PRACTICAL GEOMETRY
AND GRAPHICS

WORKS BY DAVID ALLAN LOW

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LONDON, NEW YORK, BOMBAY, AND CALCUTTA.

PRACTICAL GEOMETRY
AND GRAPHICS

DAVID ALLAN LOW

(Whitworth Scholar), M.I.Mech.E.

Professor of Engineering, East London College

(university of london)

WITH OVER SOO ILLUSTnA^ld^'S AA^B' OVER

700 e!xerois^^*' >'> vi X\ : : .'.

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NEW YORK, BOMBAY, AND CALCUTTA

1912

s)^v

PREFACE

An endeavour has been made to provide in this work a fairly
complete course of instruction in practical geometry for technical
students. The field covered is a very wide one, but by adopting a
concise style and by endeavouring to make the illustrations "talk " the
work has been kept of reasonable size.

A special feature has been made of the illustrations. These are
very numerous, and they have been carefully planned so that even the
most complicated of them should be distinct and clearly show the
constructions used, notwithstanding that most of them are compara-
tively small. Pictorial projections have been freely used in dealing
with solid geometry, and these in many cases are just as useful as
actual models in illustrating the relative positions of points, lines, and
planes in space. Nevertheless teachers are recommended to provide
models for class use in teaching solid geometry, and the student should
also make small cardboard models for himself in studying many of the
problems, especially in the earlier stages of his work.

The student cannot hope to master the subject of practical
geometry unless he works on the drawing board a large number of
examples. Hence, another special feature of this work is the large
collection of exercises. These exercises have been prepared and
selected with great care, and it will be found that they not only
provide ample practice for the student but in many cases they amplify
the text. About ninety per cent, of the exercises are original and
the remainder have been selected from the examination papers of the
Board of Education. To economize space many of the diagrams for
the exercises have been placed on a squared ground. It should
scarcely be necessary to state that in reproducing these diagrams the
squared ground need not be drawn, but by counting the squares the

26083 L

vi PREFACE

various points and lines may be plotted from two axes at right angles
to one another.

Great credit is due to Mr. J. W. Barrett for the oare, intelligence,
and skill which he has displayed in making the finished drawings for
the illustrations from the author's pencil drawings and sketches.

D. A. L.

East London CoLLEaE (University of London),
September, 1912.

CONTENTS

CHAP. PAGE

I. Introductory 1

II. The Circle 16

III. CoKic Sections 30

IV. Tracing Paper Problems 54

V. Approximate Solutions to some Unsolved Problems . . 61

VI. Roulettes and Glissettes 66

VII. Vector Geometry 82

VIII. Graphic Statics . 91

IX. Plane Co-ordinate Geometry 128

X. Periodic Motion 148

XI. Projection 166

XII. Projections of Points and Lines 171

XIII. Projections op Simple Solids in Simple Positions . . . 181

XIV. Changing the Planes of Projection 191

XV. Planes other than the Co-ordinate Planes 198

XVI. Straight Line and Plane 211

XVII. Sections op Solids 241

XVIII. The Sphere, Cylinder, and Cone 249

XIX. Special Projections of Plane Figures and Solids . . . 272

XX. Horizontal Projection 279

XXI. Pictorial Projections 290

XXII. Perspective Projection 300

XXIII. Curved Surfaces and Tangent Planes 313

XXIV. Developments * 349

XXV. Helices and Screws 362

XXVI. Intersection op Surfaces , 888

XXVII. Projection op Shadows 411

XXVIII. Miscellaneous Problems in Solid Geometry 423

Appendix, Mathematical Tables 436

Index 443

CHAPTER I

INTRODUCTORY

1. Lines. — Euclid defines a line as " that which has length without
breadth." A line, therefore, in the strict mathematical sense has no
material existence, and the finest " line " that can be drawn on paper
is only a rough approximation to a mathematical line. A straight line
is defined by Euclid as " that which lies evenly between its extreme
points." An important property of straight lines is that two straight
lines cannot coincide at two points without coinciding altogether.
This property is made use of in the applications of a " straight-edge."

A line is also defined as the locus or path traced by a moving point.
If the direction of motion of the point is constant the locus is a straight
line.

In geometry when the term " line " is used it generally stands for
" straight line."

2. The Circle — Definitions. — A circle is a plane figure contained
by one line, which is called the circumference,
and is such, that all straight lines drawn
from a certain point within the figure to
the circumference are equal to one another :
and this point is called the centre of the
circle.

The meanings of the terms radius, dia-
meter, chord, arc, sector, and segment of a
circle are given in Fig. 1 .

The term ''circle" is very frequently
used when the " circumference " of the circle
is meant. Euclid, for example, states that
" one circle cannot cut another at more than two points."

" A tangent to a circle is a straight line which meets the
ference, but, being produced, does not cut it" (see also Art. 13

3. Angles. — " A plane angle is the inclination of two
lines to one another, which m^et together, but are not in the same
straight line."

" When a straight line standing on another straight line makes the
adjacent angles equal to one another, each of the angles is called a
right angle." If it is required to test whether the angle BAG of a

Fig. 1.

circum-

, p. 12).

straight

the

set-square, which is nominally a right angle, i^ really a

right

angle,

PRACTIOAL GEOMETRY

Fig. 2.

apply the set-square to a straight-edge DE as shown in Fig. 2, and
draw the line AB. Next turn the set-square over into the position
AB'C and draw the line AB'. If AB' coin-
cides with AB, then the angle BAG is equal
to the angle BAG', and therefore by the
definition each is a right angle.

An angle which is less than a right angle
is called an acute angle, and an angle which
is greater than a right angle and less than
two right angles is called an obtuse angle.

If two angles together make up a right
angle they are said to be complementary, and one is called the comple-
ment of the other.

If two angles together make up two right angles they are said to
be supplementary, and one is called the supplement of the other.

In measuring angles in practical geometry the unit angle is
generally the l-90th part of a right angle and is called a degree. The
l-60th part of a degree is called a minute, and the l-60th part of a minute
is called a second. 47° 35' 23" is to be read 47 degrees, 35 minutes,
23 seconds. The single and double accents which denote minutes
and seconds respectively are also used to denote feet and inches
respectively, but this double use of these symbols seldom causes any
collusion.

Angles are also measured in circular measure. AOB (Fig. 3) is an
angle, and AB is an arc of a circle whose centre is O.
The circular measure of the angle AOB is the ratio

= 6. For two right angles the length of the

3-1416, and the circular measure
When

arc

RADIUS A

Fig. 3.

radius

arc is ttt, where tt

of two right angles is therefore equal to tt.

the arc is equal to the radius the angle is the unit

angle in circular measure and is called a radian.

If « is the number of degrees in an angle whose circular measure

n 6 .
is $ radians, then y^Q = -, since each of these is the fraction which

the angle is of two right angles.

180<9
Hence n = = 57-2958^, and 6 =

180

= 0-01745W.

The instrument most commonly
used for measuring angles in degrees
is the protractor, portions of two forms
of which are shown in Fig. 4. The
angle AOB on the paper beneath the
protractor is seen to be 55°. The
protractor may be made of boxwood,
ivory, celluloid, or cardboard.

Fig. 4.

INTRODUCTORY

SCALE OF CHORDS

Fig. 5.

A scale of chords is a convenient instrument for measuring angles.
The way in which the scale is constructed is shown in Fig. 5. ADF is
a right angle, and the arc ACF is de-
scribed from the centre D. The arc
ACF is shown divided into 18 equal
parts, each part therefore subtends at
D an angle of 5°. On the scale of
chords AB, lengths are marked off
from A equal to the chords of the
arcs of the different angles. For ex-
ample, AE is the chord of the arc
which subtends an angle ADE of 40"^
at the centre D, and this length is
marked off on AB by describing the
arc EH from the centre A. It will
be found that the length of the chord AC for 60° is equal to the radius
of the arc ACF Hence in using the scale of chords the radius of the
arc is made equal to the distance between the points marked and 60.

The scale shown in Fig. 5 only shows the chords for angles which
are multiples of 5° up to 90°, but by making the scale large enough,
single degrees, and even fractions of a degree, may be shown. It may
also be extended beyond 90°. But when an angle is greater than 90°
its supplement, which will be less than 90°, should first be determined.
It should be noticed that the divisions on the scale of chords are
unequal.

By using the ordinary 45° and 60° set-squares separately and
together angles which are multiples of 15° may be constructed with
i^reat accuracy. Angles may also be measured and constructed by
making linear measurements taken from tables of the trigonometrical
ratios of angles.

4. Trigonometrical Ratios of Angles. — The angle A in the
diagrams, Fig. 6, is contained by the lines OP and OX. The line OP

is supposed to start from the position OX and sweep out the angle A
by moving round O as centre in the direction opposite to that of the
hands of a watch. An angle so described is called a positive angle.
If the motion takes place in the opposite direction the angle described
is called a negative angle.

PN is perpendicular to OX. PN is positive (-f ) when it is above

4 PRACTICAL GEOMETRY

OX, and negative (~) when it is below OX. ON is positive when it
is to the right of O, and negative when it is to the left of O. OP is
always positive.

The most important trigonometrical ratios of an angle are, the smcj
the cosine^ the tangent^ and the cotangent. The sine of the angle A is

the ratio of PN to OP. Thus, sine A = ^. If PN is half of OP,

then sine A = i If PN = 2, and OP = 3, then, sine A = |. '

ON A I'N

The other important ratios are, cosine A = ^yp , tangent A = qv^>

. * ON
and cotangent A = pv^.

In addition to the above the following ratios are also used :

cosecant A = p^, secant A = -^^^ co versed sine A = 1 — sine A, and

versed sine A = 1 — cosine A.

The names of these ratios are abbreviated as follows : sin, cos, tan,
cot, cosec, sec, covers, and vers.

The following results are obvious from the definitions : —

cot A = r A , tan A = -— — . , cosec A = , — - , sec A = .

tan A cot A sm A cos A

The complement of A is 90^ — A, and the supplement of A is
180^ - A.

The following relations between the trigonometrical ratios of an
angle and those of its complement and supplement are easily proved :—
sin (90^ - A) = cos A. cos (90° - A) = sin A.

tan (90° - A) = cot A. cot (90° - A) = tan A.

sin (180° - A) = sin A. cos (180° - A) = - cos A.

■ tan (180° - A) = - tan A. cot (180° - A) = - cot A.

It may be noticed that the sine and cosine of an angle can never
be greater than 1, but the tangent and cotangent can have any
magnitudes.

The problem of constructing an angle which has a given trigono-
metrical ratio evidently resolves itself into the simple one of con-
structing the right-angled triangle OPN (Fig. 6) having given the ratio
of one side to another. But in general there are two angles less than
360° which have a given trigonometrical ratio.

5. Accuracy in Drawing. — A good serviceable line by an
ordinary draughtsman is about 0*005 inch wide, but an expert
draughtsman can work with a line of half that width. A good
exercise for the student is to try how many separate parallel lines he
can draw between two parallel lines which are 02 inch apart, as shown
in Fig. 7, which is twice full size for the sake of clearness. When the
lines are drawn as close as possible together, but distinctly separate, it
will be found that the distance between them is about equal to the
width of the lines. Hence, if the lines and spaces are about 005 inch
wide, there will be about 20 lines in a width of 02 inch.

The probable error in measuring the distance between two points

INTRODUCTORY

on a straight line may be as small as the thickness of the division lines
on the measuring scale, or say O'OOo inch.

The error e in the position of the point of intersection of two
straight lines OA
and OB (Fig. 8),
when the error in
the lateral posi-
tion of one of them

is iT, IS ~. — ;;, where
sin

6 is the angle between

when 6 = V, e = 0-057

= 10^

Using a radius of 10 inches, an angle may be constructed by
marking off the chord from a table of chords with a minimum probable
error of from 1-lOth to I- loth of a degree.

Great accuracy in drawing can only be obtained by exercising
great care, and by having the best instruments kept in the best possible
condition.

Fig. 8.

the lines. If a; = 0*005 inch, e = 0-286 inch
inch when = 5°, and e = 0029 inch when

inches, an angle

Exercises Ta

To test accuracy in drawing ayid measuring

1. On a straight line about 7 inches long mark a point near one end. ]\Iark
off the following lengths on this line : OA = 1-20 inches, AB = 0-85 inch,
BC = 1-35 inches, CD = 1-55 inches, and DE = 0*95 inch, all measured in the
same direction along the line. Measure the length OE, which should be
5-90 inches.

2. Take twenty strips of paper about 2J inches long, and number them 1 to 20.
On an edge of No. 1 mark off a length of 2 inches. On an edge of No. 2 mark off
the length from No. 1. On an edge from No. 3 mark off the length from No. 2,
and so on to No. 20. Then compare
the length on No. 20 with the length
on No. 1.

3. Draw a straight line AB (Fig. 9)
and mark two points A and B on it
2 inches apart. Draw BC, making the
angle ABC = 60°. Make BC 4 inches
long. Draw CD, making the angle
BCD = 30°. Make CD = CA. Draw
DE parallel to AB. Make DE 4 inches
long. Draw EF, making the angle
DEF = 30°. Make EF = CA. Let
DE cut BC at 0. A circle with centre
and radius OB should pass through
the points D, F, C, E, and A. Also the
chords of the arcs BD, DF, FC, CE,
and EA should each be 2 inches
long.

_ 4. On a straight line AB (Fig. 10)
2 inches long construct an equilateral

triangle ABC, using the 60° set-square. Produce the sides of this triangle as
shown. Through the angular points of the triangle ABC draw parallels to the
opposite sides forming the triangle DEF. Using the 30° and 90° angles of the
set-square, draw lines through D, E, and F, to form the hexagon DHEKFL as

Fig. 9.

PRACTICAL GEOMETRY

shown. Join FA, EB, and DC. If these lines are produced, they should pass
through the points H, L, and K, respectively. Complete the figure by drawing
the circles as shown.

5. Draw two lines OA, OA, each
10 inches long, and containing an
angle AOA of 88°. Measure accu-
rately and state the distance apart
of A, A. What should be the exact
distance AA ? State any special
precautions you may have taken to
ensure accuracy in your construc-
tion. [B.E.]

6. Take from the tables the
chord and tangent of 22°. Construct
an angle of 22° by using the value
of the chord, and a second angle of
22° by using the tangent. From
each of these angles determine the
sine of 22°, and compare the mean
of the two values with the true
value of sine 22°. [b.e.]

7. By aid of your protractor,
and without using the tables, find
the value of

sin 23° + tan 23° + cos 23°.
Now take out the true values of the sine, tangent, and cosine from the tables,
add them, and calculate the percentage error in your first answer. [b.e.]

8. In working this question employ a decimal scale of ^ inch to 1 unit.
Draw a circular arc, radius 10 units, centre 0. Mark a chord AB of this

arc 347 units long, and draw the radii OA, OB. Measure the angle AOB in
degrees.

From B draw a perpendicular BM on OA, and at A draw a tangent to
meet OB produced in N. Measure carefully B]\I, AN and the arc AB (on the
above unit scale) and calculate the sine and tangent of the angle, and the angle in
radians.

Give the correct answers for the degrees, sine, tangent, and radians, the
numbers being taken directly from
the tables. [b.e.]

9. On the base AB (Fig. 11)
construct the polygon ABCDEF to
the dimensions given. Then mea-
sure the side AF and the angles 6
and <f).

[The answers obtained by cal-
culation are : AF -- 2-65 inches,
e = 84° 56', and (p = 72° 4'.]

Straight
through

6. Some Terms used
Modern Geometry. —
lines which pass
a point are said to
be concurrent. Points which
lie on a straight line are said

to be collinear. Points which lie on the circumference of a circle are
said to be concycUc. A straight line drawn across a system of lines
is called a transversal.

The three straight lines which join the angular points of a triangle
and the middle points of the opposite sides are called the medians of

INTRODUCTORY 7

the triangle, and the point at which the medians intersect is called the
centroid of the triangle.

The three perpendiculars drawn from the angular points of a
triangle to the opposite sides meet at a point called the ortliocentre of
the triangle, and the triangle formed by joining the feet of the per-
pendiculars is called the pedal triangle, or orthocentric triangle of the
original triangle.

If a straight line he be drawn to cut the sides AC and AB of a
triangle ABC (Fig. 12) at & and c respectively, so
that the angle Acb is equal to the angle ACB,
then he with reference to AB and AC is said to
be antlparallel to BC. An antiparallel he is
parallel to the tangent AD to the circumscribing
circle at A. The middle points of the antiparallels
to one side of a triangle lie on a straight line
which passes through the opposite angular point of
the triangle, and this line is called a symmedian ^^^ -j^2

of the triangle. The three symmedians of a
triangle intersect at a point called the symmedian point of the triangle.

7. Equality of Two Triangles. — A triangle has six parts, three
sides and three angles, and, except in two cases, a triangle is completely
determined when three of its six parts are known. Another way of
stating the above is that if there be two triangles having three parts in
the one equal respectively to the corresponding parts in the other, then,
except in two cases, the two triangles are equal in every respect. The
different cases are as follows : —

Two triangles are equal in every respect when —

(1) The three sides of the one are equal to the three sides of the
other, each to each.

(2) Two sides and the included angle in one are equal to two sides
and the included angle in the other, each to each.

(3) Two angles and the side adjacent to them in one are equal to
two angles and the side adjacent to them in the other, each to each.

(4) Two angles and the side opposite to one of them in one are
equal to two angles and the side opposite to one of them in the other,
each to each, the equal sides being opposite to equal angles.

These four cases are illustrated in Fig. 13 ; the corresponding
equal parts in the two triangles have similar marks on them.

/X'/x ^t\ /^'/x i:i'/\

Fig. 13.

Of the two exceptional cases mentioned above, one is that in which
the three angles of one triangle are equal to the three angles of the
other. In this case, the two triangles have the same shape, but
not necessarily the same size. The other case is that in which two
sides and an angle opposite to one of them in one triangle are equal to

PRACTICAL GEOMETRY

two sides and an angle opposite to one of them in the other, each to
each, the equal angles being opposite to equal sides. In this case the
triangles may be equal, but they are not necessarily
so, as is shown in Fig. 14, where ABCj is one
triangle and ABCo the other.

8. To bisect the angle between two
straight lines. — First, let the two straight lines
OA and OB intersect at a point O which is accessi-
ble (Fig. 15). With O as centre and any radius
describe arcs to cut OA at C and OB at D. With centres C and D
describe intersecting arcs of equal radii. The straight line OE, which
joins with the point of intersection of these arcs, will bisect the
angle AOB.

An important property of the bisector of an angle is that the
perpendiculars, such as FH and FK, drawn from any point on it to the
lines containing the angle are of equal length.

Fig. 16.

Next, let the given lines PQ and RS (Fig. 16) meet at a point which
is inaccessible. Draw any straight line MN to cut PQ at M and RS
at N. Bisect the angles PMN and RNM by straight lines which
intersect at L. Through M and N draw perpendiculars to LM and
LN respectively, and let them meet at T. The straight line LT will
if produced meet PQ and RS produced at the same point, and will
bisect the angle between PQ and RS. It is easy to see that MT and
NT bisect the angles NMQ and MNS respectively, which suggests
another way of drawing these lines.

9. To find a point on a given straight
line such that the sum of its distances
from two given points shall be the least
possible. — Let LN (Fig. 17) be the given
straight line and A and B the two given points.
The given points are on the same side of LN.
Draw ACD at right angles to LN, cutting LN
at C Make CD equal to AC. Join DB,
cutting LN at E. E is the point required.
JoiaAE. P'°"-

Comparing the triangles ACE and DCE, it is obvious that AE is
equal to DE. Hence AE + BE = DB. If any other point F in LN

INTRODUCTORY

be taken and F be joined to A, B, and D, it is evident that AF is
equal to DF, and AF + BF = DF + BF. But DF+ BF is greater
than DB, therefore AF + BF is greater than AE + BE.

It is evident that the angles AEL and BEN are equal, and the
problem may be stated in another way, namely, to find a point E in a
given line LN such that the lines joining E to two given points A and
B shall be equally inclined to LN.

If LN represents a mirror, and a ray of light from A is reflected
from the mirror and passes through B, then the incident ray will strike
the mirror at the point E found by the above construction.

If the points A and B are on opposite sides of LN it is obvious
that E will be the point where the straight hne AB cuts LN. In this
case, however, there is no corresponding case of reflection from a
mirror.

Referring to Fig. 17, and considering LN to be a mirror, D is the
image of the point A in the mirror.

10. Similar Rectilineal Figures.
— Two rectilineal figures are said to be
similar when they have their several
angles equal, each to each, and the
sides about their equal angles propor-
tionals. For example, the figure ABCD
(Fig. 18) is similar to the figure ahcd
when the angles A, B, C, and D of the
one are equal to the angles a, 6, c, and

..1^1 , 1 ., . DA AB BC 1 CD

(i, respectively, of the other, and the ratios -pp., y.^, ^y and ^^ are

equal to the ratios — ? , f, and ^^ respectively. Stated in another
a/>' he cd da

way, the similar figures have the same sha^pe but have different

dimensions.

When the two figures are triangles, they are similar if the angles
of the one are equal to the angles of the other, each to each. The
equality of the ratios of the sides about the equal angles follows as a
consequence of the equality of the angles.

It is important to remember that the ratio of the areas of two
similar figures is equal to the ratio of the squares on their correspond •
ing sides. For example, referring to Fig. 18,
area of ABCD : area of ahcd : : AB^ : ah^

11. Through a given point to draw a
straight line which shall be concurrent
with two given straight lines when the
point of concurrence is inaccessible. —
AB and CD (Fig. 19) are the given lines, and
P is the given point. Draw a triangle PEF
having one angular point at P and the other
angular points one on AB and one on CD.

Draw HK parallel to EF, meeting AB at H and CD at K. Througji
H and K draw parallels to EP and FP respectively, and let them

10 PRACTICAL GEOMETRY

meet at Q. A straight line through P and Q will, if produced, meet
AB and CD produced at the same point.

12. Graphic Arithmetic. — The solution of ordinary arithmetical
problems by geometrical constructions is of some interest when con-
sidered as a section of practical geometry, but graphic arithmetic, when
regarded as a means to an end, is in general a poor substitute for the
ordinary method by calculation. The subject of graphic arithmetic
will therefore be treated somewhat briefly here.

Bepresentation of niimhers by lines. If the line AB, Fig. 20 (a), be
taken to represent the number owe, then a line CD n times the length
of AB will represent the number n. Again, if AB represents the
number owe, the number represented by the line EF will be the number
of times that EF contains AB. In the above examples AB is called
the unit.

Ordinary drawing scales may be used in marking off or measuring
lengths which represent numbers. Scales which are decimally divided
are the most convenient for this work.

Addition and sidHraction. To add a series of numbers together.
Draw a straight line OX, Fig. 20 (b), of indefinite length. Fix upon a
unit, that is, decide what length shall represent the number owe.
Mark a definite point O on OX. Make OA = the first number,
AB = the second number, BC = the third number, and so on. From
O to the last point determined in this way will be the answer
required.

t W^ X — S 1

(aH £ i (J

k — ; i' n — s r—^

(d)—^ i i X g 6 X

Fig. 20.

From one number to subtract another. Make OA, Fig. 20 (c),
measured to the right of O, equal to the first number. Make AB,
measured to the left of A, equal to the second number which is to be
subtracted from the first. From O to B is the answer required. If
OB is to the right of the answer is positive ( + ), and if OB is to the
left of O, the answer is negative ( — )•

To find the value of such an expression as a-\-b — c + d — e.,
where a, ?>, c, d, and e represent numbers, make (Fig. 20 (d)) OA = a,
AB = 5, BC = c, CD = d, and DE = e, then 0Y. = a -^b - c + d - e.
Note that in adding the lengths are measured to the right, while in
subtracting they are measured to the left.

Proportion. In Figs. 21 and 22, OX and OY are two straight
lines making any convenient angle with one another. AB and CD
are parallel lines meeting OX at A and C, and OY at B and D. The
triangles AOB and COD are similar, and therefore OA : OB : : OC : OD,

INTRODUCTORY H

and if any three of the terms of this proportion are known the fi£?ure
can be drawn and the fourth term found.

Arithmetical, Geometrical, and Harmonical means of two numbers.
Let M and N be two numbers. On a straight line AB (Fig. 23)
make AO = M and OB = N. Bisect AB at C. With centre C and
radius CA describe the semicircle ADB. Draw OD at right angles to
AB. Join CD. Draw OE at right angles to CD. Then

CB = arithmetical mean of M and N = i(M + N),

OD = geometrical mean of M and N = VM x N,

2M X N
DE = harmonic mean of M and N = ^>— r ^t .

M + N

OD, the geometrical mean of M and N, is also called a mean pro-
portional between M and N. If OD = \/M~x^, then OD" = M x N,
and M : OD : : OD : N.

Multiplication. In Eigs. 21 and 22, since OA : OB : : OC : OD, it
follows that OA X OD = OB X OC. Make OB equal to the unit, then
OA X OD = 1 X OC = OC. Hence the construction for finding the
product of two numbers. Make OA = one of the numbers, OD = the
other number, and OB = the unit. Join AB, and draw DC parallel
to BA, then OC is the answer.

If the given numbers are large, it is better to first divide them by-
some power of 10 so as to avoid using a very small unit. For example,
let it be required to find 485 X 363. Taking the unit as 1 inch find
the product x of 4-85 and 3-63, then 485 x 363 = 4-85 x 3-63 x 10,000
= 10,000a;. After the value of x has been found, 10,000a; is found by
simply changing the position of the decimal point.

Division. In Figs. 21 and 22, since OA : OB : : OC : OD, it follows

that QT^= QTj. Make OB = the unit, then oD~ ^ = ^-^- Hence

the construction for finding the quotient of one number divided by
another. Make OC = the dividend, OD = the divisor, and OB = the
unit. Join CD, and draw BA parallel to DC, then OA is the answer.

Square r oot. In Fig. 23, OD^ = OA X OB, therefore OD =
VOA X OB. Hence if OB is made equal to the unit, OD = VOA.

If the given number is a large one it is better to make OB = «

times the unit, and OA = - of the given number, where w is a con-

n
venient whole number. For example, to find the square root of 48,

PRACTICAL GEOMETRY

etc., may be obtained by

make O B = 10 times the unit and OA = 4*8 times tiie unit, then
0D=: v/4-8 X I0= Vis,

If a be the hypotenuse of a right-angled triangle (Fig. 24)
which h and c are the sides, then a- = W + c^ therefore a = V?>M-
also c^ = d- - 6^ therefore c = ^/ai' — lr.

Involution. The values of a^ d\ «^
multiplication as already explained. For
a? =z a X a = h, and a" = a'- x a = ^ X a.
The following construction is, however,
sometimes more convenient. Draw two
axes XOXi and YOY^ (Fig. 25) at right
angles to one another. Make OA = the
unit, and OAj = «, the given number. Join
A^A and draw AjAo at right angles toAjA.
Then OA2 = (T.

Draw A2A0 at right angles to A^A,.
Then OA3 = a'.

Draw A3 A 4 at right angles to AgA^.
Then OA4 = a^, and so on.

If AA'i be drawn at right angles to
AAj, and A'^A'a be drawn at right angles

to AA\, and so on, then 0A\ = - , OA'o = .;

OA'3 = "3

and so on.

13. Definitions relating to CuvweB.— Tangent and Normal.
If a straight line PQ (Fig. 26) cuts a curve at two points P and Q.
and if the straight line be turned about the point
P so as to cause the point Q to approach nearer
and nearer to P, then the ultimate position of the
straight line is the tangent to the curve at P.

The normal to the curve at P is a straight line
through P at right angles to the tangent to the
curve at P.

A good practical method of drawing a tangent
to a curve from an external point T (Fig. 26) may
be given here. Place a straight-edge on the paper and adjust it until
the edge passes through T and touches the curve, then draw the tangent.
This method would not be recognized by the mathematician, but as a
practical method it is as good as any p

other, and much simpler. The exact
point of contact must be obtained by
some other construction depending on
the nature of the curve.

Circle of Curvature. If a circle be
drawn through three points Q, P, and
R in a curve APB (Fig. 27), and if the ^^«' ^'^^

points Q and R be moved so as to approach nearer and nearer to P,
then in the limit the circle becomes the circle of curvature of the curve
APB at P. The centre of the circle of curvature will evidently lie on

INTRODUCTORY

curve.

Fig. 28.

and a curve is an

the normal to the curve at P. The centre of the circle of curvature is
called the centre of curvature, and the radius of the circle of curvature
is called the radius of curvature.

Evolute and Involute. The locus of the centre of curvature of a
curve is called the evolute of the
A curve has only one evolute.

The locus of a point on a straight line
which rolls Without sliding on a curve is
called an involute of the curve. A curve
has any number of involutes.

In Fig. 28, ABC is a curve and abc
is its evolute. Again, abc is a curve and
ABC is an involute of ahc. The curves
PQR and STU are also involutes of
abc.

A curve is the evolute of each of its involutes
involute of its evolute.

The normals to a curve are tangents to its evolute, and the tangents
to a curve are normals to its involutes.

Envelope. If a curve moves in a
definite manner the curve which it
always touches is the envelope of the
moving curve. Fig. 29 shows the en-
velope of a circle which moves so that
its centre traces the curve ABC.

The evolute of a curve is also the
envelope of the normal to the curve.

Cusp. If two branches BA and
CA of a curve (Fig. 30) terminate at a point A on a common tangent
AD, the terminating point A is called a
cusp. There are two kinds of cusps ; in
the first kind the branches of the curve are
on opposite sides of the common tangent,
and in the second kind the two branches of
the curve are on the same side of the com-
mon tangent.

Node or Multiple Point. A point
branches of a curve pass is called a
node or multiple point. The node
shown at (a), Fig. 31, is a double
point, and the node shown at (6) is
a triple point. At the node (a) two
tangents can be drawn to the curve.
At the node {b) three tangents can
be drawn to the curve.

Point of Inflexion. If the tangent "^
to a curve at a point also cuts or crosses the curve at that point (Fig.
32), then thtf point is appoint of inflexion.

Asymptotes. When a curve approaches nearer and nearer to a

through

which two or more

(I).

PRACTICAL GEOMETRY

Fig. 32.

straight line, but never actually meets the line however far the line and
curve be continued, the line and curve are said to
be asympioticj and the line is called an asymptote
to the curve.

Curves which approach nearer and nearer
to one another but never actually meet however
far they are continued are said to be asymptotic, and each curve is an
asymptote to the other.

Similar Curves. Two curves, PQ and PjQj (Fig. 33), are said to be
similar when any radius vectors (or
radii) OP, OQ in the one, drawn from
a fixed point O, bear the same ratio to
one another that radius vectors O^Pj,
OiQi drawn from a fixed point O^, and
including the same angle 0, bear to
one another.

If, in addition to fulfilling the
above conditions, OP and OQ are
parallel to O^P^ and OjQ^ respectively,
then the curves are said to be similar, and similarly situated.

The fixed points O and O^ are called centres of similarity, and when
these centres coincide the point is called a centre of similitude of the two
curves.

All conies having the same eccentricity are similar curves.

Fig. 33.

Draw a straight line

Exercises lb

1. Draw Fig. 34 full size to the dimensions given
bisecting the angle between AB and CD,
and through the points P and Q draw
straight lines concurrent with AB and CD
without using the point of concurrence.

2. From the points P and Q (Fig. 34)
draw two straight lines to meet on the line
passing through A and the middle point of
CD and be equally inclined to that line.

3. If a line 3^ inches long represents
26, what does a line 1{^ inches long repre-
sent, and what is the unit ? If a line If^
inches long represents If, what is the unit ?

4. Work the following exercises, taking
for the unit in each case a line ^ inch PiG. 34.
long : -

(a) 1^ + 1-^ + 5-11+21. (c) 2-4 + 0-8 - 1-9 - 2 + 4-2 - 1-6.
(6) 1-5 - 2 + 3 - 4 + 3-5. (d) ^^ + \l - 1^ + § - 21.

5. Taking a line 1^ inches long to represent the area of a square of 1 inch side,
determine a line which shall represent the area of a rectangle 2>J inches long and
I inch broad.

6. A is a line 1| inches long, and B is a line 2,'8 inches long. Determine a
line which will represent the product of A and B, the unit being a line 1^ inches
long.

7. A is a line 2^ inches long, B is a line 1^^, inches long, and C is a line \l inch
long. If A represents the product of B and C, draw a line which will represent
the product of A and C.

INTRODUCTORY 15

8. Find a line representing the volume of a rectangular solid whose dimensions
are 3 inches x 1"75 inches x 1*25 inches. Unit = 0*5 inch.

9. The unit being a line 1 inch long, find a line to represent the cube of 1-25.

10. M and N are straight lines 1*25 inches and 1-5 inches long respectively,

M 4- N
Find a line representing ^ — r;^, the unit being 1 inch.

,__ 20

11. Draw lines representmg V15 and 7/ y^. Unit = 0*25 inch.

12. Find a line to represent cc^ when x = 2V2 - /v/3. Unit one inch.

13. A, B, C, and D are lines 3fg inches, 2^ inches, 2f inches, and 1/g inches

A X B
long respectively. If ^^ ^ = C, find the unit.

I'l /2"3^
14. Using as unit a length of 1 inch, find a line to represent o.q\/ y^r

32 + 1-42
102"

CHAPTER II

THE CIRCLE

14. Properties of the Circle— Simple Problems.—

(1) In a segment ABGD of a circle (Fig. 35) the angle AOD at the
centre is double of the angle ABB at the circumference.

(2) Angles ABB and ACD in the same segment of a circle (Fig. 35)
are equal to one another.

(3) The angle in a semicircle is a right angle. The carpenter, in
cutting a semicircular groove in a piece of wood, makes use of this
property when he tests it with a square, as shown in Fig. 36.

Fig. 35.

Fig. 36.

Fig. 37.

(4) A straight line lohich bisects a chord at right angles passes through
the centre of the circle. This suggests a method of finding the centre of
a circle to pass through three given points or to circumscribe a given
triangle.

To describe an arc of a circle through three given points, A, B, and C,
when the centre of the circle is inaccessible (Fig. 37). With centres A
and C describe arcs CH3 and ASK. Join AB and CB, and produce
these lines to meet the arcs at H and K. Mark off short equal arcs HI
and Kl. The intersection P of the lines Al, CI, is another point on
the arc required. In like manner other points may be found, and a
fair curve drawn through all of them is the arc required.

The foregoing construction is based on the fact that the three angles
of a triangle are together equal to 180°, and that the angles in the same
segment of a circle are equal to one another ; also that in equal circles
equal arcs subtend equal angles at their centres. The student should
have no difficulty in showing that the construction makes the angle
APC equal to the angle ABC. For another method see Art. 59, p. 56.

THE CIRCLE 17

•

(5) A tangent to a circle is at right angles to the radius or diameter
drawn to the j)oint of contact.

To draw a tangent to a circle from an external point. A good
practical method is to place a straight-edge on the paper and adjust
it so that the edge passes through the point and touches the circle,
then the tangent may be drawn. The point of contact must, however,
be obtained by dropping a perpendicular from the centre of the circle
to the tangent.

The following construction is recommended when the one just given
is objected to. With P (Fig. 38), the given external point as centre,
describe the arc OSR, passing through O
the centre of the given circle. With centre
O and radius equal to the diameter of the
circle describe an arc to cut the arc OSR
at R. Draw OR cutting the circle at T.
PT is the tangent required and T is the
point of contact. The student should
satisfy himself that this construction
makes the angle OTP a right angle. ^^^' ^^■

Another method is to describe a semicircle on OP as diameter, to
cut the given circje at T, which will be the point of contact of the
tangent required. In this case the angle OTP is obviously a right
angle because it is the angle in a semicircle.

It is obvious that two tangents may be drawn to a circle from an
external point, and that they are equal in length.

To draw a tangent to two given circles. This may be done very
accurately by placing a straight-edge on the paper, adjusting it so that
the edge touches the circles, and then drawing the tangent. The points
of contact must however be obtained by dropping perpendiculars from
the centres of the circles to the tangent.

When the preceding construction is objected to, the following may
be used : A and B (Figs. 39 and 40) are the centres of the given circles.
Join AB cutting the circles at C and D. Make DE equal to BC. Draw /

Fig. 39

a circle with centre A and radius AE. Draw BF a tangent to this
circle, F being the point of contact. Draw the line AF meeting the
given circle, whose centre is A, at H. In Fig. 39, H is on AF produced.
Draw BK parallel to AH, meeting the circle whose centre is B, at K.

PRACTICAL GEOMETRY

The straight line joining H and K is a tangent to both the given circles,
and H and K are the points of contact.

If the circles lie outside of one another four common tangents may
be drawn to them.

(6) The angle ABC (Fig. 41) between the chord AB and the tangent
BC to the circle at B is equal to the angle ADB in the alternate
segment.

To inscribe in a circle ABD (Fig. 41) a triangle equiangular to the
triangle EFH. Draw the tangent KBC. Draw the chord AB, making
the angle ABC equal to the angle F. Draw the chord DB, making the
angle DBK equal to the angle H. Join AD. The triangle ABD is
the triangle required.

^^ -' ^- ^

Fig. 43.

On a given line ^5 (Figs. 42 and 43) to describe a segment of a circle
itlltich shall contain an angle equal to a given angle G. Draw AD, making
the angle DAB equal to the angle C. Draw AO at right angles to
AD. Bisect AB at E. Draw EO at right angles to AB, meeting AO
at O. O is the centre, and OA is the radius of the circle of which the
segment AFB contains an angle equal to the angle C.

(7) In a quadrilateral ABOD (Fig. 44) inscribed
in a circle the opjwsite angles ABC and ADC are
together equal to two right angles. Also the other
pair of opposite angles are together equal to two
right angles.

(8) In Fig. 44

AC X BD = AB X CD 4- AD X BC.

(9) AC and BD are any two chords of a circle
intersecting at F. AF x CF = BF x DF. In
Fig. 44 the point F is within the circle, but it may be outside the
circle.

(10) EH is a tangent to the circle ABCD (Fig. 44), H being the
point of contact. EKL is a line cutting the circle at K and L.
M^ = EK X EL.

To draw a tangent to a given arc of a circle, through a given point,
without using the centre of the circle.

Fig. 45. The given point P is on the given arc APB but not near

Fig. 44.

THE CIRCLE

either of its extremities. With centre P describe arcs of a circle to
cut the given arc at A and B. A line through P parallel to the chord
AB is the tangent required.

Let C be the middle point of the chord AB. If d is the diameter
of the circle of which APB is^n arc, then PC (d - PC) = AC x CB = BC^.

^ BC^-fPC^ PB2 ^, ,
Hence d = — p^ = p^. The lengths of PB and PC may be

measured and d found by arithmetic.

Fig, 45.

Fig. 46.

Fig. 46. The given point Q is on the given arc and near one of
its extremities. Draw chords QP and QB. Join PB. Draw QE
making the angle PQE equal to the angle PBQ. QE is the tangent
required.

Fig. 47. The given point S is outside the given are. Draw the

line SAB cutting the arc at A and

Find a length / such

Fig. 47.

that
/•' = SA.SB (see Art. 12, p. 11).
With S as centre and radius equal
to /, describe an arc of a circle to
cut the given arc at T. A line ST is the tangent required and T is
the point of contact.

15. To construct a triangle having given the base, the
vertical angle, and the length of the bisector of the vertical
angle. — First method (Fig. 48). On AB the given base describe a

Fig. 48.

Fig. 49.

segment of a circle ACB containing an angle equal to the given
vertical angle. From O the centre of this circle draw OF at right
angles to AB and produce it to meet the circle at D. On any straight
line make ce equal to the given length of the bisector of the vertical
angle. Bisect ce at g. Draw eh at right angles to ce and make eh
equal the chord DB. With centre g and radius gh describe an arc of
a circle to cut ce produced at d. With centre D and radius equal to

20 PRACTICAL GEOMETRY

dc describe an arc of a circle to cut the circle ACB at C. Join AC and
BC. ACB is the triangle required.

The construction for finding the length of DC is based on the fact

that DC X DE = DBl

Second method (Fig. 49). Determine the circle ACB and the
point D as in the first solution. Through D draw a straight line DQP
cutting AB at Q. Make QP equal to the given length of the bisector
of the vertical angle. Repeat this construction several times so as to
determine a sufficient number of points on the locus of P. The inter-
section of the locus of P and the circle ACB determines the vertex of
the required triangle.

This second method is a very good illustration of the use of a locus.
It is quicker and probably more accurate and certainly much easier to
discover.

16. To find the locus of a point which moves so that the
ratio of its distances from two given points shall be equal to
a given ratio. — Let A and B
(Fig. 50) be the two given points,
and let P be one position of the
moving point so that the ratio
of AP to BP is equal to a given
ratio. In Fig. 50 this ratio is
2 : 1.

Draw PD bisecting the angle
APB and meeting AB at D.
Draw PDi bisecting the angle
between AP produced and BP.
Bisect DDi at O. A circle with
centre O and radius OD will be Fig. 50.

the locus required.

Draw BE parallel to PD to meet AP produced at E. Draw BEj
parallel to PD^ to meet AP at E^.

It is easy to show that PE = PE^ = PB.

Also, AD : DB : : AP : PE, therefore AD : DB : : AP : PB, and
D is a fixed point.

Again, AD^ : D,B : : AP : PE„ therefore AD^ : D,B : : AP : PB,
and Di is a fixed point.

The angle DPD^ is obviously a right angle. Hence a circle
described on DDj as diameter will pass through P.

If the figure be rotated about ADj as an axis, the circle will describe
the surface of a sphere, and the surface of this sphere will evidently
be the locus of a point moving in space so that the ratio of its distances
from the two fixed points A and B is equal to a given ratio.

17. The Inscribed and Escribed Circles of a Triangle.— The
inscribed circle of a triangle is the one which touches each of the three
sides. An escribed circle touches one side and tlie other two sides
produced. There are three escribed circles to a triangle. The con-
structions for drawing the inscribed and escribed circles are based on
the fact, that when a circle touches each of two straight lines, its centre

THE CIRCLE

lies on the line bisecting the angle between them. In Fig. 51, O is the
centre of the inscribed circle of the triangle ABC, and Oj is the centre
of one of the escribed circles.

The following results are not difficult to
prove : —

(1) The sides of the triangle formed hy
joining the centres of the escribed circles pass
through the angular points of the original
triangle.

(2) The line joining the centre of one of the
escribed circles and the opposite angle of the

triangle passes through the centre of the inscribed circle and is perpen-
dicular to the line joining the centres of the other two escribed circles.

(3) Referring to Fig. 51, AE = AF = half the perimeter of the
triangle ABC. This suggests the construction for the solution of the
following problem.

Given the perimeter ^ and one angle of a triangle, also the radius of the
inscribed circle : to construct the triangle. Make the angle EAF (Fig.
51) equal to the given angle. Draw a circle having a radius equal to
the given radius to touch AE and AF. This will be the inscribed
circle of the required triangle. Make AE and AF each equal to half
the given perimeter. From E and F draw perpendiculars to AE and
AF respectively to meet at O^. A circle with O^ as centre and O^E or
OjF as radius will be an escribed circle of the triangle. A tangent
BC to the two circles now drawn will complete the triangle required.

18. Circles in Contact. — In considering problems on circles in
contact with one another two simple facts should be kept in view, viz.

(1) J Then two circles touch one another the straight line ichich joins their
centres or that line produced passes through the point of contact.

(2) When two circles touch one another the distance between their centres
is equal to either the sum or difference of their radii.

To draw a circle of given radius to touch two given circles. Three cases
are shown in Figs. 52, 53, and 54. A and B are the centres of the given

Fig. 53.

Fig. 54.

circles. AB or AB produced cuts the given circles at C and F. Make
CD and FE each equal to the given radius. With centre A and radius
AD draw the arc DO. With centre B and radius BE draw the arc

The perimeter of a triangle is the sum of its sides.

PRACTICAL GEOMETRY

Fig. 65.

EO, cutting the former arc at O. Join O with A and B, and produce
these lines if necessary to meet the given circles at H and K. O is the
centre of the required circle, and H and K are the points of contact.

19. To draw a series of circles to touch one another and
two given lines.- -AB and CD (Fig. 55)
are the two given lines. Draw EF, bisecting
the angle between AB and CD. Let E be
the centre of one circle : its radius is EA, the
perpendicular on AB from E. Draw HK
perpendicular to EF. Make KL equal to KH.
Draw LM perpendicular to AB to meet EF
at M. M is the centre and ML is the radius
of the next circle.

20. To draw a circle to pass through a given point and
touch two given lines. — AB and AC (Fig.
56) are the two given lines and D is the
given point. Draw AE bisecting the angle
BAC. Join AD. Take any point F in AE.
Draw FH perpendicular to AB. With F as
centre and FH as radius describe a circle.
This circle will touch the lines AB and AC.
Let this circle cut AD at K. Draw DO
parallel to KF, meeting AE at O. O is the
centre and OD the radius of the circle required.

21. To draw a circle to touch two given lines and a given
circle. — AB and CD (Fig. 57) are the given ^
lines and EF is the given circle, N being its
centre. Draw HK and LM parallel to AB
and CD respectively, and at distances from
them equal to the radius EN of the given
circle. Draw, by preceding problem, a circle
to pass through N, and touch the lines HK
and LM. O the centre of this circle is the
centre of the circle required, and OE is its
radius, OEN being a straight line.

22. To draw a circle to touch a
through two given points. — AB (Fig.
58) is the given line, and C and D are
the given points. Draw CD and produce
it to meet AB at E. If the required circle
touches AB at K, then EK^ = ED x EC,
or EK must be a mean proportional to
EC and ED. Hence the following con-
struction. Produce CE and make EF
equal to ED. On CF describe a semi-
circle. Draw EH perpendicular to CF
to meet the semicircle at H. Make EK
equal to EH. Draw KO perpendicular to AB, and draw LO, bisecting
CD at right angles to meet KO at O. O is the centre of the circle re<j[uired.

Fig. 57.
given line and pass

Fig. 58.

THE CIRCLE

Fig. 59.

23. To draw a circle to touch a given circle and a given
line at a given point in it.— ABC (Fig. 59) is the given circle, DE
is the given line, and D the given point in it.
Through F, the centre of the given circle, draw
FE perpendicular to DE and produce it to
meet the circle at C. Draw DO perpendicular
to DE. Draw CD cutting the circle at B.
Draw FB and produce it to meet DO at O.
O is the centre and OD the radius of the circle
required. There are two solutions, the second
being obtained in the same way by joining A
with D instead of joining C with D.

24. To draw a circle to pass through two given points
and touch a given circle. — A and B (Fig. 60) are the given points
and CDE is the given circle. Draw a circle
CABE through A and B, cutting the circle CDE
at C and E. Join CE, and produce it to meet
AB produced at F. Draw FD, touching the
given circle at D. Join H, the centre of the
given circle, with D, and produce it to meet
a line bisecting AB at right angles at O. is
the centre of the required circle. There are two
solutions. The second is obtained by drawing
the other tangent to the given circle from F and
proceeding as before.

Considering a straight line to be a circle of infinite diameter, the
student should endeavour to deduce from the foregoing solution the
construction for drawing a circle to pass through two given points and
touch a given line.

25. To draw the locus of the centre of a circle which
touches two given circles.— A and B (Fig. 61) are the centres of

cutting the cir-
Mark

Fig. 61.

the given circles. Join AB

cles at C and D. Bisect CD at E.

off from E, above and below it, on AB a

number of equal divisions. With centres A

and B describe arcs of circles through these

divisions as shown. A fair curve drawn

through the intersections of these arcs as

shown is the locus required. The curve is

an hyperbola whose foci are the points A

andB. ...

This locus may be used in solving problems on the drawing of a circle
to touch two given circles and to fulfil some oilier condition.

When the given circles are external to one another, as in Fig. 61,
four different loci may be drawn in the manner described above,
because there are four positions of the point E. If AB produced cuts
the upper circle again in F and the lower one again in H, then the
remaining positions of the point E are, the middle point of FH, the
middle point of FD, and the middle point of HC.

PKACTICAL GEOMETRY

Fig. 62.

26. To draw the locus of the centre of a circle which
touches a given line and a given circle. — xiB (Fig. 62) is the
given line and CDE is the given circle, O
being its centre. Through O draw OH
perpendicular to AB, meeting AB at H,
and the given circle at D. Bisect DH at F.
Mark off from F above and below it, on
OH, a number of equal divisions. With
centre O draw arcs through the divisions
above F to meet parallels to AB through
the corresponding divisions below F, as shown. A fair curve drawn
through the intersections of the arcs and parallels, as shown, is the
locus required. • The curve is a parabola.

This locus may he used in solving problems on the drawing of a circle
to touch a given circle and a given straight line, and to fuljil some other
condition.-

Two different loci may be drawn in the manner described above,
because there are two positions of the point F. If the line OH
produced cuts the circle again at K, then the second position of F is the
middle point of HK.

27. Pole and Polar. — If through a given point P (Figs. 63 and
64) any straight line be drawn to cut a circle NQR at Q and R,
tangents to the circle at Q and R will intersect on a fixed straight line
LM. Conversely, the chord of contact of tangents from any point in
LM, outside the circle, will pass through P. This lixed line LM is
called the polar of the point P with respect to the circle, and the point
P is called the pole of the line LM with respect to the circle.

Fig. 64.

LM will be found to be perpendicular to OP, O being the centre of
the circle, and if OP or OP produced cuts LM at T, and the circle
at S, then OP X OT = OkS^.

When the pole P is outside the circle (Fig. 64) the polar is the
chord of contact of the tangents from P to the circle.

If chords QR and NKof a circle meet (produced if necessary) at P,
then the chords NQ and KR will intersect on the polar of P, also the
chords NR and QK will intersect on the polar of P. This suggests a
construction for drawing the polar of a point P by using a pencil and

THE CIRCLE 25

straight-edge only ; and, having found the polar, lines from P to where
the polar cuts the circle will be the tangents to the circle from P.
Tangents to a circle from an external point may therefore be drawn,
and their points of contact determined, by using a pencil and straight-
edge only.

28. Centres of Similitude.— CD and EF (Figs. 65 and 66) are
two parallel diameters of two circles whose centres are A and B. The

Fig. 65. Fig. 66.

straight lines joining the extremities of these diameters intersect in
pairs at fixed points S and S^ on the line joining the centres of the
circles. The points S and Sj are called the centres of similitude of the
two circles, and SA : SB : : AC : BE, also SiA : SjB : : AC : BE.

When the circles are external to one another, as in Fig. 66, one pair
of the common tangents will intersect at S and the other pair will
intersect at Sj.

When the two circles touch one another one of the centres of
similitude will be at the point of contact of the circles.

29. The Radical Axis. — The locus of a point from which equal
tangents can be drawn to two given circles is a straight line called the
radical axis of the two circles.

If the circles touch one another the common tangent at their point
of contact is their radical axis, and if the circles cut one another their
common chord produced is their radical axis.

