 60.0 6.0
Shank Length
Pitch of Thread CAP CSK Set Screw Button
M 36 4. 0 32.0 3. 0 33.0 39.0 72.0 7.0
M3 0.5 8 2.5 2.0 1.5 2.0
M 42 4. 5 37.5 3.0 39.0 45.0 84.0 8.0
M4 0.7 10 3.0 2.5 2.0 2.5
M 48 5.0 43.0 3. 0 43.0 52.0 96.0 9.0
M 5 0.8 12 4.0 3.0 2.5 3.0
B S 3643
M6 1.0 16 5.0 4.0 3.0 4.0

M8 1.25 20 6.0 5.0 4.0 5.0

M 10 1.5 25 8.0 6.0 5.0 6.0

259
M 12 l.75 30 10.0 8.0 6.0 8.0
BS 4168
 260

Basic S I Units Selection of SI Units

Quantity Name of Unit Unit Symbol Physical Quantity SI Unit Unit Symbol
length metre m area square metre m2
mass kilogramme kg volume cubic metre m3
time second s frequency cycle per second s- l
electric current ampere A density kilogramme per cubic metre kg m - 3
thermodynamic temperature degree Kelvin OK velocity metre per second m s- 1
luminous intensity candela cd angular velocity radian per second rad S- I
acceleration metre per second squared m s- 2
Physical Quantity SI Unit Unit Symbol angular acceleration radian per second squared rad s - 2
force newton N = kg m s- 2 pressure newton per square metre N m-2
work, energy, surface tension newton per metre N m- 1
quantity of heat joule J = Nm newto� second per metre
power watt W = J s- 1 dynamic viscosity square N s m-2
electric charge coulomb C = As kinematic viscosity metre squared per second m2 s- 1
electric potential volt Y = W A- 1 diffusion coefficient metre squared per second m2 S - I
electric capacitance farad F = A s y- i thermal conductivity watt per metre degree Kelvin W/(m ° K)
electric resistance ohm Q = Y A- 1 electric field strength volt per metre V m- 1
magnetic flux weber Wb = V s magnetic flux density weber per square metre Wb m - 2
inductance henry H = Y s A- 1 magnetic field strength ampere per metre A m-1
luminous flux lumen Im = cd sr luminance candela per square metre cd m - 2
illumination lux Ix = lm m - 2
Multiples and sub-multiples of SI units
Useful data
Prefix Symbol Factor
Acceleration metre per second squared m s- 2 Tera T 1 01 2
Velocity metre per second m s- 1 Giga G 1 09
Angular velocity radian per second rad s - 1 Mega M 1 06
Force-Newton unit mass x acceleration kg m s - 2
Momentum mass x velocity kg m s - 1
kilo k 1 03
Moment of Inertia mass x gyration radius hecto h 1 02
squared kg R 2 deca da 10
Angular acceleration torque + mqment of inertia unit as list I
Pressure, stress,-pascal force + unit: area d
deci 10- 1
Strain stress + Young's modulus
10 - 2
= J s = Nm s- 1
Extension strain x length. centi C