For any other case the following general construction may be used
for finding the position of the radical axis of two circles. A and B
(Figs. 67 and 68) are the centres of the given circles ECF and HDK.
Describe any circle to cut the former circle at E and F and the latter
at H and K. Draw the chords EF and HK and produce them to
meet at P. A line PO at right angles to AB is the radical axis
required. The proof is as follows : Since E, F, H, and K are on the
same circle, PE X PF = PH X PK. But PC being a tangent to the
circle ECF, PC^ = PE X PF. Also PD being a tangent to the circle
HDK, PD^ = PH X PK ; therefore PC^ = PD^ and PC = PD. Hence
P is a point on the radical axis.

It can be proved that AO^ _ BO^ = AC^ - BD^.

If a circle be described with its centre at any point P on the radical

PRACTICAL GEOMETRY

axis of two given circles (Figs. 67 and 68) and having a radius equal
to PC the tangent from P to one of the given circles, then this circle
will cut the given circles orthogonally and will intersect AB at two fixed
points L and M.

[Two curves are said to cut each other orthogonal It/ when their

Fig. 67.

Fig. 68.

tangents at the point of intersection of the curves are at right angles
to one another.]

If the radical axis of each pair of a system of three given circles
be drawn, it will be found that the three radical axes will intersect at
a point which is called the radical centre of the system.

30. To draw a circle to touch two given circles and pass
through a given point. — A and B (Fig. 69) are the centres of the
given circles and C is the given
point. The line passing through A
and B cuts the circles at D, E, H,
and K as shown. Find S, one of
the centres of similitude of the
given circles (Art. 28). Draw a
circle through the points D, E, and
C, or through the points H, K, and
C, cutting the line CS produced at
F. Draw a circle to pass through
the points C and F, and touch one
of the given circles (Art. 24). This circle (centre O) will also touch
the other given circle.

There may be as many as four solutions to this problem, two for
each centre of similitude. In using the other centre of similitude (S^)
the construction is the same as described above, except that the point F
is found at the intersection of S,C and a circle through H, E and C,
or through D, K and C.

Considering a straight line to be a circle of infinite diameter, the
student should endeavour to deduce, from the foregoing solution, the
construction for drawing a circle to touch a given circle and a given
line and also pass through a given point.

Pig. 69.

THE CIRCLE 27

31. To draw a circle to touch three given circles.— There
may be as many as eight solutions to this problem. One solution is
shown in Fig. 70. A, B, and C are the centres of the given circles.
Let i\, r.>, and r.^ be their radii respec-
tively, and let r^ be the radius which is
not greater than r.^ or r.j. With centre
B and radius equal to ^g — r^ describe
the circle DE. With centre C and
radius equal to r., 4- 9\ describe the circle
FPH. Draw the circle APL which
touches the circle DE with internal con-
tact, and the circle EPH with external
contact and passes through the point A
(Art. 30). O, the centre of this circle,
is the centre of the required circle, and
ON is its radius. The points of contact „

with the given circles are M, K, and

N. The construction obviously makes AM, PN, and LK equal to one
another.

A second solution is obtained in the same way by making the circle
APL touch the circle DE with external contact, and the circle FPH
with internal contact.

Two solutions are obtained by making the radius of the circle DE
equal to r.^ -f r^ and the radius of the circle FPH equal to r., — r^.

Two solutions are obtained by making the radius of the circle DE
equal to r.^ — r^, and the radius of the circle FPH equal to r.^ — 7\ ; but
for these solutions the circle APL must have either internal or external
contact with both of the circles DE and FPH.

The remaining two solutions are obtained in the same way as the
previous two except that the radii of the circles DE and FPH are
r., + rj and r.^ -f- r^ respectively.

Remembering that a straight line is a circle of infinite radius, the
method illustrated by Fig. 70 may be easily modified to draw a circle to
touch a given line and two given circles, or two given lines and a circle.

Exercises II

Note. The points of contact of circles which touch one another, and the points
of contact of tangents to circles, must he shown distinctly.

1. ABC is a triangle. AC = 1'7 inches. BC = 1 inch. Angle C = 90°.
Draw the circle which touches AC at A and passes through B.

2. A line OA, I'b inches long, is a radius of a circle whose centre is 0. AB is
a line 2-2 inches long making an angle of 120° with OA. Draw the circle which
touches the given circle at A and passes through B.

3. On a straight line 3-5 inches long describe a segment of a circle to contain
an angle of 150° without using the centre of the circle.

4. In a circle 3 inches in diameter inscribe a triangle whose angles are to one
another as 2 : 3 : 4.

5. Draw an arc of'a circle of 6 inches radius, the chord of the arc being 4
inches long. Then draw the arc of a circle concentric with the first and of 7
inches radius without using the centre. The two arcs are to subtend the same
angle at the common centre.

PRACTICAL GEOMETRY

6. OA and OL are two straight lines. Angle AOL = 80°. OA = 2-5 inches.
OL is of indefinite length. B is a point in OA. OB = 1 inch. Find a point P
in OL such that the angle APB is the greatest possible.

7. The centres of two circles are 1-75 inches apart. The radius of one is 1*2
inches, and the radius of the other is 0-9 inch. P is one of the points of intersec-
tion of the circles. It is required to draw two straight lines APB and CPD, 3
inches long, and terminated by the circles. [Hmt. If Q is the middle point of
AP, and R is the middle point of BP, then QR is half of AB.]

8. From a point 2-2 inches distant from the centre of a circle of 1 inch
radius draw a straight line to cut the circle so that the part of it within the circle
shall be 1 inch long.

9. Two straight lines include an angle of 60°. Draw a circle 3 inches in
diameter, cutting one of the lines at points 1*7 inches apart, and the other at
points 2-3 inches apart.

10. The centres of two circles are 3 inches apart. Thh radius of one is 0*6
inch, and the radius of the other is I"! inches. Draw the four common tangents
to these circles.

11. In a circle 3 inches in diameter draw a chord dividing the circle into
two segments, one of which shall contain an angle of 65°. In the smaller segment
inscribe a circle ^ inch in diameter, and in thje larger inscribe a circle which shall
pass through the centre of the original circle.

12. ABC is a triangle. AB = 2 inches. BC = 1'75 inches. CA = 1*5 inches.
Draw the inscribed and escribed circles of this triangle.

13. Draw a sector of a circle. Radius of circle, 2 inches. Angle of sector,
60°. In this sector inscribe a circle.

14. In a circle 4 inches in diameter draw eight equal circles, each one to
touch the original circle and two of the others.

15. Draw an equilateral triangle of 3 inches side and in it place three
equal circles, each one touching one side of the triangle and the other two circles.

16. The vertical angle of a triangle is 50°, its altitude is 1*5 inches, and its
perimeter is 6*75 inches. Construct the triangle.

17. AB (Fig. 71) is an arc of a circle of 2 inches radius
circle of 1-75 inches radius. The centres of these circles are
1*25 inches apart. ADC is an arc of a circle of 0-625 inch
radius which touches the arcs AB and BC. Draw the figure
ABCD.

IS. Draw a triangle ABC. AB = 3 inches. BC = 2-5
inches. CA = 2 inches. Draw a circle of 1 inch radius to
touch the side AB and pass thro-ugh the point C.

19. Draw two circles having their centres 3 inches apart,
one circle (A) to be 2 inches, and the diameter of the other (B) to be 3 inches.
Draw a circle (C), 4 inches in diameter, to touch the circles A and B so that A is
inside, and B outside C.

20. Within a circle 3 inches in diameter draw another, 1-8 inches in diameter,
touching it. Next draw a third circle, 1 inch in diameter, inside the first circle,
outside the second, and touching both. Then draw a circle to pass through
the centre of the second circle, touch the first circle internally, and the third
externally.

21. The section of a hand-rail is shown in
Fig. 72. Draw this figure to the given dimen-
sions, which are in millimetres. Show the con-
structions by which the centres of the circular
arcs are determined and mark the junctions of
the arcs. [b.e.]

22. ABC is a triangle. AB = 1*5 inches.
BC = 1-4 inches. CA = 1 inch. C is the centre
of a circle of 1*6 inches radius. Draw two circles
to touch this circle and pass through the points
A and B.

23. Draw the locus of the centre of a circle which touches a fixed circle of

is an arc of a
A

C -B

Fig. 71.
The diameter of

H 58 ■

• Fig. 72.

^"^

THE CIRCLE

dla^

Fig. 73.

1 incli radius and passes through a fixed point 2 inches distant from the centre
of the fixed circle.

24. AB, 2 inches long, is a diameter of the circle ACB (Fig. 73). BDE is an
arc of a circle of 4 inches radius which touches the
circle ACB at B. AE is at right angles to AB.
Draw the figure ACBDE and add the circle F which
touches the circle ACB, the arc BDE and the
straight line AE.

25. Draw a circle 2*5 inches in diameter, and
take a point P at a distance of 2*25 inches from its
centre. From P draw a tangent to the circle and
determine the exact point of contact, using a pencil
and straight-edge only.

26. LjNI is a straight line of indefinite length, A and B are two points which
are 2 inches apart and are on the same side of LM. The perpendicular distances
of A and B from LM are 1 inch and 1-5 inches respectively. B is the centre of a
circle of 0*75 inch radius. Draw the two circles which have their centres on LM,
pass through A, and touch the circle whose centre is B.

27. The polar of a point P with respect to a circle 2-75 inches in diameter is
a straight line at a perpendicular distance of 0*75 inch from its centre. Determine
the position of the point P, using a pencil and straight-edge only.

28. Draw three circles and find the centres of similitude of each pair. Then
show by using a straight-edge that the line joining any two of the six centres of
similitude either passes through a third or through the centres of two of the
circles.

29. ABC is a triangle. AB = 3 inches. BC = 4 inches. CA = 3-5 inches.
A, B, and C are the centres of three circles whose radii are 1 inch, 1-25 inches,
and 2 inches respectively. Find the radical centre of the three circles, and draw
the circle which cuts the three circles orthogonally.

30. Draw all the circles which touch each of the three circles given in the
preceding exercise.

CHAPTEK III

CONIC SECTIONS

32. The Conic Sections. — The curves known as the conic
sections or conies are, the ellipse^ the hyperbola, and the parabola.
They are plane curves, and they may be defined with reference to their
properties as plane figures, or they may be defined with reference to
the cone of which they are plane sections.

33. Conies defined without Reference to the Cone. — If
F (Fig. 74) is a fixed point and XM a fixed straight line, and if a point
P move in the plane containing F and XM in such a
manner that the distance FP always bears the same
ratio to the perpendicular PM to the fixed line, then
the curve traced out by the point P is called a conic
section or conic.

The fixed point F is called the focus, and the fixed
straight line XM is called the directrix of the conic.

A straight line through the focus at right angles
to the directrix is called the axis, and the point A
where the axis cuts the curve is called the vertex of Fig. 74.

the conic.

P being any point on the curve, the constant ratio of FP to PM is
called the eccentricity of the conic.

AVhen FP is less than PM the conic is an ellipse.

When FP is equal to PM the conic is a parabola.

When FP is greater than PM the conic is an hyperbola.

A conic is therefore an ellipse, a parabola, or an hyperbola according
as the eccentricity is less than, equal to, or greater than unity.

34. To construct a Conic having given the Focus,
Directrix, and Eccentricity. — In Fig. 75, F is the given focus
and XM the given directrix. Take a point D on the axis, and draw
DE perpendicular to XD, and of such a length that DE : XT) is equal
to the given eccentricity. For example, if the eccentricity is 2 : 3 or jf,
make XI) equal to 3 to any scale, and make DE equal to 2 to the same
scale. Join XE. Draw any straight line N?i parallel to the given
directrix XM, cutting XE at n. With centre F and radius equal
to Nn describe arcs of a circle to cut wN and mN produced at P and P.
P, P are points on the required conic. In like manner any number of

CONIC SECTIONS

points may be determined, and a fair curve drawn through them is the
conic required.

It is obvious that the ratio of Nw to XN is the same as the ratio
of DE to XD, and therefore the ratio of FP to PM is the same as the
ratio of DE to XD which was made equal to the given eccentricity.

The line XE will touch the conic at a point R obtained by drawing
FR perpendicular to the axis to meet XE at R.

Fig. 75.

In Fig. 75, three different cases are shown. In the ellipse the
eccentricity is 1:^3, in the parabola the eccentricity is of
course 1 : 1, and iri the hyperbola the eccentricity is ^3:1. These
eccentricities make the angle DXE equal to 30^ for the ellipse, 45 for
the parabola, and 60° for the hyperbola.

It is obviouslv unnecessary to draw the line XE in the case of the
parabola. The radius for the arc through P may be taken at once
from XN in this case, but it is instructive to notice that the construction
given applies to all the conies. .

If lines FY and FYi be drawn making 45° with the axis and

32 PRACTICAL GEOMETRY

meeting XE at Y and Y^, then perpendiculars from Y and Yj to the
axis determine the points A and A^ where the conic cuts the axis. In
the case of the parabola there is only one such point, or to put it in
another way, the point A^ for the parabola is at an infinite distance
from A.

It will be seen that the ellipse is a closed curve w^ith two vertices,
and that the hyperbola has two separate branches each with its own
vertex.

If AjFi, measured to the left of A^, be made equal to AF, and if
AiXi measured to the right of Aj be made equal to AX, and if XiM^ be
drawn perpendicular to the axis, then the ellipse may be constructed
from the focus F^ and directrix X^Mi, both to the right of the figure, in
the same way as from the focus F and directrix XM, using the same
eccentricity. Also the hyperbola may be constructed from the focus F^
and directrix X^^Mi, both to the left of the figure, in the same way as
from the focus F and directrix XM, using the same eccentricity.

Referring to Fig. 75, it will be seen that as the eccentricity increases
or diminishes the angle DXE increases or diminishes, and therefore as-
the eccentricity of the ellipse increases and approaches to unity the
ellipse approaches to the parabola, and as the eccentricity of the hyper-
bola diminishes and approaches to unity the hyperbola also approaches
to the parabola. A parabola is therefore the limiting form of an ellipse
or an hyperbola. Remembering this fact, many of the properties of the
parabola may be deduced at once from those of the ellipse or hyperbola.
For example, the tangent to an hyperbola at any point P on the curve
bisects the angle between the focal distances FP and F^P. Now in the
parabola the focus F^ is at an infinite distance from the focus F,
therefore F^P is parallel to the axis, and the tangent to a parabola at
any point P on the curve bisects the angle between the focal distance
FP and the perpendicular PM on the directrix.

35. Conies defined with Reference to the Cone. — In Figs.
76, 77, and 78^ vtu is the projection of a right circular cone on a plane
parallel to its axis, vs being the projection of the axis of the cone. In
each Fig. xxi represents a plane which cuts the cone and is perpen-
dicular to the plane of projection.

If xx^ cuts vt and vu below the vertex of the cone (Fig. 76), the
section is an ellipse. If xx^ cuts vt above and vu below the vertex
(Fig. 78), the section is an hyperbola. If xx^ is parallel to vt (Fig. 77),
the section is a parabola.

In each Fig. the true shape of the section is shown, and is obtained
by the rules of solid geometry. The axis XX^ of the true shape of the
section is drawn parallel to .tx'j, and any point P on the curve is
determined as follows : — Through any point n within the projection of
the cone and on xx^ draw wN perpendicular to xx^ meeting XX^ at N.
Through n draw a line perpendicular to vs and terminated by vt and vu.
On this line as diameter describe a semicircle. Through n draw np
parallel to vs to meet the semicircle at p. On the line wN make NP
equal to np. P is a point on the true shape of the section. By
repeating this construction any number of points may be obtained,

CONIC SECTIONS

and a fair curve drawn through them is the curve required. The
theory of the above construction will be understood after Art. 226,
Chap. XVITI, has been studied.

Fig. 78.

To determine the positions of the directrices and foci of the conies,
draw spheres inscribed in the cone and touching the plane of section.
These spheres are represented by the circles whose centres are at s and
Si on the projection of the axis of the cone. These spheres will touch
the cone in circles whose projections are the chords of contact of the
circles which are the projections of the spheres and the lines vt and vu.
The planes of these circles of contact intersect the plane of section in
lines which are the directrices of the sections, and the projections of
these lines are the points x and Xi. (In the case of the parabola the

PRACTICAL GEOMETRY

point Xi is at an infinite distance from x.) Hence if perpendiculars be
drawn from x and x^ to XXi the directrices XM and XiM^ of the
true shape of the section are ob-
tained.

The inscribed spheres touch the
plane of section at the foci of the
section, and the projections of these
points are at / and f^. Hence
perpendiculars from / and /^ to xx^^
to meet XX^ determine the foci F
and Fj of the true shape of the
section.

A reference to the sketch shown
in Fig. 79 will perhaps make the
meaning of the foregoing state-
ments a little clearer.

As in Art. 34, it is instructive
to observe that the parabola is the
limiting form of an ellipse or an
hyperbola, as the plane of section •^^^- '^^•

(Figs. 76 and 78) is turned round so as to come nearer and nearer to a
position (Fig. 77) in which it is parallel to vt. Also, when the hyper-
bolic section is taken through the vertex of the cone, the hyperbola
becomes two straight lines, and when the plane which gives the elliptic
section is turned round so as to be perpendicular to the axis of the
cone the ellipse becomes a circle. Again, if the plane of the parabolic
section be moved parallel to itself nearer and nearer to vt, the ultimate
form of the parabola will be a straight line. Studying Figs. 76, 77,
and 78 still further, it will be seen that when the hyperbola becomes
two (straight lines, the directrices will coincide, and the foci will
coincide at the point where the axis of the conic cuts the directrix,
and where the two straight lines which form the hyperbola intersect.
Again, when the parabola becomes a straight line, that line will be the
axis of the conic, and the focus will be on the directrix. Lastly, when
the ellipse becomes a circle the foci will coincide at the centre of the
circle, and the directrices will move off to infinity.

36. Additional Definitions relating to Conies. — A perpen-
dicular PN (Fig. 80) from a point P on a conic to the
axis is called the ordinate of the point P, and if PN be
produced to cut the conic again at P', the line PP' is
called a double ordinate of the conic.

RFR', the double ordinate through the focus, is called
the latus rectum of the conic.

In the ellipse and hyperbola the point which is
midway between A and Ai, the points where the conic
cuts the axis, is called the centre of the conic, and the
ellipse and hyperbola are called central conies.

A straight line joining two points on a conic is called
a chord of the conic.

Fig. 80.

CONIC SECTIONS

37. General Properties of Conies.— Following from the
general definition of a conic (Art. 33), there are many properties
possessed by all conies which may be demonstrated. A few of the
more important of these general properties will now be given.

(1) If a straight line cuts the directrix at D and the conic at P and Q
(Figs. 81 and 82), then, F

/F \ /\.^^\^

Fig. 81.

Fig. 82.

the focus, the straight
line DF ivill bisect, either the
exterior or the interior angle
between PF and QF. Only
in the case of the hyper-
bola, and only when P and
Q are on different branches
of the curve (Fig. 82) is it
the interior angle PFQ
which is bisected by DF.

If the straight line FP'
be drawn bisecting the angle PFQ (Fig. 81) and meeting the conic at
1^', then the angle DFP' is evidently a right angle, and if the line DPQ
be turned about D so as to make P and Q approach nearer and nearer
to one another, P' will always lie between P and Q, and in the limit
when P and Q coincide they will coincide at P' and a straight line
through D and P' will be a tangent to the conic at P'. Hence the next
general property of conies.

(2) The portion of a tangent intercepted between its point of contact
and the directrix subtends a right angle at the focus.

(3) Tangents at the extremities of a focal chord intersect on the
directrix. This follows at once from f>i
the preceding property. It is evi-
dent, conversely, that if tangents
be drawn to a conic from a point on
the directrix, the chord of contact
passes through the focus.

(4) If PFP and QFQ (Fig.
83) be two focal chords, the straight
lines P'Q and Q'P intersect at a
point on the directrix; also the
straight lines PQ and Q'P' intersect
at a point on the directrix, and the
portion of the directrix DD' between the points of intersection subtends a
right angle at the focus. The above follows very
easily from the first property given in this article.

(5) Tangents TP and TQ (Fig. 84) from any
point T subtend equal angles at the focus F.

(6) If the normal at P (Fig. 84) meet the axis at
G, the ratio of FG to FP is equal to the eccentricity
of the conic.

(7) PL (Fig. 84) the projection of PG on FP
18 equal to the semi-latus rectum.

Fig. 83.

Ftg. 84.

PRACTICAL GEOMETRY

(8) The locus of the middle points of a system of parallel chords is a
straight line. This straight line is called a diameter of the conic, and
the point where a diameter cuts the conic is called the vertex of the
diameter.

(9) All diameters of a central conic pass through the centre of the
conic, and all diameters of a parabola are parallel to the axis.

(10) Tangents at the extremities of a chord PP (Fig. 85) intersect
at a point T on the diameter TW which bisects the
chord.

(11) The tangent at the vertex of a diameter is
parallel to the system of chords bisected by that
diameter.

(12) TVW(Fig. 85) being a diameter, and TP
and TP' being tangents to the conic, EE', the portion
of the tangent at }^ intercepted between TP and TP,
is bisected at V the vertex of the diameter.

38. Properties of the Parabola. — A knowledge of the following
properties of the parabola will enable the student to solve a considerable
number of problems.

(X) If from the ends of a focal chord PPy (Fig. 86) perpendiculars
be drawn to the directrix KXK^, then KK^ subtends a right angle at the
focus F.

(2) The tangent PT (Fig. 87) at any point P bisects the angle
between the focal distance FP and the perpendicular PK to the directrix.
From this it follows that the tangent is equally inclined to the focal
distance of the point of contact and the axis. Also FT = FP

Fig. 85.

Fig. 86.

Fig. 87.

Fig. 88.

(3) If the tangent at any point P (Fig. 87) meets the axis at T, and
PN be the ordinate of P, then AT = AN. This follows easily from the
preceding property thus : FT = FP = PK = XN. But AF = AX,
therefore AT = AN.

(4) Y(Fig. 87), the foot of the perpendicular from the focus F on the
tangent at any point P, lies on A Y the tangent at the vertex.

(5) PN'^ = 4AF . AN (Fig. 87).

(6) Tangents PS and P^S at the ends of a focal chord PP^ intersect
at right angles at a point S on the directrix (Fig. 87).

CONIC SECTIONS

Fig. 89.

Definition. If PG (Fig. 87) the normal at P meets the axis at G
then NG is called the suhnormal.

(7) The suhnormal is constant and equal to 2AF. For, since PT
bisects the angle FPK and PG is perpendicular to PT it follows that
PG bisects the angle FPL.^ Hence, angle FPG = angle LPG = angle
FGP, therefore FG = FP = PK = NX, and NG = FX = 2AF.

(8) The angles SFP and SF1\ (Fig. 88) subtended at the focus F hy
tangents SP and SF^ are equal to one another and to the angle LSP. •

(9) The triangles FPS and FSP^^ (Fig. 88) are similar and
FS' = FP.FP,.

(10) The focus F (Fig. 89) and the
points of intersection S^, S2, and S3 of three
tangents to a parabola lie on the same circle.

Definition. A straight line YU (Fig.
88) drawn parallel to the axis AF through
any point V of a parabola is called a
diameter and the point V is called the
vertex of the diameter YU.

(11) /Sf (Fig. 88), the point of inter-
section of two tangents SP and SP^, is
equidistant from the diameters through the
points of contact P and Pj.

( 1 2) The diameter SVU(Fig. 88) through
the point of intersection of tangents SP and SPi bisects the chord, of contact
PP^. Also SV = VU, and PP^ is parallel to the tangent at V.

(13) A diameter bisects all chords parallel
to the tangent at its vertex.

Definition. The focal chord parallel to
the tangent at the vertex of a diameter is
called the parameter of that diameter.

(14) The parameter of any diameter is four
times the focal distance of the vertex of that
diameter.

(15) The area enclosed hy the parabola AQP
(Fig. 90), the ordinatePN, and the axis AN, is
two-thirds of the area of the rectangle ANPL.

It follows from this that the area of
the figure AQPL is one-third of the area
of the rectangle ANPL.

39. To draw the Tangents to a
Parabola from an external point.—
Let A (Fig. 91) be the vertex and F
the focus of the parabola and let S be
the external point from which the tan-
gents are to be drawn.

Join SF and on SF as diameter describe
a circle. Draw YYj the tangent at the
vertex cutting the circle at Y and Yi.

1 PL is KP produced (Fig. 87).

Fig. 90.

Fig. 9].

38 PRACTICAL OEOMETRY

Join S to Y and Y^ and produce them. These are the tangents
required,

Y and Y^ are evidently the feet of the perpendiculars from F on
SY and SYj and these are on the tangent at the vertex. Hence by-
Art. 38 (4) SY and SYj are tangents from S.

40. To construct a Parabola, having given the Vertex,
the Axis, and another point on the Curve.— Let A (Fig, 92)
be* the vertex, AN the axis, and P
another point on the parabola. This
is a problem of very frequent occur-
rence and the following construction,
is the most convenient.

Draw the ordinate VN and complete
the rectangle ANPL. Divide AL into
any convenient number of equal parts,

1' 2' 3'

/ /a.^--'

r iw \h

and number the points of division 1, -pia 99

2, 3, etc. from A to L. (In Fig. 92,

four equal parts have been taken.) Divide LP into the same number
of equal parts, and number the points of division, 1', 2', 3', etc. from
L to P. From the points 1, 2, 3, etc. on AL draw lines parallel to
AN. Join A to the points 1', 2', 3', etc. on LP. The lines having
the same numbers at their ends intersect at points on the para-
bola.

Consider one of these points, Q, where A3' cuts the line through
3 parallel to AN. Draw QM and 3'H perpendicular to AN. The
ordinate PN is i of the ordinate QM. Again AN is J of AH and
AH is I of AM. Therefore AN = f x |AM = if AM.

But [Art. 38, (5)] PN^ = 4AF . AN

therefore (iQM)^ = 4AF x ^M^, and QM' = 4AF • AM,
which shows that Q is a point on the parabola.

41. Definitions relating to Central Conies. — A central conic
cuts the axis at two points A and A^, and there are two foci F and F^.
There are also two directrices, one belonging to each focus. In
referring to the focus and directrix of a conic it is understood that if
the conic is a central conic the focus and directrix belong or correspond
the one to the other.

In a central conic the line AAj terminated by the two vertices of
the curve is called the transverse axis.

In the ellipse, Fig. 93, the straight line BCB^ drawn through the
centre at right angles to AA^ and terminated by the curve is called
the conjugate axis but more generally the minor axisj and the transverse
axis is generally called the major axis.

In the hyperbola, Fig. 94, there is also a conjugate axis BCBj
passing through the centre at right angles to AA^, but it does not
meet the curve. The length of the conjugate axis of the hyperbola is
obtained by the following construction. With centre A and radius
equal to CF describe an arc of a circle to cut the line through C
perpendicular to AAj at B and Bj. Then BBi is the length of the
conjugate axis. The conjugate axis of the hyperbola is not necessarily

CONIC SECTIONS

a minor axis, as it may be either less than, equal to, or greater than
the transverse axis.

Fig. 93.

Fig. 94.

The circle described with the transverse axis AA^ as a diameter is
called the auxiliary circle.

The circle described on the minor axis of an ellipse as a diameter
is sometimes called the minor auxiliary circle.

If an hyperbola be described having BCBj for its transverse axis,
and ACAi for its conjugate axis, this hyperbola is called the conjugate
hyperbola.

Diameters PCPi and QCQi are called conjugate diameters when QQ^
is parallel to the tangents at P and Pj, or when PPj is parallel to the
tangents at Q and Q^.

42. General Properties of Central Conies. — The ellipse and
hyperbola have many properties in common, and there are other
properties possessed by one which when stated in a slightly modified
form are also possessed by the other. A number of these properties
will now be given.

(1) The sum or difference of the focal distances of any point P
(Figs. 95 and 96) on the curve is equal to the transverse axis, or
PF ± PF^ = AA^. In the ellipse it is the sum and in the hyperbola
it is the difference which must be taken.

(2) The tangent and normal at any point P (Figs. 95 and 96) on the
curve bisect the angles between the focal distances of the point. In the
ellipse it is the exterior angle between PF and PFj which is bisected
by the tangent, while in the hyperbola it is the interior angle between
these lines which is bisected by the tangent.

(3) A circle drawn through a point P (Figs. 95 and 96) on the
curve and the foci F and F^ cuts the conjugate axis at the points t and g,
ivhere the tangent and normal to the curve at P cut that axis. (In Fig. 96
the point g falls outside the limits of the figure.)

(4) Perpendiculars FY and F^Y^ (Figs. 95 and 96) on the tangent at
any point P meet that tangent on the auxiliary circle, and BC, the semi-
conjugate axis, is a mean proportional between them or BO^ = FY . FiY^.

(5) Tangents OP and OQ (Figs. 95 and 96) are equally inclined to
the focal distances OF and OFi of the point 0.

PRACTICAL GEOMETRY

(6) The locus of the intersection (Figs, 95 and 96) of pairs of
tangents FO and QO, which are at right angles to one another, is a circle
ivhose centre is C, and whose radius is equal to ^ AC^ ± BC\ This
circle is called the director circle. In determining the radius the plus
sign must be taken in the case of the ellipse and the minus sign in the
case of the hyperbola. In the case of the hyperbola when EC = AC,
the radius of the director circle is zero, and when BC is greater than

Fig. 95.

Fig. 96.

AC there is no director circle, and no pairs of tangents can be drawn
which are perpendicular to one another.

In Fig. 96 the tangents OP and OQ are shown touching different
branches of the hyperbola, but for certain positions of O they will
touch one branch only.

(7) If the tangent at P meet the transverse axis at T (Figs. 95 and
96) and the conjugate axis at t, and ifN and n he the feet of the perpen-
diculars from P on the transverse and conjugate axes respectively, then
CN . CT = AC^ and Cn.Ct = BCl

(8) PN' : AN . AiN : : BC"^ : ACl

(9) CP and CQ (Figs. 93 and 94) being conjugate semi-diameters,
CP-^ ± CQ2 = CA^ ± CBl The plus sign is taken for the ellipse and
the minus sign for the hyperbola.

(10) The area of the parallelogram formed by the tangents at the
ends of conjugate diameters is equal to the rectangle contained hy the
transverse and conjugate axes.

43. To construct a Central Conic having given the Foci
and the Transverse Axis. — F and F^ (Figs. 97 and 98) are the
given foci and AAi is the given transverse axis. For the ellipse take
any point a in AA^ and for the hyperbola take any point a in AjA
produced. With centres F and F^ and radius equal to Aa describe
arcs of circles. With centres F and Fj and radius equal to A^a describe
arcs of circles to cut the former arcs at P, Q, P^ and Qi. P, Q, P^, and
Qi are points on the conic required, and by taking other positions of
the point a and repeating the above construction any number of points
on the conic may be determined and a fair curve drawn through them
will complete the construction.

CONIC SECTIONS 41

It will be observed that the property of central conies given in
Art. 42 (1; is made use of in the above construction.

Fig. 97.

Fig. 98.

Fig. 99.

44. To construct an Ellipse from the Auxiliary Circles.— It
has already been given [Art. 42, (8)J that PN^ ; AN . AjN : : BC^ : AC^
Let the ordinate PN (Fig.
99) be produced to meet
the auxiliary circle at a,
then by a property of the
circle, «N^= AN . A^N, there-
fore PN2 : aW : : BC^ : AC^,
or PN : aN : : BC : AC.
Draw aC cutting the minor
auxiliary circle at b, then,
since «G = AC and hC = BC,
PN: aN:: 6C: aC, therefore
Fh is parallel to AC. Hence
the following construction for finding points on the ellipse. Draw a
radius aC of the auxiliary circle cutting the minor auxiliary circle at h.
Through a draw aN parallel to the minor axis, and through h draw 6P
parallel to the major axis to meet aN at P. P is a point on the ellipse.
In like manner any number of points on the ellipse may be found, and
a fair curve drawn through them completes the construction.

It may be noted here that the tangent to the auxiliary circle at a,
and the tangent to the ellipse at P meet at a point T on the major axis
produced. Also the tangent to the minor auxiliary circle at h and the
tangent to the ellipse at P meet at a point t on the minor axis pro-
duced.

45. The Trammel Method of Drawing an Ellipse.— Re-
ferring to Fig. 99, if Fn be drawn parallel to aC, meeting AC at m and
BC produced at w, then Pw = aC = AC, and PN : aN : : Pm : aC ;
but it has been shown that PN : aN : : BC : AC, therefore
P»i : aC : : BC : AC, and consequently Pm = BC. Hence if a
straight line be drawn across an ellipse, cutting the curve at P, the
major axis at m, and the minor axis at n, and if Tn = AC, then
Vm = BC. Conversely it follows that if a straight line Pmw in
which Pw = AC and Vm = BC be placed so that n is on the minor
axis and m is on the major axis, then P will lie on the ellipse.

PRACTICAL GEOMETRY

Fig. 100.

On the straight edge of a strip of paper, Fig. 100, mark points P,
m, and n, such that P?i is equal to the semi-major axis, and Pwi is equal
to the semi-minor axis of the
ellipse. Place this paper trammel
on the paper so that m is on the
major axis and n is on the minor
axis, a dot made on the paper at
P will be on the ellipse. By
moving the trammel into a number
of different positions a sufficient
number of points on the ellipse
may be obtained and a fair curve
is then drawn through them.

Referring again to Fig. 99, if m^Vrii be drawn making the angle
Vn^n equal to the angle Pwrii, then Tn^ will be equal to P?i, and Fm^ will
be equal to P?». Hence the points m and n on the trammel may be on
opposite sides of P as shown on the trammel m^Frii, Fig. 100. The latter
form of the trammel should be used when the difference between the
major and minor axes of the ellipse is small.

The trammel method of drawing an ellipse is most convenient as it
keeps the paper free of construction lines other than the axes.

46. Given a Pair of Conjugate Diameters of an Ellipse,
to find the Axes.— Let PCPj and QCQ, (Fig. 101) be the given
conjugate diameters. Draw PD
perpendicular to CQ. Make PH
and PHi each equal to CQ. Join
CH and CHp The major axis
ACAi bisects the angle HCHi, and
the minor axis BCB^ is of course
perpendicular to ACA^. Join P to
the middle point of CH, cutting the
major axis at m and the minor axis
at 11. P« is the length of the semi-
major axis, and Pjh is the length of the semi-minor axis.

47. To construct an Ellipse having given a Pair of
Conjugate Diameters.— Let PCP^ and QCQ^ (Figs. 102 and 103) be
the given conjugate diameters.

First Method. Through P,
Q, P„ and Qi (Fig. 102) draw
parallels to the given diameters
forming the parallelogi'am RT.
Divide CP into a number of
equal parts, and divide RP into
the same number of equal parts.
Join Q to 1' the first point of
division on RP. A line joining
Qi to 1 the first point of division
on CP will when produced cut
Ql' at a point on the .ellipse required. Repeating this construction

Fig. 102.

CONIC SECTIONS

Fig. 103.

with the other points of division on RP and CP other points on the arc
PQ are obtained. In like manner points on the arc PQ^ are obtained.
Points on the other half of the ellipse may be found by the same
method, or by a construction depending on the fact that all diameters
are bisected at the centre C, or by making use of the fact that chords
parallel to a diameter are bisected by its conjugate.

Second Method. Find the axes by the construction described in the
preceding article and then use a paper trammel as explained in Art. 45.

Third Method. By a triangular trammel. From P, Fig. 103, draw
Pm perpendicular to CQ. From Q draw QD perpendicular to CP and
make Pw equal to QD. Join mn. If
the triangle Pmn be drawn on a
strip of paper, mn being on one edge
and P on the opposite edge, and if
this strip be moved into different
positions, m being on QQ^ and n on
PPi, then P will be on the ellipse.
A second position of the trammel
is shown to the right of the figure.
Instead of using a strip of ordinary
paper, the points m, w, and P may
be marked on a piece of tracing paper, a needle hole being made in
the tracing paper at P, through which points on the ellipse may be
marked by a sharp round-pointed pencil.

48. The Ellipse as the Projection of a Circle.— Many of
the properties of the ellipse and many of the constructions connected
with it may be
readily demonstrated
by considering the
ellipse as the projec-
tion of a circle. A
few of these will be
considered here, it
being assumed that
the student has al-
ready studied some
of the later chapters
in this work relating
to projection.

Referring to Fig.
104, ABAiBi is a
circle lying in a hori-
zontal plane, AAj
and BBi being dia-
meters at right an-
gles to one another. . ., .,
Imagine the circle ABA^B^ to revolve about AA, as an axis until its
plane is inclined at an angle to. the horizontal, and let it then be
projected on to the horizontal plane containing AA^. ihe point r on

Fig. 104.

44 PRACTICAL GEOMETRY

the circle in its new position will have the point p for its plan, the
point p being determined by the construction shown. Other points on
the circle may be treated in the same way, and all points such as p
being joined by a fair curve the ellipse ApA-J)^ is determined.

On hhi as diameter describe the circle ahaj)^. This circle will have
its centre at C, the centre of the other circle. Now imagine the ellipse,
which is horizontal, to revolve about hhi as an axis until its plane is
inclined at the angle 6 to the horizontal and let it then be projected
on to the horizontal plane containing hh^. The point p on the ellipse in
its new position will have the point pi for its plan, and it is easy to
show that the point p^ is on the circle abaj)i, also that the points P, j?i,
and C are in the same straight line. It will be seen that the circles
ABAjBi and aba-^^ are the auxiliary circles of the ellipse AfeAjbi
and the theory of the construction given in Art. 44 is further ex-
plained.

It is easy to establish the following theorem in projection : —
" The 'projection of the tangent to a curve at any point is the tangent to
the projection of the curve at the projection of the point.""

Referring again to Fig. 104, TP is a tangent to the circle ABAiBj
at P, and when the circle is turned about AA^ as an axis and pro-
jected into the ellipse as already described, the point T which is on
that axis is stationary and therefore Tp will evidently be the tangent
to the ellipse at p. Let Tp produced meet BBj at t. Then when the
ellipse is turned about BB^ as an axis and projected into the circle
aba^i as already described, the point t which is on that axis is
stationary and tp^ will evidently be the tangent to the circle dba^^
at pi.

Another theorem in projection which is easily proved is as follows : —
The projections on a plane of two intersecting straight lines ivill intersect
at a point which is the projection of the point of intersection of the original
lines, and the point of intersection of the projections will divide the pro-
jections into segments which are to one another as the corresponding segments
of the original lines.

This enables a simple demonstration to be given of the construction
for drawing an ellipse having given a pair of conjugate diameters
which has been described in Art. 47 and illustrated by Fig. 102.

Referring to Fig. 105, PPi and QQi are two diameters of a circle
at right angles to one another. The circle is supposed to be lying on
the horizontal plane. A square RSTU is shown circumscribing the
circle, the points of contact being P, Qi, Pi, and Q. The radius CP is
divided into any number of equal parts, in this case three, at the
points 1 and 2, and RP is divided into the same number of equal parts
at the points 1' and 2'. If the lines joining Q.^ to the points 1 and 2
be produced it is easy to show that they will intersect the lines joining
Q to 1' and 2/ respectively at points on the circle as shown.

Now imagine the circle, with all the lines connected with it, to
revolve about an axis MN in its plane into an inclined position, and
let a projection of the whole be then made on the horizontal plane as
shown. The square RSTU projects into the parallelogram rstUj and

CONIC SECTIONS

the circle projects into an ellipse which touches the sides of the
parallelogram at^?, q^, p^, and q. Also pp^ and qq^ will evidently be
conjugate diameters of the ellipse. Lastly, the lines through Q^ and Q
which intersect on the circle project into lines which intersect on the

ellipse and divide cp and rp into three equal parts. Hence the con-
struction for drawing the ellipse having given two conjugate diameters.

49. Properties characteristic of the Hyperbola. — It has
already been pointed out that the hyperbola has two separate branches
and that there is a conjugate hyperbola also having two separate
branches. Also the transverse and conjugate axes of the hyperbola
are the conjugate and transverse axes respectively of the conjugate
hyperbola.

Referring to Fig. 106, F and F^ are the foci and AA^ is the trans-
verse axis of the hyperbola whose branches are Q and Qi. BB^ is the
conjugate axis and Q' and Q', are the branches of the conjugate hyper-
bola. A construction has already been given (Art. 41, p. 38) for
finding the conjugate axis, and the following construction will evidently
give the same result. With centre C, the middle point of AAj, and
radius CF describe a circle. Draw the tangent to the hyperbola at A
to cut this circle at L and L^. Through C draw BCBj at right angles
to AA^. Parallels to AAj through L and L^ will cut BCBj at B and
Bi the extremities of the conjugate axis. The circle whose centre is
C and radius CF cuts BBj produced at F' and F'l the foci of the
conjugate hyperbola.

PRACTICAL GEOMETRY

The lines CL and CLi produced both ways are the asymptotes of
the hyperbola. It will be seen that tlie asymptotes are the diagonals
of the rectangle formed by the tangents at the vertices of the hyper-
bola and its conjugate. The asymptotes are tangents to the hyperbola
and its conjugate at an infinite distance from the centre C.

The following properties should be specially noted : —

(1) Perpendiculars from the foci to the asymptotes are tangents to one
or other of the auxiliary circles. Referring to Fig. 106, FD is a perpen-
dicular to the asymptote CL and FD is a tangent to the auxiliary
circle described on AA^ as diameter and D is the point of contact.
F'D' is a perpendicular to CL and F'D' is a tangent to the auxiliary
circle described on BBj as diameter and D' is the point of contact.

(2) The auxiliary circles intersect the asymptotes at points on the

k^x' ^:>

fli-A

f] 1

"ki^

Fig. 106.

Fig. 107.

directrices. DX (Fig. 106) is a du'ectrix of the hyperbola whose
transverse axis is AA^ and D'X' is a directrix of the conjugate
hyperbola.

(3) The portion HR^ (Fig. 107) of a tangent lying between the asymp-
totes is bisected at the point of contact P, and HHi is equal to the diameter
EGEy ichich is parallel to HHi.

This suggests a method of drawing the tangent to the curve at a
point P on it. Draw PM parallel to CL^ to meet CL at M. Make
MH equal to CM. Join HP and this will be the tangent required.

(4) If from any point P on the curve (Fig. 107) PM and PN be
drawn parallel to the asymptotes, meeting them at M and N respectively,
then the product PM . PN is constant.

This last property may be used to construct an hyperbola when
the asymptotes and a point on the curve are given as follows. Referring
to Fig. 108, CL and CL^ are the given asymptotes and P is a given
point on the hyperbola. Draw PM and PN parallel to CLj and CL
respectively and produce PM and PN both ways. Through C draw a
number of radial lines to cut PM at points 1, 2, 3, etc., and PN at
points r, 2', 3', etc. Parallels to CL through the points 1, 2, 3, etc.

CONIC SECTIONS

will intersect parallels to CLj
points on the hyperbola as shown.

through

r, 2', 3', etc. respectively at

Fig. 108.

50. The Rectangular Hyperbola. — When the transverse and
conjugate axes of an hyperbola are equal the asymptotes are at right
angles to one another, and the hyperbola is then said to be equilateral
or rectangular.

In Fig. 109, OX and OY, at right angles to one another, are the
asymptotes of a rectangular hyperbola and P is a point on the curve.
Other points on the curve are
obtained by the construction
already given and illustrated by
Fig. 108. PM and FN being
parallel to OX and OY respec-
tively, it will be seen that for the
rectangular hyperbola (Fig. 109)
PMON is a rectangle.

If distances from OY parallel
to OX represent to scale the
volume «; of a given weight of a

gas,

and if distances from OX

parallel to Y represent to scale the corresponding pressure ;p of the
gas, then if the gas is expanded or compressed and the pressure is
inversely as the volume, the product pv is constant and the co-ordinates
of points on a rectangular hyperbola will show the relation between
the pressure and volume of the gas as it is expanded or compressed.
The hyperbola is then the expansion curve or the compression curve for
the gas.

51. Centre of Curvature of a Conic— The following con-
struction is applicable for finding the centre of curvature of any conic.

PRACTICAL GEOMETRY

P (Fig. 110) is a point on the curve, F the focus, FG the direction of
the axis, and PG the normal to the curve at P. Draw GH at right
angles to PG meeting PF or PF produced at H.
Draw HS at right angles to PH meeting PG pro-
duced at S. S is the centre of curvature of the
conic at P.

In the case of the parabola the point H may be
obtained by making FH, on PF produced, equal
toPF.

The above construction fails when the point
P is at A the vertex of the conic. It will be
observed that as the point P approaches nearer
and nearer to A, the points G, H, and S approach nearer and
nearer to one another and in the limit they will coincide at a point on
the axis. Now in the parabola FH is equal to PF ; hence the centre
of curvature of the parabola at A, its vertex, is on the axis at a
distance from A equal to 2AF.

In the ellipse and hyperbola PF : PFj : : FG : FjG, and when P
coincides with A, this becomes AF : AF^ : : FS : FiS. Hence the
following construction for the centre of curvature at A. Draw FjD
(Figs. Ill and 112) inclined to ¥,¥. Make F^D equal to AF^, and

Fig. 110.

Fig. 111.

Fig. 112.

DE equal to AF. Join EF, and draw DS parallel to EF meeting the
axis at S. S is the point required.

Another construction which gives the centre of curvature of a
central conic at an extremity
of the transverse axis, and at
the same time gives the centre
of curvature at an extremity
of the conjugate axis, is shown
in Figs. 113 and 114.

Referring to the ellipse
(Fig. 113) CA is the semi-
major axis and CB is the semi-
minor axis. Complete the
rectangle ACBL. Join AB. Fig. 113. Fig. 114.

Draw LSS' at right angles to

AB to cut AC at S and BC produced at S' ; then S and S' are the
centres of curvature of the ellipse at A and B respectively.

CONIC SECTIONS

The ellipse may be quickly drawn with sufficient accuracy for
many purposes by describing circular arcs through the extremities of
the axes from the corresponding centres of curvature and then joining
these arcs with fair curves to please the eye

Referring to the hyperbola (Fig. 114) CA is the semi-transverse
axis and CB is the semi- conjugate axis. Complete the rectangle
ACBL. Join CL. Draw SLS' at right angles to CL to cut CA
produced at S and CB produced at S'. S is the centre of curvature of
the hyperbola at A, and S' is the centre of curvature of the conjugate
hyperbola at B. Note that CL is an asymptote of the hyperbola. •

52. Evolute of a Conic. — The evolute being the locus of the
centre of curvature, if the construction of the preceding article be
applied to a sufficient number of points on the conic, a fair curve
drawn through the centres obtained will be the evolute of the conic.
Fig. 115 shows the evolute of a parabola, and Fig. 116 shows the

Fig. 115,

evolute of an ellipse. In the parabola the evolute cuts the conic at
two points and if the ordinate PN of one of them be drawn, then
AN =8AF. In the ellipse when OB the semi-minor axis is equal to
OF the points S^ and §2 will coincide with B^ and B respectively.
When OB is greater than OF, Sj and 83 will lie within the ellipse as in
Fig. 116, and when OB is less than OF the points Sj and So will lie
outside the ellipse.

53. Pole and Polar.— If through a given point P (Figs. 117 and
118) any straight line be drawn to cut a conic at Q and R, tangents to
the conic at Q and R will intersect on a fixed straight line LM. Con-
versely, the chord of contact of tangents to the conic from any point
in LM will pass through P.

The fixed line LM is called the ;polar of the point P with respect to
the conic, and the point P is called the pole of the line LM with respect
to the conic. The point P may be within or without the conic.

In the case of a central conic (Fig. 117) if C the centre be joined
to P and produced, if necessary, to cut the conic at S, then LM is
parallel to the. tangent to the conic at S. In the case of the parabola
(Fig. 118) if PS be drawn parallel to the axis to meet the curve at S,
then LM is parallel to the tangent to the parabola at S.

50 PRACTICAL GEOMETRY

If CP or CP produced cuts the conic at S and LM at T, then

CP X CT = CSl

If chords QR and KN of a conic (produced if necessary) intersect
at P, then the chords QK and NR (produced if necessary) will intersect
on LM the polar of P. Also the chords QN and KR (produced if
necessary) will intersect on LM the polar of P. This suggests the
simplest construction for finding the polar of a given point P.

Fro. 117. Fig. 118.

A more general way of stating the preceding property is as follows.
If a quadrilateral QNRK be inscribed in a conic, the opposite sides and
diagonals will (produced if necessary) intersect in three points such
that each is the pole of the line joining the other two.

Since the circle is a particular form of a conic it follows that all
that has been said about the pole and polar with respect to a conic will
be true for the pole and polar with respect to a circle. The student
should therefore compare this article with Art. 27, p. 24.

Exercises III

1. Draw an ellipse, a parabola, and an hyperbola as in Fig. 75, p. 31, having
given, FX = 1'5 inches, eccentricity of ellipse = ^, and eccentricity of hyper-
bola = |.

2. Using the curves of the preceding exercise, work out on each, the fol-
lowing : —

(a) P is any point on the conic. PL is the tangent at P, L being on the
directrix. G is the point where the normal at P cuts the axis, and F is the focus.
PG and LF are joined and produced to meet at p. Construct the locus of p.

(&) PQ is a. chord of the conic subtending an angle of say G0° at the focus F.
Determine the locus of the intersection of the tangents at P and Q, and also the
envelope of the chord PQ. [It will be found that the curves required are conies
having the same focus and directrix as the given conic]

3. Using the drawing of exercise 1, take a point T on the directrix XM and
0-5 inch from X, and from T draw all the possible tangents to each of the three
conies, by the method given in Art. 13, p. 12 ; then determine the points of con-
tact by Art. 37, (2).

CONIC SECTIONS 51

4. Construct Figs. 76, 77, and 78, p. 33, as explained in Art. 85, to the follow-
ing dimensions : —

For Fig. 76, vertical angle of cone tvu = SO"*, va = 2-5 inches, va^ = 1-25
inches.

For Fig. 77, vertical angle of cone tva = 60°, va = 1 inch, xXi parallel to vt.

For Fig. 78, vertical angle of cone tvtc = 80°, va = 1 inch, va^ = 0*75 inch.

The eccentricity of the conic being the ratio of AF to AX, construct each
conic separately on another part of the paper by the method of Art. 34 and
Fig. 75. Make a tracing of each conic obtained in this way and test whether it
agrees with the conic determined as a section of the cone.

5. PTQ is a triangle, PT = 2-75 inches, TQ = 1-75 inches, and PQ = 2 inches.
R is a point in PT 1 inch from P. RF is perpendicular to PT, and RF = 1-25
inches. F and Q are on the same side of PT.

PT is a tangent to a conic, P being the point of contact. F is the focus, and
Q is another point on the curve. Find the directrix and draw the conic. [Art.
87, (1) and (2).]

6. PFNQ is a quadjnlateral. The angles at F and N are right angles. PF
= 1-3 inches, FN = NQ = 1*6 inches. P and Q are points on a conic of which F
is the focus and FN the direction of the axis. Construct the conic.

7. F is the focus and S is any fixed point on the axis of a conic. From S a
perpendicular is drawn to the tangent at a point P on the curve meeting FP at Q.
Show by actual drawing that the locus of Q is a circle.

8. A focal chord of a parabola is 2-3 inches long and it is inclined at 30° to
the directrix. The middle point of the chord is at a perpendicular distance of
1'2 inches from the directrix. Draw the parabola.