Power-watt work done+ unit time milli m 10 - 3

I kgf = 9 . 8 1 N
Moment of force Newton metre N x lever length micro u 10- 6
nano n 10 - 9
gravity = 9.8 1 m s - 2 pico p I0 - 1 2
f
L9ad on beam = total mass x acceleration due to gravity 5
Maximum shear force at each support equal to reaction at support femto 10 - 1
atto a 10 - 1 8
Bending moment at centre of simple beam= length x force/8
B S 3763
Selection of British Standards Pipe Flanges. Pressures up to 689.476 kNm - 2 Pipe Flanges. Pressures up to 3. I 03 MNm - 2
Flange Thickness Bolts Flange Bolts
BS I 0. Flanges for pipes, valves and Tube I Ext. Tube Ext.
fittings Bore I Dia Dia PCD CI Steel �o. Dia Bore Dia Dia PCD Thick No. Dia
2 1 . Pipe threads 20 4 16
I5 21.3 96 66 13 10 4 12 15 21.3 1 14 82
46. Keys and keyways
I 08. Graphical symbols (power and 20 26.9 1 02 72 13 10 4 12 20 26.9 114 82 20 4 16
lighting)
292. Ball and parallel roller bearings 25 3 3 .7 1 14 82 13 10 4 12 25 33.7 127 95 22 4 16
308. parts 1 .2.3. Engineering drawing 40 48.3 1 34 98 16 13 4 12 40 48.3 1 52 114 25 4 20
practice
309. Whiteheart malleable iron castings 50 60.3 153 1 14 20 15 4 16 50 60. 3 165 127 25 8 16
3 5 1 . Rubber transmission belting 80 88.9 203 165 32 8 16
80 88.9 185 1 46 20 15 4 16
436. Spur and helical gears
545. Bevel gears BS I O : A M D 585 : 1970 B S IO
64 1 . Small rivets
72 1 . Worm gearing
1 775. Steel tubes
1936. I S O metric screwthreads
2035. Cast iron flanged pipes and fittings
2470. Hexagonal socket keys
Steel Tubes Medium Pipe Threads
25 19. Terms for toothed gearing
3027. Dimensions of worm gear units Approx Min Threads
3643. 1 .2.3 I S O metric screwthre.ids Nominal Outside Wt Socket Nominal
3939. Graphical symbols for electrical Bore Dia Thickness kg m Length Bore Pitch Depth
diagrams 17 6 0.907 0. 5 8 1
6 10.2 2.0 0.4 1 0
4 1 83. Machine nuts and screws
4 1 86. Clearance holes for met ric bolts 8 1 3.5 2.35 0.654 25 8 1 . 337 0.856
4 1 90. I S O metric black bolts and nuts
4229. I .2. Sizes of non- and ferrous bars 10 1 7.2 2.35 0.858 26 10 1 . 337 0.856
439 1 . Sizes for metal wire sheet and strip 15 2 1 .3 2.65 1 .23 34 15 1 .8 1 4 1 . 1 62
4439. Screwed studs
4500. l S O limits and fits 20 26.9 2.65 1.59 36 20 1.814 1 . 162
4504. Flanges and bol ting for pipes and 25 2. 309 1 .479
25 33.7 3.25 2.46 43
valves
45 3 3 . Electric lighting fittings 32 42.4 3.25 3. 1 7 48 32 2.309 1 . 479
4620. Rivets
4622. Pipes and fittings grey/iron 40 48.3 3.25 3. 65 48 40 2. 309 1 .479
British Standards Handbook No. 1 8 50 60.3 3.65 5.17 56 50 2.309 1 .479
British Standard Sales B ranch, 2 Park
Strcet, London W l 80 88.9 4.05 8.64 71 80 2. 309 1 . 479
1 00 1 1 4.3 4.5 I 2.4 83 JOO 2.309 1 . 4 79
125 1 39.7 4.85 1 6. 7 92 125 2.309 1 .479
150 1 65.1 4.85 19.8 92 150 2. 309 1 . 479
B S 1 387. I S O Rec R50/R65 B S 2 1 . I S O Rec R7
26 1
 262

Hydraulic Couplings. Pressures up to 10.342 MNm - 2

H Y D R A, U L I C C O U P L I N G S

Flange Bolts
Tube Ext
Bore Dia L w Th. PCD No. Dia

'8 1 3.5 82 38 16 50 2 12

10 1 7.2 82 45 20 54 2 12

15 21.3 108 50 22 70 2 16

20 26.9 108 58 25 72 2 16 ...J

25 33. 7 1 34 70 28 88 2 20

40 48.3 1 53 88 35 108 2 22

50 60.3 1 78 1 02 45 1 27 2 24

See No. 3 3 1 .

� Th
·I
Hydraulic Couplings. Pressures up to 15.5 MNm - 2 PC D
Flange Bolts
Tube Ext
Bore Dia L Th. PCD No. Dia

8 1 3.5 70 20 50 4 12

10 1 7.2 70 20 54 .4 12

15 2 1 .3 70 22 58 4 12
20 26.9 76 22 64 4 16
25 33.7 95 25 82 4 20
40 48.3 1 14 28 1 02 4 20

50 60.3 1 27 28 1 14 4 20

i-
See No. 332. �
L Th
 Metric Spur Gears ISO Screwed Studs Sunk Keys Metric Spur Gears