9. AP, a chord of a parabola, is 1-8 inches long and is inclined at 50° to the
axis. The point A being the vertex of the parabola, draw the curve.

10. TP, a tangent to a parabola from a point T on the axis, is inclined at 30°
to the axis. P is the point of contact, and TP is 3 inches long. Draw as much
of the parabola as lies between the vertex and a double ordinate whose distance
from the vertex is 2*2 inches.

11. Construct a triangle FPQ. FP = 1*5 inches, PQ = 3-4 inches, and QF
= 2-6 inches. Draw a parabola whose focus is F and which passes through
P and Q.

12. Make a careful tracing of the parabola of the preceding exercise without
any lines other than the carve ; then determine the axis, focus, and directrix of
the curve by constructions on the tracing.

13. Draw a line FS 1-3 inches long and a line SP making the angle FSP 45°.
SP is a tangent to a parabola of which F is the focus and S a point on the
directrix. Find the point of. contact of the tangent and draw the parabola.

14. PNPj, a double ordinate of a parabola, is 3-6 inches long. A being the
vertex of the parabola, the area bounded by the curve PAPi and the double ordi-
nate PNPj is 4-8 square inches. Draw the curve PAP,.

15. Construct a triangle RST ; RS = 1-7 inches, ST = TR = 2-6 inches. ST
and SR, both produced, are tangents to a parabola and TR is parallel to their
chord of contact. TR contains the focus. Draw the parabola.

16. ABC is a triangle. AB = 2-2 inches, BC = 2-3 inches, and CA = I'l inches.
D is a point in BC 0-9 inch from B. Draw the parabola which touches BC at D
and AB and AC produced.

17. The normal PG to a parabola at a point P on the curve is 1-9 inches long,
the point G being on the axis. The parameter of the diameter through P is 4-8
inches long. Construct the parabola.

18. The major and minor axes of an ellipse are 4 inches and 3 inches long
respectively. Construct the curve by the trammel method and find the foci .

19. The major axis of an ellipse is 3-5 inches long and the distance between
the foci is 2*5 inches. Draw the ellipse.

20. ABC is a triangle. AB = 2-3 inches, BC = I'l inches, and CA = 2-2
inches. Draw an ellipse whose foci are A and B and which passes through C.

21. The minor axis of an ellipse is 2-2 inches long and the distance between

52 PRACTICAL GEOMETRY

the foci is 2 inches. Draw the ellipse, and construct the locus of the middle
points of all the chords through one focus.

22. The major and minor axes of an ellipse are 3 inches and 2 inches long
respectively. Draw a half of the ellipse which lies on one side of the major axis.
Divide the curve into twelve parts whose chords are equal, and from the points of
division draw normals to the ellipse, each normal to project 0*5 inch outside the
ellipse. Lastly, draw a fair curve through the outer extremities of the normals.

23. The distance between the foci of an ellipse is 2 inches. A tangent to the
ellipse is inclined to the major axis at an angle of 30^^ and cuts that axis produced
at a point 2*5 inches from the centre of the ellipse. Draw the ellipse.

24. SABT is a quadrilateral. The angles at A and B are right angles. SA =
0'75 inch, AB = 3 inches, and BT = 2*25 inches. The points S and T are on the
same side of AB. ST is a tangent to an ellipse of which AB is the major axis.
Construct the ellipse.

25. Two conjugate diameters of an ellipse are 3 inches and 3-5 inches long,
and the angle between them is 60°. Draw the ellipse by each of the three
methods described in Art. 47, p. 42.

26. CO is a straight link, 2 inches long, which revolves about a fixed axis
at C. PON is another straight link, 4 inches long, jointed at its middle point O
to the outer end of CO. N is constrained to move in a straight line which passes
through C. Draw the loci of the middle points of OP and ON and also the locus
of P.

27. Draw a quadrilateral FPOQ. FP = 3-2 inches, angle PFQ = 70°, angle
FPO - 56°, FQ = 1-2 inches, and PO - 1-6 inches. Draw an ellipse touching
OP at P and OQ at Q, and having F for one focus.

28. The transverse axis of an hyperbola is 2 inches long and the distance
between the foci is 2-7 inches. Draw the hyperbola.

29. The transverse and conjugate axes of an hyperbola are 2-2 inches and 1*7
inches long respectively. Draw the hyperbola and the conjugate hyperbola.

30. The foci of an hyperbola are 2-8 inches apart. One point on the curve is
2-9 inches from one focus and 1-1 inches from the other. Draw the hyperbola.

31. The distance between the foci of an hyperbola is 2*9 inches. A tangent
to the hyperbola is inclined to the transverse axis at 48° and cuts that axis at a
point 0"6 inch from its centre. Draw the hyperbola.

32. The transverse axis of an hyperbola is 2 inches long. A tangent to the
hyperbola is inclined at 58° to the transverse axis and cuts that axis at a point
0*8 inch from its centre. Draw the hyperbola.

33. PMMiPj is a quadrilateral. IMM^ = 2-5 inches, the angles at M and Mj
are right angles, PM = 0*9 inch, P,Mi = 1'2 inches. P and P^ are points on an
hyperbola of which MMj (produced both ways) is a directrix, and whose eccen-
tricity is ^. Find the foci and draw the hyperbola.

34. The asymptotes of an hyperbola are at right angles to one another and
one point on the curve is 1 inch from each asymptote. Construct the two
branches of the curve by the method illustrated by Fig. 108, p. 47.

35. One cubic foot of air at a pressure of 100 lbs. per square inch expands
until its volume is 10 cubic feet. The relation between the pressure p and volume
V is given by the formula 2>v = 100. Construct
the expansion curve. Pressure scale, 1 inch to
20 lbs. per square inch ; volume scale, 1 inch
to 2 cubic feet. Draw the tangent and normal
to the curve at the point where the volume is
3 cubic feet.

36. AB and CD (Fig. 119) are two straight
lines of unlimited length. AB revolves with
uniform angular velocity about the centre P,
and CD revolves with the same angular ve-
locity, but in the opposite direction, about
the centre Q. PQ = 2-5 inches. O is the Fig. 119.
middle point of PQ. XiOX and YOY, are two

lines at right angles to one another, the angle POX being 30°. The initial positions

P B

/36^

'.-^

X
P

CONIC SECTIONS 53

of AB and CD are parallel to XjOX. Show by actual drawing that the locus of the
point of intersection of AB and CD is a rectangular hyperbola of which X^OX
and YOYi are the asymptotes and P and Q points on the curve.

37. Draw the complete e volute of an ellipse whose major and minor axes are
4 inches and 2-75 inches long respectively.

38. The focal distance of the vertex of a parabola is 0-3 inch. Draw that part
of the evolute of the parabola which lies between the vertex and a double ordinate
whose distance from the vertex is 5 inches.

39. The angle between the asymptotes of an hyperbola is 60° and the vertex
is at a distance of 1 inch from their intersection. Draw that part of the evolute
of one branch of the hyperbola which lies between the vertex and a double ordi-
nate whose distance from the vertex is 5 inches.

40. Draw an ellipse, major axis 3 inches, minor axis 2 inches. Take a point
P within the ellipse 0-8 inch from the centre C and lying on a line through C
inclined at 30° to the major axis. Through P draw a number of chords of the
ellipse and at their extremities draw tangents to the ellipse. Find the pbint of
intersection of eadh pair of tangents and see whether the points thus obtained
are in one straight line. [By a pair of tangents is meant the tangents at the
ends of a chord.] Eepeat the construction for a point Q lying on CP produced,
CQ = 2 inches.

CHAPTER IV

TRACING PAPER PROBLEMS

54. Use of Tracing Paper in Practical Geometry. — Fre-
quently draughtsmen have to make geometrical constructions on com-
plicated drawings in order to determine some point, line, or figure,
and in such cases the fewer the construction lines the better. By
using a piece of tracing paper in the manner explained in t .ds chapter
the desired result may be obtained very accurately in many cases
without making any construction lines whatever on the drawing
paper, and some problems can be easily solved by this method whigh
would be impossible by ordinary geometrical methods, or which would
otherwise involve very complicated constructions.

55. To find the Length of a Given Curved Line. —Let
ABCD (Fig. 120) be the given curved line. On a piece of tracing
paper TP draw a straight line 1 1,
Mark a point A on this line and
place the tracing paper on the
drawing paper so that this point
coincides with one end A of the
curved line to be measured. Put a
needle point through the tracing
paper and into the drawing paper
at A. Now turn the tracing paper
round until the line 1 1 cuts the
curve at a point B not far from A.
Remove the needle point from A

to B, taking care that the tracing paper does not change its position
during the operation. Next turn the tracing paper round until the
line on it takes up the position 2 2, cutting the curve at a point C not
far from B. The needle point must then be moved to C and the
operations continued until a point on the straight line coincides with
the last point on the curve. The last point obtained on the straight line
must be marked distinctly. The distance between the first and last
points marked on the straight line will be approximately equal to the
length of the curved line. The approximation will be closer the
shorter the steps AB, BC, etc. When the curve has a larger radius
of curvature the steps such as AB and BC may be longer than when
the radius of curvature is smaller. In Fig. 120 the steps, for the

Fig. 120.

TRACING PAPER PROBLEMS

sake of clearness, are of greater length than would be adopted in
practice.

The foregoing method is equivalent to stepping off the length of
the curve with the dividers, but the tracing paper method has the
advantage that the lengths of the different steps may be made to suit
the variations of curvature when the curve is not an arc of a circle.

It is obvious that this method may also be used to mark off a
portion of a given curved line which shall be of a given length.

56. To draw an Involute of a Given Curved Line. — Let
ABCD (Fig. 121) be the given curved line. On a piece of tracing
paper TP draw a straight line 11.
Mark a point A on this line. This
point will be called the tracing point.
Place the tracing paper on the drawing
paper so that the tracing point coincides
with the point A on the curve from
which the involute is to start. Put a
needle point through the tracing paper
and into the drawing paper at A.
Now turn the tracing paper round until
the straight line 1 1 cuts the curve at
a point B not far from A. Remove
the needle point from A to B, taking
care that the tracing paper does not Fig. 121.

change its position during the opera-
tion. Next turn the tracing paper round until the straight line on
it takes up the position 2 2, toucJiing the curve at B, and with a
sharp round-pointed pencil make a mark on the drawing paper
through the needle hole at the tracing point. If these operations be
continued, a number of points are obtained and a fair curve drawn
through them will be an approximation to the involute required.
The approximation will be closer the shorter the steps AB, BC, etc.

57. To draw a Straight Line to pass through a Given
Point and cut two Given Lines so that the Portion inter-
cepted between them shall have a
Given Length.— Let AB and AC (Fig.
122) be the given lines and D the given
point. On a piece of tracing paper TP
draw a straight line EF, and mark two
points H and K on this line such that
HK is equal to the given length. Move
the tracing paper into a number of dif-
ferent positions on the drawing paper, the
point K being on the line AC and the
line EF passing through D. A position Fig. 122.

will quickly be reached in which the point

H is also on the line AB. Now make a mark on the drawmg paper

at F ; a line joining this mark with D will be the line required.

Instead of using tracing paper for this problem, the points H and

PRACTICAL GEOMETRY

K may be marked on the straight edge of a strip of paper. This
strip of paper may then be moved on the drawing paper until a position
is found where H and K lie, one on AB and the other on AC. An
ordinary drawing scale may be used in the same way.

58. To draw the Path traced by one Angular Point of
a Given Triangle while the other Angular Points move,
one on each of two Given
Lines.— Let ABC (Fig. 123) be
the given triangle. Let A be the
tracing point, and let B move on
the given line DE while C moves
on the given line FH. Draw the
triangle ABC on a piece of tracing
paper TP. Make a small hole in
the tracing paper at A with a
needle. Place the tracing paper
on the drawing paper so that B is
on DE and C on FH. With a
sharp round-pointed pencil make a mark on the drawing paper through
the needle hole in the tracing paper at A. This will be one point in
the path required. By moving the tracing paper into other positions
other points may be obtained, and a fair curve KAL drawn through
them will be the required path.

59. To draw an Arc of a Circle through Three Given
Points without using the Centre of the Circle.— Let A, B,
and (Fig. 124) be the given
points. Place a piece of
tracing paper TP on the draw-
ing paper, and draw on the
former
and AE

two straight lines AD
passing through B

Fig. 124.

and C respectively. Make a
small hole in the tracing paper
at A with a needle. Move the tracing paper round into different
positions so that the lines AD and AE always pass through B and C
respectively. For each position of the tracing paper make a mark on
the drawing paper with a sharp round-pointed pencil through the hole
at A. A fair curve drawn through the points obtained in this way
will be the arc required. This construction is based on the fact that
all angles in the same segment of a circle are of the same magnitude.
Another form of this problem is — To describe on a given line BC a
segment of a circle which shall contain an angle equal to a given angle
BAC.

60. To draw any Roulette.— Let AHB (Fig. 125) be the
directing line or base, and let CDE be the rolling curve. (In Fig. 125
the base is a straight line, but it may be any curved line ; and the
rolling curve is a circle, but it may also be any curved line.) The base
is drawn on the drawing paper and the rolling curve is drawn on a
piece of tracing paper. The tracing point P is also marked on the

TRACING PAPER PROBLEMS

Fig. 125.

tracing paper by two lines at right angles, and by a small needle hole.

Place the tracing paper on the

drawing paper so that the { "^ ^ .j^-j^"^^ P5

rolling curve CDE touches

the base, say at C. Make a

mark on the drawing paper

through the needle hole in the

tracing paper at P. Place a

needle through the tracing

paper and into the drawing

paper at C. Turn the tracing

paper round about the needle

at C until the rolling curve

cuts the base at a near point F.

Transfer the needle from C to F and turn the tracing paper until the

rolling curve touches the base at F. The tracing point will now have

moved from P to Pj. Mark the drawing paper at Pj. Again turn

the tracing paper until the rolling curve cuts the base at another

near point H. Transfer the needle from F to H and turn the tracing

paper until the rolling curve touches the base at H. The tracing

point will now have moved to Po. Mark the drawing paper at Pg , and

continue the process, obtaining the points P3, P4, etc. A fair curve

drawn through the points P, Pj , P2, P3, etc., will be the roulette

required. In Fig. 125 the roulette is a trochoid.

61. To inscribe in a Given Figure a Figure similar to
another Given Figure.— Let ABCD (Fig. 126) be the given figure
in which it is required to inscribe
a figure similar to a second given
figure EFHK. Draw the figure
EFHK on a piece of tracing
paper. Take a point on the
tracing paper as a pole (prefer-
ably one angular point of the
figure EFHK, say E) and join
this point to each of the angular
points of the figure on the tracing
paper. Graduate these lines, the
divisions being proportional to
the distances of the pole from
the angular points of the figure.
In the example illustrated EF,

EH, and EK are each divided into two equal parts, and the gradua-
tions are extended on these lines produced. Place the tracing paper
on the drawing paper and move the former about until the sides of the
figure ABCD on the drawing paper cut the lines through the pole
on the tracing paper at corresponding points L, M, and N; then
prick the drawing paper through the tracing paper at these corre-
sponding points and the positions of the angular points of the figure
required are obtained.

Fig. 126.

PRACTICAL GEOMETRY

It may happen that the corresponding points mentioned are not
at points of graduation on the lines through the pole. In that case
the graduations must be made finer in the neighbourhoods where they
appear to be required. If the corresponding points required are still
not at points of graduation, it may be possible, without further sub-
division, to judge by the eye whether the points where the polar
lines cut the sides of the first given figure are corresponding
points.

62. Drawing Symmetrical Curves. — When a curve is sym-
metrical about an axis only the part on one side of that axis need be
constructed, the part on the other side may be quickly and accurately
drawn by means of a piece of tracing paper as follows.

Referring to Fig. 127, at (a) is shown one half of a curve which is
symmetrical about the axis YY. A piece of tracing paper TP is

Fig. 127.

Fig. 128.

placed over this figure and the curve and the axis YY are traced in
pencil. The tracing paper is then turned over and placed as shown at
(h) and the curve on the tracing paper is traced with the pencil. It
will then be found that the second half of the curve has appeared on
the drawing paper in pencil, the lead for this curve having come
from the lead which was put on the tracing paper when it was in the
position (a). After removing the tracing paper the curve traced
should be lined in to make it more distinct.

When a curve is symmetrical about two axes at right angles to one
another only one quarter of the curve need be constructed, the other
three quarters may be drawn, one quarter at a time, by means of a
piece of tracing paper as just described.

Fig. 128 shows the method applied to an ellipse. At (1) is shown
one quarter of the ellipse constructed, say, by the trammel method. A
piece of tracing paper TP is placed over this and on it are traced the
semi-axes and the curve lying between them. Turning the tracing
paper over and placing it as shown at (2) the second quarter of the
curve is transferred to the drawing paper, the lead coming from the
first tracing. Turning the tracing paper over again and placing it as
shown at (3) the third quarter of the curve is transferred to the draw-
ing paper, the lead coming from the second tracing. Turning the

TRACING PAPER PROBLEMS 59

traciDg paper over once more and placing it as shown at (4) the last
quarter of the curve is obtained.

Exercises IV

The foUowing exercises are intended to he luorked out by the tracing
paper method

1. Find the circumference of a circle 3 inches in diameter, and compare the
result with that got by calculation.

2. The sides of a figure ABC are arcs of circles. Radius of AB = li inches,
radius ©f BG = 2 inches, radius of CA = 2J inches. The arcs touch one" another
in pairs at A, B, and G. Find the perimeter of the figure ABC.

3. Draw the involute of a circle 2 inches in diameter. Also draw a tangent
to the circle to meet the involute, the length of this tangent between the involute
and the point of contact with the circle to be 5 inches. Find also the length of
the involute between the starting point and the point where the forementioned
tangent meets it.

4. Draw an ellipse, major axis 2 inches, minor axis 1| inches, and draw that
involute of the ellipse which starts from one extremity of the major axis. Find
the circumference of the ellipse.

5. ABC is an equilateral triangle of 3 inches side. D is a point in AB 1 inch
from A. Draw through D a straight line to cut the side AC at E, and the side
CB produced at F, such that EF = 4J inches.

6. Draw an ellipse, major axis 3 inches, minor axis 2 inches. Take a point P,
2 inches from one extremity of the minor axis, and 1^ inches from one extremity
of the major axis. Through P draw a straight line to cut the ellipse in two points
which shall be 2 inches apart.

7. X'OX and Y'OY are two straight lines intersecting at 0. Angle XOY =
60°. A straight line AB, 3 inches long moves with the end A on X'OX, and the
end B on Y'OY. Draw the complete curve traced by a point P in AB which is 1 J
inches from A. Draw also the curve traced by the middle point of AB.

8. ABC is an equilateral triangle of 3 inches side. The triangle moves with
the point A on the circumference of a circle 2| inches in diameter, and the point
B moves on a diameter of that circle produced. Draw the path traced by the
point G. Draw also the path traced by the middle point of AB.

9. On a straight line 3 inches long descriljp a segment of a circle containing
an angle of 120°, without using the centre of the circle.

10. is the centre and OP a radius of a circle 2 inches in diameter. Q is a
point in OP f inch from O, and R is a point in OP produced, and IJ inches from
O. Draw the cycloid described by P, also the trochoids described by Q and R, as
the circle rolls on a straight line. Draw as much of each curve as is obtained by
a little more than one revolution of the circle.

11. Taking the same rolling circle and the same points Q and R as in the pre-
ceding exercise, draw the epicycloid described by P, and the epitrochoids
described by Q and R as the circle rolls on the outside of a base circle 3 inches in
diameter; draw also the hypocycloid described by P, and the hypotrochoids
described by Q and R as the circle rolls on the inside of the same base circle.
Draw as much of each curve as is obtained by a little more than one revolution
of the rolling circle.

12. Draw the roulette described by one extremity of the major axis of an
ellipse (major axis, 2^ inches, minor axis, If inches) which rolls on the outside of
another ellipse (major axis, 3 inches, minor axis, 2 inches). The roulette to start
from one extremity of the minor axis of the fixed ellipse. ^ Draw also, on the
same figure, the roulette described by one extremity of the minor axis of the roll-
ing ellipse while the first roulette is being described.

13. AB is a straight line 3 inches long. BC is an arc of a circle whose centre

60 PRACTICAL GEOMETRY

is in AB and 2i inches from B, and whose chord is 4 inches long. AC is an arc
of a circle of 3 inches radius whose centre is on that side of AG which is remote
from B. Draw the largest possible equilateral triangle which has its angular
points, one on each of the sides of the figure ABC.

14. ABCD is a quadrilateral. AB = 2i inches, BC = If inches, CD = 1^
inches, DA = 2 inches, and AC = 2f inches. In this quadrilateral inscribe a
square.

CHAPTER V

APPROXIMATE SOLUTIONS TO SOME UNSOLVED PROBLEMS

63. Rectification of Circular Arcs. — The ratio of the cir-
cumference of a circle to its diameter cannot be expressed exactly, in
other words the two are incommensurable. The symbol tt is always
used to denote the ratio of the circumference to the diameter and its
approximate value is 3-1416 or nearly 3h

The best geometrical constructions hitherto given for finding
approximately the length of a circular arc, or for marking off an arc
of given length, are those due to Rankine,^ and are as follows : —

(a) To draw a straight line approximately equal to a given circular
arc AB (Fig. 129), Join BA and produce it to D making AD

Fig. 130.

= iAB. With centre D and radius DB describe the arc BC cutting
at C the tangent AC to the arc at A. AC is the straight line
requij-ed.

Tie error varies as the fourth power of the angle AOB, where O
is the centre of the circle of which AB is an arc. When the angle
AOB is 30°, AC is less than the arc AB by about -^^ of the length
of the arc.

(h) To marJc off" on a given circle an arc AB approximately equal to
a given length (Fig. 130). Draw a tangent AC to the circle at A,
and make AC equal to the given length. Make AD = ^AC. With
centre D and radius DO describe the arc CB to cut the circle at B.
AB is the arc required.

1 A Manual of Machinery and Millwork.

PRACTICAL GEOMETRY

The error in (h) as a fraction of the given length is the same as in
(a), and follows the same law.

If in Fig. 129- the angle AOB is gi-eater than one right angle and
less than two right angles the length of one half of the arc AB should
be determined by the construction and the result doubled. If the
angle AOB is greater than two right angles the length of one quarter
of the arc AB should be determined by the construction and the result
quadrupled.

If in Fig. 130, AC is greater than one and a half times the radius
OA the arc equal to one half of AC should be determined by the
construction and this arc should then be doubled. If AC is greater
than three times OA the arc equal to one quarter of AC should be
determined by the construction and this arc should then be quadrupled.

The construction in (h) follows easily from that in {a), for if the
construction in Fig. 129 be performed and CD be joined (Fig. 131),
and the angle CDB be bisected by DE meeting AC at E, a circle with

Fig. 131.

centre E and radius EC will pass through B. Since CD = DB = 3AD,
and since DE bisects the angle CDB, it follows (Euclid VI, 3) that
CE'=3AEor AE = iAC.

The following slight modification of Rankine's first construction
(a) gives a more approximate result, and is to be preferred, especially
when the angle AOB is greater than 60°. Instead of making AD
equal to half the chord AB make it equal to the chord AF of half the
arc AB (Fig. 132) and proceed as before.

For the case where the angle AOB is 90° the error in Rankine's
construction is about 1 in 170 while in the modified construction the
error is only about 1 in 2300.

64. To draw a Straight Line whose Length shall be
approximately equal to the Circumference of a Given Circle.
— Draw a straight line whose length shall be approximately equal to
a quarter of the circumference by the modification of Rankine's con-
struction explained in the preceding article ; a line four times this in
length will be the line required.

The following construction given by the late Mr. T. H. Eagles ^

^ Constructive Geovietry of Plane Curves, p. 267.

APPROXIMATE SOLUTIONS

gives a very close approximation. O (Fig. 133) is the centre and
AOB a diameter of the given circle. Draw the tangent AC and
make AC equal to three times AB. Draw a radius OD making the

Fig. 133.

angle BOD = 30^. Draw DE at right angles to AB meeting the
latter at E. Join EC. The length of EC is very nearly equal to
the circumference of the circle. EC is a little longer than the true
circumference, the error being about 1 in 21,700.

65. To draw a Straight Line whose Length shall repre-
sent approximately the Value of tt. — The circumference of a
circle of radius r is 27rr, and therefore a quarter of the circumference is

-—. If r = 2, then a quarter of the circumference is equal to tt.
'A

Hence if a quarter of a circle be drawn with radius = 2, the length

of the arc will be equal to tt, and this length may be determined by

one of the constructions of Art. 63. Instead of taking a quarter of

a circle with a radius = 2, a sector whose angle is 60° and radius = 3

may be used, or generally a sector whose angle is n°, and radius

= may be taken, and the length of its arc will be equal to tt.

The circumference of a circle whose radius is 0*5 is equal to tt and
TT may therefore be found by the construction given in the latter part
of Art. 64, by making r equal to 0*5.

66. To find the Side of a Square whose Area shall be
approximately equal to that of a Given Circle.— Solving this
problem is known as " squaring the circle." O is the centre and AB
a diameter of the given circle.

Fir^t Method (Fig. 134). Draw AC at right angles to AB and

Fig. 134.

Fig. 135.

equal to AO. Draw BC cutting the circle at D. Join BD. BD is
the line required. The error in this construction is that BD is too

64 PRACTICAL GEOMETRY

long by an amount equal to 0"0164r, where r is the radius of the
circle.

Second Method (Fig. 135). Produce AB to E, and make BE equal
to three times BO. With centre A and radius AO describe the arc
OC. With centre E and radius EA describe the arc AC to cut the
former arc at C. Draw CE cutting the circle at D. Join BD. BD
is the line required. The error in this construction is that BD is too
short by an amount equal to O'OOOTr, where r is the radius of the
circle.

67. To draw a Straight Line whose Length shall repre-
sent approximately the Value of the Square Root of tt.—
The area of a circle whose radius is r is 7rr^, and if s is the side of a
square whose area is equal to that of the circle, then s^ =7rr'-, or
8 = TaJ TT. Ifr=l, then s = /J tt, and this may be found by one of
the constructions in the preceding article.

68. To find the Side of a Square whose Area shall be
approximately equal to that of a Given Ellipse.— If a and h
are the semi-axes of an ellipse its area is irob.. If r is the radius of a
circle equal in area to the ellipse, then -n-r = Trafc, or r^ = ah, and r is
a mean proportional between a and h and may be found as in Art.
12, p. 11. The side of a square whose area is approximately equal
to that of the circle may then be found by the construction of
Art. 66.

69. To inscribe in a Given Circle a Regular Polygon
having a Given Number of Sides. — AB (Fig. 136) is a diameter
and O the centre of the given circle. With
centre A and radius AB describe the arc
BC. With centre B and radius BA de-
scribe the arc AC cutting the former arc
at C. Divide the diameter AB into as
many equal parts as there are sides in the
polygon. D is the second point of division
from A. Draw CD and produce it to cut
the circle at E. The chord AE is one
side of the polygon, and the others are
obtained by stepping the chord AE round
the circle.

The above construction is exact for an
equilateral triangle, a square, and a hexagon.
For a pentagon the central angle AOE is Fig. 136.

too small, and when the chord AE is stepped

round from A five times, the last point will fall short of A by an
amount which subtends at an angle of nearly a quarter of a degree.
For a heptagon the central angle AOE is too large, and when the
chord AE is stepped round from A seven times, the last point will
be beyond A by an amount which subtends at an angle of a little
more than five-eighths of a degree.

APPROXIMATE SOLUTIONS 65

Exercises V

1. Draw an arc of a circle of 2 inches radius subtending an angle of 60° at
the centre of the circle ; then draw by Rankine's construction, a straight line
equal in length to the arc.

2. From the circumference of a circle of 2 inches radius cut off, by means of
Rankine's construction, an arc equal in length to the radius.

3. Find, by construction, and by calculation, the circular measure of an
angle of 45°.

4. Construct an angle whose circular measure is 1*2.

5. Find, by construction, and by calculation, the circumference of a circle
whose diameter is 2'75 inches.

6. Find, by construction, and by calculation, the diameter of a circle whose
circumference is 6 inches.

7. Find, in the simplest possible way, the diameter of a circle whose circum-
ference is equal to thei sum of the circumferences of two circles one of which is
1-75 inches, and the other 1*25 inches in diameter.

8. Draw a square whose area shall be equal to that of a circle whose radius
is 1*5 inches.

9. Draw a circle having an area equal to that of a square of 2-25 inches
side.

10. Construct a square having an area equal to that of an ellipse whose major
and minor axes are 3-5 inches and 2-5 inches respectively.

11. In a circle of 2 inches radius inscribe a regular heptagon.

CHAPTEE VI

ROULETTES AND GLISSETTES

70. Roulettes. — When one curve rolls without sliding on another
curve, any point connected with the first curve describes on the plane
of the second a curve called a roulette. The curve which rolls is called
the rolling curve or generating curve ^ and the curve on which it rolls is
called the directing curve or base. The directing curve is generally
assumed to be fixed, and is sometimes called t\iQ fixed curve.

When the rolling and directing curves are circles the roulette
becomes a cycloidal curve.

71. General Construction for Drawing a Roulette. — The
best practical method of drawing any roulette is the tracing paper or
transparent templet method which is described in Art. 60, p. 56.

The following is a general construction which may be used for
drawing any roulette. ABCD (Fig. 137) is the base, Abed is the rolling
curve, and P is the tracing point. Take
a number of points h, c, d, etc. on the
rolling curve and determine points
B, C, D, etc. on the base such that the
arcs AB, BC, CD, etc. are equal to
the arcs A?>, 6c, cd, etc. respectively.
If the points are sufficiently near to
one another the arcs may be assumed
equal to their chords. Draw tangents
to the base at B, C, D, etc. and tan-
gents to the rolling curve at h, c, d,
etc. From P draw Pm perpendicular
to the tangent at h. On the tangent
at B make BM = hm, and draw MP,
perpendicular to BM and equal to Pm.
Pi will be the position of P when the
rolling curve touches the base at B,

and will therefore be a point in the roulette described by P.
manner other points may be determined.

72. General Construction for the Centre of Curvature
of a Roulette. — Three cases are illustrated in Figs. 138, 139, and
140. The description which follows applies to each.

Oi is the centre of curvature of the base line AQB at Q. (If AQB

Fig. 187.

In like

ROULETTES AND GLISSETTES

is an arc of a circle then Oj is the centre of the circle, and if AQB is
a straight line (Fig. 140), then O^ is at an infinite distance from Q in
a straight line through Q at right angles to AQB.)

Fig. 139.

Fig. 140.

O.^ is the centre of curvature of the rolling curve CQD at Q. (If
CQD is an arc of a circle, then Og is the centre of the circle.)

P is the position of the tracing point when the rolling curve is in
the position shown.

Join PQ. Draw QR at right angles to PQ to meet POg or POg
produced at R. Join ROj and produce it if necessary to meet
PQ or PQ produced at S. S is the centre of curvature of the roulette
at P.

When P, O2, and Q are in a straight line the above construction

T ^i,- ark QO1XQO2XPQ ^, , . ^.

In this case, SQ = ==: ^=—4=—-^^^, the plus sign being

fails.

QOi X POo^t QO2 xPQ
taken when 0^ and O2 are on opposite sides of Q, and the minus
sign being taken when O, and O2 are on the same side of Q. PQ
= PO2 ± QO2.

If AB is a straight line then QO^ is infinite and

SQ =

If PO2 = QO2, then SQ =

QO2 X PQ

PO2 *

QOi X PQ
QOi ±Fq'

73. The Cycloid. — The cycloid is the curve described by a point
on the circumference of a circle which rolls on a straight line, the
circle and straight line remaining in the same plane. The ordinary
geometrical construction for drawing the cycloid is shown in Fig. 141.
mo'e' is the straight line upon which the circle "Pho'q rolls and P is the
point which describes the cycloid. Tho'q is the rolling circle in its
middle position, the diameter POo' being at right angles to mo'e'.
Make o'e' equal to half the circumference of the rolling circle. Divide
o'e' into a number of equal parts at a', h', c', etc. (preferably six, but for

PRACTICAL GEOMETRY

the sake of a clearer figure o'e' in Fig. 141 has been divided into five
equal parts). Divide the semicircle Pbo' into the same number of
equal parts at a, 6, c, etc. Through O draw OE parallel to o'e'. From
rt', b', c', etc. draw perpendiculars to o'e' to meet OE at A', B', C, etc.
With centres A', B', C, etc. describe arcs of circles touching o'e\
and through a^ 6, c, etc. draw parallels to o'e' to cut these arcs at
A, B, C, etc. as shown. The points A, B, C, etc. are points on the
half of the cycloid traced by the point P as the circle P6o' makes half
a revolution to the right. Points on the other half of the cycloid
may be obtained in a similar manner, or, since the curve is sym-
metrical about Po' the part to the left of Po' may be copied from the
part to the right.

If from a point q on the circle Vbo'q a line be drawn parallel to
the base line mo' to meet the cycloid at Q then QR parallel to qo' is
the normal and QT parallel to qV is the tangent to the cycloid at Q.

■

►

— ^

^^-"^

'^f/

"^'^^s. /^ /

>v / A/"

A 1

B' ] C'/N

. E

\. I J^^^

1 v

/ /

\\/

^rS^v,^^

^^^

/ 1//

Y ^\::\

^^v ^

r^^

^.^

& a! /ycj.

Fig. Ml.

Again, if PT be drawn parallel to md to meet at T the tangent to the
cycloid at Q, the length of the arc PQ of the cycloid will be twice the
length of the tangent QT or twice the length of the chord g-P.
Hence the total length of the cycloid is four times the diameter of the
rolling circle.

If QR be produced to S and RS is made equal to QR then S is the
centre of curvature of the cycloid at Q. The locus of S is the evolute
of the cycloid. If the rolling circle be drawn below the base line
md and touching that line at R it is obvious that this circle will pass
through S. Also if the rolling circle be drawn above the base line and
touching it at R this circle will pass through Q and since the chord
RS is equal to the chord QR the arc RS must be equal to the arc QR.
But the arc QR is equal to mR, and md is equal to half the circum-
ference of the rolling circle, therefore if RU is a diameter of the
circle RSU, the arc S CJ is equal to Ro'. Hence if a line UL be drawn
pai'a-Uel to md and the circle RSU be made to roll on this line the

ROULETTES AND GLISSETTES

The

point S will describe a cycloid equal to the original cycloid,
evolute of a cycloid is therefore an equal cycloid.

74. The Trochoid.— When a circle rolls on a straight line and
remains in the same plane, a point in the plane of the circle, con-
nected to the circle but not on its circumference, describes a trochoid.
If the describing point is outside the rolling circle the trochoid is
called a superior trochoid. If the describing point is inside the rolling
circle the trochoid is called an inferior trochoid. A superior trochoid
is also called a curtate cycloid and an inferior trochoid is also called a
prolate cycloid.

The geometrical construction for finding points on a trochoid is
similar to that already given for the cycloid and is shown in Fig. 142.
mo'^n' is the base line. P is the position of the describing point when it
is furthest from the base line and O is the corresponding position of

the centre of the rolling circle, o'm' is made equal to half the circum-
ference of the rolling circle. The points a', b', c', etc. and A', B', C, etc.
are determined as for the cycloid.

With centre and radius OP describe a circle, and divide the half
of this circle which is to the right of OP into as many equal parts as
o'm' is divided into. This determines the points a, h, c, etc. Through a, b, c,
etc. draw parallels to the base line and with centres A', B',C',etc. andradius
equal to OP describe arcs of circles to cut these parallels at A, B, C,
etc. as shown. A, B, C, etc. are points on. the half of the trochoid
described by the point P as the rolling circle makes half a revolution
to the right. Points on the other half of the trochoid may be obtained
in a similar manner, or, since the curve is symmetrical about OP, the
part to the left of OP may be copied from the part to the right.

If from a point q on the circle qFb a line be drawn parallel to the
base line to meet the trochoid at Q, then QR parallel to cjo' is the

PRACTICAL GEOMETRY

normal and QT perpendicular to QR is the tangent to the trochoid
at Q.

Draw RV perpendicular to the base line to meet the parallel
through to the base line at V. V is the position of the centre of
the rolling circle when the describing point is at Q.

To find S the centre of curvature of the trochoid at Q, draw
RN perpendicular to QR to meet QV produced at N" and draw
NS perpendicular to the base line«to meet QR produced at S.

The evolute of the trochoid is the locus of the centre of curvature
S. The evolute of the superior trochoid is shown in Fig. 142. The
evolute of the inferior trochoid is partly on one side and partly on the
other side of the curve, and the normals which have the least inclina-
tion to the base line are asymptotes of the evolute.

75. The Epicycloid. — When a circle rolls on the outside of a
fixed circle, the two circles being in the same plane, a point on the
circumference of the rolling circle describes an epicycloid.

The ordinary geometrical construction for finding points on an
epicycloid is shown in Fig. 143. P is the position of the tracing point

Fig. 143.

when it is furthest from the fixed or base circle, is the corresponding
position of the centre of the rolling circle and o' is the corresponding
point of contact of the rolling and base circles. Oj is the centre of the
base circle. Make the arc om equal to half the circumference of the
rolling circle. Join O^m and produce it to meet at M the circle through
O concentric with the base circle. Divide the arc OM into a number
of equal parts (preferably six or eight) at A', B', C, etc. and divide the
semicircle Pco' into the same number of equal parts at a, h, c, etc.

ROULETTES AND GLISSETTES

With centre d, describe arcs of circles through a, h, c, etc. and with
centres A', B', C, etc. and radius equal to OP describe arcs of circles
to cut the former arcs at A, B, 0, etc. as shown. A, B, C, etc. are
points on the half of the epicycloid described by P as the rolling circle
rolls from o' to m. Points on the other half of the epicycloid may be
obtained in a similar manner, or, since the curve is symmetrical about
OiP, the part to the left of OiP may be copied from the part to the
right.

The centre of curvature of the epicycloid at any point Q is found as
follows. Draw the rolling circle in the position which it occupies
when the tracing point is at Q as shown. V is the centre of the
rolling circle in this position and R is its point of contact with the
base circle. Join QV and produce it to meet the rolling circle again at
N. Join OiN. Join QR and produce it to meet OjN at S. QR is the
normal and S is the centre of curvature of the epicycloid at Q.
QT at right angles to QR is the tangent to the epicycloid at Q.

The locus of S is the evolute of the epicycloid.

Join OiR and draw SU perpendicular to QS to meet O^R at U. On
UR as diameter describe the circle RSXJ, and with centre Oi and
radius OiU describe another circle. If the circle RSU be made to roll
on the outside of the circle whose centre is Oi and radius OiU, the
point S will describe an epicycloid which will coincide with the
evolute of the original epicycloid. This epicycloid which is the evolute
of the original epicycloid is similar to it. Two epicycloids are similar
when the ratio of the radii of their rolling circles to one another is the
same as the ratio of the radii of their base circles.

It may be pointed out here that a given epicycloid on a given
base circle may be described by a point on the circumference of either
of two different rolling circles. Referring to Fig. 144, APB is an

Fig. 144.

Fig. 145.

epicycloid on the base circle LMN. This epicycloid may be described
by a point on the circumference of the rolling circle PQN or by a
point on the circumference of the rolling circle CPD. The three
circles are such that the diameter of the base circle is equal to the
difference between the diameters of the two rolling circles. It will

PRACTICAL GEOMETRY

be observed that while both rolling circles roll on the outside of the
base circle the base circle is inside the larger rolling circle.

An interesting case of the epicycloid is that in which the rolling
circle is equal to the base circle. This is shown in Fig. 145, where
OLN is the base circle and OQP the epicycloid. PMN is the position
of the rolling circle when the tracing point P is furthest from the base
circle. The two ends of the epicycloid meet at forming a cusp at
that point.

If a straight line OLQ be drawn cutting the base circle at L and
the epicycloid at Q then it is not difficult to prove that LQ is equal to
ON the diameter of the base circle. This property suggests a simple
method of finding points on this particular form of epicycloid. This
form of the epicycloid is known as the cardioid. If the angle which
OQ makes with OP be denoted by ^ then 0Q = QL + L0 = QL4-0N cos ^
or r = tZ (1 + cos 6) where r = OQ and d is the diameter of the circle
OLN". This is the polar equation to the cardioid.

The normal to the curve at Q may be found by the construction
already given or more simply as follows. Through N draw NRj
parallel to OQ to meet the circle again at Rj. QRj is the required normal.
The tangent QT at Q is of course perpendicular to QR^.

76. The Epitrochoid. — When a circle rolls on the outside of a
fixed circle, the two circles being in the same plane, a point in the
plane of the rolling circle, connected to it but not on its circumference
describes an epitrochoid. The epitrochoid is called a superior epitrochoid
or an inferior epitrochoid according as the describing point is outside or
inside the rolling circle.

Fig. 146.

ROULETTES AND GLISSETTES

The geometrical construction for finding points on an epitrochoid
is shown in Fig. 146. The construction is so like that already used
for the cycloid, the trochoid, and the epicycloid that no detailed
description of it need be given here.

The centre of curvature of the epitrochoid at any point Q is
found as follows. Draw the rolling circle in the position which it
occupies when the tracing point is at Q as shown. V is the centre of
the rolling circle in this position and R is its point of contact with the
base circle. Join QV and produce it. Join QR and produce it.
Draw RN at right angles to QR to meet QV produced at N. Join
OjN cutting QR produced at S. QR is the normal and S is the centre
of curvature of the epitrochoid at Q.

77. The Hypocycloid. — When a circle rolls on the inside of a
fixed circle, the two circles being in the same plane, a point on the cir-
cumference of the rolling circle describes a hypocycloid.

The geometrical construction for finding points on a hypocycloid is
shown in Fig. 147. After what has been done in preceding articles
no detailed description of this construction is necessary.

.^'

<f'

MLb'W

pT/ A

Fig. 147.

PRACTICAL GEOMETRY

The construction for finding S, the centre of curvature of the
hypocycloid at any point Q, is the same as that already given for the
epicycloid except that S is on O^N produced instead of on O^N.

Join O^R and draw SU perpendicular to QS to meet O^R produced
at U. On UR as diameter describe the circle RSU, and with centre
Oi and radius OiU describe another circle. If the circle RSU be made
to roll on the inside of the circle whose centre is Oj and radius OiU,
the point S will describe a hypocycloid. This hypocycloid which is the
evolute of the original hypocycloid is similar to it.

A given hypocycloid on a given base circle may be described by a
point on the circumference of either of two different rolling circles.
Referring to Fig. 148, ACBD j^^

is a base circle of which COD -^--^

is a diameter. The hypo-
cycloid APB may be described
by a point on the circumference
of the circle CP or by a point
on the circumference of the
circle DP. The three circles
are such that the diameter of
the base circle is equal to the
sum of the diameters of the
two rolling circles.

Fig. 148 also shows the
variation in the form and size
of the hypocycloid as the
diameter of the rolling circle
is altered.

A very important case of Fig, 148.

the hypocycloid is that in

which the diameter of the rolling circle is equal to the radius of the
base circle. In this case the hypocycloid becomes a straight line EOF
(Fig. 148) which is a diameter of the base circle.

78. The Hypotrochoid. — When a circle rolls on the inside of a
fixed circle, the two circles being in the same plane, a point in the
plane of the rolling circle, connected to it but not on its circum-
ference describes a hypotrochoid. The hypotrochoid is called a superior
hypotrochoid or an inferior hypotrochoid according as the describing
point is outside or inside the rolling circle.

The geometrical construction for finding points on a hypotrochoid
is shown in Fig. 149.

The construction for finding the centre of curvature at any point
is similar to that for the ej)itrochoid. The point S corresponding to
the point Q on the inferior hypotrochoid (Fig. 149) falls outside the
lower limit of the figure.

A special and important case of the hypotrochoid is that in which
the diameter of the rolling circle is equal to the radius of the base
circle. In this case the hypotrochoid is an ellipse. Referring to Fig.
150, AMLN is the base circle whose centre is O, and AHO is the
initial position of the rolling circle. C is the centre of the circle

ROULETTES AND GLISSETTES

. 76

EVOLUTE OF
SUPERIOR
HYPOTROCHOID

Fig. 150.

AHO, B is a point in AO and D and E are points in AG produced.
aho is another position of the rolling circle. The points A and G are
the initial positioDS of the points on the rolling circle which describe
the straight lines AGL and MGN respectively, AGL and MGN being

PRACTICAL GEOMETRY

diameters of the base circle at right angles to one another. The centre
of the rolling circle will obviously describe a circle whose centre is O
and radius OC. When the rolling circle has moved into the position
aho^ the line whose initial position is AOE will now be in the position
aoe^ and if ah, ad, and ae be made equal to AB, AD, and AE re-
spectively the points h, d, and e will be points on the hypotrochoids
described by the points whose initial positions are B, D, and E
respectively.

The construction for finding the points h, d, and e is obviously the
same as in the trammel method for drawing an ellipse described in
Art. 45, p. 41.

79. The Involute of a Circle. — The involute of a circle is the
roulette described by a point on a straight line which rolls on a fixed
circle. The involute of a circle is therefore a special case of the epi-
cycloid, being the epicycloid when the rolling circle is of infinite
diameter.

The geometrical construction for finding points on the involute is
shown in Fig. 151. oP is a diameter of the circle. Draw the tangent
am and make cm equal to half
the circumference of the circle.
Divide the semicircle P^o into
a number of equal parts, say
six, at a, h, c, etc. continuing
the divisions on to the other
half of the circle if necessary,
and divide om into the same
number of equal parts at
a', h\ c', etc. continuing the
divisions beyond m if necessary.
At the points a, h, c, etc. draw
tangents a A, 6B, cC, etc. to the
circle and make a A, 6B, cC, etc.
equal to oa', oh', oc', etc. respec-
tively. P, A, B, C, etc. are
points on the involute which
starts at P and may be con-
tinued to any length. As the involute gets further and further from
the circle the distance between the points as found by the above con-
struction gets greater and greater, but intermediate points may be
found by subdividing as shown for the points U and V.

The normal to the involute at any point Q is the tangent QS to the
circle and S the point of contact of this tangent and the circle is the
centre of curvature of the involute at Q. The tangent QT to the
involute at Q is perpendicular to QS.

80. The Catenary as a Roulette. — The curve known as the
catenary will be referred to here as a roulette since it may be described
by the focus of a parabola rolling on a fixed straight line.

Referring to Fig. 152, RAL is the fixed straight line, and BAC
is the position of the parabola when its axis AY is perpendicular to

Fig. 151.

ROULETTES AND GLISSETTES 77

RAL. P is the focus and A the vertex of the parabola BAG When
the parabola has rolled into the position bac the focus has moved to

Fig. 152.

Q and has traced the arc PQ of the catenary. R being the point of
contact of the parabola hac and the straight line RAL, the straight
line QR is the normal and QT perpendicular to QR is the tangent to
the catenary at Q.

81. Envelope Roulettes. — The roulettes so far considered have
been curves traced by points and these may be called point roulettes.
If the rolling curve
carries with it a
straight or a curved
line the envelope of
the carried line is
called an envelope
roulette. -

An example of
an envelope rou-
lette is illustrated
by Fig. 153. AB
is a fixed circle on
the outside of which
another circle of the
same diameter rolls.
The rolling circle
carries a tangent,
the envelope of
which is to be
drawn. When the
rolling circle is in
its initial position
CD the carried tan-
gent is in the position EF being then furthest from the base circle.
The rolling circle in successive positions may be drawn as in

Fig. 153.

PRACTICAL GEOMETRY

determining points on an epicycloid and the corresponding positions of
the carried tangent can then be added. In Fig. 153 the carried
tangent is shown in thirteen positions, numbered from to 12. The
construction lines for position 4 of the carried tangent are completely
shown. The fair curve which touches the carried tangent in its
successive positions is the envelope roulette of that tangent.

It will be observed that this roulette has two cusps. By mathe-
matical analysis it can be shown that the tangents to the roulette at
the cusps pass through P and touch a circle concentric with AB and
having a radius equal to one-third of the radius of AB.

When the carried line is not a straight line or a circle or when
the base line or the rolling curve is not a circle, the tracing paper
method should be used. Successive positions of the carried line are
then best found by placing a piece of carbonized paper between the
tracing paper and the drawing paper, under the carried line, and going
over the latter with a hard sharp-pointed pencil or a style.

82. Glissettes. — When a line is made to slide between two fixed
points, or between a fixed point and a fixed line, or between two fixed
lines, a point carried by the sliding line describes a point glissette. The
envelope of the sliding line or of a line carried by it is an envelope
glissette. The lines referred to may be either straight or curved.

Examples of glissettes are shown in Figs. 154 and 155. In Fig.
154 OX and OY are fixed straight lines at right angles to one another.

AB is a straight line of definite length which slides between OX
and OY. The middle point Q of AB describes a circle whose centre
is at O and whose radius is equal to half the length of AB. Any
point P in AB describes an ellipse, whose centre is at O and whose
axes lie on OX and OY. The semi-axis on OX is, ^c^ual to BP and
the semi-axis which is on OY is equal to AP.

ROULETTES AND GLISSETTE 79

The envelope of AB is a hypocycloid described by a point on the
circumference of a circle, whose radius is equal to a quarter of AB,
rolling inside a circle whose centre is at O and whose radius is equal
to AB as shown.

In Fig. 155, P and Q are fixed points. ABC is a triangle whose
sides AB and AC, or these sides produced, always pass through P and
Q respectively. The locus of A is obviously a circle PQEA. Let ABC
be one position of the triangle. Draw AD parallel to BC to meet the
circle PQEA at D. Since the angle DAQ is equal to the angle ACB
the arc DEQ is of constant length and D is a fixed point. The
envelope glissette of BC or BC produced is a circle whose centre is D
and whose radius is equal to the perpendicular distance of A from BC.
The triangle is shown in three different positions.

Exercises VI

1. Draw the curve traced by a point on the circumference of a circle 2 inches
in diameter as the circle rolls on a fixed straight line and makes one revolution.
Draw also the evolute of the curve.

2. ABC is an equilateral triangle of 3 inches side. P is the middle point
of AB. PQ is a circle 2 inches in diameter outside the triangle and touching
AB at P. Draw the path traced by the point P on the circumference of the
circle as the latter rolls round the triangle. Draw also the evolute of the path
of P.

3. A circle PQ, 1*8 inches in diameter, touches a straight line PR at P. The
length of PR is equal to the circumference of the circle PQ. The circle rolls with
uniform velocity on the straight line from P to R while the straight line turns
with uniform velocity about P through an angle of 270°. The angular motion of
the line is in the same direction as the angular motion of the circle. Draw the
curve traced by a point on the circumference of the circle, the initial position of
the tracing point being P.

4. A circle 3 inches in diameter rolls on a straight line. P and Q are two
points carried by the circle. P is 2 inches and Q is 1 inch from 0, the centre of
the circle, and O, P, and Q are in the same straight line. Draw the trochoids
traced by the points P and Q. Draw also the evolutes of these trochoids.

5. Draw the epicycloids traced by a point on the circumference of a circle
2 inches in diameter which rolls on the outside, (1) of a circle 6 inches in
diameter, (2) of a circle 4 inches in diameter, and (3) of a circle 2 inches in
diameter. Draw also the evolutes of these epicycloids.

6. Draw the epicycloid traced by a point on the circumference of a circle
6 inches in diameter which rolls on the outside of a base circle 4 inches in
diameter, the base circle being inside the rolling circle. Make a tracing of this
epicycloid and apply it to the epicycloid (2) of the preceding exercise to show that
the two curves are identical.

7. Take the rolling circle and carried points P and Q as in exercise 4 and
draw the epitrochoids traced by P and Q when the base circle is 6 inches in
diameter. Draw also the evolutes of these epitrochoids.

8. A rolling circle 3 inches in diameter carries a point P 2^ inches from its
centre. This circle rolls on the outside of a base circle 2| inches in diameter,
the base circle being inside the rolling circle. Draw the epitrochoid traced by
the point P. Draw also the evolute of this epitrochoid.

9. Draw the hypocycloid traced by a point on the circumference of a circle
2 inches in diameter which rolls on the inside of a circle 6 inches in diameter.
Draw also the evolute of this hypocycloid.

PRACTICAL GEOMETRY

10. Draw the hypocycloid traced by a point on the circumference of a circle
4 inches in diameter which rolls on the inside of a circle 6 inches in diameter.
Make a tracing of this hypocycloid and apply it to the hypocycloid of the preced-
ing exercise to show that the two curves are identical.

11. A rolling circle 2*5 inches in diameter carries two points P and Q which
lie on a straight line passing through its centre 0. OP = 2 inches, and OQ
^ 0-75 inch. Draw the hypotrochoids traced by the points P and Q, the base
circle being 9 inches in diameter. Draw also the evolutes of these hypotrochoids.

12. The same as exercise 11 except that the base circle has a diameter twice
that of the rolling circle.

13. A circle A (Fig. 156), of diameter EF = 3| inches, rolls on the line CD with
uniform velocity from left to right, starting from E. Another circle B, whose
diameter is half that of A, rolls inside the circumference

of A, also with uniform velocity, but from right to left,
starting at E when A begins to move. Circle B is in con-
tact with circle A at F at the same time that F reaches
the line CD. Draw the curve described by the centre of
the circle B. [b.e.]

14. AB, 2 inches long, is: a diameter of a circle. BC,
4 inches long, is a tangent to the circle. Draw that
involute of the circle which passes through the point C.

15. Draw the roulette described by one extremity
of the major axis of an ellipse (major axis 2J inches,
minor axis If inches) which rolls on the outside of
another ellipse (major axis 3 inches, minor axis 2 inches),
the roulette to start from one extremity of the minor
axis of the fixed ellipse. Draw also, on the same figure,
the roulette described by one extremity of the minor axis of the rolling ellipse
while the first roulette is being described.

16. Draw the roulette described by one focus of an ellipse (major axis 2^
inches, minor axis 1 f inches) while the ellipse rolls on a straight line.

17. A parabola, focal distance of vertex 1*2 inches, rolls on a straight line.
Draw the roulette traced by the focus of the parabola.

18. APB is a circle 2 inches in diameter. PCD is an involute of this circle.
Draw the curve traced by the point P on the circumference of the circle as the
latter rolls on the outside of the involute PCD, the point P returning to PCD.
Draw also the evolute of the path of P.

19. A circle 2 inches in diameter rolls on the outside of a fixed circle 4 inches
in diameter. Draw the envelope roulette of a tangent carried by the rolling
circle.

20. XOX' and YOY' are two fixed straight lines at right angles to one another.
PQR is a straight line. PQ = QR = 1*5 inches. The line PQE, moves so that Q
is always on XOX' and E- is always
on YOY'. Draw the complete path
traced by the point P. Draw also
the envelope glissette of the moving
line.

21. Same as exercise 20 except
that the angle between the fixed
lines is 60° instead of 90°.

22. ABC (Fig. 157) is an equi-
lateral triangle of 2 inches side. P is
the centre of a circle of 0-5 inch
radius. Q is the centre of a circle of
1 inch radius. PQ = 3 inches. The
triangle ABC moves so that the sides
AB and AC, or these sides produced,
touch the circles whose centres are at P and Q as shown. Draw the glissette of
the point A and the envelope glissette of the side BC of the triangle ABC.

23. dPb (Figs. 158 and 159) is a moving variable triangle. The sides aP, P6,
and ba always pass through the fixed points A, B, and C respectively. The points

Fig. 157.

ROULETTES AND GLISSETTES

a and h move on the fixed lines YOY' and X'OX respectively. The angle YOX is
a right angle. The dimensions given are in inches. Draw the locus of the point

\,-

^r-

_________/

l^r^

7p~

1 \

i -1- -
1

1 X

,_^4-

ih-

---/

1 ^
1

x'i

)

1
\ X

Fig. 158.

Fig. 159.

P In case I (Fig. 158) the locus is an ellipse. In case II (Fig. 159) the locus is
an hyperbola.

CHAPTER YII

VECTOR GEOMETRY

83. Scalars, Vectors and Rotors or Locors. — A quantity
which may be completely specified by stating the kind of quantity and
its magnitude is called a scalar quantity or scalar. Thus the weight of
a body is a scalar quantity. The wfeight of a body may be, say, 2 tons
or 4480 pounds. The area of a plane figure is a scalar qfuantity. An
area may be, say, 10 square feet or 1440 square inches.

In general the kind of quantity referred to is known from the name
of the unit used in specifying its magnitude. Thus if a scalar quantity
is 10 square feet it is known that the kind of quantity is area.

Time, temperature, volume, and energy are a few other examples of
scalar quantities.

A scalar quantity is not associated with any definite direction in
space but has magnitude only.

A scalar quantity may be represented by the length of a straight
line drawn to scale, such line being drawn anywhere and in any
direction. Thus an area of 30 square feet may be represented by a
line 3 inches long. In this case the scale would be 1 inch to
10 square feet.

A quantity which, while having magnitude like a scalar quantity,
is also associated with a definite direction is called a vector quantity or
vector. Displacement, velocity, acceleration, and force are vector
quantities because they have magnitude and direction. A displace-
ment is referred to when it is stated that a body is moved 10 miles in the
direction from west to east. A velocity is referred to when it is stated
that a body is moving at the rate of 10 miles an hour in the direction
from north to south.

All vector quantities have magnitude and direction but they may
have other qualities which distinguish them from one another and it
will presently be seen that there are important propositions relating
to vectors apart from qualities other than magnitude and direction
which they may possess.

A vector may be represented by a straight line AB drawn to scale
in a definite direction. The length AB of the line represents the
magnitude of the vector and the direction of the line represents the
direction of the vector, provided it is made clear vvhether the direction
is from A to B or from B to A. A line evidently has two directions,
one being exactly opposite to the other and the distinction between

VECTOR GEOMETRY 83

them is the sense of the direction. The sense of the direction of a
vector which is represented by a line is best shown by an arrow head
placed on the line. If the line which represents a vector is lettered
at its extremities, it might be agreed to place the letters so that in
reading them in the order in which they occur in the alphabet the
sense of the direction would be given. For example if the letters at
the extremities of the line are A and B then the sense of the
direction would be from A to B.

A quantity which in addition to having magnitude and direction
also has position is called a rotor or locor. A rotor or locor is
therefore a localized vector. A locor has the qualities of a vector in
addition to its quality of having position. A force acting at a definite
point of a body is an example of a locor. A displacement of a body
from one definite position to another is another example.

It has been seen that the vector part of a locor may be represented
by a straight line and in order that this straight line may also
represent the locor all that is necessary is to place the straight line
in the proper position.

Absolute position in space cannot be defined and the position of a
point or line can only be fixed in relation to other points or lines. In
considering problems on locors it is only necessary to know their
relative positions.

Since vectors have direction but not position, lines which represent
vectors may be placed anywhere, provided that they have the proper
directions.

The following is a convenient way of specifying a vector. Let OX
(Fig. 160) be a fixed direction of reference, say from west to east, and
let OA be a vector whose sense is from O to A and
whose magnitude OA is equal to a. Also let the
angle XOA, measured in the anti-clockwise direction,
be denoted by 6. Then if the vector OA is referred
to as the vector A it may be specified by the equation
A = a. An extension of this method to locors is
described in Art. 91, p. 91.

84. Addition of Parallel Vectors. — Let A, B, C, and D
(Fig. 161) be parallel vectors, it is required to find a single vector
which is the sum or resultant of A, B, C, and D.

Draw a straight line XiX parallel to A, B, C, and D. Take a
point O in XjX. Mark off
on XjX a distance Oa equal Ai — >— < (I d b

to A. Observe that as the Bi * ' Xj — j I I h~^^

sense of A is from left to right C i * •

Oa is measured to the right of Di * ' ^^ *

0. Make ah equal to B, Fig. 161.

measuring to the right of a

because the sense of B is from left to right. Make he equal to C,

measuring to the left of h because the sense of C is from right to left.

Make cd equal to D, measuring to the right of c because the sense of

D is from left to right. Then Od is a vector which is the sum or

84 PRACTICAL GEOMETRY

resultant of the vectors A, B, C, and D. If R is the resultant of the
vectors A, B, C, and D, then R = A-|-B + C + D.

If a vector whose sense is from left to right is said to be posi-
tive, then a vector whose sense is from right to left would be said
to be negative. Then the result of the above example would be
written R = A + B-C + D. Opposite signs represent opposite
senses.

Subtraction of parallel vectors is converted into addition by
changing the signs of the vectors to be subtracted and then pro-
ceeding as in addition.

It should be observed that in the addition of vectors the order in
which they are taken does not aftect the result. For A -h B -f C + D
= A + B + D-fC = B-fD4-A + C.

85. Addition of Inclined Vectors. The Vector Polygon.

Let A, B, and C (Fig. 162) be three given vectors, it is required to
find a single vector which is the sum or re-
sultant of the vectors A, B, and C. b > r

Take a point o and draw the vector oa C \ > i /\\ /^
parallel and equal to A. The sense of oa is ^ / \ \^' \

from to a the same as that of A. The \ya^ \ f^/'^^
vector A is now represented by oa. From >fX ^v I '/' ' '
a draw the vector ah parallel and equal to B. /^ V ^J/' '

The sense of ah is from a to h the same as ° ^ £, ""/l

that of B. The vector B is now represented Fig. 162.

by ah. From h draw the vector he parallel

and equal to C. The sense of he is from 6 to c the same as that of

C. The vector C is now represented by he. Join oc, then the vector

oe whose sense is from o to c is the sum or resultant of the

vectors A, B, and C. The polygon oahe is called a vector polygon.

The vector polygon is of very great importance in vector geometry
and it has numerous applications in mechanics.

It will be useful to consider the vector polygon as applied (1) to
displacements, (2) to velocities, (3) to accelerations, and (4) to
forces.

(1) Suppose a man to be standing on the deck of a ship which is
sailing through water which is itself moving in relation to the earth.
Let the man walk across the deck along a straight line whose length
and direction in relation to the ship are given by the vector A
(Fig. 162). Let the vector B represent the direction and distance
which the ship sails through the water in the time that the man
walks the distance A. Lastly let the vector C represent the
direction and distance which the water moves in the same time.

Let the vector polygon oahe (Fig. 162) be drawn. While the man
is walking along oa that line is carried parallel to itself by the ship
into the position mh. Consequently, neglecting for the moment the
motion of the water, the man travels along the imaginary line oh over
the water. But while this is happening the imaginary line oh is
travelling parallel to itself with the water and reaches the position nc
when the man has finished his walk. The actual displacement of the

VECTOR GEOMETRY

man in relation to the earth is therefore represented by the line oc.
That is, oc is the sum or resultant of the three displacements A, B,
and C.

(2) Since velocity is displacement or distance moved in unit time

it follows that if t is the time taken by the man in walking the

distance A, t will also be the time of the displacement B of the ship

suad of the displacement C of the water, and if each displacement is

divided by t the results are the several velocities. Hence the

. . oa ah he oc

velocities are y, —, — , and —^ and the polygon oahc, measured with

a suitable scale, will be a polygon of velocities in which oa represents
the velocity of the man's walking on the ship, ah represents the
velocity of the ship through the water, he represents the velocity of
the water over the earth, and oc is the resultant velocity of the man
or his velocity in relation to the earth.

(3) Let / denote the uniform acceleration or uniform rate of
increase of the velocity of a moving body, then in time t the increase
in velocity is ft. If A, B, and (Fig. 162) represent three accelera-
tions simultaneously imj^ressed on a body these will also represent
three corresponding increases in velocity and the resultant increase in
velocity will be represented by oc, and oc will therefore represent the
resultant acceleration, oahc is therefore an acceleration polygon.

(4) A force acting on a body causes it to move with a uniformly
increasing velocity or acceleration and the magnitude of the acceleration
is proportional to the magnitude of the force and takes place in the
same direction as that of the force. Hence if oahc (Fig. 162) is an
acceleration polygon it is also a force polygon.

If R is the sum or resultant of the vectors A, B, and C, then
R = A -f- B + and the order in which the vectors are taken in
performing the summation is immaterial. That is, if R = A + 1^ + CJ,
then, also, R = B + A-fC = C + A4-B. The addition is per-
formed by drawing the vector polygon, three sides of which are
A, B, and C, and the fourth or closing side is R.

Observe that if R = A -f B + C, then A + B + C-R = 0,
which shows that if the sense of R is reversed the sum of the four
vectors A, B, C, and R is zero.

When the sum of a number of vectors is zero the vector polygon
closes without the use of
another vector. / / \C

86. Subtraction of / \ A/fTn)

Vectors.— At (/) Fig. 163, V -' v ^^ /

three vectors A, B, and C
are given, B and C being
equal and parallel but the
sense of C is opposite to
that of B. Hence C = -B.

If R = A - B then R = A + C.

The solution of R = A + C is shown at (m) and the solution of
R = A — B is shown at (n).

R=A+C

Fig. 163.

PRACTICAL GEOMETRY

The rule for subtraction is evidently, change the sign of the vector
to be subtracted and then proceed as in addition.

87. Example. — A, B, C, and D are four vectors specified as
follows. A = 2-460O, B = 1-5
tudes being in inches. It
is required to find the value

330O, C = 2-09^0, and D = 2-li8oo, the magm-
as

ScaLe^

R=A+B-C+D

Fig. 164.

A + B-C + D = r0.

of K = A + B-C + D.

The magnitudes and direc-
tions of A, B, C, and D
are shown to the right in
Fig. 164. Take a point 0.
Draw oa parallel and equal
to OA. From a draw ah
parallel and equal to OB,
then oh is equal to A + B.
From h draw he parallel
to DC reversed and equal
to 00, then oc is equal to
A + B — C. From c draw
cd parallel and equal to OD, then od is equal to R
R will be found to be equal to 0*79;5oi^.

88. Resolution of Vectors. — If R is the sum or resultant of any
number of vectors A, B, C, D, etc., then A, B, C, D, etc., are called
the components of the vector R. The operation of finding R when
A, B, C, D, etc., are given is called .the summation of vectors or
composition of vectors. The converse operation of breaking up a
vector into a number of components is called the resolution of a
vector. It is evident that a given vector may be resolved into any
number of components by constructing a polygon on the given vector.

Generally when a vector has to be resolved into components only two
components are required the directions of which are given. Let R or ah
(Fig. 165) be a given vector and let OX and OY be
two given directions ; it is required to resolve R
into two components P and Q whose directions shall
be parallel to OX and OY respectively. Through
a draw ac parallel to OX and through h draw he
parallel to OY to meet ac at c, then ac and ch will
be the required components P and Q. A common
case in practice is that in which the angle XOY
is a right angle.

89. Example. — The thick curved line shown in Fig. 166
represents the section of a vane of a water wheel or turbine which
revolves about an axis at O. A jet of water impinges on the vane
at A with a velocity v-^. The jet is then deflected and flows over
the vane leaving it at B. The wheel revolves in the direction of
the arrow R. The radius of the wheel at A is r^, and the linear
velocity of the vane at A is c^. The radius of the wheel at B
is r^ and the linear velocity of the vane at B is c^. Obviously
Cg : Ci : : r2 : r^.

Fig. 165.

VECTOR GEOMETRY

C--

.'<-^'

Fig. 166.

Consider the water in the jet at A where it comes in contact
with the vane. This water is beginning to slide along the vane
with a velocity s in a
direction tangential to the
vane at A. This water is
also carried round with
the wheel with a velocity
Cj in the direction of the
tangent to the wheel
circle at A, and in order
that there shall be no
shock the resultant of
these two velocities s and
c, should be Vj^. Of the
three velocities v^, c^, and
s, if one be known com-
pletely and the directions
of the other two are given,
their magnitudes are
readily found. Or if two of the velocities be known completely, the
magnitude and direction of the other is readily found.

Assume that the water slides along the vane with a constant
velocity s relatively to the vane and consider the water in the jet at
B. This water has a velocity 8 in the direction of the tangent to
the vane at B and also a velocity Cg in the direction of the tangent
to the wheel circle at B. Hence BD or v^ the resultant or absolute
velocity of the water at B is the resultant of the velocities s and c^.

Draw BC parallel and equal to v^ and join CD; then CD is the
vector change Vg — v^ in the velocity of the water in passing over
the vane.

The actual path of a particle of water in passing through the
wheel may be found as follows. Make the arc AE equal to a definite
fraction of s. With centre O and radius OE describe the arc EF.
Make the arc EF equal to the same fraction of c, the velocity of
the wheel at E, that the arc AE is of s. Then F is a point in the
actual path AFK of a particle of water which enters the wheel
at A.

Exercises VII

1. Three vectors, A, B,.and G, acting in a horizontal plane are defined in the
following table :— See also Fig. 167.

Vector.

Magnitude.

Direction

A
B
C

1-23 units
1-95 „
2-60 „

Eastwards.

33-2° northwards of east.

112° northwards of east.

A + B + C

Fig. 167.

PRACTICAL GEOMETRY

The angles of 33-2° and 112° are to be set off with the protractor, and not by
copying the diagram.

Determine the resultant or vector sum A + B + C, using a scale of 1 inch to
1 unit. Measure and tabulate the results (thus completing the above table).

Theorem : — A vector sum is the same in whatever sequence the vectors are
added. Verify this principle by actual drawing in the following case : — Show that
A + B + C = A + C + B. [B.E.]

2. Three coplanar vectors A, B, C are given as follows : —

Vector.

Magnitude.

Direction.

A
B
C

37-2 units
59-5 „
88-0 „

23-6°
115-5°
238-0°

A + B + C
A-B + C

Fig. 168.

In defining direction, the vectors are supposed to act outwards from a point O
(Fig. 168), and the angles are measured anti-clockwise from a fixed line OX.
These angles must be set of! with a protractor, and not copied from the diagram.
Find A + B + C and A — B + C. Measure and record the results (in the
form required to complete the foregoing table).

Verify by drawing that A - (B - C) = A - B + C. '

Use a scale of J inch to 10 units. [b.e.]

3. Find the vector sum A + B + C + D + E, having given A = 2-15qo

l-60„o, C = 2-25,300,

being 1 inch.

4. Four coplanar forces P, Q,
follows :— P = 15„o, Q = 9,

D = 2-80qo, and E = 5 35,o„o, the unit of magnitude

R, and S, acting at a point, are specified as
S = 21, --0, the magnitudes of

Find the resultant of these forces, and specify it in

0"' "* — "21;
the forces being in pounds,
the form T = t^o- Use a scale of 1 inch to 10 pounds.

5. A locomotive engine A represented by a point is approaching a level cross-
ing C from the south at a speed of 15 miles an hour (22 feet per second), that is
velocity of A = 22gQO f.s., directions being measured anti-clockwise from the east.
An engine B is approaching from the W.S.W., its velocity being 44.22io f.s. They
a,rrive together at C.

(a) Show their positions one second before the collision, scale 1 inch to 10
feet. Measure their distance apart and the direction from A to B.

(6) What is the relative speed of the two engines when the accident
occurs? [b.e.]

6. An aeroplane is headed due west, and is propelled at 50 miles per hour
relatively to a steady wind which is blowing at 20 miles per hour from the north-
west. Find the actual direction and speed of flight as regards the earth.

If the pilot wishes to travel westward, in what direction must he apparently
steer, and what will be his speed to the west ? [b.e.]

7. A ship is sailing eastwards at 10 miles an hour. It carries an instrument
for recording the apparent velocity of the wind, in both magnitude and direction.

(a) If the wind registered by the instrument is apparently one of 20 miles per
hour from the north-east, what is the actual wind ? Give the answer in
miles per hour and degrees north of east of the quarter from which the
wind comes.

(6) If a wind of 15 miles per hour from the north-east were actually blowing,
what apparent wind would the instrument on the vessel register ? State
this answer in miles per hour and degrees north of east as before.

Use a scale of J inch to 1 mile per hour, [b.e.]

YECTOR GEOMETRY

8. A weight of 15 pounds is supported by two cords. One cord is inclined at
40° and the other at 55° to the horizontal. Determine the tensions in the cords.

9. A wheel weighing 100 lb, rolls at a certain speed on a horizontal rail. The
wheel is out of balance to the extent that at the speed of rolling there is a radial
centrifugal force of 20 lb. On a base 6 inches long, representing one revolution
of the wheel, plot, for every 1-1 2th of a revolution, the pressure exerted by the
wheel on the rail. Use a force scale of 1 inch to 20 lb.

10. A, B, C, D, E, and F are six forces acting in a plane at a point 0. The
magnitudes of the forces are 125, 1*10, 2*25, 1-75, 2-45, and 1-35 pounds re-
spectively. The forces act outwards from in lines inclined to a fixed line OX at
the following angles, 30°, 60°, 135°, 180°, 225°, and 300° respectively. Find R the
resultant of these forces. If these forces are balanced by a force P acting in the
line OX and a force Q acting in a line at right angles to OX, determine P and Q.

11. A jet of water having a velocity of 353^0 feet per second passes over a fixed
vane as shown in Fig. 169, and is deflected, leaving the vane with a velocity of
35jjjQO feet per second. If w^ denote the change of velocity that has occurred,
find u and 6 ; that is, solve the vector equation

Un = 35

100°

30°'

The mass m of water passing per second being 2-5 units {= 2-5 X 32*2 lb.), calcu-
late muQ, the magnitude and direction of the change of momentum per second ; find
also — mUg, the magnitude and direction of the force acting on the vane, [b.e.]

Fig. 170.

Fig. 171.

12. A stream of water flowing in the given direction AB (Fig. 170) impinges
on a succession of moving vanes, one of' which is shown. BT, CT are tangents to
the curve of the vane at its ends. The vector V represents the velocity of the
vane to a scale of 1 inch to 20 feet per second.

(a) Find and measure the speed of the water along AB in order that the water

may come on to the vane at B tangentially, or without shock.

(b) Suppose the water to flow over the vane without change of relative speed,

represent graphically to scale the absolute velocity of the water as it
leaves the vane at C. , i. xu 4. i,

occurred owing to its passage over the vane. Find and measure tue com-
ponent of U in the direction of V. .

(d) Determine, either the actual path of the water as it flows over the moving

vane, or the line of the resultant force on the vane, due to the change of
momentum of the water. [b.e.J

PRACTICAL GEOMETRY

13. A jet of water having a velocity Vi of SSg^o feet per second passes over a
succession of curved vanes (one of which is given in Fig. 171) moving with a
velocity V of 23^,o feet per second. Find Vi - V, the velocity of the water rela-
tively to the vane ; that is, find s and o in the vector equation

35,

30°

23oo = Yx-Y,

s being the speed along the vane, and a the direction of the vane at entrance, the
water coming on tangentially.

The water leaves the vane with a
velocity relatively to the latter of s^g^o-
Find U(= Uq), the change of velocity that
has occurred.

The mass m of the water flowing per
second being 2-5 units (= 2*5 X 32-2 lb.),
calculate mU, the magnitude and direc-
tion of the change of momentum per
second. Find also the power developed,
which is equal to the scalar product
- mlJV. [B.E.]

14. A jet of water passes over a suc-
cession of curved vanes (one of which is
shown in Fig. 172), entering tangentially
with a velocity Vj of 353QO feet per second,
and leaving with a velocity Vg of 8gQ0 feet
per second. The vanes have a velocity V
of Vqo feet per second, and the speed s of
along the vanes is assumed

Fig. 172.

the water
constant.

Find V, the speed of the vane. Find also a and j8 the directions of the vane
at entrance and exit. The mass m of the water flowing per second being 2'5
units (= 2*5 X 32*2 lb.), and U denoting the vector change of velocity of the
water, find wU, the magnitude and direction of the change of momentum per
second. Find also the power developed, which is equal to the scalar product
- mUV. [B.K.]

(

CHAPTER Vlir

GRAPHIC STATICS'

90. Coplanar Forces. — Unless otherwise stated when a system
of forces is considered it will be assumed that the lines of action of all
the forces are in the same plane. In the great majority of problems
in statics with which engineers have to deal the forces are in the same
plane.

The composition and resolution of forces acting at the same point
have been considered in Chapter VII. as particular cases of vectors.

91. Specification of Forces in a Plane. — A force is specified
when its line of action, its sense, and its magnitude are known. Let
OX (Fig. 173) be a fixed line or axis in the plane in which a force A
acts, and let O be a fixed point in OX. Let the line of action of the
force A cut OX at L, and let x denote the length of the intercept OL.
Let tt denote the angle, measured anti-clockwise, which the line

F^^

" " '» %;

Fig. 173.

Fig. 174.

of action of the force makes with OX, the force being supposed to
act outwards from L. Lastly let a denote the magnitude of the force.
Then the force is specified by giving x, a, and a.

A convenient way of stating the quantities x, a, and a in speci-
fying a force A is, A =,««• For example, if the magnitude of the
force A is 19-5 units of force, and the intercept x is 0*7 unit of length
and the angle a is 35 degreed, then A = 07^^'^35O'

This method of specifying a force fails when the line of action of
the force is parallel to OX, but this difficulty may be overcome by
giving the intercept OM = y on an axis OY at right angles to OX.
To show clearly that it is ij and not x that is given when specifying

92 PRACTICAL GEOMETRY

the force, the value of y may be placed above instead of below the
level of the value of a. Thus the force C (Fig. 173) whose line of
action is parallel to OX would be given by the equation C = %o. The
use of the value of y instead of the value of x is also desirable when
the intercept x is large because the point L may be practically inac-
cessible although the line of action of the force may be at quite a
convenient distance from O. The force B (Eig. 173) would therefore
be specified by B = ^6/3.

The intercept x is positive when it is to the right of O and nega-
tive when to the left. Also, the intercept y is positive when it is
above O and negative when below.

A number of forces are represented in Fig. 174. Without a
drawing these forces would be specified as follows, the unit of force
being the pound, and the unit of distance the inch.

-^ — 0-6 17330

B — i-ilOigoo

C = _o.7l 3*5900

D = -vM:

E = 1-4300

F = U9,so.

H = -"Hi,^

K = -o-^iO,

92. Bow's Notation. — In Fig. 175 the diagram (m) shows the
lines of action of a number of forces which are in equilibrium. The
diagram (n) is the corresponding
polygon of forces. In one system
of lettering, each force is denoted
by a single letter, as P. In Bow's
notation, each force is denoted by
two letters, which are placed on
opposite sides of the line of action
of the force in diagram (?«), and ^ ' ^
at the angular points of the poly-
gon in diagram (w). In Bow's notation the force P is referred to as
the force AB. In like manner the force Q is referred to as the force
BC. The diagram (m), which shows the lines of action of the forces,
is called the space diagram or frame diagram, and the diagram (n)
which shows the polygon of forces is called the force diagram.

93. The Funicular Polygon or Link Polygon.— Let P, Q, R,
and S (Fig. 176) be four forces, acting on a rigid body, but their lines
of action do not meet at one point. The forces P, Q, R, and S are
balanced by a fifth force T which is at present unknown. Draw the
polygon of forces ahcde, that is, draw a polygon whose sides are
parallel to the lines of action of the forces and of lengths equal to
the magnitudes of the forces as in the case of forces acting at a point.
ea the closing side of the polygon will represent in magnitude and
direction the fifth force T. It now remains to find a point in the line
of action of the force T.

Take any point and join it to a, h, c, d, and e. Take any point 2
in the line of action of P and draw the line 2B3 parallel to oh to
meet the line of action of Q at 3. Draw 3C4 parallel to oc to meet
the line of action of R at 4. Draw 4D5 parallel to od to meet the
line of action of S at 5. Draw 5E1 parallel to oe to meet 2A1

GRAPHIC STATICS

parallel to oa at 1. Then 1 is a point in the line of action of T
which may now be drawn parallel to ea.

Conceive that the lines A, B, C, D, and E represent bars jointed
to one another at the points 1, 2, 3, 4, and 5. Then these bars may
be supposed to take the place of the rigid body upon which the five
forces P, Q, R, S, and T are supposed to act. In the case under con-
sideration (Fig. 176) it is obvious that the bars A, B, C, D, and £
are subjected to tension. Consider the point 2. Here there are
three forces acting which balance one another, namely, the force P
and the tensions in the bars A and B, and these three forces are repre-
sented in magnitude and direction by the three sides of the triangle
abo. Again, the three forces acting at the point 3 are represented by
the sides of the triangle hco, also the three forces acting at the point 4
are represented by the sides of the triangle cdo, and the three forces
at 5 by the sides of the triangle deo. Now in order that the tensions
in the bars E and A may be balanced by the force T, the force T

Fig. 176.

Fig. 177.

must act at the point of intersection of the bars E and A. The point
1 is therefore a point in the line of action of T.

The polygon 1 2 3 4 5 is called the funicular polygon or link polygon
of the forces P, Q, R, S, and T with reference to the point o, which
is called the pole.

Since the pole o may have an infinite number of positions, there
is an infinite number of link polygons to any system of balanced
forces.

If the diagrams (F) and (/) (Fig. 176) be compared it will be seen
that each line on the one is parallel to a corresponding line on the
other. Also, if a system of lines on the one meet at a point the
corresponding lines on the other form a closed polygon. From these
properties the diagrams (F) and (/) are called reciprocal figures.

No reference has yet been made to Fig. 177, but all that has
been said with reference to Fig. 176 will also apply to Fig. 177, where
the given forces are parallel to one another, except that the bars E
and A are in compression, the remaining bars B, C, and D being in
tension.

PRACTICAL GEOMETRY

An examination of Figs. 176 and 177 will show that the simple
rule to be remembered in drawing the link polygon is, that any side
of that polygon has its extremities on the lines of action of two of the
forces, and that that side is parallel to the line which joins the pole to
the point of intersection of the lines which represent these two forces
on the polygon of forces.

Referring to Figs. 176 and 177, it may be noted that the equili-
brant of P and Q is represented in magnitude and direction by ca,
and that the point of intersection of. the sides A and C of the link
polygon is a point in the line of action of this equilibrant. Also the
equilibrant of P, Q, and R is represented in magnitude and direction
by da, and the point of intersection of the sides A and D of the link
polygon is a point in the line of action of this equilibrant.

Having shown that the link polygon together with the polygon of
forces may be used to determine the equilibrant of any system of
forces in a plane, it is obvious that the same construction will also
determine the resultant of that system of forces, since the resultant
acts along the same line and has the same magnitude as the equilibrant,
but acts in the opposite direction or has the opposite sense.

Again referring to Figs. 176 and 177, it will be observed that the
letters A, B, C, D, and E on the space diagrams are situated in the
spaces between the forces P, Q, R, S, and T, and the letters P, Q, R,
S, and T may be omitted and the forces referred to as AB, BC, CD,
DE, and EA. Also the same forces in the force polygons are
lettered ah, he, cd, de, and ea respectively, which is Bow's notation.
It will further be noticed that the sides A, B, C, D, and E of the link
polygon are parallel to the polar lines oa, oh, oc, od, and oe respectively
of the force polygon.

94. Examples. — The following two examples illustrate the
application of the link polygon to the solution of problems on forces
whose lines of action are either parallel or do not all meet at the same
point.

(1) Using Bow's notation, AB, BC, and CD (Fig. 178) are three
vertical forces acting on a horizontal beam. These forces are balanced
by the vertical forces DE and

EA whose magnitudes and t 23li. 20W. ZSW. |
senses are required.

Since the forces are all
parallel, the polygon of forces
will be a straight line ahcdea,
the position of the point e being
as yet unknown.

Choose a pole o, and join oa,
oh, oc, and od. Draw OA, OB,
OC, and OD parallel to oa, oh,
oc, and od respectively as shown.

These lines OA, 01>, OC, and OD will form four sides of the link
polygon of which OE will be the closing side. Draw oe parallel to OE
to meet ac at e. This completes the solution. It will be found that

Fig. 178.

GRAPHIC STATICS

Also, EA is 18-9 lb.

DE is 13*9 lb. and that it acts downwards,
and acts upwards.

(2) LMN (Fig. 179) is a framed structure supported at M and N,
MN being horizontal. There is a vertical load AB of 1100 lb at L
and a load BC of 1520 lb.
at the middle point of LN
and at right angles to LN.
The supporting force CD at
N is known to be vertical,
but its magnitude is un-
known. The supporting
force DA at M is unknown
both as regards magnitude
and direction. It is re-
quired to complete the de-
termination of the forces
CD and DA.

The triangle of forces
ahc determines the direc-
tion of the resultant of the given forces AB and BC and a line through
R, the point of intersection of AB and BC, parallel to ac is the line of
action of that resultant. Replacing AB and BC by their resultant
there are now only three external forces acting on the frame and as
they are not parallel they must meet at a point which is obviously
the point S where the resultant of AB and BC meets CD. This
determines the line of action of DA which passes through M and S.
The polygon of forces ahcd may now be completed. It will be found
that CD = 1380 lb., DA = 1340 lb., and the inclination of DA to
the horizontal is 36*5 degrees.

This example may be worked without using the point S, as follows.
After drawing the sides ah and he of the force polygon choose a pole
0, and join oa, oh, and oc. Next proceed to draw the link polygon
starting at the point M which is the only point in the line of action of
DA which is known. Draw Ml, across the space A, parallel to oa.
Draw 1 2, across the space B, parallel to oh. Draw 2 3, across the
space C, parallel to oc. Then M3 is the closing line of the link
polygon. Draw od parallel to M3 to meet a vertical through c s.t d
which is the remaining angular point of the force polygon. DA may
now be drawn parallel to da.

95. The Centre of Parallel Forces. — If a system of parallel
forces acts at fixed points, the resultant will act through another fixed
point called the centre of the system. This centre is independent of the
direction of the forces so long as the sense of each in relation to the
sense of one of the forces is unaltered.

In Fig. 180, P, Q, R, and S are parallel forces acting at the fixed
points A, B, C, and D respectively in a plane. By means of the force
and link polygons the line of action LK of the resultant is determined.
Let the direction of the forces be changed so that they act as shown
by F, Q', R', and S'. The line of action MK of the resultant is

PRACTICAL GEOMETRY

determined as before. The point K, where LK and MK intersect, is
the centre of the parallel forces P, Q, R, and S acting at the fixed
points A, B, C, and D respectively. If the construction be repeated
with the forces acting in any other direction, it will be found that the
new resultant will act through the same point K.

In Fig. 180, the forces P, Q, R, and S have all the same sense, and
therefore P', Q', R', and S' must have the same sense. But if the sense

Fig. 180.

of Q, say, were opposite to that of P, then the sense of Q' would be
opposite to that of P'.

In applying the above method to the determination of the centre
of a system of parallel forces, it is usually most convenient to take
the two directions of the forces at right angles to one another.

96. Centres of Gravity or Centroids. — The particles of which
any body is made up are attracted to the earth by forces which are
proportional to the masses of these particles. For all practical purposes
these forces may be considered to be parallel, and their resultant will
pass through the centre of these parallel forces. In this case the
centre of the parallel forces is called the centre of gravity or centroid
of the body, and the determination of a centre of gravity resolves into
finding the centre of a system of parallel forces.

The centre of gravity of a body may also be defined as that
point from which if the body is suspended it will balance in any
position.

When the term centre of gravity is applied to a line, the line is
supposed to be an indefinitely thin wire ; and when the centre of
gravity of a surface is spoken of the surface is supposed to be an
indefinitely thin sheet of material.

The following results, which are not difficult to prove, should be
noted : —

The centroid of a straight line is at its middle point.

The centroid of a triangle is at the intersection of its medians.

The centroid of a parallelogram is at the intersection of its
diagonals.

GRAPHIC STATICS

If a plane figure is symmetrical about a straight line, the centroid
of the figure is in that straight line.

To find the centroid of a line made up of a number of straight lines.
At the centres of the straight lines apply parallel forces whose
magnitudes are proportional to the lengths of these lines. The centre
of these parallel forces is the centroid required.

To find the centroid of a curved line. Divide the line into a number
of parts, preferably of equal length. At the centres of these parts
apply parallel forces whose magnitudes are proportional to the lengths
of the parts. The centre of these parallel forces is approximately the
centroid required. Theoretically the approximation is closer the more
numerous the parts into which the curved line is divided, but practically
when the parts are very numerous or very short the drawing of the
link polygon becomes less accurate.

To find the centroid of any irregular figure. Divide the figure into
parts whose centroids and areas are known or easily found. At the
centroids of these partial areas apply parallel
forces whose magnitudes are proportional to
these areas. The centre of these parallel forces
is the centroid required. If the given figure has
an irregular curved boundary line such as is
shown in Fig. 181, divide the figure into a number
of parallel strips as shown by the full straight
lines. Draw the centre lines of these strips,
shown dotted. The centroids of these strips may

be taken at the middle points of their centre lines, and the areas of
the strips may be taken as proportional to the lengths of their centre
lines.

To find the centroid of a quadrilateral. Let ABCD (Fig. 182) be
the quadrilateral. Draw the diagonals AC and BD, intersecting at
O. Let OA be less than OC. Make CE equal to AO. Join BE and
DE. G, the centroid of the triangle BDE is also the centroid of the
quadrilateral ABCD.

To find the centroid of an arc of a circle. Let ABC (Fig. 183) be
the arc of a circle of which is the centre. Draw OB at right angles

Fig. 181.

Fig. 182.

to the chord AC. Draw BD at right angles to OB and make BD
equal to the arc BC. Join OD. Draw CE parallel to OB to meet
OD at E. Draw EG parallel to AC to meet OB at G. G is the
centroid of the arc ABC.

98 PRACTICAL GEOMETRY

To find the centroid of a sector of a circle. Let OABC (Fig. 184)
be the sector of a circle of which O is the centre. Draw OB at right
angles to the chord AC. Draw BD at right angles to OB and make
BD equal to the arc BC. Join OD. Find a point F in OC such that
OF is two-thirds of OC. Draw FH parallel to OB to meet OD at H.
Draw HG perpendicular to OB to meet OB at G. G is the centroid
of the sector OABC.

To find the centroid of a figure considered as part of another figure.
Frequently the addition of a simple figure to a given comparatively
complicated one will make a simple figure.
For example the figure ABCD (Fig. 185)
only requires the addition of the triangle
OCD to convert it into the sector of a
circle whose centre is O. G^ the centroid
of the sector O^B is readily found and
so 'is G2 the centroid of the triangle OCD.

Let G be the centroid of the figure
ABCD. Then if at G^ and G parallel ^'^- ^^^•

forces P and Q be applied, the magnitudes of P and Q being propor-
tional to the areas of OCD and ABCD, the centre of these parallel forces
will be at Gj the centroid of the sector OAB.

Hence, if parallel forces P and R be applied at Go and G^, hut in
opposite directions, the magnitudes of P and R being proportional to
the areas of OCD and OAB, the centre of these parallel forces will be
at G the centroid of the figure ABCD.

The area of a sector of a circle is equal to half the product of the
arc and the radius.

97. Centre of Pressure and Centre of Stress. — If a plane
figure be subjected to fluid pressure, the point in the plane of the
figure at which the resultant of the pressure acts is called the centre
of pressure. If the plane figure is a section of a bar or part of a
structure which is subjected to stress, the point in the plane of the
section at which the resultant of the stress acts is called the centre of
stress.

In what follows " pressure " will be taken to include " stress."

If the pressure be uniform over the figure, then the centre of
pressure coincides with the centroid of the figure.

A general construction for determining the centre of pressure of
any plane figure when the pressure varies uniformly in one direction
is illustrated by Fig. 186. ABCD is a plane figure supposed to be
vertical, and AB and CD are horizontal. AAj is the altitude of the
figure, and the pressure is supposed to vary uniformly from an amount
represented by AP at the level AB to an amount represented by A^Q
at the level CD. AP and A^Q are horizontal.

Join QP and produce it to meet A^A at O. Draw any horizontal
SRMN to cut the given figure. Draw the horizontal OF, and the
verticals MMj and NNj. Through K, the middle point of MN, draw
the vertical KF. Join FMj and FN^ cutting MN at m and n. If this
construction be repeated at a sufficient number of levels, and all points

GRAPHIC STATICS

correspondinrr to m be joined, also all points corresponding to n, a
figure abnCDma is obtained and the centroid of this figure will be the
centre of pressure of the original figure.

The proof is as follows. Suppose that the line MN is the centre
line of a very narrmo horizontal strip of the original figure, and let the
width of this strip be denoted by w. The magnitude of the resultant
pressure on this strip is equal to MN X w; X KS, and it will act at K
the middle point of MN. '

Since SRMN is parallel to QA^DC,

MiNi : mn:: OAj
and, MN : mn : : A^Q

therefore, MN X RS = mn x A^Q,

and MN XwxUS = mnXwx A^Q,

that is, the resultant of the pressure on the strip of length mn when
subjected to a pressure A^Q will have the same magnitude as the
resultant of the pressure on the strip of length MN when subjected
to a pressure RS, and it will act at the same point K which is also
the middle point of mn.

OR,
RS,

It follows that the resultant of the pressure on the figure abnCDma
when subjected to a uniform pressure A^Q will be the same as the
resultant of the varying pressure on the original figure. But when
the pressure on a plane figure is uniform, the centre of pressure is at
its centroid. Therefore the centroid of the figure abnCDma is the
centre of pressure of the original figure.

The figure abnCDma is called a modulus figure of the original
figure.

A common and important practical case is that in which the given
figure is a rectangle and the pressure on it varies uniformly in a
direction parallel to a side of the rectangle.

In Fig. 187, ABCD is a rectangle on which the pressure varies
uniformly from nothing at A to an amount DT at D. Applying the

100

PRACTICAL GEOMETRY

construction just proved the student will have no difficulty in seeing
that the modulus figure for the rectangle ABCD is the, isosceles
triangle OCD and that the centre of pressure for the rectangle ABCD
is at Ci on the vertical centre line of the rectangle and at a distance
from AB equal to |H where H is the height of the rectangle. Also
the magnitude of R the resultant of the pressure on ABCD is equal
to the area of the rectangle multiplied by the half of DT, where DT
is the intensity of the pressure or the pressure per unit area at the
level CD.

If the rectangle ABCD (Fig. 187) be divided into two rectangles
by the line EF, then, Cg, the centre of pressure of the rectangle ABFE

is at a distance \h from AB, where h is the

height

of the rectangle

ABFE, and the magnitude of P the resultant of the pressure on ABFE
is equal to the area of ABFE multiplied by the half of ES.

The centre of pressure of the rectangle CDEF will be at Cg the
centroid of the quadrilateral CDe/ and the magnitude of Q the
resultant of the pressure on CDEF will be equal to the area of CDEF
multiplied by half the sum of DT and ES. Q may however be
determined both as regards magnitude and position by considering
it as the equilibrant of the parallel forces P and R, the sense of R
being opposite to that of P.

98. Masonry Dams. — A masonry dam is a wall for holding back
the water at the end of a natural reservoir.

One form of dam section is shown in Fig. 188, A V being a vertical line.

The principal problems connected with dams are: (1) the de-
termination of the line of resistance when the reservoir is empty ;
(2) the determination of the line of resistance when the reservoir
is full, and (3) the determination of
the stresses at various horizontal
sections of the dam.

Reservoir empty. When the reser-
voir is empty the stresses in the dam
are those due to its weight. Refer-
ring to Fig. 188. Consider a portion
of the dam lying between two vertical
cross sections one foot apart, then the
weight of a part of this between two
horizontal section planes will be equal
to the area, in square feet, of the
cross section between these planes
multiplied by the weight of a cubic
foot of the. material of the dam. |

Let horizontal sections BC, EF,
GH, and KL be taken. The resultant
M?i of the weight of the top portion

ADLK acts vertically through Cj the Fig. 188.

centroid of ADLK. w^ cuts KL at

Cj which is the centre of pressure or centre of stress for the horizontal
section KL.

Irmer
Face}

GRAPHIC STATteS •- ' ^ ' , ' > .|*qi

The resultant w.^ of the weight of the portion KLHG acts vertically
through Ca the centroid of KLHG. The resultant of the load on GH
is Wg which is the resultant of w^ and w^. Wg cuts GH at Co. which
is the centre of stress for the horizontal section GH.

Continuing in the same way, Wg, the resultant of w^, W2, and w.^,
cuts EF at C3 the centre of stress for EF, and W4, the resultant of
iDi, W2, Ws, and w;4, cuts BC at c^ the centre of stress for BC.

A fair curve drawn through the centres of stress of the various
horizontal sections is called the line of resistance for the cross section
of the dam.

Beservoir full. When the reservoir is full the gravity forces which
have just been considered will still act but there will now be in addition
the pressure of the water on the inner face.

The intensity of the pressure of the water per square foot at any
depth h feet from the upper or free surface of the water is equal to h
multiplied by the weight of a cubic foot of water or say 62 -SA.

Still considering a portion of the dam one foot long. The resultant
force on the horizontal section BC is the resultant ofc W^ and P4, where
P4 is the resultant of the pressure of the water on the face of the
one foot length of dam from A to B. P4 acts at two-thirds of AB
from A and at right angles to AB. The magnitude of P4 is the area
of the face AB, one foot long, multiplied by half the intensity of the
water pressure at B. Let P4 and W4 intersect at O. B4 the resultant
of P4 and W4 acts through O in a direction determined by the triangle
of forces rpw. The line of action of R4 cuts BC at s^ which is the
centre of stress for BC.

In a similar manner «3, s^, and s^ the centres of stress for EF, GH,
and KL may be determined. A fair curve through these centres of
stress of the various horizontal sections determined as above is the line
of resistance for the cross section of the dam when the reservoir is full.

If each of the horizontal lines BC, EF, etc. be divided into three
equal parts then each of the middle parts is called a middle third. If
the corresponding extremities of the various middle thirds be joined
by fair curves ^ these curves will enclose the middle third of the cross
section of the dam, and if a dam is pro-
perly designed the lines of resistance
should fall within this middle third
whether the reservoir is empty or full.

If R is the resultant force on any hori-
zontal section XXj (Fig. 189) of the dam,
then S the vertical component of R causes
a normal stress on XX, and the horizontal
component T causes a tangential or shear
stress on XXj.

The normal stress produced by S on XX^ is not uniformly dis-
tributed but varies uniformly from 5 - p at X which is furthest from
S to 5 -f jp at Xj.

1 If the contour of the cross section of the dam is made up of straight lines
these fair curves will become a series of straight lines.

102

FIIACTICAL GEOMETRY

Let C be the middle point of XX^. Let XX^ = 2a, and let the

distance of S from C be x, then, still considering 1 foot length of the

dam, a = <T- and p = -^^.
' ^ '2a ^ a

If a; = ^ then p = q and there is no stress at X. If x is greater than

a .

o then j9 is greater than q and there is a tension in the dam at X

which should be avoided.

The maximum compressive stress p ■}- q should not exceed 7 tons
per square inch.

The average weight of masonry dams is about 150 lb. per cubic
foot.

Exercises Villa

1. The horizontal distance between the axles of a bicycle is 3 feet 6 inches, and
its weight is 28 lb. Assuming that the weight on the saddle is vertically above a
point 9 inches in front of the rear axle, and that a greater pressure than 200 lb.
must not be brought on either wheel, find by a funicular polygon, and mark
distinctly, the greatest weight the bicycle will bear. Scale of weight, 60 lb. to 1
inch. Scale of length, Jth full size. [b.e.]

2. Using a funicular polygon, determine the resultant of the five parallel
forces given in Fig. 190. The magnitudes of the forces are given in pounds.
Use a force scale of 1 inch to 20 lb.

3. Determine the parallel forces, which, acting through the points L and M
(Fig. 191), should balance the given forces. The magnitudes of the forces are
given in pounds.

8 18 12 105

140

70 340

Fig. 190. Fig. 191. Fig. 192.

In reproducing the above diagrams take the small squares as of 0-5 inch side.

4. Six forces are given in Fig. 192, the magnitudes being in pounds. Find the
resultant of these forces.

5. Find the resultant of the following three given forces, that is, find and
measure r, 0, and a in the vector equation

^ = Je- I'^^iiP + 4'33-jo 4- 8'l^ii5° pounds.
Employ scales of i inch to 1 foot, and 1 inch to 10 pounds. [b.e.]

6. Five coplanar forces P, Q, R, S, T are given as follows : —

(To explain this manner of

defining a force see Art. 91,
p. 91.)

Find the resultant of the
given system of forces. Mea-
sure its intercept, direction,
and magnitude.

Find also and measure the
resultant of Q, R, and S.

Employ a linear scale of
} full size, and a force scale
of J inch to 10 lb. [b.e.]

Force.

Intercept.

Direction.

Magnitude.

Q
R
S
T

10*5 inches
6-4 „
2-8 „
4-2 „

-4-8 „

32°

71°

153°

291°

196°

1
65-1 lb. 1
28-5 „
56-3 „
15-2 „
601 „

GRAPHIC STATICS

103

7. Find the resultant of the forces given in Fig. 174, p. 91. The intercepts
are in inches and the magnitudes of the forces are in pounds.

8. Determine r, x, and d in the equation

^ = x'-9= Ol65oo + 0-7355285O + 44IO330 + 2.7323260° + ol62i,oo + 4-32052400.

Force unit, 1 pound. Linear unit, 1 inch.

9. Given the equation of equilibrium of five coplanar forces

0^90° + 10 '^e ~ 12OO3050 + 3I5O27QO + S^OOgjgO

find r, s, and d. The unit of force is the pound and the unit of length is the foot.
Employ a linear scale of J inch to 1 foot, and a force scale of 1 inch to 100
pounds.

10. Find the centre of gravity of a piece of wire bent to the form shown at (a)
Fig, 193.

(ft)

(c)

(d)
Fig. 193.

(e)

if)

hi reproducing the above diagrams take the small squares as of 0-5 inch side.

11. Determine the centroids of the figures (6), (c), (d), [e] and (/) Fig. 193.

12. The triangle ABC (Fig. 194) is a diagram showing the intensity of earth
pressure on a retaining wall, its area representing the total pressure P to the same
scale that the area of the cross section of the

wall represents the weight W of the wall.
Determine the lines of action of P and W, the
face AB of the wall being vertical. Find the
centre of pressure on the base of the wall, this
being the point where the resultant of P and
W intersects the base.