Module Module Module Nominal Overall Plain Metal Nut Shaft Dia Key PCD
Module =
Length Length Length end end Section
111
-y-
l 4 16
3 1 5. 75 20.25 3. 75 4.5 12 8 IO 3x3 Pitch Circle Dia PCD = T x m
1.25 5 20
4 19 25 5 6 14 IO 12 4x4 PCD
Teeth
1.5 6 25
T
5 22.25 29.75 6.25 7.5 16 12 I7 5 x5
2 8 32 Addendum A = m
6 25.5 34.5 7.5 9 I8 I7 22 6x6
2.5 IO 40 Clearance C = 0. 157 111
8 32 44 IO 12 22 22 30 8x7
3 12 50 Dedendum D = 1 . 157 m
12.5 15 26 30 38 JQ X 8
=
IO 38.5 53.5
B S 436 Outside Dia OD (T + 2) m
I2 45 38 44 12 X 8
=
63 15 18 30
Centre Distance CD HT + t) m
16 58 82 20 24 38 44 50 14 X 9
Root Dia RD = (T - 2. 3 14) 111
20 7I IOI 25 30 46 50 58 1 6 x IO
B S 436. B S 4582. B S 2 5 19
25 87.25 144.75 3 1.25 37.5 56 58 65 1 8 x 11

30

B S 4439
1 03.5 148.5 37.5 45 66
I 65 75 20 X 12

I S O TC ! 6. B S 46
Designation of Metric
Screwthreads
Selected Hole and Shaft Fits Material
Head Shape
Clearance Transition Interference Bolt (or Nut)
Diameter
Hole Shaft Hole Shaft Hole Shaft Pitch
Length
Average HS f7 H 7 n6 H7 s6 Type of Fit
Fine H 7 g6 H7 k6 H7 n6 Bolts and Nuts Example:
Class of Fits-general use Steel
Hexagonal Head
I Clearance Transition Interference Bolt Nut Bolt
Examples Fit H 8-f 7 Fit H 7-k 6 Fit H 7-s 6
I M 10 x 1.5 x 30 8 g
Close 4h 5H
B S 4500: I A, B Medium 6g 6H Nut (or internal thread)
Free 8g 7H M 1 0 x 1.5 x 35 7 H

B S 3643 B S 3643 263

 Limits of Tolerance for Selected Holes (Upper and Lower derh·ations) 264
Unit = 0.001 mm

Nominal Sizes
C: C:

H ll
0
H9 ·">g
0
H7 H8
Up to
and EI
CJ
u
.,,..." 'f.,,
..."
ES
...
ES EI ES EI El ES
� "
C:

Over including + + + +
:Ir
.2
Q.
0
mm mm -
3 10 0 14 0 25 0 60 0

3 6 12 0 18 0 30 0 75 0
-�" I
I t "'
'&. � -� 'S. �
1/1
6 10 15 0 22 0 36 0 90 0 SHAFT f u

� I � �
X
z < 0

� � I
10 18 18 0 27 0 43 0 1 10 0

18 30 21 0 33 0 52 0 1 30 0

30 50 25 0 39 0 62 0 1 60 0 HOLE & SHAFT TOLERANCES,

L I M ITS & F ITS.
50 80 30 0 46 0 74 0 1 90 0 - DIAGRAMMATIC -
80 1 20 35 0 54 0 87 0 220 0

ES = Upper deviation. EI = Lower deviation

BS 4500 : A, B, Table 5, 6
Limits of Tolerance For Selected Shafts (Upper and Lower deviations)

Nominal Sizes ! C1 1 d\O I cg f7 g6 h6 k6 n6 p6 s6

- - - - -
es
- -
ei es
- -
es
-
es
-
Up to cs ci es ei es ei es ei es ei es ei ei ei ei
and + + + + + + + +
Over including

-
mm mm
3 60 1 20 20 60 14 39 6 16 2 8 0 6 6 0 10 4 12 6 20 14

3 6 70 145 30 78 20 50 10 22 4 12 0 8 9 I 16 8 20 12 27 19

6 JO 80 1 70 40 98 25 61 13 28 5 14 0 9 IO l 19 10 24 15 32 23

10 18 95 205 50 1 20 32 75 16 34 6 17 0 11 12 l 23 12 29 I8 39 28

30 I IO 240 65 1 49 40 20 41 7 20 0 13 15 2 28 15 35 22 48 35
18 92

30 40 1 20 280
80 1 80 50 1 12 25 50 9 25 0 16 18 2 33 17 42 26 59 43
40 50 1 30 290

65 1 40 330 100 220 60 1 34 30 60 10 29 0 19 21 2 39 20 51 32 72 53

50

Up -' via( - -
Selection of Materials. FERROUS HS Steel Co 60, Tu 25, C r 1 5 Hiduminium R R A l 94. 1 , Cu 2, Ni 470
' Stellite ' 1 . 3, M g 0.8, Si 0.7,
Material Composition /� Tensile Fe 1.0, Ti 0. 1
MNm - 2 Fe 1.0, Ti 0. 1