Also find the centres of pressure on the
sections D, E, and F. [b.e,]

13. A circle 4 inches in diameter is sub-
jected to pressure which varies uniformly
from 10 pounds per square inch at one end of
a diameter to 5 pounds per square incli at the
other end of that diameter. Find the centre
of pressure and the magnitude of the resultant
pressure.

14. A rail section is given in Fig. 195.
Find the centroid of this section. Through
the centroid draw a line parallel to the
bottom line of the section ; this is the "neutral
axis " of the section. This section is subjected
to stress which is normal to the section and
varies uniformly from the top to the bottom,
being zero at the neutral axis where it changes
sign. Find the centres of stress of the portions of the section above and below
the neutral axis. If the stress at the lower edge of the section is 4 tons per square
inch what is the stress at the upper edge, and what are the magnitudes of the
resultants of the stresses on the portions of the section above and below the
neutral axis ?

104

PRACTICAL GEOMETRY

15. A section of a large dam is given in Fig. 196. The two curved faces are
plotted from the vertical line shown. The weight of the masonry composing the
dam is 125 lb. per cubic foot. The weight of the water may be taken as 62-5 lb.
per cubic foot.

Draw the lines of resistance for this dam, (a) when the reservoir is empty, and
(&) when the top surface of the water is 10 feet below the top of the dam.

double

this

sice

1 17-4

165-i

Fig. 195.

Fig. 196.

Draw also the "middle third" lines of the section of the dam, and the
diagrams showing the distribution of the vertical stress on the lowest horizontal
section when the reservoir is empty and when the reservoir is full.

Linear scale, say, 10 feet to 1 inch. [u.l. modified.]

99. Moment of a Force. — The moment of a force about a
point or about an axis perpendicular to its line of action, is the
measure of its turning power round that point or axis. The magnitude
of the moment (generally called the moment) is the product of the
magnitude of the force and the perpendicular distance of its line of action
from the point or axis. For example, the moment of the force AB
(Fig. 197) about the point M is equal to the magnitude of the force
AB multiplied by MN, the perpendicular distance of M from the line
AB. If the unit of force is the pound, and the unit of distance is the
inch, then the unit of moment is the inch-pound or pound-inch. Other
units of moment in common use are the foot-pound or pound-foot, the
foot-ton or ton-foot, and the inch-ton
or ton-inch.

The construction shown in Fig.
197 is a very convenient one for
determining graphically the moment
of a force about a point. AB is the
line of action of the force, and M is
the point. The construction is as
follows. Draw ah parallel to AB,
and make the length of ah to repre-
sent the magnitude of the force.
Through M draw a'Mh' parallel to
AB. Choose a pole o. .Join oa and oh.

Fig. 197.

Take any point o' in AB.

GRAPHIC STATICS

105

Draw o'a! parallel to oa to meet oJWj! at a', and draw o'h' parallel to
oh to meet a!Mh' at h'. Then a'h' measured with a suitable scale
will be the magnitude of the moment of the force AB about the
point M.

Draw oil perpendicular to ah, and o'Ji! perpendicular to a!h'. The
triangles oah and o'a'h' are obviously similar, and ah : a!h' :\ oh: o'h'.
Hence ah x o'h' = a'h' X oh. But ah is the magnitude of the force
AB, and o'h', which is equal to MN, is the perpendicular distance of
M from AB. Therefore ah X o'h' is equal to the moment of AB about
M, and therefore a'h' X oh is equal to the moment of AB about M.

If oh is made equal to the linear unit, then a'h' measured with the
force scale will give the moment required. For example, if oh is 1
inch and a'h' measures 20 lb. on the force scale, then the required
moment is 20 inch-pounds. It is not always convenient to make oh
equal to the unit of distance, but it should be made a simple multiple
or sub-multiple of it.

The following is the simple rule for determining the moment scale.
Let oh be m times the linear unit, and let the force scale be n units of
force per inch. Then the moment scale will be w X n units of
moment per inch. For example, let the linear unit be one foot, and
suppose that oh, measured with the linear scale, is 4 feet. Let the force
scale be 100 lb. per inch, then the moment scale will be 100 X 4 = 400
foot-pounds per inch.

It may be pointed out that the figure a' o'h' is the link polygon of
the force AB with reference to the pole o.

The following problems are important.

(1) An unknown force P acting through the point A (Fig. 198)

Fig. 198.

Fig. 199.

has a moment m^ about the point O^ and a moment mg about the point
Og : it is required to determine the force P.

Join O1O2. Draw dMi and O.Ma at right angles to OjOa. Make
OiMi = mi to any convenient scale and make O2M2 -■ m^ to the same
scale. OiMi and O2M2 are on the same or on opposite sides of OjOg
according as Wj and m^ have the same or opposite signs. Join MjMg
and produce it if necessary to cut O1O2, or OiOg produced at B. Then
B is another point in the line of action of P. The sense of P *-

106

PRACTICAL GEOMETRY

readily determined by inspection. To find the magnitude of P, draw
OjN at right angles to AB. Then, magnitude of P

(2) An unknown force P has moments Wj

7)12, and mg about the

points Oi, O2, and O3 (Fig. 19^) respectively : it is required to
determine the force P.

Draw OiMi and O2M2 at right angles to OiOg. Make OiM^ = mj
to any convenient scale, and make OgMg = 171.2 to the same scale.
OiMi and OgMg are on the same or on opposite sides of O1O2 according
as m^ and m.^ have the same or opposite signs. Join M^Mg and produce
it if necessary to cut OjOa, or OiOo produced, at A. Then A is a
point in the line of action of P. In like manner the point B is found
as shown, and AB the line of action of P is determined. The sense of

P is found by inspection, and the magnitude of P is equal to Ty^, where

OiN is the perpendicular from Oi on AB.

100. Resultant Moment of a System of Forces. — The
resultant moment of a system of forces about a point is equal to the
algebraical sum of the moments of the separate forces about that point,
and it is obvious that this sum must be equal to the moment of the
resultant of the system about the same point. Hence the graphical
determination of the resultant moment of a system of forces about a
point resolves into constructing the resultant of the system, and the

Fig. 200.

Fig. 201.

determination of the moment of this resultant about the given point
by the construction of the preceding article. The two constructions
may, however, be combined in one, as shown in Figs. 200 and 201.
AB, BC, and CD are three given forces, and M is a given point. It

GRAPHIC STATICS 107

is required to determine the resultant moment of the given forces
about the given point.

The polygon ahcda is the force polygon ; ad, the closing line, gives
the magnitude and direction of the resultant of the three given forces.
A pole is taken at a perpendicular distance oh from ad, which is a
simple multiple or sub-multiple of the linear unit. The link polygon
of the forces with reference to the pole o is next drawn, and the
intersection of the closing sides OA and OD determines a point on
the line of action of the resultant force AD. A line through M
parallel to ad intersects the closing sides OA and OD of the link
polygon at a' and d'. The moment required is equal to a'd' X oh.
The triangle o'a'd' is obviously similar to the triangle oad, and therefore,
as shown in the preceding article, the moment of AD about M is
equal to a'd' X oh.

It may be observed that the moment of any one of the forces, say
BC, is obtained by drawing through M a parallel to BC to intersect
the sides of the link polygon which meet on BC at h' and c' ; h'c! X oh
is the moment of BC about IVL

101. Principle of Moments. — When a number of forces acting
on a rigid body are in equilibrium, then the moments of all the forces
about any given axis being taken, the sum of the moments of those
forces which tend to turn the body in one direction about the axis is
equal to the sum of the moments of those forces which tend to turn
the body in the opposite direction about the same axis.

102. Couples. — A couple consists of two equal parallel forces
acting in opposite directions. The arm of a couple is the perpendicular
distance between the lines of action of the two forces. The moment of
a couple is the product of the magnitude of one of the forces and the
arm of the couple. A couple tends to cause a body upon which it acts
to rotate.

A couple cannot be balanced by any single force.

Two couples acting on a body will balance one another when (I)
they are in the same plane or in parallel planes, (2) they have equal
moments, and (3) their directions of rotation are opposite.

103. Bending Moment and Shearing Force Diagrams for
Beams. — When a horizontal beam is acted on by vertical forces or
loads, these forces tend to bend the beam, and the bending action at
any transverse section is measured by the algebraical sum of the
moments of the forces on one side of the section about a horizontal
axis in that section. For example, the beams shown in Fig. 202 are
acted on by forces P, Q, and R to the right of the transverse section
XY, and the bending moment at XY is equal to

P X /+ Q X m- R X «.
The loads on a beam also tend to shear the beam transversely, and
the shearing action at any transverse section is equal to the resultant
of the transverse forces on one side of the section. For example, the
shearing action at the section XY of the beams shown in Fig. 202 is
equal to the resultant of the forces P, Q, and R which act to the
right of the section, and this resultant is equal to P + Q — R.

108

PRACTICAL GEOMETRY

The drawing of the bending moment diagram for a beam is simply
the application of the construction explained in Art. 99. In Fig. 203
is shown a horizontal cantilever carrying vertical loads AB, BC, and
CD. abed is the line of loads, or polygon of forces. A pole o is chosen,
so that the pole distance oh is a simple multiple or sub-multiple of the
linear unit. The link polygon a'd'n' is then drawn. It is easy to show,
as in Art. 100, that the bending moment at any section XY is equal
to a^d^ X oh, that is, the depth of the link polygon under the section
multiplied by the pole distance. The depth of the link polygon is
measured by the force scale, and the pole distance by the linear scale.

Fig. 202.

Fig. 203.

It follows that, since the pole distance is the same for all parts of the
link polygon, the depth of the link polygon under any section of
the beam is a measure of the bending moment on the beam at
that section, the scale for measuring the bending moment being found
as explained in Art. 99.

The shearing force diagram is constructed by drawing horizontals
across the spaces A, B, C, and D at the levels a, h, c, and d respectively.
The depth of this diagram under any section of the beam, measured
with the force scale, gives the vertical shearing force on the beam at
that section. For example, at the section XY the shearing force is
the resultant of the forces to the right of XY, and is equal to
BG-^CD = bc + cd = bd.

Another example is illustrated by Fig. 204. Here a beam AB
rests on supports at A and C and carries a distributed load, the
intensity of which at any p^int is proportional to the height of the
shaded diagram above A13 at that point. The load diagram is divided

GRAPHIC STATICS

109

into a number of vertical strips of equal width as shown by the
vertical dotted lines. If these strips are sufficiently numerous it will
be sufficiently accurate to assume that the load which any one of these
strips represents acts at the middle of its width and has a magnitude
proportional to its mean height. Figures instead of letters have been
used to denote the forces in Bow's notation.

The line of loads rs is drawn and a pole o chosen, and the link
polygon (a) is determined, the closing line being mn. A line, shown
dotted, through o parallel to mn to cut rs at the point 12 determines
the reactions 11-12 and 12-1
at the supports.

The diagram (a) is the
bending moment diagram.
This has been redrawn at (6)
on a horizontal hasm line.
The heights in {b) and (a)
are the same on the same
vertical lines. It will be ob-
served that the bending
moment changes sign near
to the line of action of the
force 7-8.

The stepped diagram is
the shear force diagram and
is obtained by drawing the
horizontal lines across the
spaces between the forces
and projected from the corre-
sponding points on the line
of loads. The shear force
changes sign at a point
between the supports, and
again at the support C as

shown by the part (d) of the Fig. 204.

shear force diagram, shown

dotted. The part (d) has been transferred to (d') so as to show the
whole of the shear force diagram on the same zero base.

The shear forces as determined above are correct for the middle
vertical lines of all the spaces between the forces except the end
ones, and in the true shear force diagram the steps would be replaced
by the dotted curve shown.

The link polygon is- not quite exact because the forces will not
act along the vertical middle lines of the strips but will act through
the centroids of the strips which will be slightly to the right of
the middle positions. The point 12 on the line of loads is therefore
not quite exact but should be slightly higher.

If the forces be made to act through the centroids of the strips
then the point 12 on the line of loads will be determined exactly.
Also the link polygon then determines exactly the bending moments at

110

PRACTICAL GEOMETRY

the middle vertical lines of all the spaces between the forces except
the end ones, and the true bending moment diagram is obtained by-
drawing a fair curve through the points thus determined and
continuing the curve to the verticals through A and B where the
bending moments are zero.

104. Moment of Inertia. — The sum of the products of the mass
of each elementary part of a body and tlie square of its distance from
a given axis is called the moment of
inertia of the body about that axis.
Thus, if mi, Wg, mg, etc., be the masses
of the parts of the body, and r^, r^, r.j,
etc., be the distances of these parts
respectively from the axis, then the

moment of inertia
moTo^ 4- m^n^ + etc. . .

m,r| +

*2'2

. = %mr^.
The moment of inertia of an area
and the moment of inertia of a line are
defined in a similar manner by sub-
stituting area or length for mass. But
since areas and lines have no inertia,
they have, strictly speaking, no moment
of inertia.

The moment of inertia of a force about an axis perpendicular to the
line of action of the force is the product of its magnitude and the
square of the distance of its line
of action from the axis.

The graphic method of de-
termining the moment of inertia
of a plane area, or of a system
of parallel forces, will be under-
stood from the two examples
worked out in Figs. 205 and
206.

Fig. 205 shows the applica-
tion of the method to finding
the moment of inertia of a force
AB about a point M or about
an axis through M and perpen-
dicular to the plane 9£ the
paper. Through M draw MY
parallel to AB. Draw MN
perpendicular to AB. Apply-
ing the construction explained
' in Art.59, a'6'xo/i=ABxMN.
Choose a pole o' at a distance
o'h' from a'h', which is a simple
multiple or submultiple of the

linear unit. From a point n" in AB draw n"a^' parallel to o'a' and n"h"
parallel to o'h'. Since the triangle a"b"n" is similar to the triangle

GRAPHIC STATICS

111

a'h'o\ it follows that a!'h" y o'h' = a'b' x MN, and therefore
a"b" y o'h' X oh = a'h' X oh X MN. But a'b' x oh = ABxMN. There-
fore a"b'' X o'h' X oh = AB x MIP = moment of inertia of AB about
M. a'b'n' and a"b"n" are link polygons, of which the first determines
the moment AB X MN, and the second determines the moment of
this moment, namely, (AB x MN) x MN. The lengths a'b' and a"b"
must be measured with the force scale, and the lengths oh and o'h' with
the linear scale.

Fig. 206 shows the application of the method to the determination
of the momen*t of inertia of the shaded figure about an axis a'a" in
the plane of the figure. The area is divided into parallel strips, and
parallel forces AB, BC, CD, DE, and EF are supposed to act at the
centres of gravity of these strips, the magnitudes of the forces being
proportional to the areas of the strips. The sum of the moments of
these forces about the given axis is equal to a'f x oh, and the sum of
their moments of inertia is equal to a"f" x o'h' x oh. The lengths a'f
and a"f" must be measured with the area scale and the lengths oh and
o'h' with the linear scale.

Exercises Vlllb

1. Three coplanar forces P, Q, and E (Fig. 207) act on a rigid body which is
capable of rotation about an axis through O at right angles to the plane of the
forces. P = 8 lb., Q = 10 lb., and R = 6 lb. Find the resultant moment of these
forces about 0. Determine the magnitude and sense of the force S which will
prevent the rotation of the body.

2. AB (Fig. 208) is a lever whose fulcrum is at A. Forces P and Q act on the
lever as shown. P = 10 lb., and Q = 15 lb. Rotation of the lever about its ful-
crum is prevented by the force E.. Find the magnitude of the force R and
determine the reaction of the fulcrum on the lever.

-?::?

'1,— s— Z.

Fig. 209.

Fig. 210.

Fig. 207. Fig. 208.

Reproduce the above diagrams three times this size and then consider that the
reproduced diagrams are drawn to the scale of \ full size.

3. A bar AB (Fig. 209) is acted on by forces P and Q as shown. P = 20 lb.,
and Q = 25 lb. A couple whose moment is 30 inch-pounds also acts on the bar in
the same plane as P and Q and tends to give clockwise rotation to the bar. Find
the additional force acting on the bar which will produce equilibrium.

4. A bent lever AOB (Fig. 210) is capable of rotation about a pin at 0. The
lever is at rest under the action of the forces P, Q, and R, and the reaction S of
the pin on the lever. P = 15 lb., and Q = 20 fb. Find the magnitude and sense
of R and determine S.

5. Draw a square ABCD of 2 inches side, the lettering being in the clockwise
direction. A force P acts through A in the plane of the square. P has a clock-
wise moment of 11 inch-pounds about B and an anti-clockwise moment of 19
inch-pounds about D. Deterinine the force P.

Another force Q also acts through A in the plane of the square. Q has a

112

PRACTICAL GEOMETRY

clockwise moment of 24 inch-pounds about B and a clockwise moment of 8 inch-
pounds about G. Determine the force Q.

6. ABC is a triangle. AB = 8 inches, BC = 9 inches, and CA = 10 inches.
The lettering is clockwise. A force K acting in the plane of the triangle has a
clockwise moment of 40 inch-pounds about A, an anti-clockwise moment of 30
inch-pounds about B, and a clockwise moment of 12 inch-pounds about C.
Determine the force R, using a linear scale of J.

7. ABCD is a square of 2 inches side, lettered clockwise. A force P acts along
the side AB and another force Q acts along the diagonal AG. The resultant
moment of the forces P and Q about D is 10*4 inch-pounds, anti-clockwise. The
resultant moment of P and Q about the middle point of BG is 25-5 inch-pounds,
clockwise. Determine the forces P and Q.

8. The cantilever (Fig. 211) carries five loads each of 100 lb. as. shown. Draw
the bending moment and shear force diagrams.

9. The cantilever (Fig. 212) is loaded £us shown, the loads being in tons. Draw
the bending moment and shear force diagrams.

[

5 ;

(

^-;

~i 1

u 1

■

<T^r

^ _

ll ^

, ,

"ll

3^ 1

^2 \

±_

-_2

U \

^-U

Fig. 211.

Fig. 212.

Fig. 213.

Fig. 214.

10. The beam (Fig. 213) rests on supports 16 feet apart and is loaded as
shown, the loads being in tons. Draw the bending moment and shear force
diagrams.

11. The beam (Fig. 214) rests on supports 10 feet apart and is loaded as
shown, the loads being in tons. Draw the bending moment and shear force
diagrams.

12. A beam is built firmly into a wall at one end and projects 24 feet from the
face of that wall. The other end rests freely on a support. The beam carries a
uniformly distributed load of 1 ton per foot run. It may be proved that the
reaction at the support is three-eighths of the total load on the beam. Draw the
bending moment and shear force diagrams.

13. A beam (Fig. 215) rests on supports 16 feet apart. The load at A is 2 tons
and each of the other loads is 1 ton. Draw the bending moment and shear force
diagrams.

14. A log of timber 24 feet long and of uniform cross section floats in still
water. On the top of the log at points 6 feet from its ends are placed loads of
200 lb. each. Draw the bending moment and shear force diagrams.

[Note that the bending moments and shear forces on the log are those due to
the given loads and a uniform upward water pressure whose total is 400 lb.]

15. The beam A (Fig. 216) rests freely on two cantilevers B and G as shown.

i: I

±2. u u

.Ah

, i

"" "

U- -

-IG'-U-

iitnT

Z 3 i

1— tii

:ts —

■ '

rsjE

U^nl _

Fig. 215.

Fig. 216.

The given loads are in tons. Draw the diagrams of bending moment and shear
force for this structure.

16. A beam rests on two supports 12 feet apart. This beam carries a dis-
tributed load which varies uniformly in intensity from nothing over one support
to a maximum over the other support. The total load is 15 tons. Draw the
bending moment and shear force diagrams.

GRAPHIC STATICS

113

17. petermine the moment of inertia of each of the figures given in Fig 217
about a horizontal axis through its centroid. ^6- ^-»-<

Fig.. 217.
In reproducing the above figures take the small squares as of half inch side.

105. Stress Diagrams for Framed Structures.— It will be
assumed that the framed structures considered are made up of bars
which are connected by frictionless pin joints at their ends. It will
also be assumed that the loads on the structure are concentrated at
the joints. If a bar carries a load uniformly distributed over its
length this load is divided into two equal parts, and one part is placed
at each end of the bar. If a bar carries a load concentrated at an
intermediate point, this load is divided into two parts, which are to
one another as the distances of the load from the ends of the bar ;
these parts are then placed one at each end of the bar, the greater
part being at that end of the bar which is nearest to the original load.

In studying the equilibrium of a structure, two kinds of forces
have to be considered, (1) the external forces, which for the whole
structure must balance one another, and (2) the internal forces. As
a consequence of the two assumptions mentioned at the beginning
of this article, the bars forming the structure are subjected either
to direct compression or to direct tension under the action of the
external forces. It follows, therefore, that the lines of action of the
internal forces are the lines which represent the bars on the diagram
of the structure (called the frame diagram). At any joint, there-
fore, the forces acting are the internal forces acting along the bars
meeting at that joint, and the external forces, if there are any, acting
at that joint.

If a sufficient number of the forces acting at any joint are known,
the polygon of forces for that joint can be drawn and the unknown
forces determined.

The general method of drawing the complete stress diagram for
a framed structure will be understood by reference to the example
worked out in Fig. 218. A simple roof truss is shown carrying a
load AB at its apex. The other external forces are the reactions
BC and CA at the supports. The internal forces are the forces
acting along the bars AD, BD, and CD. The lines of action of all
the forces are known, but AB is the only force whose magnitude is
known as yet.

At each joint there ajre three forces acting, and the polygon of forces

114

PRACTICAL GEOMETRY

for each joint is therefore a triangle. The triangle of forces for the
joint 2 or for the joint 3 cannot yet be drawn, because the magnitudes
of all the forces at these joints are as yet unknown, but the triangle
of forces for the joint 1 may be drawn, and this is shown at (m).
This triangle determines the magnitudes bd and da of the internal
forces in the bars BD and DA respectively. The sense of these forces
is also determined, and it will be observed that the internal forces in
the bars BD and DA both act towards the joint 1, therefore these bars
are in compression. In drawing the triangle (m) the forces have been
taken in the order in which they occur in going round the joint 1
in the watch-hand direction, beginning with the known force AB.
Beginning with BA, and going round the joint in the opposite direc-

Fig. 218.

tion, the triangle (s), which i^ similar to (m) but differently situated,
is obtained.

Passing next to the joint 2, the three forces acting there are known
in direction, and the magnitude of one of them, BD, has been deter-
mined by the drawing of the triangle (m) or the triangle (s). Begin-
ning with DB, and taking the forces in the order in which they occur
in going round the joint in the watch-hand direction, the triangle of
forces (n) is drawn. If the forces be taken in the order in which they
occur in going round the joint in the opposite direction, beginning with
BD, the triangle (u) is obtained. Proceeding next to the joint 3, the
triangle (o) is obtained when the forces are taken in the watch-hand
order, and the triangle (v) is obtained when the forces are taken in the
opposite order.

The construction of the three triangles (m), (n), and (o), or the
three triangles (s), (m), and (c), determines the magnitude and sense
of each of the three internal forces, and also the magnitudes and
sense of the external forces BO and OA.

GRAPHIC STATICS 115

It is obvious that the triangles (w) and (o) may be applied to the
triangle (m) so as to form the figure (r), and this figure gives all the
results which were found from the separate triangles (m), (n), and (o),
and this figure (r) is the complete stress diagram for the given framed
structure. The figure (r) may of course be drawn at once without
drawing the triangles (m), (n), and (o). It should, however, be noticed
that in order that the force polygons for the diflferent joints may be
combined into one diagram, these polygons must be drawn by taking
the forces in the order in which they occur in going round each joint
in the same direction, (r) is the form of the stress diagram when the
forces are taken in the order in which they occur when going round
each joint in the watch-hand direction, and (w) is the form of the
diagram when the order is reversed.

It is important to observe that in going from the joint at one end
of a bar to the joint at the other end the sense of the force in that
bar is reversed, because if a bar is in tension it must exert a pull at
each of the joints at its ends and if it is in compression it must exert
a thrust at each of the joints at its ends.

The sense of the force in any bar may be determined from the
polygon of forces for the joint at either end and from the sense of one
of the forces represented by one side of that polygon because the
polygon of forces represents forces which are in equilibrium, and if
arrow heads be placed on the sides of the polygon to show the senses
of the forces these arrow heads must follow one another round the
polygon.

106. Examples of Stress Diagrams. — (1) The frame diagram
for a " saw-tooth" roof truss is shown at (u), Fig. 219. The truss is
loaded at the joints as shown, the loads being in pounds.

The first step is to draw the line of loads ah ... g. To determine
the reactions GH and HA at the supports, which are obviously
vertical, choose a pole o and draw the link polygon shown at («;).
The closing line of this polygon is shown dotted. From the pole o
draw oh parallel to the closing line of the link polygon to meet the
line of loads at ^, then gh and Jia are the magnitudes of the reactions
GH and HA respectively.

Commencing at the joint ABKHA at the left-hand support the
polygon of forces ahhha is completed. Proceeding to the joint BCLKB
the polygon of forces hdkh is completed. Proceeding from joint to
joint round the frame diagram the various polygons of forces are com-
pleted, the whole of the polygons of forces making up the complete
force diagram for the truss as shown at {w).

The magnitudes of the forces in the various bars may now be
scaled off from the diagram {w). The senses of the forces at any joint
are determined by inspecting the polygon of forces for that joint and
the sense of the force in a bar at a joint determines whether that bar
is in tension or compression. Lines on the frame diagram which
represent members in compression have been made bold.

(2) The lower part of Fig. 220 shows a roof truss under two
systems of loads. One system is vertical and is due to the weight of

116

PRACTICAL GEOMETRY

Fig. 219.

/■

yT^

' ^-^'''''

' \

^^\^^^--'-_

"" ^:^^ ^s">^^» ~ — -

_ _

>Ei ^^^^v^^::^^

l'^

\di

-\f

c\P,

1050

2100

1050

2' L

40 -HH

Fig. 220.

GRAPHIC STATICS

117

the roof covering. The other system is at right angles to the left-
hand surface of the roof, and is due to wind pressure. All the loads
are in pounds. The truss is supposed to be fixed at the right-hand
support u but at the left-hand end v it is merely supported ; hence the
line of action of the reaction at v must be vertical. The direction of
the reaction at u is unknown and the magnitudes of both reactions
are unknown.

The part abcdefgJik of the polygon of external forces acting on the
truss is first drawn. The next step is to determine the reactions KL
and LA at the supports.

Choose a pole o and draw the link polygon shown by dotted lines
on the frame diagram, starting at the point u. By starting at the point
u the space K, which is as yet not completely defined, is eliminated.
The closing line of the link polygon is wu. A line through the pole o
parallel to wu to meet the vertical through a determines the point /
and therefore also Ik. This completes the polygon of external forces.

Starting at the joint at one of the supports the complete stress
diagram may now be drawn as in the preceding example.

107. The Method of Sections. — A little more than one half of
a roof truss is shown in Fig. 221. Conceive that this truss is divided
into two parts by a plane of
section XX which cuts three
bars FG, GH, and HA. Next
suppose that the part of the
structure to the right of XX
is removed and that external
forces are applied at the
sections of the bars cut so as
to balance the internal forces
in these bars. These external
forces will take the place of
the forces exerted on the part
of the structure to the left
of XX through the parts of
the cut bars which have been
removed. The part of the
structure to the left of XX will now be in equilibrium under the
action of the given system of loads and the three external forces FG,
GH, and HA. It will now be shown how these three applied external
forces may be found. The determination of these three forces is
evidently the determination of the forces in three of the members of
the original structure.

Draw the line of loads ahcdef. To complete the polygon of ex-
ternal forces on the structure to the left of XX requires the locating
of the points g and k Choose a pole o and, starting at the point w,
where the lines of action of two of the unknown external forces inter-
sect, draw the link polygon of which vu is the closing line. A line
through o parallel to vu to meet a line through a parallel to AH
determines the point h. The intersection of hg parallel to HG and fg

118 PRACTICAL GEOMETRY

parallel to FG determines the point g. The senses of the external
forces FG, GH, and HA are found at once by going round the com-
pleted polygon of external forces in the direction fixed by the sense of
any one of the known external forces. It will be seen that the bar
FG is in compression while the bars GH and HA are in tension.

This method of sections is extremely useful, for it may be applied
to determine directly the forces in three members of a structure with-
out drawing the stress diagram for any other part of the structure.

In drawing the stress diagram for the truss, a part of which is
shown in Fig. 221, the polygon of external forces would be first drawn,
then the stress diagram would be built up polygon by polygon com-
mencing at one of the supports where there are only two forces whose
magnitudes have to be determined. In passing from joint to joint if
a joint is arrived at where the magnitudes of three forces have to be
determined the polygon of forces for that joint becomes indeterminate.
This happens when the joint AMNHA or the joint DEPNMLD is
reached. If however a section be taken as shown in Fig. 221 and
the force AH determined as explained above the polygon of forces
for the joint AMNHA may be completed and the stress diagram may
then be proceeded with in the ordinary way.

108. Deflections of Braced Frames. — The bars which make
up a braced frame are generally in tension or compression and the
forces acting along the bars may be found as described in the preceding
articles of this chapter. The bars have then to be proportioned to
resist these forces. The pull or push on a bar will cause it to lengthen
or shorten by an amount x and will produce a stress / in it. If E is
the modulus of elasticity of the material in tension or compression

and I is the length of the bar, then E = . = ttz ~ ' *^^

"'- E-

For steel E may be taken as 30,000,000 pounds per square inch

and if /is, say, 15,000 pounds per square inch, then x = oTT aoo 00 ~

9000' *^^* ^^ *^® ^^^ ^^^^ lengthen or shorten by an amount which
is the l-2000th part of its length. A bar 100 inches long
would therefore lengthen or shorten 05 inch under these con-
ditions.

It is outside the scope of this work to discuss further the
design of structures or the determination of the strains produced in
their members by given systems of loading, but it will now be
shown how the deflections of braced frames may be determined
when the alterations in the lengths of the members due to loading
are given. •

The lower part of Fig. 222 shows the frame diagram of a pent-roof
truss loaded at the joints B, C, and D. It is given that the various
bars of the truss alter in length by the amounts stated on the lines

GRAPHIC STATICS

119

which represent the bars. + denotes increase and - denotes
decrease in length. These alterations in length are so small that
the change m the form of the truss due to the loads cannot be
clearly shown on an ordinary scale drawing of the truss. The
displacements of the various joints may however be determined as
follows.

The joints A and E are supposed to be fixed, that is these joints
are not displaced when the truss is loaded. Choose an origin and
draw horizontal and vertical axes OX and OY. is the origin and
OX and OY are the axes of a displacement diagram which is to be
drawn to a large scale. In Fig. 222 the displacement scale is five
times full size, but on a drawing board
the displacement scale would not be
less than ten times full size.

Consider the joint B. Through
draw Obi parallel to AB and make
Obi = 0*08 inch on the displacement
scale. Draw Ob.^ parallel to BE and
make Ob^ = 0*06 inch on the dis-
placement scale. Draw b^b perpen-
dicular to 0^1 and ^g^ perpendicular
to 0^2 and let these, perpendiculars
meet at b. If is the original posi-
tion of B then b is its new position
when the bar AB has lengthened 008
inch and the bar BE has shortened
0-06 inch.

To be strictly accurate b^b should
be an arc of a circle whose centre is in
h^O produced at a distance from O
equal to the length of AB measured on the displacement scale ; but
the radius of this arc is so large compared with the length of the arc
that the arc is practically a straight line at right angles to ^^O. So
also bj) should be an arc of a circle whose centre is in 0^2 produced
at a distance from O equal to the length of BE measured on the dis-
placement scale.

Consider next the joint C. C has a downward displacement whose
vertical component is the vertical component of the displacement of B
plus the displacement 0*04 inch due to the lengthening of the bar BC.
The former is represented by Ocq and the latter by c^Ci. The horizontal
component of the displacement is 0*05 inch due to the shortening of the
bar CE and this is represented by Oc^. The horizontal c^c to meet the
vertical c^c determines c the new position of C on the assumption that
was its original position.

Lastly consider the joint D. D has a displacement whose com-
ponent in the direction BD is made up of the component of the
displacement of B in that direction, that is Obj, and the amount
0'04 inch by which the bar BD lengthens and which is represented by
b^di. The total component of the displacement of D in the direction

Fig. 222.

120

PRACTICAL GEOMETRY

BD is therefore Od^. The horizontal component of the displacement
of D is made up of the horizontal component of the displacement of
C, that is Ocg, and 0*05 inch the amount by which the bar CD
shortens and which is represented by Cg^g- ^^^ total horizontal
component of the displacement of D is therefore Oc^a- ^i<^
perpendicular to Od^ meeting the vertical through ^2 determines
d the new position of D assuming that its original position was 0.

The angular deflection of any bar is measured by the ratio y/l
where y is the relative displacement of the ends at right angles to
the bar and I is the length of the bar. For example, consider the bar
BD. The component of B's displacement at right angles to BD is h\
or d^di where bd.^ is parallel to Od^. The component of D's displace-
ment at right angles to BD is dd^^. Hence dd^ is the relative
displacement of D and B at right angles to BD, and the angular

If BD is 14 feet and dd^ is

0-292

= 0-00174. Whether

displacement of BD is dd^ -4- BD.

0*292 inch, the angular displacement Ois ^ ^ ,-,

this be taken as the circular
measure, or the sine, or the
tangent of the angle, the angle
being very small, the angle
is 6 minutes or 1-1 0th of a
degree.

In the foregoing example
one member AE (Fig. 222)
was supposed to be fixed and
to remain of the same length.
An example will now be con-
sidered in which all the
members are displaced. Fig.
223 shows a roof truss ADF
in which the joint A is fixed
and the joint F is free to
move horizontally. The truss
supports dead loads and in
addition a wind load on the
left-hand side. These loads
cause alterations in the
lengths of the bars which are
stated on the lines which re-
present them.

Assume in the first place
that, say, the bar CD re-
mains vertical and that C is
fixed in position. Choose
an origin and axes OX and
OY and determine the dis-
placements of the various
joints, as in the previous example, the original position of each joint

Fig. 223.

GRAPHIC STATICS 121

being taken as at O. Then on the assumption that C is fixed and CD
remains vertical, A suffers a horizontal displacement 0«i to the left and
an upward vertical displacement a^a. Also F suffers a horizontal dis-
placement O/i to the right and an upward vertical displacement//.
But under the actual conditions A is fixed. Let, therefore, the
whole truss be now lowered an amount aa^, or, which is the same
thing, let the axis OX be raised to O'X' the level of a. Let, also,
the whole truss be moved to the right a distance equal to Oaj, or,
which is the same .thing, let the axis OY be moved to the left so as
to pass through a. The axes are now O'X' and O'Y' and a which
represents the position of the joint A is at the origin 0'. It will be
seen that the joint F has now a downward vertical displacement
equal to /'/. To bring F to its original level, which is the level of
A, the whole truss must now be rotated about A until F rises
through a distance equal to ff\ This angular movement will cause
the various joints to travel at right angles to the lines joining
them to A distances proportional to the lengths of these lines.

Consider the joint D. Referred to the origin 0' and the axes
O'X' and O'Y', d shows the displacement of D when A is fixed and
F is at the distance // below its proper level. It is now required
to find the final displacement of D when the truss is turned about
A to bring F to its proper level. With A as centre and AD as
radius describe the arc DDj to cut the horizontal AF at Dj. Draw
the vertical D^HK to cut 0/ at H and 0'/' at K. Then HK is
the amount that D will travel at right angles to DA when the
truss is turned about A to raise F the distance ff. Draw dd'
perpendicular to AD and make dd! equal to HK, then d! shows
the final displacement of D referred to the origin 0' and axes O'X'
and O'Y'. In like manner the points V, c', and e' are found.

109. The Three-Hinged Arch. — If the ends of a roof or bridge
truss are secured to foundations by hinged joints, and there is another
hinged joint at an intermediate point, say, at the middle of the truss,
such a truss is known as a three-hinged arch, and it is said to be con-
structed on the three-hinged system. The determination of the stresses
in the various bars of such a truss may be proceeded with as in an
ordinary truss as soon as the reactions at the hinges are determined.

One method of finding the reactions at the hinges is as follows.
Fig. 224 shows a truss hinged at A, B, and C. The resultant load on
the part AB is the force P, and the resultant load on the part BC is
the force Q. First neglect the load acting on the part BC. The
part BC is then under the action of two forces only, namely, the
reactions at B and C, and these forces must balance one another, and
will therefore act in opposite directions along the straight line BC.
The truss as a whole is now under the action of three forces, namely, the
force P, the reaction Tj at C, which acts along CB, and the reaction Si
at A. Since these three forces are in equilibrium, and since the lines
of action of two of them, Ti and P, meet at m, therefore the line of
action of the third one, Sj, must be Am. By means of the triangle of
forces the magnitudes of Si and Tj can be determined.

122

PRACTICAL GEOMETRY

Fig. 224.

Next neglect the load on the part AC, and consider the load Q on

the part BC. This load Q will cause reactions Sg and Tg at A and B

respectively, and these

reactions may be found

in the same way as Sj

and Ti were found.

When both loads P

and Q act, it is evident

that the reaction at A

will be the resultant of

Si and Sjj, and the re-
action at B will be the

resultant of Ti and Tg.
The reaction of the

part AB on the part CB

at B will be the force

which will balance the

force Q and the reaction

at C, and the reaction

of the part CB on the

part AB at B will be the force which will balance the force P and

the reaction at A. These two reactions will, of course, be equal and

opposite.

When the truss is symmetrical about a vertical centre line, and is

symmetrically loaded, the reactions at B will be horizontal, and the

line of action of the reaction at A will be the line joining A with the

point of intersection of the line of action of the resultant load on the

half truss AB with the horizontal line through B. The direction of

the reaction at C is found
in like manner.

The two parts of a
three-hinged arch may be
considered as two girders
or two beams loaded ob-
liquely. When these
beams are solid or built
up of plates and angles
instead of being open
brace work they are called
arched ribs.

An arched rib of a
three-hinged arch is shown
in Fig. 225, the joints
being at 1 and 4. This
rib carries vertical loads

AB and BC. The reactions CO and OA at the joints may be deter-
mined in the manner already explained, and oafec, the polygon of external
forces, may then be drawn. If the link polygon 1 2 3 4 for the external
forces be drawn so as to pass through the joints 1 and 4 this polygon

Fig. 225.

GRAPHIC STATICS

123

is called the linear arch for the rib. The sides of the linear arch are the
lines of action of the thrusts which the rib has to support. The
magnitudes of these thrusts are given by oa, oh, and oc. It will be
observed that each of these thrusts has a horizontal component whose
magnitude is given by oh, the perpendicular from o on abc.

Consider the normal section LM of the rib. Let G be the centroid
of this section. The thrust 11 on LM has the magnitude oh and
since R does not pass through G there is a bending moment on the
rib at LM whose magnitude is R X GN, where GN is the perpen-
dicular from G on 2 3. Draw the vertical GE to meet 2 3 at E. Com-
paring the triangles GNE and ohh it will be seen that they are similar,
and that GE : GN : : oh : oh, or GE x oh = oh x GN = R x GN.
But R X GN is the bending moment at the section LM, therefore
GE X oh is also the bending moment at LM. Hence the vertical
intercept between the locus of G and the link polygon 1 2 3 4 at any
point on that locus is a measure of the bending moment at that point,
since oh is constant. The shaded figure is therefore the bending
moment diagram for the rib.

Let 2 3 meet ML produced at F. Resolve R into components P
and Q perpendicular and parallel respectively to LM. It is easily
seen that P x GF = R x GN ; therefore the bending moment on the
rib at G is equal to P X GF.

In addition to the bending moment on the rib at the section LM
there is a shear force equal to Q and a normal thrust whose resultant
passes through G and is equal to P

Exercises VIIIc

In working the following exercises on forces in framed structures it is not
sufficient to draio the force diagrams. The forces should he scaled off from the
force diagrams and the results stated on the lines of the frame diagram. The lines
of 4he frame diagram which represent bars which are in compression should be
lined in considerably thicker than the others after the forces have been determined.

1. A crane frame is shown at (a) Fig. 226. The external forces are, the load

■ -

k+A

—

— I—

20'^

• •

-»-

J'a]

^\0>

.;w

2000

400

(

»,

IFT

■^

^^1

151

—

innn

?m^

Fig. 226.

W of 2 tons, the horizontal force R and the reaction at the footstep at the lower
end of the post. The lower end of the post is hemispherical and the reaction
there may be assumed to pass through the centre of the sphere, the axis of the

124

PRACTICAL GEOMETRY

lower sloping member also passes, when produced, through that point. Draw the
frame diagram four times the given size and then determine the reaction at the
footstep and the forces in the bars B, C, D, and E, (1), assuming that the load W
is hung directly from the joint A, and, (2) assuming that the load W is suspended
by a chain which passes over a pulley at A as shown by the dotted lines.

[Note that when a chain passes over a pulley the resultant force on the axle
of the pulley is the resultant of two forces each equal to the tension in the chain
and acting along the straight parts of the chain which proceed from the pulley.]

2. A crane frame is shown at (6) Fig. 226. The load W is 5 tons. Draw the
frame diagram, say, three times the given size. Determine the balance weight L
so that there shall be no bending action on the lower part of the crane post.
Determine the forces in the various members of the frame, (1) assuming that the
load W is applied directly at the point of the jib, (2) assuming that the load W is
suspended by a chain which passes over two equal pulleys as shown by the dotted
lines. [See note to preceding exercise.]

3. A projecting truss for the roof of an open shed is shown at (c) Fig. 226.
The truss is loaded as shown, the loads being in pounds. Draw the frame
diagram, say, three times the given size and then determine the forces in the
various members.

4. A pent roof truss is shown at {d) Fig. 226. The truss is loaded as shown,
the loads being in pounds. Draw the frame diagram three times the given size
and then determine the forces in the various members.

5. The given roof truss (Fig. 227) is loaded as shown. Find and measure the

double this size
boolb.

Fig. 227.

vertical supporting forces at A and B. Determine the thrust in each of the two
upper horizontal members of the frame, writing the answers on the bars
themselves. [b.e.]

6. A curb roof truss is shown at (a) Fig. 228. The truss is loaded as shown,

—

-1

r—

|—

M-

p-j

1 1

''N

<**

,.>

^»^

(d)

y •

t^ -

j^-

ib)

»^

«^

i£

^ —

■^

—%

)

. _

—

wmT

I4i:

—

Fig. 228.

the loads being in pounds. Draw the frame diagram three times the given size
and then determine the forces in the various members.

GRAPHIC STATICS

125

7. The diagram (6) Fig. 228 represents a girder supported and anchored down
by vertical forces Rj and Rg respectively. The girder carries loads of 60 tons at
each end as shown. Draw the frame diagram three times the given size and then
determine the forces Rj and Rg and the forces in the various members.

8. At (c) Fig. 228 is shown the truss for the roof of an island platform of a
railway station. The loads (in tons) due to the roof covering and wind pressure
are given. Draw the frame diagram double the given size and determine the
forces in the members of the frame.

The scale of the figure, when drawn double the given size, being J inch to 1
foot, what are the bending moments on the supporting pillar at A and
at B ? [B.E.]

9. In the crane shown at (d) Fig. 228 the horizontal member has a pin
joint at the left-hand end, and is supported at the other end by an inclined tie
bar as shown. The load W of 1 ton is suspended from the middle of the hori-
zontal member as shown. Draw the frame diagram to the scale of J inch to 1
foot and then determine, (1) the magnitude and line of action of the resultant
shear force on the pin at the left-hand end of the horizontal member, (2) the
tension in the tie bar, (3) the thrust in the horizontal member, (4), the total bend-
ing moment at the middle of the horizontal member.

10. The diagram. Fig. 229, represents a swivel bridge, its pivot placed at AA.

I-2S

1-2$

Scale 0-75mh=Wfed

10 I

10 A

Fig. 229.

The loads, in tons, are placed as indicated. Draw the diagram double the given
size. Determine the equal loads on the counterpoise to the right of AA placed as
shown, then find the forces in the various members of the bridge. [b.e.]

11. You are given in Fig. 230 a bowstring truss under the action of wind

Loads irv tons
Brow double this size

Fig. 230.

forces, and supported at the ends A and B. Determine the supporting forces at
the ends, that at A acting in a vertical line. Also draw the force diagram or
reciprocal figure for the truss. [b.e.]

12. Determine the reactions at the supports and draw the stress diagram for
the truss of example (2) p. 115 with the same loads except that the wind forces
act on the right-hand instead of on the left-hand side of the truss,

13. The given crane (Fig. 231) is set out to a scale of' 1 cm. to 3 feet. A load
W causes the bar AD to lengthen, and the three bars radiating from C to shorten,
by the amounts written on the bars. Assuming AB to be rigid, find the vertical

126

PRACTICAL GEOMETRY

component of the deflection of D. Find also the angular deflection of

CD. [B.E.]

14. A crane has the form shown in Fig. 232, A load W causes the bars to
alter in length by the amounts written thereon, the minus sign denoting shorten-
ing. The joint A being fixed and the joint B being free to move horizontally,
find the deflection of D and measure its vertical and horizontal components.

What is the angular deflection of the bar CD, if its actual length is 100
inches ? [b.e.]

15. Forces, not shown, acting on the given roof truss (Fig. 233), cause the

Fig. 231.

Fig. 232.

Fig. 233.

bars to alter in length by the amounts written thereon, the minus sign denoting
shortening. If the end A is fixed, and B is free to move horizontally, find the
displacement of B.

What is the angular displacement of the horizontal bar CD if its actual length
is 110 inches ? [b.e.]

16. The diagram (Fig. 234) shows a simple roof truss, resting on walls and
loaded as shown, the span being 15 feet and the inclination of the rafters 39°.

(a) Find and measure the horizontal thrust on the walls.

(6) Draw diagrams showing the thrust, shearing force and bending moment
throughout the length of one of the rafters. Measure the maximum values of
these quantities. [b.e.]

Fig. 234.

Fig. 235.

17. A braced arch, hinged at the crown and at each springing, is loaded as
shown in Fig. 235. Determine the horizontal thrust on the abutments. Find
the forces in all the members of the structure. What is the shearing force on the
pin at the crown ? [b.e.]

18. An arched rib in the form of a circular arc ACB, is hinged at each end A,
B, and at the crown C. The span AB is 150 feet, and the rise 30 feet. Draw the
curve of the arch to a scale of J inch to 10 feet.

A load of 10 tons passes over the arch. Confine your attention to one position
only of this load, that for which its horizontal distance from A is 50 feet.
Determine and measure the horizontal thrust of the arch in tons. Draw a
diagram of bending moment on the rib, and measure the maximum bending
m,oment in ton-feet. Determine also and measure the greatest thrust and the

GRAPHIC STATICS

127

greatest shear in the rib, and state the horizontal distances from A of the places

where these occur

[B.E.]

19. The form of an arched rib is a circular arc of 100 feet span and 16 feet rise
the supports being at the same level. It is hinged at the ends and loaded with a
weight of 12 tons at a horizontal distance of 80 feet from one end. The hori-
zontal thrust of the arch is known to be 12-5 tons.

Draw a diagram of bending moment for the arch. Indicate the places
where the shearing force, thrust, and bending moment on the rib have their
maximum values, and give these values. rB.B.l

20. A flying buttress has the form and weight shown in Fig. 236. It is sub-

Draiv double
this size.

^»- '

-■^

—

40'--_

1 ^

20'

Fig. 236.

7-0 6-6 6-3 6-0 76 5*3 5
Loads in tons. Draw double tMs size.
Scale ZOFeet to an incfv.

Fig, 237.

ject to a thrust P as indicated. Show the lines of the forces acting across the
several joints. State the magnitude of the force on the bed joint BB. Adopt a
force scale of ^ inch to 1 ton. [b.e.]

21. ABDC, Fig. 237, represents the section of a half-arch. The span of the
half -arch is 40 feet, rise 20 feet, thickness of key-stone 6 feet, thickness of arch-
ring at abutment, AB, 9 feet.

The loads on the half-arch are supposed concentrated as shown.

The dotted arcs bd and ac represent the limits of safety, and are at one-third
of the thickness of the arch-ring from the extrados and intrados respectively.
The horizontal thrust at the crown is assumed to be along the line dt, and the
curve of thrust is further assumed to pass through the point e where the vertical
through the abutment FA meets the dotted arc ac. Determine and draw the
curve of thrust between the points d and e, by means of a funicular polygon, and
give the value of the horizontal thrust at d.

Scale of lengths, 10 feet to an inch.

Scale of loads, 10 tons to an inch. [b.e.]

CHAPTER IX

PLANE CO-ORDINATE GEOMETRY

110. Co-ordinates of a Point. — The position of a point P in a
plane (the plane of the paper) may be fixed by giving its distances
X and y from two fixed intersecting straight lines OX and OY (Figs.
238 and 239), the distances x and y being measured parallel to OX
and OY respectively as shown.

The fixed lines OX and OY are called the axes. The axes are of
unlimited length and each extends both ways from O. The point O
where the axes intersect is called the origin. The distance x is called
the abscissa and the distance y the ordinate of the point P, and these
two distances together are called the co-ordinates of the point P. A
point P whose co-ordinates are x and y may be referred to as the
point x,y.

*-x — ^

Fig. 238.

Fig. 239.

Fig. 240.

When the axes are at right angles to one another (Fig. 238) x and
y are the rectangular co-ordinates of the point P, and when the angle
between the axes is not a right angle (Fig. 239) x and y are the oblique
co-ordinates of the point P.

The abscissa x is positive ( 4- ) or negative ( — ) according as the
point P is to the right or left respectively of the axis OY, and the
ordinate y is positive or negative according as the point P is above or
below the axis OX.

The position of a point P in a plane (the plane of the paper) may
be fixed in the manner illustrated by Fig. 240. O is a fixed point
called the pole, and OX is a fixed straight line called the initial line,
or line of reference. The point P is joined to O and the line OP
whose length is r is called the radius vector. The angle which OP
makes with OX, measured from OX in the anti-clockwise direction is
called the vectorial angle, r and 6 are called the polar co-ordinates of

PLANE CO-ORDINATE GEOMETRY

129

the point P, and these co-ordinates fix the position* of the point P.
As defined above r and are both positive. If is measured in the
clockwise direction from OX then it is negative. The angle whether
it is positive or negative fixes the position of the positive radius vector
OP. If the radius vector is negative then produce PO to F and make
OP' equal to OP, then OP being the positive radius vector, OP' is the
negative radius vector. A point P whose polar co-ordinates are r and
may be referred to as the point r,0.

Rectangular co-ordinates are the most common in practical problems
and when co-ordinates are referred to, without qualification, rectangular
co-ordinates will be understood.

111. The Straight Line.— AB (Figs. 241 and 242) is a straight

line which intersects the axis OY at C. Let 00 = c. Take any point

P in AB. Draw the ordinate PM and through C draw CL parallel to

PL
OX to meet PM at L. Then ^^ is a measure of the slope of AB to OX,

and since the slope of AB is independent of the position of the point P,

~ = constant = m, say. But PL = PM - LM = PM - 00 = y - c,
CL

and

CL = iK. Therefore ^- ^ = w?, or y = mx -\- c, and this is

the

EiQ. 242.

equation to the straight line AB. The meaning of this equation
is that, the constants m and c being known, the co-ordinates of
every point in the
straight line satisfy
the equation. The
distance OC = c is
called the intercept of
AB on the axis of y.