Cast Iron Fe 94, Si 2.2, Mn 0.8 Birmabright A l 96.75-93.25, 160-190

C3 Mg 3 --:6, Mn 0.25-
0.75
Cast Iron Fe 89.3, C3, Si 2.2, 470
M alleable Mn 1.0 Ni 3, Cr 1 .0, Selection o f Materials NO!'I-FERROUS Al. Magnesium Al 94. 1, Mg 1.5, 270-160
Mo 0.5 Alloy Si 0.7, Fe 0. 7
Material Composition Tensile
Steels. B S 970 C 0.07-0. 15, Mn 1.4, 360-420 MNm- 2 Al. Silicon alloy Al 87, Si 13 220-150
Low carbon S 0.3-0.6, P 0.06
Free machining Brass Std Cu 70, Zn 30 620-530 Die-casting Zn 92.95, Cu 3, 3 10
Alloys Al 4, M g 0.05
Steel Medium C 0.5-0.6, Mn 0.5- 700-850 Brass Turning Cu 58, Zn 40, Pb 2 330-540
Carbon 0.8 Babbitt Metal Sn 86, Cu 3.5, Sb I 0 --
Brass pressing Cu 60, Zn 40 330-530
Steel High Carbon C 0.9-1.2. Mn 0. 3 - 700- 1 000
0. 75 Cr I .0- I . 6 Brass Casting Cu 85, Zn 5, Sn 5, 300
Pb 5
Steel Nickel- C 0.3 5-0.45, M n 780-1500
chrome 0.45-0. 7, Ni 1.3- Bronze Phosphor Cu 85, P 0.9, Sn 14. 1 280-880
1.8. Cr 0. 9-1 .4,
Mo 0.2-0.35 Bronze Silicon Cu 96, Si 3, (Mg, 420- 1 100
Zn, Fe, A l , small
Steel Nickel- C 0.74- 0.84, Mn 780-1500 %)
chrome 0.2-0.6, Ni 1. 1 5-
1.65, Cr 19-20.5, Bronze Lead Cu 65, Pb 30, Sn 5 1 50
Si 1.75-2.25, S & P
0.03 Gun metal Cu 86, Sn 12, Zn 2 280-320

Stainless Steel C 0 . 1 , Cr 1 8, Ni 8 700 Beryllium Copper Cu 98, Be 2 940- 1230

Steel Chrome C 0.45-0.55, Mn 0.5, 800 Nickel Alloys Cu 26- 30, Ni 65-70, 820-1000
Vanadium 0.8, Cr o:8-1.2, Mone! Metal Mg 1.5, Si 0.25,
V 0. 15-0.25, Si 0.5 C 0.25, Fe 3

High Speed C 0.6-0.8, Mn O.], - Nimonic Alloys C 0. 1 , Ti 1.8-2.7, 370

Cutting Steels Si 0.2-0.3, Tu 6, No. 80 Cr 18-2 1 , Al 0.5-
Tungsten Cr 3-5. V 0.9-2.0, Mo 5 1.8, Si 1.0, Mn 1.0,
Fe 5.0, Co 2.0, Ni
H S Steel Cobalt c 0.8-0.85. Tu 20.0. - remainder
Co 12. JV! o 1 . 0. Cr
4.5, V 1 . 5, Mn & Si Aluminium Alloys Al 95, Mg 0.5, Mn 390-970
traces Duralumin 0.5, Cu 4

265
 266

APPENDIX II 6. The upper portion of a radial cam is given. The lower 14. The framework, loaded as shown, is supported at
portion is to be of such shape as to give UV lift from A to R L and R R. Determine the magnitude and direction
EXAMINATION TYPE QU ESTIONS B. Draw the cam twice the printed size, together with its of the two reactions. Obtain graphically the force and
performance graph base 120 mm representing one R P M types of stress in each member of the structure, tabulate
of the cam shaft. State the rotation. (See 9 1 .) the results. Draw out the given diagram twice the printed
size, the scale of the load line is discretionary. (See No.
7. Draw three teeth of an involute pinion meshing with 147.)
four straight rack teeth to suit the following data. Module
l. An arm A B, 75 mm effective length, turns on a vertical I O mm. Pinion teeth I 0. Pressure angle 20°. PC D I 00 mm. 1 5. The lifting crane is pivoted at B and anchored at A.
shaft through 90c and also slides up and down the shaft Addendum as module. Dedendum I . I 57 module. Clear­ Draw the diagram twice the printed size. Obtain graphi­
for a distance of 50 mm. Plot the locus of A, the free end, ance 0. 1 57 module. (See 1 1 3- 1 16.) cally the direction and magnitude of the reactions at A
through the following motion: I 0 ° , lift 10 mm ; I O ° , lift and B for the load given. Tabulate the force. and type of
8 mm ; 10°, lift 4 mm ; l 0°, lift 3 mm ; 1 0°, lift 2 mm ; 1 0°, 8. Draw one complete tooth of an involute gearwheel to stress in each member of the frame. (See No. 1 5 1 B.)
lift O ; I 0°, lift 2 mm; l 0°, lift 5 mm ; 1 0°, lift 1 6 mm. The satisfy the following data . P C D 240 mm. Module 12 mm.
return locus is an elliptical curve falling to zero with the Teeth 20. Pressure angle 20 ° . Addendum as module. 16. The wall framework is loaded as shown. Obtain
limits of the locus graph as half axes. (See 1 88.) Dedendum 1 . 1 57 x module. Clearance 0 . 1 5 7 x module. graphically the magnitude and direction of the reactions
(See Nos. 1 1 3-1 1 6.) A and B. Tabulate the force and stress in each member of
2. Draw the hyperbolas by constructing the hyper­ the framework. State the scale adopted for the load vector
boloid given the following data. Height I 00 mm ; genera­ 9. A slider crank mechanism is shown. Draw out twice diagram (See No. 1 49.)
ting circle 40 mm dia ; asymptotes at 60°. Find the foci, the printed size and using the instantaneous centre
draw any suitable tangent, and a centre of curvature from method, find the angle of velocity of B when the R P M of 17. Draw the laminae shown, height 50 mm, other sizes
the point of tangency. (See 67, 70, 72, 73, 76A.) the crank is 600. Plot the locus of P when B P equals are discretionary. Find the position of the centroid in the
BC. (See No. 1 75.) figure by link polygons. (See Nos. 1 59-1 6 1 . )
3. A channel of section as shown is soldered to a tube
40 mm in diameter and follows a L H helical downward I 0. A four bar chain is shown. Draw out twice the printed 1 8. Draw the laminae, which is symmetrical, proportion­
path for 240° in 1 00 mm height. Draw the plan and ele­ size. Using the instantaneous centre method, find the ately from the one dimension given. Find the position of
vation of the assembly, show hidden edges. (See Nos. velocity of D when the crank revolves at 500 R P M. the centroid in the figure by graphical integration. Find
100-1 04.) (See No. 177.) also the second movement of area about X X . The figure
may be turned to a more convenient position before
4. A spring is made from 20 mm dia metal, has a height 1 1 . In the Watt's quick return motion, plot the movement integration. (See Nos. 1 63-1 69.)
of 1 20 mm, lower overall diameter of 90 mm, top overall of P during one revolution of the crank AB. Draw out
diameter of 75 mm, and is composed or two R H turns. twice the printed size. (See No. 180.) 19. Two right cones stand on the H P and interpenetrate
Draw the elevation-no hidden lines arc required. (See as indicated in the given views. Draw the views twice the
102.) 1 2. The diagram shows a profile round which the wheel printed size, add a plan, project the lines of intersection
can roll without slipping from D to E. The slotted link is in the views. A plane A/A inclined to the H P cuts the
5. Design a radial plate cam to fulfil the following con­ pivoted at A, and can move about the fixed pin B. Plot cones, portion B is removed. Project an auxiliary view in
ditions. Shaft diameter I O mm; base circle diameter the locus of C as the wheel rolls from D position to the which the true shape of the cut surfaces may be seen.
25 mm; rotation clockwise; vertical knife follower I O mm stop E. (See Nos. 1 9 1-202.)
offset. Performance: 0° to 125 ° UV lift 20 mm; 125° to
260 ° S H M lift 20 mm; 260° to 295° rest; 295° to 360° 1 3. The beam shown carries a varied load. Draw twice 20. A ring of circular cross section is cut by two planes
U Ace and Rtd fall 40 mm. Draw also the performance the printed size, find graphically the resultant and the A/A and B/B, both inclined to the H P, portion C is re­
graph, base length 1 20 mm for one revolution of the equilibrant-magnitude and position. Draw also the shear moved. Draw the .diagram twice the printed size as a
camshaft. (See Nos. 90-93.) force and bending moment diagrams to a suitable scale. front elevation, project a plan in third angle. Project also
State the magnitude of the two reactions. (See No. 1 39.) the view seen from position D. (See Nos. 192-202.)
2 1 . A right cone and cylinder stand on the H P and inter­ 28. Two views of a folded metal plate are shown. Draw 35. M ake a freehand isometric sketch of a capstan lathe
penetrate. Draw the two views 1.5 times the printed size, l .5 times the printed size the two views. Find the true toolbox, assembled. Also make a detail sketch-ortho­
complete the lines of intersection. The solids are cut by shape of the metal before it was folded and also the di­ graphic-showing a section of the spring loaded locating