For the line shown
in Fig. 241, m = J and
c = 5, hence the equa-
tion to AB is ?/= ^ a; -f- 5.
It will be found*^that for the point Q, re = 6, and ^ =8. Insertmg
these values in the equation, 8 = l X 6 -h 5 = 8, and the equation is
satisfied. Again, for the point R, a; = -3, and i/ = SJ. Inserting
these values in the equation, 3J = -J X 3 -f 5 = -1^ + o = 3^, and
the equation is again satisfied, and so it will be found for every point

in AB. , ^ .1 i.-

The line AB (Fig. 241) is called the graph of the equation
y = \x + b. Conversely if any pair of values of x and y which satisfy
the equation y = ^x + b he taken as the co-ordinates of a point, and
this point be plotted, all such points will lie in the straight line AB
which is the graph of the equation.

An equation containing the variables x and y in the first power
only is called an equatior, of the first degree. The general form of _the
equation of the first degree, with two variables, is Ax + ^y + ^ - ^,
where A, B, and C are constants which are finite or zero and may be
positive or Negative. It may be proved that the graph of an equation

130

PRACTICAL GEOMETRY

of the first degree is a straight line, and since a straight line is fixed
by two points in it, the graph of such an equation may be drawn as
soon as two values of x and the two corresponding values of y are
known. For example, take the equation —2x-\-^y — ^ = 0. Put
X = 0, then ^ = 3. Put a; = 10, then ?/ = 9|. Plotting the points
a; = 0, 2/ = 3j ^^d ic = 10, 2/ = 9|, and joining them determines the
graph of the equation — 2.t; + 3^/ — 9 = 0.

If a straight line is pa'rallel to OX and at a distance h from it,
then for every point in the line y = h, and this is the equation to the
line. The distance h is positive or negative according as the line is

Fig. 243.

above or below OX. If a straight line is parallel to OY and at a
distance a from it, then for every point in the line x = a, and this is
the equation to the line. The distance a is positive or negative
according as the line is to the right or left of OY.

For problems in co-ordinate geometry " squared paper " will be
found very convenient. Fig. 243 represents a piece of squared paper
on which are drawn eleven straight lines numbered 1 to 11.

The equations to the lines shown in Fig. 243 are given below and
the student should carefully study these, comparing them with their
graphs. The unit used is the length of side of the small squares.

1. y z= ^x -^ 5, or 2y = X -{- 10. 3. y z= ^x — 7, or 2?/ = £c - 14.

2. 2/ = 2^ + 0, or 2y = x, i. y = — 2x -\- 5, or 2x -{- y = 5,

PLANE CO-ORDINATE GEOMETRY

131

5. y = — JiT + ^> or a; + 2?/ = 8.

6. 2/ = -^x - 5, or a; + 3?/ + 15 = 0.
j^ y = —X — b, or X -{- y -{- 5 = 0.

= 5,

-5 = 0.

X =

or a; - 7 =

10.

X =

-10,

or a; + 10

= 0.

11.

y =

-7,

or 2/ 4- 7 =

Comparing the lines 1, 2, and 3 it will be seen that they are
parallel, and when their equations are written in the form y = mx + c,
the coefficient of x is the same for each line. Line 4 is perpendicular
to lines 1, 2, and 3, and the " m" for 4 is the reciprocal of the " m"
for 1, 2, and 3, with the sign changed.

If the units for x and y are represented by the same length on the
graph of the equation y = mx -\- c, then m is the tangent of 6 (Figs.
241 and 242) the inclination of AB to the axis of x, 6 being measured
in the anti-clockwise direction. But in many practical problems in
co-ordinate geometry the co-ordinates x and y represent different kinds
of quantities and even when they represent the same kind of quantity
the scales used for measuring them on the graphs may be different,
and the slope of the line which is the graph of the equation y = mx-\-c,
where wi is a definite number, may be very different in different cases
depending on the scales used for measuring x and y. These remarks
will be better understood after studying the following two examples.

Example 1. — In a certain machine for raising weights the effort
Q, in pounds, at the driving point required to raise a load W, in
pounds, was found to be given
by the equation Q = 0-2W -f- 4.
Since Q is much smaller than W
a more satisfactory graph of the
equation is obtained by using a
larger scale for Q than for W.
Fig. 244 shows a graph of the
equation with the scale for Q five
times that for W. Observe that

PI
to find m or ^ from the graph,

PL must be measured on the
effort scale and CL on the load
scale. On the effort scale PL = 80
and on the load scale CL = 400.

Therefore §^ = ^ = ^'^ ^^ ^^

the equation given. .

Example 2.— The horse-power, H, of a steam engine was varied
by altering the initial steam pressure. The weight of steam, W, in
pounds, used by the engine
per hour, was measured at
different powers and the
results were as given in
the table.

Plotting H horizontally

200
LOAD,W

Fig. 244.

H
' W

8
150

16
303

454

548

40
650

132

PRACTICAL GEOMETRY

IV\}-

- ■ 1

J D

ocno-

^-^

S^no-

_^ ^

-^^

>r^nn-

^z!

^^i"

^300 .-^ 1 1 1 1 1 1 1 1 1 1

ui200 ^^

55'°°c^^

'6 10

20 30 40

HORSE -POWER, H

Fig. 245.

to a scale of 1 inch to 20 horse-power, and W vertically to a scale of
1 inch to 500 pounds, the results are shown by the dots in Fig. 245.
It will be seen that these
dots are approximately in
a straight line, and that the
straight line passing through
the points H = 0, W = 50,
and H = 40, W = 650,
lies evenly amongst the
dots. Hence the rela-
tion between W and H
is given approximately by
an equation of the form
W = wH + c, where c = 50,

PL 600
and m= ^,^= W= ^^-

Observe that c and PL are

measured on the W scale and that CL is measured on the H scale.

Judgment is required in choosing the scales for x and y in plotting.
In general the scales should be such that, in the case of a straight
line, the line should be inclined to the axis of x at an angle lying
between 30° and 60°.

In co-ordinate geometry when the part of a line or a figure is being
dealt with which is at considerable distances from the axes, a new pair
of axes may be drawn nearer to the figure and parallel to the original
axes which may now be off the paper. The new origin will however
not be the zero point for x or y. For example
if the new axis of y is 50 units to the right
of the original axis of y the new origin will
represent 50 on the axis of x.

112. The Circle.— Referring to Fig. 246,
C is the centre of a circle, the co-ordinates of
C being a and b. P is any point on the cir-
cumference of the circle, the co-ordinates of P
being x and y. PM and CN are the perpen-
diculars from P and C on OX. CL is parallel
to OX and meets PM at L. Let CP, the radius
of the circle, be denoted by r.

In the right-angled triangle CPL, CL^ + PL^

CF.

But CL = NM = OM - OK = a; - a,

PL = PM - LM = 2/ - t, and CP = r.
Therefore (x - of + (y - hf = r"

or x^^f- 2ax - 2ay + a^ + 1'' - 7^ = 0.

This is the equation to the circle, r must always be positive, but
a and h may be either positive or negative.

Conversely the graph of an equation of the form

x'-^f + Ax-hBy + C =

PLANE CO-ORDINATE GEOMETRY

133

is a circle whose centre has for co-ordinates, a = — and 6 = — ,

n 2i

and

whose radius is r = JVa^ + B^ — 40.

113. The Farabola.~OK (Fig. 247) is a fixed straight line,
and F is a fixed point. P is a point which moves in the plane of F and
OK, so that its distance from F is always equal to its distance from
OK. The path of P is a parabola, whose axis is the line through F
perpendicular to OK. The line OK is called the directrix, and the
point F the focus of the parabola. The curve cuts the axis at A, the
vertex of the parabola. FA is equal to AO.

Draw PK perpendicular to OK, and PN perpendicular to the axis.
Draw the tangent to the parabola at A, and let it meet PK at K'.
The tangent at A is obviously perpendicular to the axis of the parabola.
Let FA = «, PN = X, and PK' = y.

Then PN^ + FN' = PF^ = ON^.

That is, x^-i-(ij - of = (y + af.

Therefore x^ = iay . , (1)

which is the equation to the parabola referred to the axis of the
parabola and the tangent at the vertex, the axis being the axis of y,
and the tangent the axis of x

Fig. 247.

Fig. 248.

Fig. 249.

If the axis of x be moved parallel to itself until it is at a distance
h from the vertex (Fig. 248), then ^ in (1) will become y — h, and the
new equation will be

x'^ = 4<x(2/ — b) = iay — iab . . . . . (2)

If the axis of y be moved parallel to itself until it is at a distance c
from the axis of the parabola (Fig. 249), then x in (2) will become x — c,
and the new equation will be

{x-cY = ia(y^b) ...... (3)

If the axes of x and y be interchanged then x and y in the above
equations will change places.

114. The Ellipse. — The equations
to the ellipse are not of great importance
in practical geometry.

If the origin O be taken at C (Fig.
250) the centre of the ellipse, and if the
major and minor axes be taken as
the axes of x and y respectively and if the
semi-major axis be denoted by a and the

- 1^

*x-^

i^'iQ. 250.

134

PRACTICAL GEOMETRY

Fig. 251.

Ou IT

semi-minor axis by 6, then the equation to the ellipse is -2 + |2= 1j

If a = h = r, then a? ^ y^ = r^, which is the equation to the circle
when the origin is at its centre.

115. The Hyperbola. — Let a denote the semi-transverse axis and
h the semi-conjugate axis of an hyperbola. Let the origin be taken^«*
at C (Fig. 251) the centre of -

the hyperbola, and let the trans- v^ vK- - X

verse and conjugate axes be
taken as the axes of x and y
respectively.

If X and y are the co-
ordinates of any point P on the

x^ iP-
hyperbola, then -^ — p = 1»

or h^^ — €?y^ = a^h\ and this is
the equation to the hyperbola.

If X and y are the co-
ordinates of any point Q on
the conjugate hyperbola, then

V X

y 2 ^ = 1, or ay — 6V = a^¥, and this is the equation to the con-
jugate hyperbola.

If 6 = a, the hyperbola is equilateral or rectangular, then x^ — y^^a^
for the hyperbola and y"^ — x^ = a^ for the conjugate hyperbola. The
rectangular hyperbola and its conjugate are identical but differently
situated. In the rectangular hyperbola the asymptotes are at right
angles to one another.

An important equation to the hyperbola is that obtained by taking
the asymptotes as the axes of x and y as shown in Figs. 252 and 253,
which show one branch only
of the curve. OA the semi-
transverse axis bisects the
angle XOY and the curve is
symmetrical about OA pro-
duced. In Fig. 252 the asymp-
totes are oblique while in Fig.
253 they are at right angles
to one another. The hyper-
bola in Fig. 253 is therefore Fig. 252.
rectangular.

-xH,

^^^

Fig. 253.

For Fig. 252, xy =
semi- axes as before.

+ h'

= constant, where a and h are the

In Fig. 253, 6 = a and xy = -^ = constant.

The curve xy = ^ = constant (Fig. 253) is an important one.

PLANE CO-ORDINATE GEOMETRY

135

the form PV = constant, the co-ordinates of this curve give the relation
between the pressure and volume of a gas when it expands or is com-
pressed isothermally, that is, at constant temperature.

A simple construction for finding any number of points on the
curve xy = constant when one point is given is described in Art. 50,
p. 47.

116. The Curve y = x^— Whatever be the value of n in the
equation y = .x" if x = 0, then y = 0. Therefore the curve whose
equation is y = x1 passes through the origin.

Let n = 1, then y = x which is the equation to a straight line.

Let n = 2, then y = x^ which is the equation to a parabola.
Whether x is positive or negative its square is always positive, there-
fore ?/ is always positive and the curve lies entirely above the axis of
X. Also since x = ± ^/y the two values of x corresponding to any
given value of y are numerically equal, but one value is positive and
the other negative. The curve is therefore symmetrical about the
axis of y. Also if a; = 4-1,?/= +1, and ii x = —l,y= -f 1.

Let n = 3, then y = x^. If x is positive x^ is positive, and if x is
negative ay^ is negative. Hence the sign of y is the same as that of ic,
and the curve lies in the first and third quadrants. If ic = -f-l>
2/ = -f-1, and ii x = —I, y = —1.

To plot the curve y = x^ for any given value of n, first select values
of X and calculate the corresponding values of y and tabulate as here

shown for n =

= 2 anc

n = 3, then plot.

X =

0-0 ±0-4

zkO-8

±1-0

±1-2

±1-4

n = 2,y =

+0-16

+0-64

+ 1-0

+ 1-44

4-1-96

n = S,y =

0-0

±0-064

±0-512

±1-0

±1-728

±2-744

Fig. 254 shows the curves (1) y = x, {2) y = x^, and (3) y = a^.
The curve y = x^ is known as the cubic parabola.

In Fig. 254, the scale for y is the same as that for x and the true
forms of the curves are shown. It will be seen that when n is greater
than 1 y increases more rapidly than ic, and for comparatively small
values of x, y becomes very large when w = -3. To show a sufficient
amount of the curve within a reasonable area when n is greater than
1 it is generally necessary to make the scale for y smaller than that
for X. This will alter the shape of the curve without altering its
character. This will be referred to again when considering the curve
y = bx\

Considering further the influence of n on the form of the curve
y = x^, it is evident that so long as w is a positive integer, y will have
the same sign as x when n is odd, and y will always be positive when
n is even whether x is positive or negative. Calling the curves having
n odd the odd curves and those having n even the even curves it
follows that all the odd curves lie in the first and third quadrants
while all the even curves lie in the first and second quadrants. Also,

136

PRACTICAL GEOMETRY

(2)

as n
nearer

but for a given

]og y = 1'3 log X. Assume values
for X, calculate and tabulate as
follows.

(3)f

IIS

-2-4

-2-a

H-6

-0-a

Fig. 254.

.'(3) ,

(I)
y=x

all the odd curves pass through the points (-f- 1, -f l)and (— 1, —1),
that is through the points P and R, Fig. 254, and all the even 'curves
pass through the points (+ 1, 4- 1)
and (—1, +1)5 that is through
the points P and Q, Fig. 254.

It will be seen that for a given
value of y, greater than ± 1
increases the curve
to the axis of y ;
value of y less than ± 1 the curve
moves further from the axis of y
as n increases.

Generally in practical problems
only positive values of x and y
have to be considered, and these
relate to the curve in the first
quadrant.

When n is positive but not a
whole number, values of y corre-
sponding to assumed values of x in
the equation y = x" are found by (D
using logarithms as follows. If
y = x^ then log y = n log x. For
example, to plot the curve y = x^"^

X -

log X =

0-3010

0-4771

0-6021

0-6990

0-7782

log y =

1-3 log X =

0-3913

0-6202

0-7827

0-9037

1-0117

y =

2-46

4-17

6 06

8-10

10-27

The curve is shown plotted in Fig. 255, and for comparison the

curves y = x, and y = x^ are also
shown.

117. The Curve y = bx'^— Let
y' = x^, then y — by'. Hence if the
curve ?/' = ic'* be plotted, the curve
y = by' = bx'' will be obtained by en-
larging (or reducing if Z^ is a fraction)
the ordinates of the curve y' = x" b
times. This is evidently equivalent to
drawing the curve y' = x'' with another
scale for the ordinates. The factor h
may therefore be called a scale factor
since it does not alter the character of the curve.

tyx^ w ^

-yx

PLANE CO-ORDINATE GEOMETRY

137

118. The Curve y = a + bx".— Let y' = hx'\ then y = a + y'.
Hence if the curve y' = hx" be plotted, the curve y = a -\- y' = a -{- bx""
will be obtained by raising (lowering if a is negative) the curve bodily-
through a height equal to a, or more simply, by lowering (raising if a
is negative) the axis of a; a distance equal to a.

An important problem of frequent occurrence is to find the equation
to a given curve which may have been obtained by plotting a number
of observations or the results of an experiment. Let AEF (Fig. 256)
be such a curve referred to the axes OX and OY. An inspection of
the curve suggests that it may be of the form y = a -\- hx^. The value
of a is found at once and is equal to OA which in this case is 80.

Take A as a new origin O' and let y denote the ordinate of any
point on the curve, measured from O'X' ; then y' =y — a=:y — 80 = hx".

Will

400

A^^=^

10 20

30 X

JD^

7 ^

- ^^

^•u-

"5^ ;

^ ^

•^ ^^

1.0 ^^

10 ^^

^ /

C? :

U r

10 l-S

logx

Fig. 257.

Fig. 256.
Values of X and y are given in the following table

114

175

255

348

455

574

y' = y - 80

175

268

375

494

logx

0-6990

1-0000

1-1761

1-3010

1-3979

1-4771

log y'

1-5315

1-9777

2-2430

2-4281

2-5740

2-6937

If the curve is of the form y' = bx"", then log y' = logb + n logx. ^

Let log y' = y", log b = c, and log x = x', then y" = c -\- nx\
which is the equation to a straight line.

Values of y' corresponding to the given values of y are inserted m
the table, also the values of log x and log y\ taken from a table of
logarithms.

Plotting the values of log x and log y', as shown in Fig. 257, it is

138

PRACTICAL GEOMETRY

PL
found that the points lie in a straight line for which n = — - = 1*5,

PL being measured on the log y' scale and CL on the log x scale.

c = OC = 0'477. Hence log b = 0477, and the equation to the

curve is, y = 80 -{- 3x^^. *

If the points (log x, log y') are not exactly in a straight line, a
mean line may be drawn amongst them, from which n and log h may
be taken to get an approximate equation to the given curve.

119. The Carve yx^ = c. — The most important application of
the equation ^a;"*= c is to the curve whose co-ordinates give the rela-
tion between the pressure and volume of a gas which expands or is
compressed. If p is the pressure of a gas when its volume is v, then
when the gas expands or is compressed the pressure and volume both
vary and in general the relation between them is given by an equa-
tion o6 the form j9y" = c. In the case where n = I, pv = c, and the
curve is a rectangular hyperbola which was considered in Art. 115,
p. 134.

In studying the curve pv'"- = c only positive values of p, Vj and n
need be considered. If y = 0, then >;" = 0, and in order that p X
may have the finite value c, p must be infinite. Again, if j? = 0, then
v'^ must be infinite, and v must also be infinite if n is finite. Hence
the curve j?y" = c approaches nearer and nearer to the axes but never
actually meets them, or it meets them at infinite distances from the
origin. The axes are therefore asymptotes to the curve.

As an example consider the curve py^^ = 292. Taking logarithms
of both sides of the equation,

logp + 1-3 log V = log 292.

Letp = 60, then

, log 292
log V = —

log 60 2-4654 - 1-778^

'& "^ —

1-3

= 5286,

and V = 3-38.
Let V = 4, then

logp = log 292 - 1-3 log 4 = 2-4654 - 1-3 x 0-6021 = 1-6827,
and p = 48-2.
In like manner for any given or assumed value of i; or j? corre-
sponding values oi p ov V may be calculated.

Various corresponding values of v and p are given in the following
table, and these are plotted in Fig. 258, giving the curve ABD.

3-38
600

4
48-2

5
36-0

7
23-3

10
14-6

15
8-6

An important practical problem is to determine whether a given
curve is of the form pv" = c, and if so to find the values of n and c.
If pv'* = c then log p -{- n log v = log c, which is the equation to a
straight line.

PLANE CO-ORDINATE GEOMETRY

139

Let EFH (Fig. 258) be a given curve which appears to be of the
fonn pv"" = c. Corresponding values of v and p for this curve are
given in the following table, also the values of log v and log p.

"nMuT

I 1 1 [■ 1 1 1 1

50 Xa

50- ^

•An 3 ^

L^^ -

30 ^ ^ -

\ ^1

9n ^F^

20- (^

_^^

10 ^^

^"*" — .^ — ~. D

°'i"" !'

10 IS

Volmne v

Fig. 258.

Cns-

Z^O" s^

B^ "^S -

^ "^^

^f :

- ^^

i*u-

^SP

__^^_^

logv

Fig. 259.

2-2

£0

7-5

4-2

log V

0-3424

0-4771

0-6990

10000

1-1761

log 2)

1-7782

1-G021

1-3010

0-8751

0-6232

The values of log v and log p are plotted in Fig. 259, from which

it is seen that the resulting points are in a straight line for which the

PL
equation is log p = — p-r log v + OC. Measuring PL on the log p

scale and CL on the log v scale ^.^ = 1-4. Also OC, measured on the

OL
log p scale, is 2-267. Hence log p = -1'4: log v + 2-267, therefore
w = 1-4, log c = 2-267 and c = 185. The equation to the curve is
therefore pi;^"^ = 185.

In a problem such as the foregoing, if the points (log v, log p) are
not exactly in a straight 'line, a mean line may be drawn amongst
them, from which n and c may be found to get an approximate
equation to the given curve.

120. The Curve y = ae^*.— In the equation y = ae^% e is the
base of the Napierian system of logarithms, its numerical value to
four places of decimals being 2-7183. a and h are constants. The
above equation is called an exponential equation because the variable x
occurs as part of the index or exponent of the power of e. The
curve which is the graph of y — ae^"" is called an exponential curve.

Let X = 0, then e*^ = e" = 1, and y = a. The constant a is there-
fore the intercept of the curve on the axis of y.

Let y = 0, then e*^ = 0, and a; = - oo . The curve is therefore
asymptotic to the axis of x.

140

PRACTICAL GEOMETRY

y=0'de^

y-o

8.^

'^N,

HiT^

Fig. 260.

Taking logarithms of both sides of the equation y = ae*%
log y =log a -f hx log e = log a + 0-434'i6a?.

From the equation
log y = \og a -\- 0-43436a;, y
may be calculated if x is given
or assumed, or x may be calcu-
lated if y is given or assumed.

As an exam pie take a = 0*8,
and 6 = 2, then y = OSe^, and
log y = i-9031 + 0-8686a;.
This curve is shown plotted
in Fig. 260. The curve
y = O'Se'^ is also shown in
Fig. 260, and it will be seen
that this second curve is the
image of the first in a mirror
represented by the axis of y.

121. The Curve y = a log« bx. — The curve y = a log, bx is called
a logarithmic curve. Referring again to the exponential curve y = ae*"",
logg y = logg a -\- bx log^ e, but log^ e = 1, therefore log^ y =loge a + bxj

1 y

and X = T loge - , which is of the form x = p log^ qy.

The logarithmic curve y = a log, bx is therefore of the same form
as the exponential curve y = ae^'' with the variables x and y inter-
changed. The logarithmic curve will therefore be asymptotic to
the axis of y while the exponential curve is asymptotic to the
axis of oa.

122. Spiral Curves. — If a straight line OP revolves in a plane
about a fixed point O in a fixed straight line OX in the plane, and if
at the same time a point P in OP moves along the line OP, the point
P describes a curve called a spiral. The form of the spiral depends
on the law connecting the displacement of the point P along OP with
the angular displacement of OP.

The point O is the pole and the fixed line OX is the initial line.
The position of OP at any instant is given by the angle 6 which it
makes with OX. 6 is the vectorial angle and r the length of OP is
the radius vector. The position of P at any instant is fixed by r
and 0.

123. The Archimedean Spiral. — In the Archimedean spiral the
displacement of P from O is proportional to 6 and the equation to
the curve is, r = a6 where a is a constant. It follows that for equal
increments of there are equal increments of r. Stated in another
way this means that if successive values of are in arithmetical pro-
gression successive values of r are also in arithmetical progression.

When ^ = 0, r = 0, and when = 27r, r = 27ra. If 6 be increased
by 30° at a time from ^ = 0, then the constant increment of r is
27ra
"12*

PLANE CO-ORDINATE GEOMETRY

141

The coDstruction for plotting the curve when a is given is shown

in Fig. 261. Radii are drawn at intervals of %r 30°. OA is the

radius vector when 6 = 27r,
or 360°. Hence OA = 27ra.

O A is divided into -^^ =12

equal parts. The remainder
of the construction is clearly
shown in the figure.

The tangent PQ at any
point P makes an angle with
the radius OP whose tangent

is equal to — -. Hence if on

a radius at right angles to

OP, OL be made equal to a,

PL will be the normal to the Fkj. 261.

spiral at P, and the tangent

PQ will be at right angles to PL.

124. The Logarithmic or Equiangular Spiral.— If the suc-
cessive values of are in arithmetical progression and the correspond-
ing values of r are in geometrical progression the curve traced by P
becomes a logarithmic spiral.

The equation to the logarithmic spiral is r = a*. Hence
log r = log a, and this logarithmic equation may be used to
calculate values of r for given or assumed values of 6 when the
constant a is known. If ^ = 0, then r = 1 .

Fig. 262 shows one and a quarter convolutions of the spiral to a
scale of I when a = 1 -2, the unit being 1 inch.

If^ =^or 30°, then

log r = g log a
3-1416

X 0-0792

= 0-04147,
and r = 1*10.

If the increments of ^ be taken equal to ^ then the increments of

log r are equal to 0-04147 and successive values of r are quickly
calculated.

The logarithmic spiral has the property that the angle <^ which the
tangent at any point of the curve makes with the radius vector at
that point is constant. On account of this property the logarithmic
spiral is also known as the equiangular spiral.

The tangent of the angle (f> is given by the formula

142

PRACTICAL GEOMETRY

tan <^

logio g _ 0-4:343
logi

log a '

For the curve shown in Fig. 262,

logr^ebgd
Scale^i

*^^^=0W92 = ^'^^-

To draw the tangent at the point L, on the radius OL produced
make LK = 1 on any scale, and make KH at right angles to OL and
equal to 5-48 on the same
scale. Join LH, then LH is
the tangent to the spiral at L.

If OP and OQ are any two
radii of the logarithmic spiral,
and if OR is a radius bisecting
the angle POQ, then OR is
a mean proportional between
OP and OQ. This property
may be used in constructing
the curve geometrically when
the pole and two points on
the curve are known.

125. Graphic Solution
of Equations. — Many equa-
tions which are difficult or
impossible of solution by ordi-
nary algebraic methods may
be readily solved by graphic
methods. On the other hand
the graphic method is gene-
rally inferior to the algebraic
method in the solution of the

simpler forms of equations. But in any case a study of the graphic
method will lead to a better understanding of the theory of equations.

Consider first the simple equation ax -\- h = 0, where a and h are
constants. To solve this equation graphically, let y = ax -j- h. This
is the equation to a straight line which may readily be drawn. This
straight line cuts the axis of x where y — 0, and the distance of this
point from the origin is the value of x which satisfies the original
equation. In this case the graphic method is obviously inferior to the
algebraic.

Next consider the pair of simultaneous equations ax -\- hy -{- c = 0^
and a^x -\- h^y + Cj = 0, where a, 6, c, a^, h^, and Ci are constants,
The graph of each of these equations is a straight line, and the lines
may readily be drawn, and the co-ordinates of their point of inter-
section are the values of x and y which satisfy both equations.

Ta-ke the equation I'Tx"^ — 0-75a; -1-5 = 0. This is not a difficult
equation to solve by the method of algebra but its graphic solution
will now be considered in order to illustrate graphic methods.

First method. Let y = ITic^ — 0-75a; — 1*5. Choose a number
of simple values of x and calculate the corresponding values of y.
Plot the points whose co-ordinates are these values of x and y. %^ The

Fig. 262.

PLANE CO-ORDINATE GEOMETRY

143

curve joining these points will cut the axis of x where y = 0. There
will be two such points and their distances from the origin are the
values of x which satisfy the original equation.

This curve is shown plotted in Fig. 263. The curve cuts the axis
of X at A, giving x = 1-18, and at B, giving x = — 0'74, and these are
the roots of the equation. The curve should first be sketched to a
small scale through a few calculated points in order to see the probable
form and position of the curve. The parts of the curve in the
neighbourhood of the axis of x should then be plotted carefully to
a large scale in order to obtain more accurately the points A and B.

Second method. Rewrite the equation in the form 1 -7;?;^ = 0-75a;4- 1 *5.
Let y' = l'7x', and let y" = 0-7b x -{- 1-b. Draw the graph of
y" = 0*75a; 4-1*5 which is a straight line. Next draw the graph of
y' = l'7u;^ which is a parabola. These two graphs are shown in

Small sduar es 0-2 unit side lY

jtx

^4^

^^7

•♦.

^ r

s ^ >

^ ^*^ •

^ t ^

1 \

? %^-.^

t ^

'sL E^

-T H

v^^

1 ?

T "^

7- "^^

-^-l!^^

Fig. 263.

Fig. 264.

Fig. 264, and they intersect at the points E and F. The values
of X for the points E and F are the required roots of the given
equation. This is evident, for at these points y' = if and there-
fore l*7a:^ = 0-7ic -|- 1*5 when for x is substituted the abscissa of E or
the abscissa of F.

The graph of y' = l'1x^ should first be sketched through a few
calculated points and the parts in the neighbourhood of the straight
line ?/" = 0'75a; -f- 1'5 should then be drawn carefully to obtain as
accurately as possible the points E and F.

In general the second method of solution just described will be
found preferable to the first as it involves less calculation.

As a final example take the equation ic^ — 0*6a; — 0*1 = 0. Re-
write the equation in the form x^ = 0'6a; -f O'l. Let y' = x^, and
y" = O'Qx -f 0-1. The graph of y" = 0'6x -f 0-1 is a straight line and
the graph of y' = x^ is a cubic parabola.

A rough sketch of the graphs to a small scale will show that their
points of intersection are within the limits x = -\-l and x = —1.
The graphs should now be drawn to a large scale between the limits
a;= -f-1 and x= —1. They will appear as shown in Fig. 265.

144

PRACTICAL GEOMETRY

agia

Their points of intersection, A, B, and C, give the roots x — 0-85,
X = — 0'18, and x = — 0*67. In Fig. 265 the small squares are of
O'l unit side.

The given equation is of
the form x^ ■\- ax -\-'b =i ^^
and is a cubic equation. In
the theory of equations it is
shown that a cubic equation
has three roots but one or
more of these roots may be
imaginary, that is, there may
be three real roots, or two
real roots, or one real root,
or no real roots. This will
be understood by reference
to Fig. 265, for a straight
line which is the graph of
y" = —ax — b may meet the
graph of y' = x^ at three
points, or two points only, or -pj^ 265

one point only, and the line

may not meet the curve at all. When the straight line does not meet
the curve there is no real value of x which will satisfy the original
equation.

Similar remarks apply to the quadratic equation which may have
two, but not more than two real roots.

Exercises IX

1. The axes being at right angles to one another and the unit of length being
1-lOth of an inch, plot the points whose co-ordinates are given in the following
table :—

Point

-17

-23

-18

- 9

-12

-8

2. Same as exercise 1 except that the co-ordinates are oblique, the angle
XOY between the axes being 60°.

3. Plot the points whose polar co-ordinates are given in the following table,
the unit of length being 1 inch.

Point

2-30
30°

1-25
90°

1-85
105°

2-65
180°

0-95.
200°

1-00
-100°

2-05
320°

-1-70
-120°

PLANE CO-ORDINATE GEOMETRY

145

4. State in tabular form the polar co-ordinates of the points given in exercise
1, r and d to be positive in each case.

5. The polar or vector co-ordinates of two points A and B are (3 inches, 30°)
and (4-2 inches, 100°). That is, it O is the pole, and OX the line of reference,
OA = 3 inches, XOA = 30°, OB = 4-2 inches, XOB = 100°.

Draw OX and plot the points A and B, Measure AB and the perpendicular
from on AB, and calculate the area of the triangle OAB. Verify your answer
by calculating the value of |OA.OB sin AOB. [b.e.]

6. Given the equation y = 0-6x + 4 calculate the values of y corresponding to
the following values of x, namely, 5, 10, 15, 20, 25, and 30. Taking the unit for
X and y as 0*1 inch plot the six points of which the

above values of x and the corresponding values of y
are the co-ordinates. Show that all these points lie on
the same straight line.

7. Draw the straight line (a) passing through the
points (10, 15) and ( — 15, —15) and find its equation.
Through the point (15, —3) draw a straight line (&) at
right angles to (a) and find its equation. Find the
co-ordinates of the point of intersection of (a) and (6).

8. Draw the graphs of the equations, x — 2y = 9,
and a: -f 3?/ = 24 and determine the co-ordinates of
their point of intersection. FiG. 266.

9. In arranging some elementary experiments in .

statics, cyclists' trouser clips are used as .spring balances. In order to be able to
measure pulls, a series of weights are hung on a clip, and the corresponding
openings AA (Fig. 266) are measured. The results are as tabulated.

Pull = P ounces . .

1 ^

Opening AA = x inches

0-51

1-15

1-83

2-53

3-24

3-89

4-44

4-96

Plot a curve showing the relation between P and x, the scale for P being I inch
to 1 ounce. Use this curve to graduate a decimal scale of ounces, which being
applied to the spring at A A should measure any pull up to 32 ounces.

10. The co-ordinates of the centre of a circle being a and b, and the radius of
the circle being r, draw the circles given in the following table. Dimensions are
in inches.

\ '

III

VII

VIII

1-5

0-0

-1-5

-0-5

1-2

-1-0

0-0

1-5

-2-0

1-0

-1-2

0-0

1-3

1 1-5

1-5

1-8

2-0

1-6

1-2

0-8

1-7

11. Draw the circles whose equations are —

I. a;2 -f ?/2 = 4. III. x'^+y^ -2x = 8.

II. ic2 -f- 2/2 - 27/ = 3. IV. x'^ + y-i -2x-y = 2-75.

12. Draw the circle which passes through the three points (9, 8), ( — 16, 13),
and (- 5, —12). The unit is to be taken as 0*1 inch. Find the equation to this
circle. Draw the straight line joining the points (10, 18), and (— 25, 8), At the
points where this line cuts the circumference of the circle draw the tangents and
normals to the circle and find their equations.

13. Given the equation x^ = 2y, take the following values of x, namely,
0, ± 0-5, ± 1-0, ± 1-5, zt 2-0, ± 2-5, and ± 3-0, Calculate the corresponding
values of y, plot the points, and draw a fair curve through them. Name the
curve. Unit = 1 inch.

146

PRACTICAL GEOMETRY

14. Given the equation y^ = 2x — 1, take the following values of y, namely,
0, ±0-5, db 1-0, ±1-5, ±20, ±2-5, and ± 3-0. Calculate the corresponding
values of x, plot the points, and draw a fair curve through them. Name the
curve. Unit = 1 inch.

15. The equation to an ellipse is 0-49x2 + y^ = 1-96. Take the following
values of x, namely, 0, ± 0-4, ± 0-8, ± 1-2, ± 1-6, ± 1-8, and ± 2-0. Calculate
the corresponding values of y plot the points and draw a fair curve through them.
Unit = 1 inch.

16. Draw the rectangular hyperbola xy = 200 between the limits a; = 5, and
X = 40. Unit = 0-1 inch.

17. A cubic foot of gas at a pressure of 100 pounds per square inch expands
isothermally until its volume is 5 cubic feet. Draw the expansion curve which
shows the relation between the pressure and volume of the gas. Scales, —
Pressure, 1 inch to 20 pounds per square inch. Volume, 1 inch to 1 cubic foot.

18. Plot the curves, y = x\ and y = x\ between the limits « = ± 2. Scale
for X, 1 inch to 1 unit. Scale for y, 1 inch to 10 units.

19. Plot the curves, y = x-, y = x^'^, and y = x^, between the limits, x =
and x = S. Scale for x, 1 inch to 1 unit. Scale for y, 1 inch to 5 units.

20. Plot the points whose co-ordinates are given in the following table, and
find the equation to the fair curve joining them.

0-40

0-80

1-20

1-60

2-00

2-40

2-80

0-10

0-57

1-58

3-24

5-66

8-92

13-13

Take the scale for a; as 1 inch to 1 unit and the scale for y a,sl inch to 5 units.
21. Plot the points whose co-ordinates are given in the following table, and
find the equation to the fair curve drawn through them.

60 78

144

266

448

694

22. Plot the curve y{x + 0-75)' -'^ = 6, between the limits x = 0-75 and cc = 4,
X and y being in inches.

23. The volumes {v cubic feet) and corresponding pressures {p pounds per
square inch) of a pound weight of steam as it expands are given in the following
table. Plot the expan»ion curve and test whether it approximately follows the
law pv'^ = c, and if it does, find the values of n and c.

3K)

3-7

4-7

6-1

7-9

10-6

14-5

20 2

153-3

118-3

90-0

67-2

49-3 ^

35-5

25-0

17-2

24. Draw the first and second convolutions of an Archimedean spiral, having
given that a certain radius vector cuts the first convolution at 0*7 inch from the
pole and the second at 2 inches from the pole. Draw the tangent to the curve at
the point which is 1*5 inches from the pole.

25. The equation to an Archimedean spiral is, r =: J0 inches. Draw two
convolutions of the spiral.

26. Draw a triangle POQ. OP = 1-1 inches, OQ = 1-4 inches, and the angle
POQ = 45°. P and Q are points on the first convolution of an Archimedean
spiral of which is the pole. Find the initial line and draw the first convolution
of the curve.

27. The equation to a logarithmic spiral is, r — 1-3^ inches. Draw one con-
volution of the spiral and draw a tangent to the curve inclined at 60° to the
initial line.

PLANE CO-ORDINATE GEOMETRY

147

28. In a logarithmic spiral the common ratio of two radii containing an angle
of 30° is 8:9. The longest radius vector in the first convolution is 3 inches.
Draw one convolution of the spiral.

29. Plot the curve, y — 6-l"e

(^-rs)

from cc = to aj = 27r.

You may obtain points on the curve by calculation, using mathematical tables ;
or by projection from the logarithmic spiral, r = ae~*^'^^^^*

Observe that in this spiral the ratio of the lengths of two radii which differ in
direction by 360° = e " ""^^^ ^ ^'^ = 2-718 " ^'^^ = 1:3 very nearly. [b.e.]

30. The curve (Fig, 267) representing the curtail of a stone step is constructed
of circular arcs, each of 90°. Set out the figure to the given dimensions, which
are in centimetres. Scale \. [b.e.)

31. The given volute (Fig. 268) of one convolution is constructed of circular
arcs each of 90°. Set out the curve, when the dimensions are as figured, in
centimetres. [b.e.]

Fig. 267.

Fig. 268.

Fig. 269.

32. Fig. 268 shows a construction for describing a volute of one convolution.
Set out a similar curve of one and a half convolutions, when the dimensions, in
centimetres, are as in Fig. 269. [b.e.]

33. Solve, graphically, the equations : —

(1) x" - 2-bx + 1-5 = 0. (3) x^ - O'bx + 0-1 = 0.

(2) «2 + 3a; + 2 = 0. (4) x^ + 2a;2 - 3a; - 3-375 = 0.

34. Solve, graphically, the simultaneous equations : —

(2y^x-l=^0. , (x^+l = 9y.

^"■f \4:X + dy + 1-8 = 0. ^^^ \x^ +x = 6y.

35. Solve, graphically, the equations : —

(1) 32=^ + 5x = 6. (2) 15a;2 - 2c2« = 0.

CHAPTER X

PERIODIC MOTION

126. Periodic Motion. — If any position of a moving body be
selected and it is found that after a certain interval of time the body
is again moving through that position in the same direction and with
the same velocity, the body is said to have periodic motion. The
interval of time between two successive appearances of the body in the
same position moving in the same direction with the same velocity is
called the periodic time or period of the motion. Periodic time is
generally measured in seconds and will be denoted by T.

The simplest case of periodic motion is that of a body revolving
about a fixed axis with uniform velocity. In this case the periodic
time is the time taken to make one complete revolution. A swinging
pendulum has periodic motion, but it has different velocities in different
positions. In general all the parts of a machine have periodic motion
although some parts may have very variable velocities and very compli-
cated motions during a period.

The number of periods in the unit of time is called the frequency
and is the reciprocal of the periodic time, or frequency = N = 1/T.

127. Simple Harmonic Motion. — P (Fig. 270) represents a
point which is moving with a uniform velocity V along the circum-

FiG. 270.

ference of the circle XYXjYi whose centre is C. M is another point
which is moving backward and forward along the diameter YY^ of the
same circle in such a manner that PM is always at right angles to
YYi, in other words, M is the projection of P on YYj. Under these
circumstances the point M has simple harmonic motion. If XX, is a

PERIODIC MOTION 149

diameter at right angles to YYi and if PN is perpendicular to XXj
then the point N also has simple harmonic motion.

CP is the representative crank of the harmonic motion of the
point M or the point N, and the circle XYXiYj is the auxiliary circle
or circle of reference.

The amplitude of a harmonic motion of a point is the greatest dis-
placement of the point from its mean position and is equal to the
radius of the representative crank.

Denote the angle PCX by 6 and let CP = r, CK = x, and CM = y.
Then, x = r cos 6, and y = r sin 6.

If the angular velocity of CP be o> = V/r, and if P moves from
X to P in the time t, then 6 = lot. Hence, x = r cos wt, and y = r
sin o)t.

If the time t be plotted along the straight line OL and the corre-
sponding displacement y be plotted at right angles to OL as shown in
Fig. 270, the harmonic curve OEFGL is obtained. The length OL
represents the periodic time and this has been divided into twelve
equal parts. The circumference of the circle has been divided
into the same number of equal parts to obtain the corresponding
values of 6.

The distance y is to be taken as positive when measured above
XXi and negative when measured below. Also, the distance x is to
be taken as positive when measured to the right of YYi and negative
when measured to the left.

Since 6 is proportional to t, lengths along OL will also represent
values of and the curve OEFGL is then called a sine curve.
If is in degrees OL represents 360°, but if is in radians OL
represents 27r.

The tangent to the harmonic or sine curve at any point Q is

determined as follows. Draw Qq parallel to OL to meet the circle at

q. Draw qn perpendicular to XX,. Make Q/= FO. Draw /S

at right angles to OL. Make /S = tt to any convenient scale and

make /H = 1 on the same scale. Make fu = Cw. Draw wK parallel

to HS to meet /S at K. KQ is the tangent required.

2r
The mean height of the curve OEF is — and the mean depth

2r
of the curve FGL is also — or 0-637r. In other words, the mean ordi-

nate for half a period is 0'637r. If s^ is the mean of the squares of
the ordinates for half a period, then s^ = Jr^ and therefore s = 0-707r.
These results are of importance in connection with alternating electric
currents.

Consider the equation y = 2 sin l"2t. This is the equation to a
simple harmonic curve for which r = 2 feet, and <o = 1*2 radians

per second. The periodic time is T = -^ = ^.^^ = 5 236 seconds. The

curve may be drawn as shown in Fig. 270. OL will represent
6'236 seconds and will be measured on a time scale. If OL and the

150 . PRACTICAL GEOMETRY

circle be divided into twelve equal parts the successive values of t will be

2 2 X Q.Q ' 12 X 0-6 ' TTx 0^6' ^*^' harmonic curve may also be

constructed by calculating values of ij from the equation ?/ = 2 sin 1-2/.
In this case values of t increasing by, say, 5 second, may be
taken. The values of t would be multiplied by 1*2 to obtain the
values of 9 which would be in radians. Taking the values of the
sines of these angles from a table of sines and multiplying them
by 2 would give the corresponding values of y. t would be plotted
along OL on a time scale while y would be plotted at right angles to
OL on a distance scale.

Consider the equation y = 2-b sin 6. This is the equation to a
sine curve which may be constructed as in Fig. 270. The radius of
the circle is 2*5 on any convenient scale and y will be measured on the
same scale. If the circle be divided into twelve equal parts the
successive values of will be 30^, 60°, 90°, 120°, etc. and these will
be plotted along OL to any convenient scale. The corresponding
values of y are projected from the circle as shown. This sine
curve may also be constructed by calculating values of y from the
equation y = 2'5 sin 0, values of sin being taken from a table of sines.

The exact position at any instant of the point M (Fig. 270)
which has simple harmonic motion is defined by the angle 6 which
is called the phase angle. The term phase is generally used to
designate the position of the point M at any instant by stating the
fraction of its period of motion which it has performed reckoned
from its middle position when moving in the positive direction. The
first or zero phase is when M is passing through C towards Y. For
this position ^ = 0. For a complete period ^ = 27r or

360° and for any position the phase is the ^ period

phase. For example, when 6 is equal to tt the phase
is the half period phase.

A well known mechanism which gives simple har-

monic motion is shown in Fig. 271. The crank pin of ^j^' ^/'
the crank works in the slot on the end of the piece ""'^ ^<^ J
which is made to reciprocate. The reciprocating piece ^ ^^\.^'

has simple harmonic motion when the crank rotates Fig. 273.

with uniform velocity.

128. Lead and Lag. — Instead of using the phase angle 6 to fix
the position of a point having harmonic motion it is frequently more
convenient to give the angular position of the representative crank
with reference to a fixed radius of the auxiliary circle which is
inclined at an angle a to the zero position of the representative crank.
In Fig. 272 CX is the zero position of the representative crank and
CA is a fixed radius inclined at an angle a to CX. CP is any position
of the crank and </> is the angle which CP makes with CA. The
angle a is positive when measured in the anti-clockwise direction and
negative when measured in the clockwise direction.

PERIODIC MOTION

151

Evidently 9 = (ft -^ a and y = r sin (cf) -\- a).

If w is the angular velocity of CP and t is the time taken by P to
travel from A to P, then (f> = wt, and y = r sin (wt + a).

When a is positive it is called the advance or lead, and when it is
negative it is called the lag.

The curve which is the graph of the equation y = r sin (^ + «) is
the same as the curve which is the graph of the equation y = r sin cf>
except that it starts at a different point ; this is clearly shown in
Fig. 272. The dotted part of the curve corresponds to the motion

Fig. 272.

of the crank from the zero position CX to the position CA which
is to be taken as the new starting position. Observe that the equal
divisions on the circle which correspond to the equal divisions on OL
start from the point A.

129. Composition of CoUinear Harmonic Motions. — A
simple way of giving a point a motion which is compounded of two
or more simple harmonic motions is shown
in Fig. 273. A and B are cranks which
give simple harmonic motions to the rods
C and D which carry pulleys E and F
at their upper ends. G and H are sus-
pended guide pulleys. A thin cord or line
wire is fixed at one end K and passes
under or over the different pulleys as
shown. The free end of the wire is loaded
and guided in a vertical direction and
carries a pen or pencil P which traces a ..>.. ^ ,
curve on a sheet of paper stretched round / (g/g. ' / ^ '
a revolving drum L. The cranks and the '\^ y ^^-...^
drum are driven so that each has a uniform "' Fig. 273.

velocity.

The vertical components of the motions of the crank pins are
communicated to the pencil but the amplitudes of these motions are
doubled by the action of the pulleys. If one of the cranks is stopped
while the other rotates uniformly, the pencil will have simple harmonic
motion of the same period as that of the rod driven by the moving
crank but the amplitude of the motion of P will be twice the radius of

152

PRACTICAL GEOMETRY

the moving crank. When both cranks rotate the pencil traces on the
moving paper a curve whose ordinates are the sum of the ordinates
of the harmonic curves due to the separate cranks. Any number of
harmonic motions may be compounded in this way by introducing
a crank for each.

The mechanism illustrated in Fig. 273 is the basis of Lord Kelvin's
tide predicting machine.

If the crank shafts be at a considerable distance from the guide
pulleys G and H compared with the radii of the cranks, the pulleys
E and F may be mounted directly on the crank pins and the mechanism
is considerably simplified but the motions communicated to the pencil
by the separate cranks are not quite simple harmonic motions.

When two collinear simple harmonic motions of the same period
and therefore of the same frequency are combined the resulting motion
is also a simple harmonic motion.

Referring to Fig. 274, OApL is the harmonic or sine curve for the
crank CP which starts from the zero position CX. The radius of the

'f^

/>\n/ X

,''

■f

....

\ ;

....

»N«

r^y4o

i;^

"■"^3

Fig. 274.

crank CP is 1-8 inches and the equation to the curve OApL is therefore
2/i = 1-8 sin B in which Q is measured from the zero position of the
crank. EFgH is the harmonic or sine curve for the crank CQ which
has a lead of 45^. The radius of the crank CQ is 1*2 inches and the
equation to the curve EFij'II is therefore 3/2 = 1*2 sin {0 + 45°) in
which B is measured from the radius which is 45° in advance of the
zero position CX.

To combine the two curves the ordinates of the one must be added
to the ordinates of the other. For example, rn =pw + 5^- In the
addition regard must of course be paid to the signs of the ordinates.

The equation to the resulting curve is ?/ = 2/1 + 2/2 = 1*8 sin ^ + 1*2
sin {B + 45°).

The resulting curve in this case may be drawn independently
as follows. Complete the parallelogram PCQR. Then the diagonal
CR is the representative crank whose harmonic curve, having an angle
of advance RCP, is the curve which has just been obtained by
combining the harmonic curves Oj^L and EF^H.

The general construction for determining the representative crank

PERIODIC MOTION

153

-Y---

which will give to a point a harmonic motion which is the resultant
of the harmonic motions of the same period due to any number of
given representative cranks is shown
in Fig. 275. CP, CQ, CR, and CS are
the relative positions and radii of the
given cranks at a given instant. Draw
P^ parallel and equal to CQ. Draw
qr parallel and equal to CR. Draw rs
parallel and equal to CS. Then CS,
the closing line of the polygon CP^rs is
the crank required. The point whose
motion is being considered is supposed
to be reciprocating in the line YCYj.
The proof of the above construction
is as follows. The distance of the
moving point from C at any instant is

the sum of the projections of the given cranks on the line YCYi at
the instant considered, regard being paid to the signs of these pro-
jections. But the sum of these projections is evidently equal to the
sum of the projections of CP, Vq, qr, and rs on YCYj, and this sum
is also equal to the projection of Gs on YCY^.

When the simple harmonic motions which have to be combined are
not of the same period the resulting motion is not a simple harmonic
motion. To construct the resulting displacement curve the separate
simple harmonic curves are first drawn and then combined as already
explained for motions of the same period (Fig. 274).

An example of the compounding of two simple harmonic motions
of different periods is illustrated by Fig. 276, where the full curve
is the graph of the equation

2/ = 2 sin (^ + 30°) 4- 1-4 sin (W + 50°)
12 3 4 5

The curve (p ) is the graph of the equation 2/i = 2 sin (0 + 30°) and
is the harmonic curve for the crank CP which starts with an advance
of- 30°. from the zero position CX.

154

PRACTICAL GEOMETRY

The curve (q) is the graph of the equation y., = I'i sin {20 + 50°)
and is the harmonic curve for the crank CQ which starts with an
advance of 50° from the zero position CX.

It will be observed that the frequency of the second motion is
double that of the first, that is, the crank CQ rotates twice as fast as
the crank CP.

130. Composition of Harmonic Motions at Right Angles
to one another. — The diagram (a) Fig. 277 shows two slotted bars
at right angles to one another and driven by two cranks so that each
of the slotted bars has simple harmonic motion. If a pencil P fitting
both slots be passed through them where they overlap, then the pencil
will have a motion which is compounded of two simple harmonic
motions at right angles to one another. The remainder of Fig. 277
shows how the curve traced by the pencil P may be drawn.