the inclined plane VT H and the upper portion removed. hedral angle between the two faces of the component. plunger and describe briefly its operation. (See Nos. 324
Project new lines of intersection to the views, project also (See No. 232A.) and 432.)
a view normal to VT on which the true shape of the cut
surfaces may be seen. (See Nos. I 9 1-202.) 29. The projections of two skew lines AB and C D are 36. Make a freehand sketch of a dovetail slide, showing
shown. Draw the diagram 1.5 the printed size. Find the the method of adjustment to obtain a close sliding move­
22. A solid of revolution is shown cut by an inclined length of the shortest distance perpendicular between the ment without lateral play. Add a further sketch showing
plane A/A portion B to be removed. Draw, as an eleva­ two lines. (See No. 227.) a detail of a spring loaded vernier collar attached to the
tion, twice the printed size, the diagram, project a plan operating screw control. If the screw had a pitch of 1 . 5
in third angle. Project an elevation to show the face at C 30. Two views of a solid are shown. Draw the diagrams mm, how would the collar be marked out t o be capable of
in correct relationship to the other views. Project an twice the printed size. Project a fresh plan below X, Y, advancing the slide 0.01 mm?
auxiliary view to show the true shape of the cut face, and from this new view project an auxiliary view as indi­
indicated by D. (See No. 258.) cated by arrow A. Imagine now that the object to be made 37. Sketch a pipe coupling to withstand heavy pressures
from sheet metal. Draw the complete development up to 1 0 MNm - 2 (ten meganewtons per square metre)
23. A hemisphere is pierced by a cylinder. Complete the enabling the object to be made from one piece by folding suitable for pipes of 25 mm internal diameter pipe. The
views, first angle, also a view indicated by arrow A. and soldering. No jointing flaps need be shown. (See diagrams should include a full size sectional orthographic
Show lines of intersection in all the views. Draw the pro­ Nos. 204-206, 275- 277.) view showing the seal used and explanatory notes of its
jections 1.5 times the printed size. Project a development function. (See Nos. 330-335.)
of the tube piece showing the true shape of the hole. 3 1 . A sheet metal hood is shown, rectangular at the base,
Twelve chordal positions may be taken. (See Nos. 250- elliptical at the top, the axes of which are AB and CD. 38. Sketch a shut down valve heavy duty type as above
254.) Draw the diagrams twice the printed size, draw the for 25 mm internal diameter pipe. The diagrams should
development by the triangulation method. (See Nos. 275- include an orthographic view showing details of the
24. The prism shown has a portion removed by an 277.) coupling, sealing, shutter, with explanatory notes on the
oblique plane. A circular hole is then drilled perpendicular function. (See No. 350.)
from the cut face through the block. Draw the block 1.5 32. A vertical elliptical tube and a parabolic reflector
times the printed size in first angle projection, two eleva­ intersect as shown. The minor axis is indicated, and the 39. Using the cast component in No. 434, make a free­
tions and plan, hidden lines required. Produce also an major axis is 1.5 times the minor axis. The point V is the hand sectional view showing the wooden pattern in place
auxiliary view showing the true shape of the face. Di­ vertex of the parabola, and F is the foci. Draw the dia­ in the cope and drag with core prints and runner provision.
mensions not shown are discretionary. (See Nos. 220- gram 1.5 times the printed size, treat this view as an Explain the need for core prints and cores for cored holes,
22 1 .) elevation, project a plan, third angle, show the line of the need for filleted corners, and the purpose of a runner.
intersection between the tube and the reflector. Project (See No. 434.)
25. An octagonal oblique pyramid stands on the H P and the development of the tube, jointing line at XX, sixteen
is cut by an oblique plane VT H. Draw 1.5 times the chordal points are allowed, show accurately the hole 40. A pressure airline in a garage has an A C motor driven
printed size the elevation and plan of the truncated where inter-section takes place with the reflector. (See pump feeding a pressure container with safety release
pyramid. (Sec Nos. 220-223.) Nos. 250- 257.) valve which also actuates an on/off switch to the pump
when the pressure falls below 700 kNm - 2 • Make a
26. An oblique cylinder stands on the H P and is cut by an 33. Using No. 39 J make an axonometric view of the diagram using B S 3939 symbols for the electrical circuit
oblique plane VT H. Draw 1 . 5 times the printed size, the component in the easiest position enabling circles to be including mains coupling cut-out switch and fusebox,
elevation and plan of the cut cylinder, the upper portion compass drawn where possible. (See No. 35.) warning light, plus the items mentioned above. Show
being removed. Project a development of the circum­ diagrammatically the mechanical details relative to the
ference wall of the cut cylinder. (See Nos. 220-223.) 34. Using No. 39 K, make an isometric projection of the function of the airline, and describe briefly how the
component. The position should be taken viewing from whole system would work. (See Nos. 424-430.)
27. Two views of a laminae are given. Draw the given the lower angle, (refer to 423 B Face Valve, position.
views 1.5 times the printed size and proceed to find the The nearest quarter of the component should also be re­ 4 1 . Draw an electrical circuit similar to the one used on
true shape of the laminae and its true angle to the H P. moved, as in 423 A. See also 26 for isometric scale.) an automobile which shows on a dial the amount of
(See No. 2 3 1 . ) fuel remaining in the petrol tank. Use B S 3939 symbols,
negative earth, battery fuse, dial float contact mechanism,
wiring. Describe briefly how the system works.
267
 268

0 0 0
42. Make a diagram, using BS 3939 symbols, of the
ignition system suitable for a four cylinder petrol engine.
Describe briefly how the system works, with especial
regard to the coil. State also the meaning of suppression
and how it is achieved in the case of this circuit. (See No.
3 7 1 , 424--429.)

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273
J
 Index
The numbers in the Index refer to sections in the text.
Acme thread, 288 Clutches, 309-3 1 1 Gib-head key, 296-300
Addendum, 1 12-123 Compound motions, 190A Glands, 325-328, 337, 364
Alloys, Appx. Compound shapes, isometric, 28-32 Glissette, I 05
Aluminium, Appx. Concurrent forces, 126, 127 Gunmetal, Appx.
Angled interpens, 243, 244, 266-270 Cone, isometric, 31, 32 Gusset, 285
Angled roof truss, 145-148 Connecting rods, 360-369
Approx. curve for involute, 22 Conventions, 374, 375 Helices, 100-104
Arc equal to line, 1 3 Co-planar forces, 125-128 Hinged frame, I 53
Areas, by integration, l 62, I 69 Cotters, 296-300 Hood, 238, 274, 275-277
Assembly solutions, 408-423 Couplings, Hooke's joint, 307
Auxiliary projections, 1 97-232 flanged, 302 Hydraulic flanges, 329-337
Axle, 31 6-328 flexible, 303, 304, 306 Hydraulic pumps, 352-359
Axonometric, 35, 423A muff, 301 Hyperbola, 66-67
rigid, 305 Hyperboloid, 73
universal, 307, 308 Hypocycloid, 78
Babbit metal, Appx. Curves in isometric, 29 Hypertrochoid, 85-87
Ball bearings, 3 1 6-320 Cycloid, 77-87
Beam moments, 1 29-135, 1 39, 140 I.C. engines, 366-373
Bearings, Dedendum, 1 12-123 Ignition, 371-373
ball, 3 1 6-320 Developments, 233-249 Inclined plane, 1 9 1 -203
footstep, 3 1 4, 328A-B Diametric projection, 40A Instantaneous centre, 174-176
halved, 3 1 3 Dihedral angle, 2 1 7 Interpenetration, 241-272
ring oiler, 3 1 5 Dimensioning, 374-3 8 1 Involute, 107-1 1 5
roller, 32 1-324 Duralumin, Appx. gears, 1 1 2- 1 22
simple, 3 1 2 Isometric assemblies, 4 1 5-423A
Bending moments, 1 30 Eccentricity, 42, 62, 66 Isometric drawing, 26-32, 40
Bevel gears, 1 1 8-1 2 1 Electric circuits, 424-430 Joints,
Bolts, 287-291 Electric symbols, 425 cottered, 296-300
Bow's Notation, 1 28-155 Ellipse, 29, 41-66, 74-76 expansion, 336, 33 7
Box framework, 1 4 1 Engine parts, 360-373 flanged, 329-337
British Standards, Appx. Enlargement, 23 Hooke's, 307
Bronze, Appx. Epicycloid, 78 pipe, 329
Butt joints, 283 Epitrochoid, 83, 84 riveted, 28 1-286
Equilibrant, 124-155 strap, 286
Cams, 88-99 Evolutes, 59, 60, 7 1 , 72, 79, 80 universal, 308
Canted truss, 145 Junction piece, 265, 273-280
Cantilever, 1 35, 149 First angle projection, 24
Carbon, Appx. Fits and tolerances, 374-3 8 1 Keys, 294
Car Brake System, 373A Followers, 89, 99 Keyways, 294
Castle nuts, 290 Frameworks, 1 36-155 Kinematics, 173-179
Centre of curvature, 57, 58, 79, 80 Freehand sketching, 431-434
Centroids, 1 56-172 Lap joints, 283, 286
Circle, isometric, 29 Gears, 1 12-123 Lead, 100-104
Circles touching, 6, 14-16 Geometric solids in isometric, 26-32 Lead, Appx.
Circular pitch, 1 1 3 Geometry, 1-23 Leaders, 374
Clearance, 1 1 2- 1 23 Gib and cotter, 296-300 Lettering, 374
275
 ......