26 w 25
l4jLl3

Fig. 277.

In the example worked out in Fig. 277 the crank at (6) has a
radius of 2 inches while that at (c) has a radius of 1 5 inches. The

PERIODIC MOTION

155

first crank makes two revolutions while the second makes three revolu-
tions. The frequencies of the two harmonic motions are therefore in the
ratio of 2 : 3. When the first crank makes 30^ with the horizontal
position CX the second makes 45° with CX. Calling these the initial
positions, if the first crank turns through an angle W from its initial
position the second crank will turn through an angle Z6 from its
initial position, in the same time.

The position of the point P is determined, with reference to the
axes OX' and OY', {d) Fig. 277, by the two equations

x = \'b cos (3^ + 45°) a-nd 2/ = 2 sin {26 + 30°)

The circle (h) has been divided into 18 equal parts, starting from
tlie initial position of its crank, while the circle (c) has been divided
into 12 equal parts, starting from the initial position of its crank.
Observe that the. numbers 18 and 12 are in the inverse ratio of the
frequencies.

The construction for determining points in the path of P is clearly-
shown and need not be further described.

taking different

A great variety of curves may be obtained by

A few of the simpler cases

The result

The result is a

frequencies and different angles of lead,
may be mentioned here.

(i) Equal frequencies and equal cranks.

(1) One crank 0° or 180° in advance of the other.

is a circle.

(2) One crank 90° in advance of the other.

straight line.

(3) For any other angle between the cranks the result is an

ellipse,
(ii) Equal frequencies and unequal cranhs.

(1) One crank 90° in advance of the other. The result is a

straight line.

(2) For any other angle between the cranks the result is an

ellipse.

It is easy to show that when the path of P is a straight line its
motion in that straight
line is a simple harmonic
motion.

131. Composition of
Parallel Harmonic
Motions. — AB (Fig.
278) is a vibrating link.
The vertical component
of the motion of A is a
simple harmonic motion
of which oa is the repre-
sentative crank. The
vertical component of the
motion of B is a simple
harmonic motion of which oh is the representative crank. The motions

156 PRACTICAL GEOMETRY

of A and B have the same frequency. The advance angles of the
cranks oa and oh are a and ^ respectively.

C is a poiQt in AB or in AB produced either way. The vertical
component of the motion of C is obviously a simple harmonic motion
and it is required to find the representative crank for this motion.

From a point o' draw o'a' parallel and equal to oa. Draw o'h'
parallel and equal to oh. Join a'h' and divide it at c' so that a'c' : c'V
: : AC : CB. Then o'c' is the crank required and y is its advance
angle. The proof is as follows. Draw c'r parallel to h'o' to meet o'a'
at r. Draw cs parallel to a'o' to meet o'h' at 8. Let the vertical motion
of B be destroyed, then the vertical motion of C will be to that of A
as CB : AB or as c'h' : a'h' or as o'r : o'a'. o'r is therefore the repre-
sentative crank for the vertical harmonic motion of C when B has no
vertical motion. In like manner when the vertical motion of A is
destroyed and B is driven, the representative crank for the vertical
motion of C will be o's. Hence, when A and B are both driven the
representative crank for the vertical motion of C will be o'c' which is
the diagonal of the parallelogram o'rc's as was proved in Art. 129.

Denoting the lengths of the cranks oa, oh, and oc by a, h, and c
respectively the equations for the vertical displacements of A, B, and C
from their mean positions are

7/i = a sin [B -f a), y^^h sin {6 + )S), and 2/3=0 sin {0 + 7)

where B is measured from the initial positions oa, oh, and oc.

The above problem occurs in connection with reciprocating steam
engine valve gears, oa and oh are eccentrics which drive the link
AB. The valve is driven from a point C in AB. o'c' is the equivalent
eccentric, that is, o'c' is an eccentric which would give the same motion
to the valve, driving it directly, as the two eccentrics and the link AB
give it.

In a vertical engine if the crank is in the vertical position oYj when
oa and oh are in the positions shown, then 180 -- a is the angle of
advance of the eccentric oa, and /3 is the angle of advance of the
eccentric oh. In the actual engine the eccentrics oa and oh are on the
same shaft.

132. Velocity and Acceleration in Harmonic Motion. — CP
(Fig. 279) is the representative crank for the
simple harmonic motions of the points M and
N which reciprocate along the vertical and
horizontal diameters respectively of the auxiliary
circle.

Let CP = r and let the position of CP be
defined by the angle B which it makes with
ex. Let 0) be the angular velocity of CP and
let V be the linear velocity of P. Then V = wr.
Let the velocities of M and N when in the Fig. 279.

positions shown be V^ and V^ respectively.

These velocities are the vertical and horizontal components of
V respectively. Hence V = V cos ^ = wr cos ^ = wr cos w/, and

\v^

PERIODIC MOTION

157

V^ = V sin 6 = cor sin = mr sin wf, where t is the time taken by
P to travel from X to P.

If a scale for velocity be chosen such that CP = V, then PM = V^,,
and PN = Y^. The auxiliary circle therefore becomes a velocity
diagram on a displacement base for the simple harmonic motion of the
point M or the point N.

If the angular position of CP be measured from a radius CA
which makes an angle a with CX and if the angle ACP = ^
then Vy = V cos (<^ + a) = wr cos ((^ + a) = wr cos {oit + a)
and V^ = V sin (</> + a) = inr sin (<^ + a) = cur cos (cu< + a)
where t is now the time taken by P to travel from A to P.

The velocities V^ and V^ may be plotted on a time t or angle B or
angle <^ base in exactly the same way as displacements were plotted in
Arts. 127 and 128.

The radial acceleration of the point P is/= — = wV. Let the

acceleration of the points M and N be fy and f^ respectively. These
accelerations are the vertical and horizontal components of / respec-
tively. Hence (Fig. 280)

f^ =fsm 6 = — ^ sin ^ = coV sin 6 = oy^r sin wt,

and

fx=f cos = — cos = ooV cos 6 = (oh cos wt,

where t is the time taken by P to travel from X to P.

Since sin 6 = -, and cos $ = - the above equation may be written,

and

r r^'

If a scale for acceleration be chosen such that CP = /, then fy = y
and/^ = X.

The acceleration of the point M is
shown plotted on the vertical diameter
YYi as a base.

If the angular position of CP be
measured from a radius CA making an
angle a with CX and if the angle
ACP = ^ then for in the above equa-
tions substitute ^ -f- a.

The accelerations fy and /, may be
plotted on a time t or angle B or angle <^
base in exactly the same way as displace-
ments were plotted in Arts. 127 and 128.

When a point has a motion which is
compounded of two or more simple har-
monic motions the resultant velocity and resultant acceleration at any

f 1

\*w*^ »

M^^

'>^ /

/\^

\ fe

ArX

\ '

\x-*

^^^Y.

Fig. 280.

158 PRACTICAL GEOMETRY

instant are determined from the component velocities and component
accelerations exactly as for displacements.

133. Harmonic Analysis. — It has been shown that the displace-
ment of a point from its mean position, when it has simple harmonic
motion in a straight line, is given by the equation y = r sin (0 -\- a)
or hj X = r cos (^ + ct).

Again, if a point has periodic motion in a straight line, and if its
motion is compounded of a number of simple harmonic motions in
which the separate displacements from the mean position are given by.
the equations —

x-^ = Vi cos (0 + tti), X.2 = rg cos (20 -f og), x^ = r^ cos (3^ -j- ag), and so on,

then the resultant displacement is given by the equation —

a; = ri cos (0 + aj) + r^ cos (20 -f- a^) + r^ cos (3^ + as) -\- . • •
and it has been shown how such a resultant displacement may be
found graphically and plotted to obtain a curve of resultant displace-
ment on a time t or angle base.

If instead of measuring the resultant displacement from the mean
position it is measured from a point at a distance c from it, then,

X = c -{- r^ COB (0 -^ tti) 4- r.2 cos (20 + o^) + rg cos (30 + a^) + . . .

The right hand side of this equation is known as a Fourier series.

Any periodic curve of displacement being given it may be resolved
into a number of simple harmonic components. The number of these
components is greater the greater the complexity of the given periodic
curve.

The process of breaking up a given periodic curve into a number
of simple harmonic curves, or the process of finding the equation to' the
curve in the form

X = c -{- Ti cos (0 + tti) + **2 COS (20 + ttg) + r. cos (3^ + as) + • • •

is called harmonic analysis.

The process of harmonic analysis is one of great importance in
connection with the study of the periodic motions of machines and of
alternating electric currents.

A graphic method of harmonic analysis which is comparatively
simple and easily applied will now be described. This method is due
to Mr. Joseph Harrison and was described by him in Engineering of
Feb. 16th, 1906.

The procedure will be illustrated by reference to a definite example.
The full curve EFGH (Fig. 281) has been plotted on the base line OL.
The base represents one complete revolution of a shaft upon which
there is an eccentric which drives a slide valve through certain inter-
mediate link-work. The ordinates of the curve represent the displace-
ments of the valve from a certain fixed position. In order that the
student may transfer this curve accurately and of full size to his
drawing paper, and work out the example for himself, the ordinates
have been dimensioned. Twelve ordinates at equal intervals will be
used and these are numbered to 11.

PERIODIC MOTION

159

In most engineering problems in harmonic analysis it will be found
that not more than three of the harmonic terms are required and in
many cases two terms are sufficient. In the example to be worked
three terms in addition to the constant will be found.

Fig. 281.
The equation is
i» — c + »'i cos (^ + tti) + r.^ cos (2^ + o.^ + **3 cos (3^ + a.^

and it is required to find c, r^, r<^ , and r.^ , also ai, a^, and a^.
Expanding the right hand side, the equation may be written

X = c -\- Vi COS tti cos — ri sin a^ sin

-\- r2 cos tta cos 20 — r^ sin a^ sin 2^
+ r^ cos ttg cos 3^ — rg sin a..^ sin 3^.

Let r^ cos a^ = a^, r.^ cos 02 = aa , and r.^ cos a^ = a-^.

160 PRACTICAX GEOMETRY

Also, let —Ti sia a^ = 6^, — rg sin 02 = ^2, and —r^ sin ag = 63.
Then ic = c + «i cos ^ 4- ^2 cos 20 + % cos 3^

+ b^ sin ^ + 62 sin 26 + 63 sin 3(9.

Since ri cos aj = a^, and — r^ sin a^ = hi it follows by division that
tan ai = \ Also rj^ = a^^ + 61^ and r^ = ± Va^^ + 6i"^. There are

similar relations for the other corresponding constants.

The constant c is the mean of all the given ordinates and in this
case it is 1*63.

From an origin O^ beginning with the zero direction draw 12 radial
lines at equal angular intervals. Along these radial lines, beginning
with the zero line, mark off in succession from Oi the ordinates to 11.
Find the horizontal and vertical projections of these radiating vectors
as shown. Find the sum of the horizontal projections having regard
to their signs. This sum is 4*00 and is equal to Qai^ the coefficient 6
of «! being half the number of ordinates used. Therefore aj = 0'67.
Find the sum of the vertical projections, having regard to their signs.
This sum is 7*26 and is equal to 66^. Therefore b^ = 1*21.

tan ai= = — 1*806 and aj = — 61° to the nearest degree

Ti = ± \ ai^ -\- bi^ = ± 1'38. Since r^ cos a, = a^ and cos a^ and a^
are here both positive it follows that rj = + 1 -38.

The first harmonic term or the fundamental term is therefore
+ 1-38 cos (^- 61°).

Next, from an origin Oo , beginning with the zero direction, draw,
at equal angular intervals, half the number of radial lines that were
drawn from O^. On these radial lines, beginning with the zero line,
mark off in succession from O2 the ordinates to 1 1, going twice round.
Find the sums of the horizontal and vertical projections of these
radiating vectors. In this case these sums are —1*09 and — 0*36
respectively.

Now, 6a, = — 1*09, therefore a^ = — 0-18.
Also, 6^2 = - 0*36, therefore ^2 = - 0-06.

tan tta = = — 0*333 and 02= — 18° to the nearest degree.

r^ = ± V a2^ + 62^ ~ i 0*19. Since Tq cos ao = a^, and since cos aa
is positive and a^ is negative r2 must be negative. Hence rg = —0*19.

The second harmonic term or the octave term is therefore
-0-19 cos (2(9- 18°).

Lastly, from an origin O3 , beginning with the zero direction, draw,
at equal angular intervals, one third of the number of radial lines that
were drawn from Oj . On these radial lines, which in this case are at
right angle intervals, mark off in succession from O3 the ordinates to
11, beginning with the zero line, and going three times round as shown.
The sums of the horizontal and vertical projections of these radiating
vectors in this case are +0*24 and -f0 04 respectively.

PERIODIC MOTION 161

Now, 6a,, = 0-24, therefore a, = 0-04.
Also, (jh, = 0-04, therefore 63 = 0-007.

tan a.. = — ■' = — 0-175 and a.^ = — 10'^ to the nearest degree.

?••; = ± ^/ a.^^ + 63'^ = ± 0-04. Since r.^ cos ao = a., and since cos a^
and «3 are both positive, r.^ = -\- 0*04.

The third harmonic term or the term of the third order is therefore
+ 0-04 cos (3^ - 10°>

The complete equation is now

fl!=l-63 + l-38 cos (6l~6rj-0-19 cos (2i9-18°) + 0-04 cos (3(9-10°).

The crank circle for the fundamental term x^ = 1-38 cos {0 — 61°)
is shown and the curve for this has been projected as explained in
Art. 128, and is shown dotted.

If the student will work out this example, step by step as described,
and also the last four exercises at the end of this chapter, he will find
that the method is simpler than it looks from the foregoing somewhat
lengthy description.

After finding the Fourier equation it is a useful exercise to con-
struct the component simple harmonic curves and then construct the
resultant curve to see how near it approaches to coincidence with the
original curve.

If the angular velocity w in the equation 6 = (ot is known, curves
of velocity and acceleration for the component harmonic motions may
be constructed, and from these the resultant velocity and acceleration
curves may be determined. Or the equations for the component
velocities and accelerations may be found, and then by addition the
equations for the resultant velocity and acceleration may be obtained
and then used to calculate the velocity or acceleration for any position
of the moving body.

Exercises X

1. A point has simple harmonic motion of amplitude 2 inches and periodic
time 2 seconds. Construct the curve which shows the relation between displace-
ment from mean position and time. Scales. — Displacement, full size ; time 3
inches to one second. Draw the tangents to the curve where the time is 1 second
and 1-3 seconds.

2. Taking the equation to a simple harmonic curve, y = r sin oot, cpnstruct
the curve for the case where r = 0-75 foot and the frequency is 2 periods
per second. Scales. — Displacement, 3 inches to 1 foot; time 10 inches to 1
second. What is the value of u in radians per second ?

3. Construct the sine curve y = 1-75 sin 6, from 6 = 0° to 6 = 360° taking
intervals of 30°, y being in inches. Scales. — For y, full size ; for 9, 1*5 inches to
90°. Find the mean of the mid positive ordinates.

4. Given the equation, y = 2 sin 0, fill up the following table, by calculation,
taking sin 6 from a table of sines.

162

PRACTICAL GEOMETRY

h°

15°

25°

36°

45°

55°

65°

75°

85°

j Meau.

Sin.e

i -

i
1

Plot e and y and 6 and y^ from = 0° to = 90°. Scales.— 7/ in inches, full
size ; Q, 5 inches to 90°.

. 5. Given the equations, y^ = 1-7 sin 6, and y^ = 2 sin {6 + 30°), plot the values
of 2/i + 2/2 and 6, and y^yo and 6 for values of 6 from 0° to 360° at 30° intervals.

6. A point M has a simple harmonic motion, in which the displacement x
from the mid position C is given in inches by x = 2 sin (1-5^ + 0*4), t being time
in seconds and the angle being in radians.

Draw a horizontal line (along the tee square) for the path of M, on which mark
the centre C and the limits.of the swing. Let positive displacements be those to
the right of C. Draw the representative crank, its rotation being clockwise.
Find the positions of M when ^ = and when t = 2 seconds and measure CM in
both cases. [b.e.]

7. A weight hangs by a spring, and has an up and down simple harmonic
motion of period T = 2 seconds and amplitude a = 2-25 inches, the advance a

being — radians (at the instant when time begins to be reckoned). The dis-

placement y from mean position at any time t is thus given by the equation
(i^t + a ) = 2-25 sin

y = asm

(^t + l)

Draw a curve showing the relation between y and t at any time from t
t = T = 2 seconds.

Adopt as the horizontal scale for time 3 inches to 1 second, and take the ver-
tical scale for y full size. Read off the displacement y when t = 025 second, [b.e.]

8. A motion in a straight line, which is compounded of two simple harmonic
motions of the same period, is itself a simple harmonic motion of that period.

If X is displacement of a point at time t, this theorem is represented by the
equation : —

jc = «! sin (qt + 61) + a-i sin {qt + e^) = A sin [qt + E).

Determine graphically and measure the amplitude or radius A, and the
advance E of the resultant motion, having given the corresponding elements of
the component motions, viz. : —

ai = 2 inches, a^ = S inches ; Ci = 0*25 radian, e, = l"! radians.

Find and measure the displacement x when ^ = 0, and also when ^ = 3 seconds,
the angular velocity q being J radian per second. [b.e.]

9. Three simple harmonic motions in a straight line are represented by the
equations : —

2/, = 1-5 sin O-U, y. = 1-2 sin (0-8^ + 1), and y, = 0-9 sin (1-6^ + 0-5),
where y^, y^, and 2/3 are displacements in inches from the mean position, the
angles being measured in radians and the time in seconds.

These three motions are combined. Plot the separate displacements and the
resulting displacements on a time base for a complete cycle.

10. The motion of a point in a straight line is compounded of two simple
harmonic motions of nearly equal periods, represented by the following
equation : —

X = 2-1 sin fdt + ij + sin 8^,
where x is displacement in inches from mean position and t is time.

PERIODIC MOTION

163

Let the complete period of vibration be divided into nine equal intervals.
Taking only the first, fourth, and seventh of these intervals, in each case draw a
curve in which abscissae shall represent time,
and ordinates the corresponding displacements
of the point.

Let the time of one of the intervals be re-
presented on the paper by a length of 8 inches.
In determining successive ordinates the method
of projection from the resultant crank may be
used with advantage. [b.e.]

11. In the equation
y = 2-6 sin [6 + 31^) + 0-33 sin (2a + 112°),

which represents a simple vibration with a
small superposed octave, the displacement y for
any value of 6 is given approximately by the
construction defined on the diagram. Fig. 282.
By means of this construction, or otherwise,
determine y for values of d of 0°, 30°, 60° . . . 360°.

Plot a curve with y as ordinate on a 6 base. Scale for 6. — 1 inch to 60°. From
your figure measure y when e = 192°, and compare this with the true value
of y as calculated from the equation, using the table of sines. [b.e.]

12. A point P in a plane has a compound harmonic motion, whose components
parallel to two perpendicular axes OX, OY are given by the equations

y = rsln (e+(i)+s sm(za-t 6)

Fig. 282.

X = a cos {ut + o) = 2*5 cos ( ^ +

7/ = 6 sin {2a>^ + )8) = 1*5 sin

i)'

inches.

Plot the complete locus of P. [b.e.]

13. Referring to Fig. 277, p. 154, draw, full size, the locus of the point P when
the frequencies of the component simple harmonic motions are as 3 : 4 instead
of 2 : 3.

14. A, B, C are three coUinear points in a vibrating link. AB : BC : AC =
5:1:6. The component motion of A in a certain direction is simple harmonic,
with half travel 4 inches, advance —90° ; that is, displacement

a?i = 4 sin ( »^ - ^ j = 4 sin (a - 90°).

The component motion of B in the same direction is given by

x^ = h sin o)t = b sin d.
Let the component motion of C be defined by the equation
X = a sin {6 -{- a).

Find the half travel a and the advance a for the following values of b, and
tabulate the results, as indicated ; —

' Values of b in inches.

0-5

1-0

1-5

Half travel a

Advance o

15. Taking the data of exercise 7 plot the velocities and accelerations of the
weight on a time base. Scales. — Time, 3 inches to one second ; velocity, maxi-
rnum^ velocity in feet per second = 2-25 inches ; acceleration, maximum accelera-
tion in feet per second per second = 2*25 inches. Determine the lengths which
represent a \velocity of 1 foot per second and an acceleration of 1 fobt per second
per second.

164

PRACTICAL GEOMETRY

16. Taking the data of exercise 9 plot the velocities and accelerations of the
component simple harmonic motions also the resultant velocities and accelera-
tions, all on the same time base, for a complete cycle. Construct the velocity
and acceleration scales.

17. The slide valve of a steam engine is actuated by a Joy gear. Fig. 283

AS 6 4

I Ll_

7 30

Copy double this size

Fig. 283.

I 2

-*- +

gives eight positions, 0, 1, 2, ... 7 of the valve, corresponding to the eight crank
positions e of 0°, 45°, 90°, . . . 815°.

Measuring from the point A, the displacement x of the valve for any crank
position Q is given approximately by the Fourier equation —

a; = c + a, cos + tto cos 20 + a^ cos 3a + a^ cos 40
+ hy sin e + hi sin 20 + 63 sin 30.

Determine the eight constants in this equation.

If the speed of the crank shaft is 10 radians per second, what is the velocity
of the valve when = 0? [b.e.]

18. One cycle of a periodic curve A is given in Fig. 284. Express 2/ as a

360*

Fig. 284.

function of 0, assuming that all the Fourier vectors of the fourth and higher
orders are negligible.

19. The curve B (Fig. 284) shows the displacement ij inches of the block in
the link of a Stephenson valve gear, for any crank position degrees, during one
revolution of the crank shaft. Express y approximately in terms of by the first
three terms of the Fourier series —

2/ = ri sin (0 + Oj) + r^ sin (20 + Uo) + r^ sin
4- a constant

Give values of r,, r,, rg, and Oi, a,, 03.

. ? IE M

Copy double this size + direction

Fig. 285.

' + ^3)

20. A point P oscillates in a straight line, the motion being repeated in-
definitely. The period of oscillation being divided into twelve equal iuteryjijs,

PERIODIC MOTION 165

beginning when t = 0, you are given, in Fig. 285, the twelve corresponding
positions of P, numbered to 11.

Suppose the displacement from mean position at any time t is given very
approximately by the first three terms of the Fourier series—

cc = a sin {wt + a) + 6 sin (2a>i + 3) + c sin (3«^ + 7).

Find the elements of the motion, that is, find the three half travels a, &, c and
the three angles of advance, a, )3, 7 for this case.

CHAPTER XI

PROJECTION

134. Descriptive Geometry. — Practical Solid Geometry, or Be-
8crij)tive Geometry is that branch of geometry which treats (1) of the
representation, on a plane surface, of points, lines, and figures in
space, in such a way that the relative positions of the points, lines,
and figures, and also the exact forms of the lines and figures are
determined, and (2) of the graphic solution of problems connected
with points, lines, and figures in space. The problems of descriptive
geometry are best solved by means of the method of projections.

135. Projection. — When an object is seen by the eye of a
spectator, rays of light come from all the visible parts of the object
and converge towards a point within the eye. Now suppose that a
flat sheet of glass is placed between the object and the eye of the
spectator, and that each ray of light, in passing through the glass
from the object to the eye, leaves a mark on the glass of the same
colour and tint as the part of the object from which the ray came.
In this way a picture would be produced on the surface of the glass,
and if the object be removed while the picture and the eye remain
stationary, the picture would convey to the mind of the spectator the
same knowledge of the object as was conveyed by the presence of the
object itself. Again, if instead of the rays of light from all the visible
points of the object leaving an impression on the glass, only those
which came from the edges of the object were to do so, an outline
would be produced on the surface of the glass which, although it would
not convey to the mind of the spectator the same impression as the
presence of the object itself might still give a good idea of its form.

The foregoing remarks are illustrated by Fig. 286, where AB
represents an object viewed by an eye at E ; CD is a plane interposed
between E and AB ; the thin dotted lines represent a few of the rays
of light passing from the edges of the object to the eye, and A'B' is
the outline obtained from the intersections of the rays of light with
the plane CD. The figure A'B' is called a projection of the object
AB on the plane CD.

The plane upon which a projection is drawn is called a plane of
projection.

The rays of light or imaginary lines passing from the different
points of the object to the corresponding points of the projection are
called projectors.

PROJECTION

167

When the projectors converge to a point the projection is called a
radial, conical, or jperspective projection.

When the point to which the projectors converge is at an infinite
distance from the object the projectors become parallel, and the
projection is called a parallel projection.

If besides being parallel the projectors are also perpendicular to
the plane of projection the projection becomes a perpendicular, an
orthogonal, or an orthographic projection.

For the purposes of descriptive geometry orthographic projections
are the most convenient and most commonly used, and when the term
projection is used without any qualification orthographic projection is
generally understood. In what follows projection will mean ortho-
graphic projection.

The projection of a point upon a plane is the foot of the perpendicular
let fall from the point on to the plane.

The projection of a line upon a plane is the line which contains the
projections of all the points of the original line.

The projecting surface of a line is the surface which contains the

Fig. 286.

projectors of all the points of that line. When the projecting surface
of a line is a plane it is called the projecting plane of the line. The
projecting surface of a straight line is always a plane, but a line is not
necessarily straight because its projecting surface is a plane. These
definitions of projecting surface and projecting plane of a line and the
statements which follow them apply to all kinds of projection.

One projection alone of a figure is not sufficient for determining its
exact form. For example if a triangle ahc drawn on a sheet of paper
be taken as the projection on the paper of a triangle ABC somewhere
above it, it is clear that the exact form of the triangle ABC will
depend on the relative distances of its angular points from the paper,
but the projection ahc gives no information about these distances. If,
however, another projection a'b'c of the triangle ABC be obtained on
a sheet of paper at right angles to the former one, then, as will be
shown later, the true form of the triangle ABC may be obtained from
these two projections.

The representation of an object by means of two projections, one
on each of two planes at right angles to one another, and how these

168

PRACTICAL GEOMETRY

projections are drawn on a flat sheet of paper will be understood by
reference to Figs. 287 and 288.

Eig. 287 is to be taken as a pictorial projection of a model. Two
planes of projection are shown one being vertical and the other
horizontal. These planes, called co-ordinate planes, divide the space
surrounding them into four dihedral angles or quadrants which are
named, first, second, third, and fourth dihedral angles or first, second,
third, and fourth quadrants. If an observer be facing the vertical
plane of projection, then the first quadrant is above the horizontal
plane of projection and in front of the vertical plane of projection.
The second quadrant is behind the first and the others follow in order

ELEVATION

PLAN

Fig. 287.

Fig. 288.

as shown. The line of intersection of the planes of projection is called
the ground line and is lettered XY.

An object is shown in the first quadrant and projections o£ it on the
horizontal and vertical planes of projection are also shown. The
projection on the horizontal plane is called a i^lan and the projection
on the vertical plane is called an elevation.

Now imagine the vertical plane to turn about XY as an axis,
carrying with it the elevation, until it is in a horizontal position. The
horizontal and vertical planes of projection will now coincide and the
plan and elevation of the object will be on one flat surface and their
exact forms may be drawn as shown in Fig. 288.

Instead of imagining the vertical plane to turn about XY until it
is horizontal, the horizontal plane may be imagined to turn about XY
until it is vertical as shown in Fig. 289.

PROJECTION

169

In British and European countries the general practice in making
working drawings is to conceive the object to be placed in the first
quadrant as shown in Fig. 287 and the working plan and elevation are
then in the positions shown in Fig. 288, the plan being below and the
elevation above XY. In the United States of America the practice of
conceiving the object to be placed in the third quadrant as shown in
Fig. 289 is now very general and the working plan and elevation are
then in the positions shown in Fig. 290, the plan being above and the
elevation below XY.

Whether the object be placed in the first quadrant or in the third
quadrant it is supposed to be viewed from above in obtaining the plan,
consequently when the object is in the first quadrant it lies between

PLAN !

—

ELEVATION

Fig. 289.

Fig. 290.

the observer and the plane of projection and the projectors from the
visible parts have to go through the object to the plan, but when the
object is in the third quadrant the plane of projection lies between
the observer and the object, hence the projectors from the visible parts
to the plan are not obstructed by the object. In like manner for the
elevation, the projectors from the visible parts of the object go through
the latter when it is in the first quadrant but are clear of it when in
the third quadrant; hence the advantage claimed for placing the
object in the third quadrant. The practice of placing the object in
the first quadrant is however so well established and the advantage
claimed for placing it in the third quadrant is so small, being probably
more imaginary than real, that it is doubtful whether a change from
the older practice should be encouraged. In any case drawings on

170 PRACTICAL GEOMETRY

either system can be made with equal facility if the principles are
understood.

In working problems in descriptive geometry by the method of
projections on co-ordinate planes, the given points and lines may be
in any one quadrant, but the lines for the solution may extend into
any or all of the other quadrants.

136. Notation in Projection. — For the purpose of reference
and for clearness, points, lines, and figures may be lettered. In
general a point in space is denoted by a capital letter, its plan by a
small italic letter, and its elevation by a small italic letter with a dash
over it. Thus P denotes a point in space, p its plan and p its
elevation. A line AB in space would have its plan lettered ah and its
elevation ah'. A point P in space may be referred to as the point P
or as the point jip'. In like manner a line AB in space may be referred
to as the line AB or as the line ah, a'h'.

The horizontal and vertical planes of projection may be referred
to by using the abbreviations TI.P. and*V.P. respectively.

CHAPTER XII

PROJECTIONS OF POINTS AND LINES

137. Rules relating to the Projections of a Point.— Fig.

291 is a pictorial projection of a model showing the horizontal and
vertical planes of projection in their natural positions together with
four points A, B, C, and D in space, A being in the first quadrant or
first dihedral angle, B in the second, C in the third, and D in the
fourth. The plans and elevations of these points as obtained by
dropping perpendiculars from them on to the horizontal and vertical
planes of projection respectively are also shown. The positions of the

Fig. 291.

Fig. 292.

BP'elevations of the points when the vertical plane of projection is
turned about XY until it coincides with the horizontal plane are indi-
cated, and in Fig. 292 the various plans and elevations are shown as
they appear when the planes of projection are made to coincide and
then placed flat on this paper.

A careful study of Figs. 291 and 292 should convince the student
of the truth of the following rules relating to the projections of a
point.

172

PRACTICAL GEOMETRY

(1) The plan of a point is beloic or above XY according as the point
is in front or hehlnd the vertical plane of projection.

(2) The elevation of a point is above or below XY according as the
point is above or below the horizontal plane of projection.

(3) The distance of the plan of a point from XY is equal to the
distance of the point from the vertical plane of projection.

(4) The distance of the elevation of a point from XY is equal to
the distance of the point from the horizontal plane of projection.

(5) The plan and elevation of a point are in a straight line
perpendicular to XY.

The term projector has already been defined and is the line joining
a point and a projection of it, but the line joining the plan and
elevation of a point is also called a projector. A plan and elevation
are also said to be projected, the one from the other.

138. True Length, Inclinations, and Traces of a Line. —

Fig. 293.

Pig. 294.

The projection of a line on a plane will be shorter than the line itself
except when the line is parallel to the plane ; in the latter case the
line and its projection have the same length.

A line, its projection on one of the co-ordinate planes, and the
projectors from its ends to that plane, form a quadrilateral concerning
which everything required for constructing it is known if the plan and
elevation of the line are given. One method of finding the true
length of a line is therefore to construct this quadrilateral.

Let the plan ab and elevation a'b' of a line AB be given as in
Fig. 294. Referring to the pictorial projection in Fig. 293, it will be
seen that the line AB, its plan ab, and the projectors Aa and B?> form
a quadrilateral, the base ab of which is given. Also Aa is equal to a'a^ ,
B6 is equal to b'b^, and the angles Aab and Bba are right an'gles.
Hence to find the true length of AB, draw (Fig. 294) oAj at right
angles to ab and equal to a^a'. Next draw bV*^ at right angles to ab
and equal to bjj'. A/Bi is the true length of AB.

PROJECTIONS OF POINTS AND LINES

173

Note that if the extremities of the line AB are on opposite sides
of the horizontal plane, the perpendiculars aAi and hB^ must be
drawn on opposite sides of the plan ah.

The inclination of a line to a plane is the angle between the line
and its projection on that plane.

Referring to Figs. 293 and 294, it is evident that the inclination
of A]^ to the horizontal plane is the angle between ah and AjBi, so
that the construction just given for finding the true length of AB also
serves for finding its inclination to the horizontal plane.

The inclination of AB to the vertical plane of projection is found
by constructing the quadrilateral a'h'B2-^2i ^^ which a'A^ and h'\% are at
right angles to a'h' and equal to a^a, and hj) respectively. A2B.2 is the
true length of AB and the angle between a'h' and A2B2 is the inclina-
tion of AB to the vertical plane of projection.

If al)., be drawn parallel to AjB^ (Fig. 294) to meet hB^ at B.(,
then aB;5 will also be the true length of AB, and the angle haB.^ will
be the inclination of A15 to the horizontal plane. Also the length of
?)B.j is equal to the difference between the distances of B and A from
the horizontal plane. Hence the true length of AB and its incli-
nation to the horizontal plane may be found by constructing the
triangle ahB.^. In like manner the true length and the inclination of
Al> to the vertical plane may be found by constructing the triangle
a'h'A.f in which a' A3 is equal to the difference between the distances
of A and B from the vertical plane.

The inclination of a line to the horizontal plane is usually denoted
by the Greek letter 6 (theta) and its inclination to the vertical plane
by the Greek letter <f> (phi). Notice that 6 is the letter O with a
horizontal line through it, while <^ is the same letter with a vertical
line through it.

The true length of a line AB and its inclinations to the planes of
projection may also be found as follows. Referring to Fig. 295,
through h draw afih^^ parallel to XY. With centre h and radius ha
describe the arc aa^, cutting a^hh^ at a^.
Draw «!«/ perpendicular to XY to meet a
line through a' parallel to XY at a/, a///
will be the true length of A15 and the angle
h'a^'a' will be its inclination 6 to the hori-
zontal plane. An inspection of the figure
will be sufficient to make clear the corre-
sponding construction for finding the incli-
nation <f> of the line to the vertical plane.

Comparing the constructions shown in
Figs. 294 and 295, it will be seen that in
both a quadrilateral is drawn having a base
equal to one of the projections of the line,
and in Fig. 294 this base is made to coincide
with that projection while in Fig. 295 the
base is made to coincide with XY.

When the inclination of a line is mentioned without reference to

a^^

a' • :
1 : Y

iJ [■

>^'

Fig. 295.

174

PRACTICAL GEOMETRY

any particular plane, inclination to the horizontal plane is generally
understood.

The trace of a line on a surface is a point where the line intersects
the surface. When the traces of a straight line are mentioned without
reference to any particular surface, the points in which the line or the
line produced cuts the planes of projection are understood.

The horizontal trace of a line is the point where the line or the
line produced cuts the horizontal plane of projection, and the vertical
trace of a line is the point where the line or the line produced cuts the
vertical plane of projection.

After making the construction for finding the true length of a
line, shown in Fig. 294, if < B^Ai be produced to meet ba at h then h
will be the horizontal trace of the line AB. In like manner if AJi,
be produced to meet a'b' at v' then v' will be the vertical trace of AIj.
The correctness of these constructions is obvious from an inspection
of Fig. 293.

If it is only the traces of a line which are required, then it is
only necessary to produce the elevation a'b' (Fig. 294) to meet XY at
h' iind then draw h'h perpendicular to XY to meet the plan ab pro-
duced at h in order to determine the horizontal trace. In like
manner, by producing the plan ab to meet XY at v, and drawing vv'
perpendicular to XY to meet the elevation a'b' produced at v', the
vertical trace is determined. This construction fails however when
the projections of the line are perpendicular to XY but the con-
struction previously given will apply in this case also, provided that
the plans and elevations of two points in the line are definitely
marked and lettered.

When the projections of a straight line are perpendicular to XY
the line itself is perpendicular to XY although it may not meet XY.

When a line is parallel to one of the planes of projection it has
no trace on that plane.

When a line is perpendicular to one of
the planes of projection it has no trace on
the other plane of projection, and its projec-
tion on the plane to which it is perpendicular
is a point.

139. True Form of a Plane Figure.
— The projection of a plane figure on a plane
will not have the same form or dimensions
as the figure itself excepting when the plane
of the figure is parallel to the plane of pro-
jection, in which case the figure and its pro-
jection will be exactly alike. To determine
the true form of any plane figure, whose pro-
jections are given, it is necessary to know
the true distances of a sufficient number of
points in it from one another ; now these
distances may be found by one of the constructions given in the
preceding article.

Fm. 29G.

PROJECTIONS OF POINTS AND LINES

175

An example is shown in Fig. 296, where ale and a'h'o' are the given
projections of a triangle. The true lengths of the sides are found by
the constructions shown, and the triangle which is the true form of
the triangle whose projections are given is then drawn.

If the figure given by its projections has more than three sides it
should first be divided into triangles ; the true forms of these triangles
are then found and assembled to give the true form of the whole figure.

In many cases the true form of a plane figure whose projections
are given is best determined by the method of rabatment described in
Art. 189, p. 221.

The traces of the sides of a plane figure whose projections are given
may be determined by one of the constructions of the preceding article,
and it will be found that all the horizontal traces will lie in one
straight line, and all the vertical traces will lie in another straight
line, and these two straight lines will intersect on XY, excepting
when they are parallel, in which case they will be parallel to XY.
This matter will be referred to again in Chapter XVI.

140. To mark off a given Length on a given Line.— Let
aJ), a'b' (Fig. 297) be the projections of the line, it is required to find
the projections of a point C
in this line so that AC shall
be a given length.

Determine A^Bi the true
length of AB. Make A,Cj
equal to the given length.
Through Cj draw C^c perpen-
dicular to ah to meet ah
at c. Through c draw cc'
perpendicular to XY to meet
a!h' at d. c and c' are the
projections required.

If the given projections
of the line are perpendicular
to XY (Fig. 298), determine c as before then make the distance of
c' from XY equal to C^c.

141. Given the True Length of a Line and the Distances
of its Extremities from the Planes of

Projection, to draw its Projections, —
First determine the projections aa' (Fig. 299)
of the end A of the line. On a^a make a^c
equal to the distance of the end B of the line
from the vertical plane of projection, and on
a^a' make a^c' equal to the distance of B from
the horizontal plane. Through c and c' draw
lines parallel to XY. These parallels to XY
will contain the plan and elevation respec-
tively of B. With a' as centre and the given
true lenojth of the line as radius describe an

Fig. 297.

arc to cut the parallel to XY through c' at

Fig. 299.

176

PRACTICAL GEOMETRY

Fig. 300.

B2. From B2 draw the perpendicular BJ)^ to XY. Then since a'B., is
the true length of the line and a^a' and h.^Bo are the distances of its
extremities from the horizontal plane it follows (see Fig. 293, p. 172)
that a^h^ is the length of the plan of the line. Hence, if with centre
a and radius equal to aji^ an arc be described to cut the parallel to
XY through c at b then ab is the required plan of the line. A per-
pendicular to XY from h to meet the parallel to XY through c'
determines b', and a'b' is the required elevation of the line.

142. Given the Projection of a Line on one of the Planes
of Projection, its Inclination to that Plane, and the Distance
of one end from it, to determine its other
Projection. — Let ab (Fig. 300) be the given
projection, the inclination to the horizontal
plane and let the distance of the end A of the line
from the horizontal plane be given.

The distance of a' from XY is equal to the
given distance of A from the horizontal plane.
At a make the angle haB^^ equal to and draw
?>Bi at right angles to ab to meet aB^ at B^. The
distance of b' from XY is equal to bBi plus the
distance of a' from XY.

If the elevation a'b' is given and the incli-
nation <^ of the line to the vertical plane of projection and also the
distance of A from the vertical plane of projection, the construction
is similar to that just given and is shown in Fig. 300.

143. Given the Inclination of a Line to one of the Planes
of Projection and the Angle which its Projection on that
Plane makes with the Ground Line, to draw
its Projections, — Let the line be inclined at an
angle to the horizontal plane, and let its plan
make an angle a with XY. From a point C in
XY (Fig. 301) draw Cb' inclined at an angle ^ to
XY. Draw b'b at right angles to XY. Then bC
is the length of the plan of the line whose true
length is Cb' and whose inclination to the hori-
zontal plane is 6. With b as centre and bC as
radius describe an arc, and from b draw ba to
meet this arc at a and make an angle a with XY.
ab is the required plan of the line. A perpendicular
from a to XY determines a' and a'b' is the required
elevation of the line. «

144. Given the Inclinations of a Line to the Planes
Projection, to determine its Projections. — Let the line be in-
clined at an angle to the horizontal plane and at an angle <^ to the
vertical plane of projection. From a point C in XY (Fig. 302) draw
Cb' inclined at an angle to XY. Draw b'b at right angles to XY.
Then bC is the length of the plan of the line whose true length is Cb'
and whose inclination to the horizontal plane is 0.

From b' draw b'T> making the angle C^'D equal to <^. Draw CD

PROJECTIONS OF POINTS AND LINES 177

perpendicular to //D. Then fe'D is the length of the elevation of the
line whose true length is C// and whose inclination to the vertical
plane of projection is cf). ^^-^-^A'

With centre h' and radius h'T> describe an arc D,,,— — -""T'J/
to cut XY at a'. With centre b and radius ^C '^ y/

describe the arc Qa to meet the perpendicular to \\ / /

XY from a' at a. ah is the plan and a'b' is the \\ / /
elevation required. \/OV/a!

Note. The sum of the angles $ and <j6 may ^'"cf [ /J ^

vary between 0° and 90°. When ^ + (^ = 0° the \ \ /

projections of the line are parallel to XY, and \ | /

when ^ + <^ = 90° the projections of the line are \,ix

perpendicular to XY. p^ ^^2

145. Projections of Parallel Lines, — The

projections of parallel lines on to the same plane are parallel. Hence
if it is required to draw the projections of a line which shall pass
through a point whose projections j?p' are given and which shall be
parallel to another line whose projections ah^ a'b' are given, draw
through the plan p a line pq parallel to ab and through p' draw p'(^
parallel to alb'. Then p</ and p'q are the projections required.

The projections on the same plane of equal parallel lines are equal.

146. Conditions that Two Lines whose Projections are
given may intersect. — If two lines intersect, their point of inter-
section is a point in each of the lines, therefore the plan of that point
must be on the plan of each of the lines, and therefore the plans of
the lines must intersect, and the point of intersection of the plans is
the plan of the point of intersection of the lines. In like manner the
elevations of the lines must intersect at a point which is the elevation
of the point of intersection of the lines. But the plan and elevation
of a point are in the same straight line at right angles to the ground
line.

Hence the conditions that two lines intersect is that their plans
and elevations respectively intersect and that the points of intersec-
tion are in the same straight line at right angles to the ground line.

There are exceptions to this rule. When the lines are perpen-
dicular to the ground line and lie in the same vertical plane or when
one of the lines only is perpendicular to the ground line the lines may
or may not intersect. In such cases an auxiliary projection will show
whether the lines intersect or not.

147. Angle between Two Intersecting Lines. — Let AC and
BC be two intersecting lines whose projections are given ; it is required
to find the true angle between these lines. Take any point D in AC
and any point E in BC. Determine by Art. 139, p. 174, the true form
of the triangle DCE. The angle C of this triangle is the angle
required.

The construction is simplified in many cases by taking for the
points D and E the horizontal or vertical traces of the lines. In
Fig. 303, D and E are the horizontal traces of the lines AC and BC
respectively. dC^e is the true angle between AC and BC.

178

PRACTICAL GEOMETRY

Figv 303 also shows how to determine the projections of a line
which bisects the angle ACB. Draw C^r
bisecting the angle dC.e and intersecting de
at r. Find / the elevation of R. Join
cr and c'r'. These are the projections of
the line which bisects the angle between
AC and BC. This also suggests the con-
struction for finding the projections of a
line which shall intersect the lines AC and
BC and make a given angle with one of
them.

148. Angle between Two Non-
intersecting Lines. — Let AB and CD be
two non-intersecting lines whose projections
are given ; it is required to find the true
angle between these lines. Through any

point P in one of the lines, say AB, draw a line PQ parallel to the
other line. The angle between these two intersecting lines PQ and
AB is the angle required.

Exercises XII

1. Show the projections of the following points, using the same ground line
for all the projections, and making the projectors 0*5 inch apart.

A, 1-2 inches in front of the V.P. and 1-8 inches above the H.P.

B, 1-9 inches in front of the V.P. and 1'7 inches below the H.P.

C, 1-4 inches in front of the V.P. and in the H.P.

D, 2 inches behind the V.P. and 1-6 inches above the H.P.

E, 2 inches behind the V.P. and 1-6 inches below the H.P.

2. Show the projections of the following points as in the preceding exercise.

A, 1'3 inches behind the V.P. and in the H.P.

B, in the V.P. and in the H.P.

C, 1-4 inches in front of the V.P. and 1-6 inches below the H.P.

D, in the V.P. and 1-2 inches below the H.P.

E, 1-5 inches behind the V.P. and 1-5 inches above the H.P.

3. State the exact positions of the points whose projections are given in Fig.
304 with reference to the planes of projection.

4. a'b', the elevation of a straight line, is
2 inches long and it is inclined at 30° to XY.
The end A is in the horizontal plane and 2 inches
from the vertical plane. The end B is in the
vertical plane and the whole line is in the first
dihedral angle. Draw the plan and elevation
of the line.

5. A triangle ABC is in the first dihedral
angle. A is in the V.P. and 2 inches above the
H.P. B and C are in the H.P. ab makes 45°
with XY and abc is an equilateral triangle of
2 inches side. Draw the plan and elevation of
the triangle ABC.

6. The same as the preceding exercise except
that while the side AB remains in the first

dihedral angle the side AC is placed in the second dihedral angle.

7. ab, be, and cd form three sides of a square of 1'5 inches side, a is on
XY, ab is inclined at 30° to XY, and abed is below XY. abed is the plan of a
piece of thin wire ABCD of which the parts AB, BC, and CD are straight. The

^IT^

a^p-

lis

Fig. 304.

4±

PROJECTIONS OF POINTS AND LINES

heights of the points A, B, C, and D above the H.P. are 0-5, 0*9, 1-2, and 1-1
inches respectively. Draw the plan and elevation of the wire.

8. a'h', 2-5 inches long, is the elevation of a straight line which is parallel to
the V.P. The end A is in the H.P. and 1 inch in front of the V.P. The end B is
1-5 inches above the H.P. Dra^V the plan and elevation of AB.

9. Draw the projections of the following lines, and then find their traces
where possible.

AB, 2 inches long, parallel to XY, 1 inch above the H.P. and 1*3 inches in
front of the V.P.

CD, 2-2 inches long, parallel to the H.P., inclined at 30° to the V.P., the end
C to be in the V.P. and 1-2 inches above the H.P.

EF, 1-5 inches long, perpendicular to the H.P., the end E to be 0*5 inch above
the H.P. and 1 inch in front of the V.P.

GH, 2 inches long, perpendicular to the V.P., 1 inch above the H.P., the end
G to be 0-5 inch in front of the V.P.

10. The plan of a line is 2 inches long and it makes 35° with XY, the eleva-
tion makes 45° with XY, and the line intersects XY. Draw the plan and
elevation of the line and then find its true length and its inclinations to the
planes of projection.

11. Find the true length, the inclinations to the planes of projection, and the
horizontal and vertical traces of each of the lines whose projections are given in

-j

—

jt_

tsj

VYVnTf

e^ y^

^ I I 1 1

r-]

r-^

—

Fig. 305. Fig. 306. Fig. 307. Fig. 308.

In reproducing the above diagrams the sides of the small sguares are to he taken

equal to half an inch.

Figs. 305-308. Show also in each case the projections of a point in the line
whose true distance from the lower end of the line is 1 inch.

12. Determine the true form of each of the plane figures whose projections
are given in Figs. 309-312. Find also for each figure the horizontal and vertical

»N^

■-:

:::

(

71'

•rf

Fig. 309. Fig. 310. Fig. 311. Fig. 312.

In repi-oducing the above diagrams the sides of the small squares are to be taken

equal to half an inch.

traces of each side, where possible, and show that the horizontal traces are in one
straight line and the vertical traces in another.

180

PRACTICAL GEOMETRY

13. A straight line 2-5 incites long has one end 0-3 inch in front of the vertical
plane and 1-5 inches above the horizontal plane while the other end is 1'75 inches
in front of the vertical plane and 0*6 inch above the horizontal plane. Draw the
plan and elevation of the line.

14. A straight line inclined at 50^ to the horizontal plane has one end 2 inches
above the horizontal plane and 0*5 inch in front of the vertical plane of projection.
The other end of the line is on the horizontal plane and 2 inches in front of the
vertical plane of projection. Draw the projections of the line.

15. A straight line is 2*75 inches long. One end is in the horizontal plane
and the other end is in the vertical plane of projection. The line is inclined at
30° to the horizontal plane and its plan makes an angle of 45° with XY. Draw
the projections of the line.

16. a'h' (Fig. 313) is the elevation of a straight line. B is in the vertical
plane of projection. The line is inclined at 35° to the horizontal plane. Draw
the plan of the line.

17. cd (Fig. 314) is the plan of a straight line whose true length is 3 inches.
The end G of the line is 0-5 inch below the horizontal plane and the end D is
above the horizontal plane. Draw the elevation of the line.

18. dbc (Fig. 315) is the plan of a triangle. The point A is on the horizontal
plane. The point B is above the horizontal plane and is higher than the point C.
The true length of AB is 2-5 inches and the inclination of BC to the horizontal
plane is 40°. Draw the elevation of the triangle and find the true length of AC.

rrl'V-

h '-
ay

" 31-3- — m

Fig. 313. Fig. 314. Fig. 315. Fig. 316. Fig. 317.

In reproducing the above diagrams the sides of the small squares are to be taken

equal to half an inch.

19. The middle point of a straight line 3 inches long is 1 inch above the hori-
zontal plane and 1*25 inches in front of the vertical plane of projection. The line
is inclined at 30° to the horizontal plane and 40° to the vertical plane of pro-
jection. Draw the projections of the line.

20. Find the real angle between the lines whose projections are given in Fig.
316. Show also the projections of the line bisecting the angle ABC.

21. Determine the angle MON of the triangle whose projections are given in
Fig. 317, and draw the projections of the line which passes through the point M
and intersects the line ON at right angles.

22. ABC is an equilateral triangle of 2-5 inches side. A is in the horizontal
plane and B is in the vertical plane of projection. AB is inclined at 45° to the
horizontal plane and 30° to the vertical plane of projection. BC is inclined
at 35° to the horizontal plane. Draw the plan and elevation of the triangle.

CHAPTER XIII

PROJECTIONS OF SIMPLE SOLIDS IN SIMPLE POSITIONS

149. Projections of Solids. — It has already been indicated
(Chap. XI.) that one object of Descriptive Geometry is to convey to
the mind an impression of the exact form and size of objects, which
have length, breadth, and thickness, by means of representations of
them on a surface, which has length and breadth only.