276

INDEX (continued)

Levers, 129, 173 couplings, 329-337 Steel, Appx.

Limits, 375-38 1 threads, Appx. Str. line to arc, 9
Linkages, 173-190 Pistons, 360-376 Studs, 287
Link polygons, 125-155, 1 6 1 Pitch, 1 0 1 , 1 13 -123 Stuffing box, 325-328, 382
Load line, 129-155 circle,. 1 13 Sun shadows, 37
Loads, 1 29-155 of rivets, 283
Loci, 41- 1 1 1 of teeth, 1 1 2, 123 Tangency, 17-21
Locking devices, 290, 291 Proportionate division, I Tangent planes, 228-230
Logarithmic progression, 5 Projection, 24, 25 Teeth, 1 1 2-123
spiral, 1 1 0 Pumps, 352-359 Third angle projection, 25
Threads,

�
Machine assemblies, 408-423 Rack, 1 1 7 pipe, 329- 337
Machine drawing, 382-407 Radius of gyration, 1 56-172 screw, 287-289
Machine parts, 295-423 Ram, 382 Tooth wheels, 1 1 2-123
M agnitude of forces, 125-1 55. Ratio of eccentricity, 42, 62, 66 Train of gears, 123
Malleable C . I., Appx. Reactions, 129-134 Transformer pieces, 273-280
Ma·nganese, Appx. Rectangle, 2 Triangles, 3
Mechanisms, 173-190 Regular polygons, 1 1 , 12 Triangulation, 273- 280
Metrication, Appx. Relative velocities, 173-175 Trimetric projection, 40A
Module, 1 1 2-123 Resolution of forces, I 24-155 Tolerances, 375-381
Moinents, 124-135 Resultant, 124-155 Trochoids, 8 1 - 84
Motion, 91-93 Rivets, 281 True lengths, 1 9 1-203
joints, 283
Non-parallel forces, 144 Roof truss, 136-155 Uniform acceleration and retard, 93
Nuts, 287, 290 Roulette, 106 Uniform velocity, 91
locking device.,s, 290, 291 Universal joint, 307, 308
Screws, 289
Oblique, Screwthreads, 287- 289
cone, 278 Seals, 325-328, 337 Valves,
plane, 208-232 Second mom_ent of area, 156-172 ball, 341
projection, 33, 34 Sections, 197-202, 220-223 check, 340
pyramid, 273 Shap�d block, 27 disc, 345
sections, 220-223 Shear force, 130 flap, �42
Offset interpens, 245-:)48, 268- 270 S.H.M., 92 poppet, 338
Overhung beam, 1 34' Si'mpson's rule, 10 release, 339
Ordinates, 10, 26, 163, 169 Skew lines, 227 safety, 347, 348
Slider, 175, 176 stop, 349-351
Packing, 325-32.Jl, 337, 364 Solids on O.P., 21 8-223
Parabola, 61-75 Spheres, 228, 25 1, 252, 257, 259-263 Warren frameworks, 142
Parts, Spirals, 109-1 1 1 Whitemetal, Appx.
assembly, 382-407 Splines, 295 Wind loads, 143, 144, 146-148
engine, 360-373 Spur gears, I 12-123 Woodruff key, 294
Perspective, 36, 37 Square roots, 4 cutters, 294, Appx.
Pictorial projection, 26-37 Square thread, ·100- 104 Worm gears, 120, 122
Pipes, 329-337 Standards, 374-381, Appx. Wormwheel, 122

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