Now, a solid may be conceived to be made up of an immense
number of small particles or material points, the relative positions of
which may be represented by means of their projections on two planes,
as explained in the two preceding chapters ; but as in looking at a
solid it is generally only the points on its external surface which are
seen, all impressions of the form and size of objects are derived from
the form and extent of their surfaces. It is therefore unnecessary in
representing an object on paper to give the projections of points in
its interior, that is, it is only necessary to represent the surface of the
object.

Again, the form and extent of a surface are generally known when
the forms, lengths, and relative positions of a suflScient number of lines
on that surface are known. But it has been seen that by the method
of projections the forms, lengths, and relative positions of lines may be
represented on a single flat surface. Hence, a' solid may be represented
on paper by the projections of a sufficient number of lines on its surface.

When the solid has plane faces, the projections of the boundary
lines of the faces are all that is necessary in order to represent
them.

150. Definitions of Solids. — There are certain solids, of definite
and simple form, which are of frequent occurrence and which will now
be defined.

A polyhedron is a solid bounded entirely by planes. The edges
of a polyhedron are the lines of intersection of its bounding planes.
The sides or faces of a polyhedron are the plane figures formed by its
edges.

A polyhedron is said to be regular when its faces are equal and
regular polygons. Generally when a polyhedron is referred to a regular
polyhedron is understood.

There are only five regular polyhedra, namely, the tetrahedron,
the cube, the octahedron, the dodecahedron, and the icosahedron.

182

PRACTICAL GEOMETRY

TETBAHEDRON.

CUBE.

The tetrahedron (Fig. 318) has four faces, all equilateral triangles.

The cube (Fig. 319)
has six faces, all squares.

The octahedron (Fig.
320) has eight faces, all
equilateral triangles.

The dodecahedron
(Fig. 321) has twelve
faces, all pentagons.

The icosahedron (Fig.
322) has twenty faces,
all equilateral triangles.

A prism is a poly-
hedron having two of its faces, called its ends or bases, parallel, and the
remaining faces are parallelograms. Those faces of a prism which are
parallelograms are generally called the sides of the prism.

. 1.

Pig. 318.

Fig. 319.

OCTAHEDRON.

DODECAHEDRON.

ICOSAHEDRON.

Fig. 320.

Fig. 321.

Fig. 322.

A parallelepiped is a prism whose bases are parallelograms.

A pyrandd is a polyhedron having a polygon for its base, and for
its sides it has triangles which have a common vertex and the sides of
the polygon for their bases. The common vertex of the triangles is
called the vertex or apex of the pyramid.

The axis of a prism is the straight line joining the centres of its
ends ; and the axis of a pyramid is the straight line joining its vertex
to the centre of its base.

A right prism has its axis at right angles to its ends ; and an oblique
prism has its axis inclined to its ends.

A right pyrarnid has its axis at right angles to its base ; and an
oblique pyramid has its axis inclined to its base.

The altitude of a prism is the perpendicular distance between its
ends ; and the altitude of a pyramid is the perpendicular distance of
its vertex from its base.

Prisms and pyramids are named from the form of their bases — as,
triangular, square, pentagonal, hexagonal, etc. Examples of prisms
and pyramids are shown in Figs. 323-326.

A cylinder resembles a prism. If the sides of the bases of a prism
be continually diminished in length and increased in number, the
ultimate form of the boundary lines of the bases will be curved lines,

SIMPLE SOLIDS IN SIMPLE POSITIONS

183

and the ultimate form of the prism will be a cylinder. A right circular
cylinder has its axis at right angles to its ends, which are equal circles.

BIGHT
SQUARE
PBISM.

OBLIQUE

PENTAGONAL

PBISM.

RIGHT OBLIQUE

HEXAGONAL HEPTAGONAL

PYRAMID PYRAMID.

Fig. 323.

Fig. 324.

Fig. 325.

Fig. 326.

BIGHT
CIRCULAR
CYLINDER.

RIGHT

CIRCULAR

CONE.

SPHERE.

A right circular cylinder may also be defined as a solid described by
the revolution of a rectangle about one of its sides, which remains
stationary. The fixed line about which the rectangle revolves is the
axis, and the circles described by the opposite revolving sides are the
bases or ends of the cylinder.

The diameter of a right circular cylinder is the diameter of its
circular ends.

A cone resembles a
pyramid. If the sides
of the base of a pyra-
mid be continually
diminished in length
and increased in
number, the ultimate
form of the boundary
line of the base will be
a curved line, and the
ultimate form of the
pyramid will be a cone.

A right circular cone has its axis at right angles to its base, which is a
circle. A right circular cone may also be defined as a solid described
by the revolution of a right angled triangle about one of the sides
containing the right angle, which side remains stationary. The fixed
line about which the triangle revolves is the axis, and the circle de-
scribed by the other side containing the right angle is the base of
the cone.

A sphere is a solid every point on the surface of which is at the
same distance from a point within it called its centre. A sphere may
also be defined as a solid described by the revolution of a semicircle
about its diameter, which remains stationary. The middle point of
the diameter of the semicircle is the centre of the sphere.

151. Projections of a Prism. — The projections of a prism are
simplest when the solid is so placed that its ends are parallel to one of
the planes of projection, and whatever projections of the prism may be

Fig. 327.

Fig. 328.

Fig. 329.

184

PRACTICAL GEOMETRY

required it is generally necessary to fii*st draw the projections of the
solid when so situated.

When the ends of the prism are parallel to one of the planes of
projection, their projections on that plane will show their true form,
and as the form of the ends is supposed to be given these projections

Fig. 330.

Fig. 331.

are drawn first. If the prism is a right prism the projections of the
two ends on the plane of projection to which they are parallel will
evidently coincide. The projections of the ends on the other plane of
projection will be straight lines parallel
to the ground line and at a distance apart
equal to the altitude of the prism.

The foregoing observations are illus-
trated by Figs. 330, 331, and 332. Fig.
330 shows the plan and elevation of a
right triangular prism when its ends are
horizontal. The plan ahc is first drawn,
its position in relation to XY being arbi-
trarily chosen unless it is specified. The
elevation is projected from the plan, as
shown. The height of the elevation of the
lower end above* XY is equal to the height
of that end above the horizontal plane,
which may be arbitrarily chosen, unless it
is specified. The distance between the
horizontal lines which are the elevations
of the ends is equal to the altitude of the
prism, which would be given.

Fig. 331 is a pictorial projection of the prism and the planes of
projection in their true positions.

SIMPLE SOLIDS IN SIMPLE POSITIONS

185

Fig. 332 shows the plan and elevation of an oblique hexagonal
prism when its ends are parallel to the vertical plane of projection,
one end being in that plane. In this case the elevation is first drawn.
The hexagons which are the elevations of the ends of the prism have
the sides of the one respectively parallel to the sides of the other, their
relative positions being otherwise determined from the specification of
the prism. It will be observed that the hexagons have been placed
with a diameter parallel to XY.

It will be noticed that in Fig. 330 the elevation of one of the edges
of the prism is shown as a dotted line and that in Fig. 332 .the
elevations of five of the edges are so shown. The reason for this is
that these edges are hidden by the solid when it is viewed from the
front. In like manner if an edge of the solid is hidden when
viewed from above the plan of that edge would be shown by a dotted line.

When a projection of an edge which should be a full line coincides
with a projection of an edge which should be a dotted line only the full
line can be shown.

152. Projections of a Pyramid. — The projections of a pyramid
are simplest when the solid is so placed that its base is parallel to one
of the planes of projection, and whatever projections of the pyramid
may be required it is generally necessary to first draw the projections
of the solid when so situated.

When the base of the pyramid is parallel to one of the planes of
projection, its projection on that plane will show its true form, which

Fig. 333.

Fig. 334.

is supposed to be given. This projection is therefore drawn first. If
the pyramid is a right pyramid the projection of its vertex on the
plane of projection to which the base is parallel will be at the centre
of the projection of the base on that plane. The projection of the
base on the other plane of projection will be a straight line parallel

186

PRACTICAL GEOMETRY

to the ground line, and the projection of the vertex will be at
a distance from the projection of the base equal to the altitude of the
pyramid.

These observations are illustrated by Eigs. 333 to 336. Fig. 333
shows the plan and elevation of a right square pyramid when its base
is horizontal. The plan is first drawn. This plan is a square abed
with its diagonals intepsecting at y, the plan of the vertex of the
pyramid. The position of the square abed in relation to XY is
arbitrarily chosen unless it is specified. The elevation is projected
from the plan, as shown. The height of the elevation of the base
above XY is equal to the height of the base above the horizontal
plane, which may be arbitrarily chosen, unless it is specified. The
distance of the elevation of the vertex from the elevation of the base
is equal to the altitude of the pyramid, which would be given.
Fig. 334 is a pictorial projection of the pyramid and the planes of
projection in their true positions.

Fig. 335 shows the plan and elevation of an oblique pentagonal
pyramid when its base is parallel to the vertical plane of projection.

ELEVATION^^^^^^^^ ^

-A/

¥^

b
a

V ^\.

plan\\

Fig. 335.

Fig. 336.

The vertex of the pyramid is in a* straight line at right angles to
the base and passing through one corner of the base. The
elevation is first drawn. The elevation of the base is a regular
pentagon ah'c'd'e which has been placed with one side a!b' at
right angles to XY. The vertex has been placed in that per-
pendicular to the base which passes through the corner whose
elevation is a! . The elevation v' of the vertex therefore coincides
with a'. Joining v' to c' and dl completes the elevation. It will be
noticed that the elevations of the edges VB and VE coincide with the
elevations of the edges AB and AE of the base. The plan is projected
from the elevation, as shown. Fig. 336 is a pictorial projection of
the pyramid and the planes of projection in their true positions.

SIMPLE SOLIDS IN SIMPLE POSITIONS

187

Fig. 337.

153. The Altitude of a Regular Tetrahedron.— Since the
regular tetrahedron has four faces all equilateral triangles, this solid
is evidently a right pyramid on a triangular base.
Its projections may therefore be drawn as described
in the preceding article if its base and altitude are
known. The altitude will evidently depend on the
size of the base and is found as follows. Eig. 337
shows the plan of a tetrahedron when one face
ABC, which may be called its base, is on the hori-
zontal plane, av is the plan of one of the sloping
edges, and since all the faces are equilateral triangles, the true
length of the edge of which av is the plan will be equal to a&.
Hence, with a as centre and a radius ah draw an arc to cut vv\ which
is perpendicular to av, at v'. vv' is the altitude required.

154. Projections of the Octahedron. — The lines joining the
opposite angular points of the octahedron are its axes. There are
three of these lines all equal in length, and bisecting one another at
right angles at the centre of the solid.

It will be found on examination that the octahedron may be
divided, in three different ways, into two right square pyramids having
a common base, the triangular faces being equilateral triangles. Two of
the axes of the octahedron are the diagonals of the square which forms
the base of the two forementioned pyramids, while the third axis is the
line joining their vertices.

If the octahedron be placed with an axis perpendicular to one of
the planes of projection its projection on that
plane will be a square with its two diagonals.
Eig. 338 shows the plan and elevation of an
octahedron when one axis is vertical. The plan
is a square ahcd with its two diagonals ac and
hd and this projection is first drawn. The
square ahcd is the plan of the common base of
two pyramids into which the octahedron may be
divided, the point v where the diagonals ac and
hd intersect being the plan of the vertices of the
pyramids, vv' the elevation of the vertical axis
of the octahedron is at right angles to XY and
has a length equal to ac or hd. a'h'c'd' the
elevation of the common base of the two
pyramids is a straight line parallel to XY and
bisecting v'v'.

155. Projections of the Cylinder, Fig. 338.
Cone, and Sphere. — When the axis of a

right circular cylinder is perpendicular to one of the planes of pro-
jection, the projection of the cylinder on that plane will be a circle
having a diameter equal to that of the cylinder. The projection of
the cylinder on the other plane of projection will be a rectangle, one
side being parallel to the ground line and equal to the diameter of
the circle, while an adjacent side is e(j[ual in length to the axis of the

188

PRACTICAL GEOMETRY

cylinder. Fig. 339 shows the plan and elevation of a right circular
cylinder when its axis is perpendicular to the vertical plane of
projection.

When the axis of the cylinder is parallel to both planes of
projections, each projection will be a rectangle equal to the one

Fig. 339.

Fig. 340.

Fig. 341.

Fig. 342.

described above, but that side which is equal to the diameter of
the cylinder will be perpendicular to the ground line. Fig. 340
shows the plan and elevation of a right circular cylinder when its
axis is parallel to both planes of projection and therefore ako parallel
to the ground line.

When the axis of a right circular cone is perpendicular to one of
the planes of projection, the projection of the cone on that plane will
be a circle having a diameter equal to that of the base of the cone.
The projection of the cone on the other plane of projection will be an
isosceles triangle, its base being parallel to the ground line and equal
to the diameter of the base of the cone, and having an altitude equal to
that of the cone. Fig. 341 shows the plan and elevation of a right
circular cone when its axis is vertical.

When the axis of the cone is parallel to both planes of projection,
each projection will be a triangle equal to the one just described, but
the base will be perpendicular to the ground line. Fig. 342 shows
the plan and elevation of a right circular cone when its axis is
parallel to both planes of projection, and therefore also parallel to the
ground line.

All projections of a sphere are circles having a diameter equal to
that of the sphere. The plan and elevation of a sphere have their
centres in a straight line at right angles to the ground line as shown
in Fig. 343.

Exercises XIII

1. The three rectangles (Fig. 344) are the plans of three right prisms the ends
of which are vertical and 1 inch square. Each prism is 3 inches long, ab, cd,
and ef, the plans of the axes of the prisms, bisect one another. The first, or
lowest, prism rests on the ground, the second rests on top of the first, and the
third rests on top of the second. Draw these plans and from them project the
elevations of the prisms.

SIMPLE SOLIDS IN SIMPLE POSITIONS

189

2. A pictorial projection of an angle bracket is given in Fig. 345. Draw the
plan and elevation of this bracket vv^hen the edge AG is in the vertical plane of
projection and the edge AB is in the horizontal plane of projection and inclined
at 30° to XY.

Fig. 346.

3. Fig. 346 shows, in pictorial projection, two flights of steps and an inter-
mediate landing. Draw the plan and elevation of the steps and landing when
the edge AC is in the vertical plane of projection, and the edge AB is on the
ground and inclined at 30° to XY. Scale, half inch to one foot.

4. A cube of 1 inch edge has two faces horizontal, the lower one being 1 inch
above the ground. The square which is the plan of the cube has one side in-
clined at 30° to XY, the corner which is nearest to XY being 1 inch distant from
it. On each face of this cube there is fixed a cube. The corners of the bases of
the added cubes are at the middle points of the edges of the original cube. Draw
the plan and elevation of this built up solid.

5. A plan of an oblique triangular prism is given in Fig. 347. The base ABC
is on the horizontal plane. The altitude of the prism is 1-5 inches. Draw this
plan and from it project an elevation on XY. Draw also the plan and elevation
of this prism when the base is on the ground and the edge AB is parallel to the
ground line.

6 . Fig. 348 shows the plan of an oblique prism when one end is on the hori-
zontal plane. The altitude of the prism is 1-5 inches. Draw this plan and from
it project an elevation on XY.

X-

■ !!■

""7

Fig. 347. Fig. 348. Fig. 349. Fig. 350. Fig. 351.

In rep'oducing the above diagrams the sides of the small squares are to be taken

equal to half an inch.

7. A plan of a right square prism is given in Fig. 349, the edge AB of one end
being on the ground. Draw this plan and from it project an elevation on XY,

8. An elevation of a right square pyramid is given in Fig. 350, v' being the
elevation of the vertex. Draw this elevation and from it project the plan.

9. An elevation of a right square prism surmounted by a right square pyramid
is given in Fig, 351. The corners of the base of the pyramid are at the middle
points of the edges of one of the square faces of the prism. Draw this elevation
and from it project the plan.

10. A cube, whose edges are 1*5 inches long, has two faces horizontal, the
lower one being 1*25 inches above the groynd. The square which is the plan of

190

PRACTICAL GEOMETRY

this cube has one side inclined at 25'^ to the ground line. On each face of this
cube is placed a right square pyramid, the corners of the square bases being at the
middle points of the edges of the cube. The triangular faces of the pyramids are
equilateral triangles. Draw the plan and elevation of this built up solid.

11. Draw the plan and elevation of a tetrahedron whose edges are 2 inches
long when one face is in the vertical plane of projection and an edge of that face
is inclined at 20° to XY.

12. Draw the plan and elevation of an octahedron whose edges are 1-5 inches
long when one axis is
perpendicular to the ver-
tical plane of projection
and the elevation of
another axis is inclined
at 60° to XY.

13. The semicircle
(Fig. 352) is the eleva-
tion of a hemisphere
whose base is on the
ground and touching the ground line. The triangle is the elevation of a right
circular cone whose base is on the ground and touching the ground line. The
circle is the elevation of a right circular cylinder, 1-5 inches long, with one end
in the vertical plane of projection and resting on the other two solids. Draw the
plan of this group of solids.

14. The semicircle a'h'c' (Fig. 353) is the elevation of the half of a right
circular cone, the base being in the vertical plane of projection. The semicircle
a'd'c' is the elevation of the half of a right circular cylinder, one end being in the
vertical plane of projection. The altitude of the cone is the same as that of the
cylinder, namely 1*75 inches. Draw the given elevation and from it project the
plan.

FiG. 352.

Fig. 353.

CHAPTEK XIV

CHANGING THE PLANES OF PROJPXTION

156. Auxiliary Projections. — In general the true form of an
object is determined by a plan and an elevation, bub there are cases
where two projections are not sufficient, and in many other cases
additional projections would make the form of the object much easier
to understand. For example, in Fig. 354 are given a plan (a) and
elevation (a') of a rectangular block having recesses in three of its
faces. It is clear from the plan (a) and elevation (a') that the recess

1 V.R

ra')

•ELEVATION

PLAN !

X _ >

•T|

fa)

H.P.

^ 1

Fig. 354.

Fig, 355.

in the top is rectangular and that the recess in the front is cylindrical,
but the recess in the left hand end might be either rectangular or
cylindrical. The form of the end recess is however determined by an
end view (a/) which is a projection of the block on a plane parallel to
the ends. This second vertical plane of projection intersects the
horizontal plane in a line X^Yi which is a second ground line. The
relative positions of the solid and the three planes of projection are
clearly shown in the pictorial projection, Fig. 355, where A.E.P. is the
additional vertical plane of projection and may be called the auxiliary
elevation plane.

157. Auxiliary Projections of a Point. — Referring to the
pictorial projection. Fig. 356, A is a point in space, a is its projection

192

PRACTICAL GEOMETRY

on the horizontal plane H.P. and a! is its projection on the vertical
plane V.P. XY is the ground line in which these two planes of pro-
jection intersect. XjYj is another ground line and is the line of
intersection of the vertical plane A.E.P. with the horizontal plane
H.P. a/ is the projection of the point A on the plane A.E.P. If these
two vertical planes of projection with the projections on them be
rabatted about their respective ground lines into the horizontal plane,
Fig. 357 is obtained. An examination of Figs. 356 and 357 will show
that the distance of the auxiliary elevation a/ from X^Y^ is equal to
the distance of the elevation a! from XY ; and just as the straight
line joining a and a' is perpendicular to XY so also is the straight line
joining a and a/ perpendicular to XiYj.

Fig. 356.

Fig. 358.

Fig. 359.

Referring next to Figs. 358 and 359, A, a, and a' and the planes
of projection H.P. and V.P. are the same as before, but XjYj is now
the line of intersection of a plane A. P. P. and the vertical plane of pro-
jection V.P., the plane A.P.P. being perpendicular to the plane V.P.
When the planes H.P. and A.P.P. (Fig. 358) with the projections on
them are rabatted about XY and XjY^ respectively into the vertical
plane V.P., Fig. 359 is obtained. An examination of Figs. 358 and
359 will show that the auxiliary plan aj is at a distance from XjYi
equal to the distance of the plan a from XY, and just as the straight
line aa' is at right angles to XY so also is the straight line a'ax at right
angles to XjYi.

It will be observed that the auxiliary plane of projection A.P.P. is
not a horizontal plane, but it is usual to speak of the projection a^ of
the point A on this plane as an auxiliary plan.

The student must carefully study Figs. 356-359 and satisfy him-
self as to the correctness of the following rules which are used in the
determination of auxiliary projections : —

(1) The plan and elevation of a point are in a straight line which
is perpendicular to the ground line.

(2) When a number of elevations are projected from the same plan,
the distances of all the elevations of the same point from tlieir corre-
sponding ground lines are the same.

(3) When a number of plans are projected from the same elevation,
the distances of all the plans of the same point from their correspond-
ing ground lines are the same.

CHANGING THE PLANES OF PROJECTION 193

X^JL.^

a\'

■ -^ I,

Fig. 360.
is the true distance of

158. Distance of a* Point from the Ground Line. — The
distance of a point aa! (Fig. 360) from the ground line XY is the
length of the perpendicular from the point
on to XY. As in general this perpendicular
will be inclined to both planes of projection,
neither its plan nor its elevation will show
its true length ; but by making a projection
of the planes of projection and the perpen-
dicular on a plane at right angles to XY the
perpendicular will then be projected on a
plane parallel to it and will therefore have
its true length shown in this new projection.
In Fig. 360, X^Yi is at right angles to XY.
a'l is the new elevation of the point A and dcC
the point A from XY.

159. Auxiliary Projections of Solids.— Having mastered the
rules for auxiliary projections of a point the student should have no
difficulty in applying them
to the drawing of auxili-
ary projections of solids.
An example is illustrated
by Fig. 361. The eleva-
tion (1) and plan (2) repre-
sent a right rectangular
prism in a simple position.
From the plan (2) an ele-
vation (3) is projected on
a ground line X^Y^. Con-
sidering the point R, the
distance of r/ from XjYj
is equal to the distance
of r' from XY; also the straight line rrl is perpendicular to XiYi.
From the elevation (3) a plan (4) is projected on a ground line XgYg.
Again considering the point R, the distance of r<^ from XgYg is equal to
the distance of r from XiYj ; also the straight line r^r^ is perpendicular

to XoYg.

160. Projections of a Solid when a given Line in it is
vertical. — A plan and an elevation of the solid are first drawn when
it is in a simple position such that the given line is parallel to the
vertical plane of projection. The plan of the given line will be parallel
to the ground line. A new ground line is then taken at right angles
to the elevation of the given line and a new plan of the solid is projected
from the elevation. This new plan will be a plan of the solid when
the given line is vertical. Two examples are illustrated by Figs. 362
and 363.

In the first example (Fig. 362) the plan of a right hexagonal pyra-
mid is required when an edge containing the vertex is vertical. The
solid is placed with its base on the horizontal plane and with a sloping
edge A^ A parallel to the vertical plane of projection ; the plan va of

<(!)

X ; •,- \

(2)j

1
1 ,

Fig. 361.

194

PRACTICAL GEOMETRY

this sloping edge is parallel to XY. Completing the plan (1) the ele-
vation (2) is projected from it. XiYj is drawn at right angles to v'a'
and from the elevation (2) the plan (3) is projected. Applying the
rules for auxiliary projections of a point it is found that the new plans
v^ and ^1 of the points V and A, coincide, which shows that the line
VA is vertical when the plane of the plan (3) is considered to be a
horizontal plane. An elevation (4) is also shown projected from the
plan (3) on a ground line X2Y.2.

€•

v<<

^^v

^v.

--,^

V— .

, "-.

A"^

'-.

Fig. 362.

Fig. 363.

In the second example (Fig. 363) it is required to draw the plan
of a cube when a diagonal of the solid is vertical. The cube is first
placed with one face on the horizontal plane. The plan of the cube in
this position is a square which is drawn with one diagonal ac parallel
to XY as shown at (1). The elevation (2) is projected from the plan
(1). ac is the plan and dc is the elevation of a diagonal of the cube.
XiYi is drawn at right angles to dd and the plan (3) is projected from
the elevation (2) as shown. The plan (3) is the one required. The
outline of the plan (3) is a regular hexagon.

The plan of a solid may be drawn when a given line in it is
inclined at any given angle to the horizontal plane by proceeding in
the manner just explained, but making the new ground line inclined at
the given angle to the elevation of the given line instead of perpen-
dicular to it. But as the solid in this case can occupy any number of
positions with reference to the horizontal plane and still fulfil the
given condition, the problem is indefinite. In the case where the
given line is vertical, however, the plan is always the same. These
remarks will be understood if the solid be imagined to turn round the
given line as an axis.

161. Projections of a Solid when a given Face is inclined
at a given Angle, the Base of that Face being horizontal. —
The solid is first placed in a simple position and having the plan of the
base of the given face at right angles to the ground line. The elevation
of the given face will then be a straight line. A new ground line is
then drawn, making with the elevation of the given face an angle

CHANGING THE PLANES OF PROJECTION

195

equal to the given inclination. A plan on this new ground line is the
plan required.

Two examples are illustrated by Figs. 364 and 365. In the first
example (Fig. 364) projections of a right hexagonal pyramid are
determined when the face VAC is inclined at a given angle 6 to- the
plane upon which the new plan is to be projected. The plan (1) and
elevation (2) are drawn representing the pyramid with its base on the
horizontal plane and with <xc, the plan of the base of the face VAC, at
right angles to XY. It will be seen that the elevation of the face
VAC is the straight line v'a'. XjYi is drawn at the given inclination
B to via\ and the plan (3) is projected as shown.

In the second example (Fig. 365) projections of a regular octahe-
dron are determined when one face VAC is horizontal. The various
projections are drawn in the order in which they are numbered. From
the plan (3) an extra elevation has been projected on the ground line

Fig. 364.

Fig. 365.

XoYg. It is of interest to observe that the boundary line of the
plan (3) is a regular hexagon, and this plan may evidently be drawn
directly without the aid of any other projections.

Exercises XIV

1. Draw the elevations of the points given in Fig. 366 on a ground line
inclined at 45° to XY.

2. Find the true distances of the points given in Fig. 866 from XY.

3. A triangle ABC is given in Fig 367. Draw an elevation of this triangle on
a ground line parallel to ac. Also, project from the given elevation a new plan on
a ground line parallel to a'c'.

4. A figure ABCDEF (not a plane figure) is given in Fig. 368. Draw an
elevation of this figure on a ground line parallel to he. Also, project from the
given elevation a new plan on a ground line parallel to a'h'.

5. A plan and an elevation of a curved line are given in Fig. 369, Draw an
elevation of this curve on a ground line inclined at 45° to XY.

6. Show the plans and elevations of the following points : — A, 1*5 inches in
front of the V.P., above the H.P., and 2 inches from XY. B, 1-2 inches above

196

PRACTICAL GEOMETRY

the H.P., behind the V.P., and 1-8 inches from XY. C, 1-6 inches behind the
V.P., below the H.P., and 2 inches from XY. D, 2 inches below the H.P., in
front of the V.P., and 25 inches from XY.

—

' —

—

■N.

,.■

-v*

V-

-''

—

/
f

Fig. 366. Pig. 367. Fig. 368. FiQ. 369.

In reproducing the above diagrams take the small squares as of 0-3 inch side,

7. The ends of a right prism are regular pentagons of 1*25 inches side, and the
altitude of the prism is 1-5 inches. This prism rests with one of its rectangular
faces on the ground and with its ends inclined at 45° to the vertical plane of pro-
jection. Draw the plan and elevation.

8. Draw the plan of a right square prism when a diagonal of the solid is
vertical. Side of base, 1*25 inches. Altitude, 2 inches.

9. A front elevation of a Maltese cross is given in Fig. 370. The thickness of
the cross is 0*7 inch. Draw the plan and from it project an elevation on a ground
line making 50° with XY.

10. Draw an elevation of the solid given in Fig. 371 on a ground line inclined
at 60° to XY.

11. Draw a plan of the solid given in Fig. 371 when the line RS is vertical.

12. Draw an elevation of the solid given in Fig. 372 on a ground line inclined
at 60^ to XY.

-—5

7^
i —

ffl

Fig. 370.

Fig. 371.

Fig. 372.

Fig. 373.

In reproducing the above diagrams take the small squares as of 0*4 inch side.

13. Draw a plan of the solid given in Fig. 372 when it is tilted about one edge
of its base through an angle of 30°.

14. The solid given in Fig. 373 is formed out of two right square prisms.
Draw an elevation of this solid on a ground line inclined at 60° to XY.

15. Draw the plan of the solid given in Fig. 373 when it is tilted about one
edge of its base through an angle of 30°.

16. Draw the plan of a right hexagonal pyramid when it rests with one
triangular face on the ground. From this plan project an elevation on a ground
line inclined at 45° to the plan of the axis of the pyramid. Side of base, 1*25
inches. Altitude, 2 inches.

17. An isosceles triangle, base 1 inch, sides 1*5 inches, has its base parallel to
XY. This triangle is the elevation of one of the triangular faces of a right

CHANGING THE PLANES OF PROJECTION

197

hexagonal pyramid. Side of base of pyramid, 1 inch. Altitude, 2 inches. Com-
plete the elevation of the pyramid and from it project the plan.

18. Two elevations of an inkstand are given in Fig. 374. From the right-
hand elevation project a plan on XY and from this plan project an elevation on a
ground line inclined at 50° to XY.

19. Draw the plan of the inkstand (Fig. 374) when the edge AB is vertical.

TTl

_ 71] 1 j:-.;

u t

.,!..i

L i :

iL :±__

...Y

- / - --r K.

/^ sX

,. - , ^z s^

/{L £_Z S_

xt:::±:__ _:i

Fig. 374.

Fig. 375.

In reprodiLcing the above diagrams take the small squares as of 0*3 inch side.

20. Two elevations of a trestle are given in Fig. 375. From the right-hand
elevation project a plan on XY and from this plan project an elevation on a
ground line inclined at 60^ to XY.

21. Draw a plan of the trestle (Fig. 375) when the line joining the points R

and S is vertical.

CHAPTER XV

PLANES OTHER THAN THE CO-ORDINATE PLANES

162. Representation of Planes — Traces of a Plane. — Planes
other than the co-ordinate planes are represented by the lines in
which they meet the latter. The lines in which a plane meets the
co-ordinate planes or planes of projection are called the traces of that
plane, the intersection with the vertical plane being called the vertical
trace, and that with the horissontal plane the horizontal trace.

The line in which one plane meets another is also called the trace
of the one plane on the other, but when the traces of a plane are
mentioned its traces on the planes of projection are generally understood.

Planes occupying various positions in relation to the planes of
projection are shown in pictorial projection in the lower parts of
Figs. 376 to 384. The upper parts of the same Figs, show how the
same planes are represented by means of their horizontal and vertical
traces when the planes of projection are made to coincide as explained
in Art. 135, p. 166.

V.T.

NO
HORIZONTAL
TRACE

VERTICAL

TRACE

X Y

H.T

>
X 1

Fig. 376.

Fig. 377.

Fig. 378.

Fig. 379.

Referring to Figs. 376 to 384 separately. Fig. 376 shows a hori-
zontal plane. . Fig. 377 shows a plane which is parallel to the vertical
plane of projection and is therefore also a vertical plane. Fig. 378
shows a plane which is perpendicular to both planes of projection and

PLANES OTHER THAN THE CO-ORDINATE PLANES 199

is therefore also a vertical plane, and it is also perpendicular to the
ground line. Fig. 379 shows a plane which is perpendicular to the
vertical plane of projection and inclined to the horizontal plane.
Such a plane is generally called an inclined plane. Fig. 380 shows a

Fig. 380.

Fig. 381.

Fig. 382.

Fig. 383.

plane which is inclined to the vertical plane of projection and is
perpendicular to the horizontal plane and is therefore a vertical plane.
Fig. 381 shows a plane which is inclined to both planes of projection
but is parallel to the ground line. Figs. 382 and 383 show planes
which are inclined to both planes of projection and to the ground line.

Fig. 384 shows a plane containing the ground line XY and inclined
to both planes of projection. The horizontal and vertix;al traces
coincide with XY and these traces do not fix
definitely the position of the plane. In this
case a trace on another plane of projection is
necessary, say on a vertical plane of which Xj Yi
is the ground line.

A perpendicular plane is one which is at
right angles to one or other or both of the
planes of projection (Figs. 376, 377, 378, 379,
and 380).

An oblique plane is one which is inclined to
both planes of projection (Figs. 381, 382, and
383).

It is obvious that in the case of a perpen-
dicular plane P, its trace T on the plane of
projection Q, to which the plane is perpen-
dicular, is an edge view of the plane, and the
projections on Q of all points and lines on P
will lie on T.

It is not difficult to see that if the traces of a plane meet

X H.T&V.T. V

.Fig. 384.

one another their point of intersection is on the ground line, and if the

200

PRACTICAL GEOMETRY

traces are parallel to one another they must be parallel to the ground
line.

163. Angle between Two Planes. — The angle between two
planes, or the inclination of one plane to another, is the angle between
two straight lines drawn from any point in their line of intersection,
at right angles to it, one in each plane.

Referring to Fig. 385, which is a pictorial projection, AB and CD
are two planes and CE is their line of
intersection. FH is a straight line in the
plane AB and at right angles to CE. FK
is a straight line in the plane CD and at
right angles to CE. If HF be produced
to L, then the angle between the planes
AB and CD is either the angle HFK or
its supplement the angle LFK.

Of the two angles which one plane
makes with another it is generally the
acute angle which is taken as the angle
between the planes. In the case of the angle between two adjacent
plane faces of a solid it is generally the angle within the solid which is
understood, whether it be acute or obtuse.

The angle between two planes is called a dihedral angle.

164. Inclinations of a Plane to the Planes of Projection. —
A plane is given by its traces in Fig. 387. The same plane and the
planes of projection in their natural positions are shown in oblique
projection in Fig. 386. Referring to Fig. 386, iVB is a line in the

Fig. 386.

Fig. 387.

given plane and at right angles to its horizontal trace H.T. AO is a
line in the horizontal plane and at right angles to H.T. By definition
the angle OAB is the angle which the given plane makes with the
horizontal plane. 0, the true value of this angle, is found by the
following construction, shown in Fig. 387. Take a point o in XY and
draw oa at right angles to H.T. With o as centre and oa as radius
describe the arc aAi to cut XY at Aj. Draw oh' perpendicular to XY
to meet V.T. at b'. Join h'Ai. The angle oAjt' is the true value of 0.

PLANES OTHER THAN THE CO-ORDINATE PLANES 201

lb is obvious that the triangle oA^h' in Fig. 387 is the true shape of
the triangle OAB in Fig. 386. A similar construction gives the angle
(j), the inclination ot the given plane to the vertical plane of projection.

The arc aA^ (Fig. 387) or the arc AAj (Fig. 386) may be looked
upon as part of the outline of the base of a right circular cone whose
axis is ob' (Fig. 387) or OB (Fig. 386). The base of this cone is on
the horizontal plane, and the given plane is tangential to the slant
surface of this cone. Hence 6 which is the inclination of the slant
surface of the cone to its base is also the inclination of the given plane
to the horizontal plane. Similarly <^ is the base angle of another right
circular cone whose base is on the vertical plane of projection and
whose axis is in the horizontal plane, and the given plane is tangential
to the slant surface of this cone. Hence <p is the inclination of the
given plane to the vertical plane of projection. It may be observed
that it is not necessary that the axes of the two cones should spring
from the same point O on XY.

The construction given above will apply whatever be the positions
of the traces of the given plane ; but when the traces are parallel it is
not necessary to make a construction for finding tp after 6 has been
found because in this case ^^ = 90° — ^.

No constiniction is necessary for or (p when the given plane is
situated as shown in Figs. 376, 377, 378, 379, or 380. In these cases
0, the inclination of the plane to the horizontal plane, is equal to the
angle which the vertical trace makes with XY, and <^, the inclination
of the plane to the vertical plane of projection, is equal to the angle
which the horizontal trace makes with XY. When there is no
horizontal trace (Fig. 376) ^ = 90°, and when there is no vertical
trace (Fig. 377) 6 = 90°.

165. Plane having given Inclinations to the Planes of
Projection. — It is required to draw the traces of a plane whose
inclinations to the horizontal and vertical planes of projection are
and <^ respectively. The sum of the angles and <^ must lie between
90° and 180°.

This problem is the converse of that discussed in the preceding
Art. and two methods of solving it will be given.

In the first method the two right circular cones referred to in the
preceding Art. are used. One cone has its base on the horizontal
plane and its axis in the vertical plane of projection, and its base angle
is equal to 6. The other cone has its base in the vertical plane of
projection and its axis in the horizontal plane, and its base angle is
equal to (p. The required plane is tangential to the slant surfaces of
both cones. The important point to observe is that the relative
dimensions and positions of the cones must be such that it is possible
for the same plane to be tangential to the slant surfaces of both. The
simplest way of ensuring that the two cones shall have a common
tangent plane is to arrange the cones so that they envelop the same sphere.

Referring to Fig. 389 a circle with centre o on XY and of any con-
venient radius oe is first drawn. This circle is both plan and elevation
of a sphere whose centre is on the ground line. Draw VA^ to touch

202

PRACTICAL GEOMETRY

this circle and make with XY an angle equal to 9, cutting XY at A^
and a line through o at right angles to XY at V. Draw dC^ to touch
the same circle and make with XY an angle equal to <j> cutting XY at
Cj and a line through o at right angles to XY at d. With centre o

Fig. 388.

Fig. 389,

and radius oA^ describe the arc A^a. With centre o and radius oC^
describe the arc Q^c' . A tangent da to the arc A^a is the required
horizontal trace, and a tangent Vc' to the arc C^c' is the required
vertical trace. If these traces meet, their point of intersection must
be on the ground line. The sphere and enveloping cones having been
drawn there will, in general, be four planes which will be tangential
to both cones. The student who has mastered the problem of the
preceding Art. should now have little difficulty in understanding the
solution just given if he carefully studies Figs. 388 and 389. It only
remains to emphasize the point that the two cones are made to envelop
the same sphere in order to ensure that they shall have a common
tangent plane.

The second method of solving this problem depends on the following
theorems. (1) The inclination of a plane to one of the planes of pro-
jection is the complement of the inclination of its normal to that plane
of projection. (2) The trace of a plane on one of the planes of pro-
jection is perpendicular to the projection of its normal on that plane
of projection. Hence the construction. — Draw (Art. 144, p. 176) the
projections of a line inclined at 90° — ^ to the horizontal plane and
90° ~ <^ to the vertical plane of projection. The horizontal and vertical
traces of the required plane will be perpendicular to the plan and
elevation respectively of this line. If the traces meet, their point of
intersection must be on the ground line. If the traces do not meet
within a convenient distance, or if they are parallel to the ground line,
care must be taken to ensure that the traces are the traces of the same
plane.

166. Given one Trace of a Plane and the Inclination of
the Plane to one of the Co-ordinate Planes, to determine the

PLANES OTHER THAN THE CO-ORDINATE PLANES 203

other Trace. — In Figs. 390 and 391, 6 the inclination of the plane to
the horizontal plane is supposed to be given.

In Figs. 392 and 393, <p the inclination of the plane to the vertical
plane of projection is supposed to be given.

Case L Figs. 390 and 391, and Figs. 392 and 393. LA is the
given trace.

From any point in XY draw OA at right angles to LA. With
O as centre and radius OA describe the arc AC to cut XY at C.
Draw CB making the angle OCB equal to the given inclination of the
plane. Draw OB at right angles to»XY to meet CB at B. Join LB.
LB is the other trace of the plane.

Fig. 390.

By the

Case II. Same figures. LB is the given trace.

From any point C in XY draw CB making the angle OCB equal
to the given inclination of the plane, and meeting LB at B. Draw
BO at right angles to XY. With O as centre and OC as radius
describe the arc CA. Through L draw LA to touch the arc CA. LA
is the other trace of the plane.

The correctness of the above constructions will be obvious, if, after
the required trace has been found in each case, the construction for
finding the given inclination be applied. It will be found that all the
lines for the latter construction are already on the figures.

167. True Angle between the Traces of a Plane.-
true angle between the
traces of a plane is
meant the angle between
them when the plane
and the planes of pro-
jection are in their
natural positions as
shown in Fig. 394 where
the angle ALB is the
angle referred to. This
angle is determined by
finding the true form of

the triangle ABL in Fig. 394. Fig. 395.

which the angle ABL is a right angle.

Referring to Fig. 395, from a point a in XY draw ah at right angles

204

PEACTICAL GEOMETRY

to the horizontal trace of the plane and draw aa' at right angles to XY
to meet the vertical trace of the plane at a'. With centre L, where
the traces meet XY, and radius La', draw the arc a'A^ to meet ah pro-
duced at Ai. Join A^L. Then a, the angle A^Lfe, is the true angle
between the traces of the plane.

The theory of the above construction is that the triangle ABL
(Fig. 394) is supposed to turn about LB as an axis until it comes into
the horizontal plane. The point A describes an arc of a circle whose
plane is perpendicular to LB and whose centre is B.

Instead of taking the triangle ABL with the right angle at B, the
triangle may be taken with the right angle at A and the triangle would
then be supposed to turn about LA as an axis into the vertical plane
of projection. The student should draw the figure for this case as
well as for the other and show that the results are the same.

168. Given one Trace of a Plane and the true Angle
between the Traces, to draw the other Trace. — Let LB (Figs.
396 and 397) be the given
trace. Draw LA^ making
the angle BLAi equal to «,
the given angle between the
traces. Draw A^B at right
angles to LB and produce it
to meet XY at O. Draw
OA at right angles to XY.
With centre L and radius
LA^ describe the arc A^A
to cut OA at A. Join LA.
LA is the other trace of the -^^^ g^g
plane.

The correctness of the above construction will be obvious, if, after
the required trace has been found, the construction for finding the given
angle be applied. It will be found that all the lines for the latter
construction are already on the figure.

169. Intersection of Two Planes. — The intersection of two

Y X

Fig. 398.

Fig. 399.

planes is a straight line which is determined when two points in it are
known or when one point in it and the direction of the line are known.

PLANES OTHER THAN THE CO-ORDINATE PLANES 205

If the vertical traces of the two planes meet one another and the
horizontal traces almost meet one another and if the meeting points
are within a convenient distance (Figs. 398 and 399) the intersection
of the planes is readily found. It is clear that the point aa\ where the
horizontal traces intersect, is a point in the intersection of the two
planes. It is also evident that the point hh\ where the vertical traces
intersect, is another point in the intersection of the two planes. Hence
the required intersection is the line ah, a'h'.

Fig. 400.

Fig. 401.

The case where the traces of the two planes on one of the planes of
projection, the horizontal plane, are parallel, is shown in Figs. 400 and
401. The point hV where the vertical traces meet is a point in the
intersection of the two planes and the line of intersection is obviously
a horizontal line, its plan being parallel to the horizontal traces of the
planes, and its elevation parallel to XY.

Figs. 402 and 403 illustrate the case where the traces of the two
planes intersect at the same point on the ground line. This point on
the ground line is obviously one point on the line of intersection of the

Fig. 402.

Fig. 403.

two planes. To find another point cut the given planes by a third
plane. In Figs. 402 and 403 this third plane is perpendicular to the
horizontal plane and inclined to the vertical plane of projection. The
lines in which the third plane intersects the first and second planes

206

PRACTICAL GEOMETRY

intersect at a point pp' which is another point in the intersection of
the two given planes.

The case where the traces of the given planes are parallel to the
ground line, and the planes themselves are therefore parallel to that
line, is shown in Figs. 404 and 405. A point pp' in the line of inter-
section of the two planes is found by means of a third plane as before
and lines through |? andp' parallel to XY are the plan and elevation
respectively of the intersection required

V.T.I

V.T.2. /

J^^

yp'

/" y

i iX

HJ.\yS^

HT.2.

-Y X

Fig. 404.

Fig. 405.

A plane may be fixed by the projections of three points in it, and
the intersection of two planes fixed in this way may be found without
using the traces of the planes on the planes of projection. Referring
to Figs. 406 and 407, aa\ hV ^,
and cc' are the projections
of three points in a plane.
The triangle a&c, dVd is ob-
viously in this plane. Jc?',
ee', and ff are the projec-
tions of three points which
are in another plane. The
triangle def^ d'e'f is obvi-
ously in this second plane.
•It is required to find the
intersection of these two
planes.

Referring to Fig. 406,
a horizontal plane at the
level of the point ff is
taken. This plane cuts
the plane of the triangle
abc, a'h'd in the line gh, g'h'.

This same plane cuts the p .-g

plane of the triangle def,
d'e'f in the line fk, fh'. pp' the point of intersection of the lines gh^

PLANES OTHER THAN THE CO-ORDINATE PLANES 207

g'h', and fh, f'Tz is obviously a point in the intersection of the two
given planes.

Fig. 407.

A vertical plane of which hi is the horizontal trace is taken. This
plane cuts the plane of the triangle abc^ a'h'c' in the line hl^ 67'. This
same plane cuts the plane of the triangle def^ d'e'f in the line ww, m'n'.
q^ the point of intersection of the lines 6/, h'V and mn, m'n' is obviously
another point in the intersection of the two given planes. That part
of the line of intersection of the two given planes which lies within
both of the given triangles is shown thicker, and this thicker line is the
intersection of the given triangles.

Fig. 407 shows the various planes and their intersections in oblique
projection.

When an auxiliary cutting plane is required in order to find the
intersection of two given planes, this auxiliary plane may be placed in
any number of positions, and the student must exercise his judgment
and ingenuity in determining a convenient position, that is, a position
which will lead to points of intersection of lines which will fall within
convenient distances and which can be found with accuracy. Lines
intersecting at very acute angles should be avoided if possible.

170. Distance between two Parallel Planes.— Parallel planes
have parallel traces, the vertical and horizontal traces being parallel to
one another respectively.

In Figs. 408 and 409, V.T.I, and V.T.2. are the vertical traces and
H.T.I, and H.T.2. are the horizontal traces of two parallel planes. It
is required to find the distance between these planes. Cut the given
planes by a vertical plane whose horizontal trace oc (Fig. 409) or
OC (Fig. 408) is perpendicular to the given horizontal traces. This
plane will cut the given planes in two parallel lines the true distance
between which will be the required distance between the given planes »,

208

PRACTICAL GEOMETRY

Referring to Fig. 409, with o as centre and radii oa and oc describe
arcs to cut XY at A^ and C^ respectively. Draw oh'd' at right angles

Fig. 408.

Fig. 409.

to XY cutting the given vertical traces at h' and d' as shown. Join
Ai6' and C^d'. The perpendicular distance between A^h' and C^d' is the
distance between the two given planes.

The theory of the foregoing construction is that the vertical plane
COD (Fig. 408) is supposed to turn about its vertical trace OD, which
is perpendicular to XY, carrying with it the lines of its intersection
with the given planes, until it is in the vertical plane of projection.
The true distance between these lines can then be measured.

The auxiliary cutting plane may be, quite as conveniently, taken
perpendicular to the given vertical traces, and the student should draw
the figure with the cutting plane so taken.

171. To draw the Traces of a Plane which shall be parallel
to a given Plane and at a given distance from it. — The solution
of this problem is so obvious, if the preceding one has been understood,
that no detailed description of it need be given. It may be pointed
out however that two planes may be placed at a given distance from a
given plane, one on each side of it.

Exercises XV

1. Planes are given by their traces at (a), (6), (c), [d), [c), and (/) in Fig. 410.
Determine in each case the inclinations of the plane to the planes of projection.

Fig. 410.

PLANES OTHER THAN THE CO-ORDINATE PLANES 209

2. What are the true angles between the traces of the planes (a), (6), and (c),
Fig 410 ?

3. Find the true angles between the traces of the planes (d), (e), and (/),
Fig. 410.

4. The vertical trace of a plane is inclined at 45° to XY and the plane is
inclined at 60° to the horizontal plane : draw both traces of the plane.

5. The vertical trace of a plane is inclined at 50° to XY and the plane is
inclined at 50° to the vertical plane of projection : draw both traces of the plane.

6. The horizontal trace of a plane is inclined at 45° to XY and the plane is
inclined at 60° to the horizontal plane : draw both traces of the plane.

7. The vertical trace of a plane is parallel to XY and at a distance of 2*5 inches
from it. The plane is inclined at 55° to the horizontal plane. D^aw the traces of
this plane. .

8. Draw the traces of a plane whose inclinations to the horizontal and vertical
planes of projection are 55° and 60° respectively.

9. Draw the traces of a plane which is inclined at 50° to the horizontal plane
and 40° to the vertical plane of projection.

10. The true angle between the traces of a plane is 60° and the vertical trace
is inclined at 35° to XY. Draw both traces of this plane.

11. Draw the traces of a plane which is inclined to the horizontal plane at 50°,
the true angle between the traces being 70°.

12. The true angle betwee.n the traces of a plane is 60° and the plane is
equally inclined to the planes of projection. Represent this plane by its traces.

13. Pairs of planes are given by their traces at (a), (6), (c), and (d) Fig. 411.
In each case show the plan and elevation of the line of intersection of the planes.

Fig. 411.

14. Find the plan and elevation of
the point which is situated in each of
the three planes given in Fig. 412.

15. Determine the intersection of
the plane triangle abc, a'b'c'- (Fig. 413)
with the plane triangle def, d'e'f. The
dimensions given in Fig. 413 are in
inches.

Fig. 412.

Fig. 413.

210 PRACTICAL GEOMETRY

16. Two parallel planes are parallel to the ground line. The horizontal trace
of the first is 1-5 inches below XY and that of the second 2 inches below XY.
The vertical trace of the first is 2 inches above XY. Find the vertical trace of
the second plane and the distance between the two planes.

17. Referring to Fig. 410, take each of the given planes and show two planes
parallel to the given plane and 0*7 inch distant from it.

18. Two parallel planes have an inclination of 50°, and their horizontal traces
are 1 inch apart. Determine a vertical plane which cuts these planes in lines
which are 1'5 inches apart.

CHAPTER XVI

STRAIGHT LINE AND PLANE

172. Lines and Points in a Plane. — Theorem I. A straight
line contained hy a given plane lias its traces on the traces of the plane.
The traces of the line are points in the planes of projection, but they
are also points in the plane containing the line, therefore the traces of
the line must lie on the intersection of the plane with the planes of
projection, that is, on the traces of the plane.

Referring to the pictorial projection, Fig. 414, LA and LB are the
traces of a plane and AB is a straight line contained by the plane
ALB. The points A and B are the traces of AB.

Theorem II. A straight line which is parallel to one of the planes
of projection and is contained hy a given plane is parallel to the trace of
that plane on the plane of projection to which the line is parallel.
Referring to the pictorial projection, Fig. 414, CD is a horizontal line
lying in the plane LAB. CD is parallel to LA the horizontal trace of
the plane. Hence cd the plan of CD is parallel to LA. c'd\ the
elevation of CD, is parallel to XY. This line has only one trace C
which is a vertical trace. EF is a line parallel to the vertical plane
of projection and lying in the plane ALB. EF is parallel to LB the
vertical trace of the plane. Hence ef the elevation of EF is parallel
to LB. ef, the plan of EF, is parallel to XY. This line has only one
trace E which is a horizontal trace. '

Fig. 415 shows the various lines referred to above by ordinary
plan and elevation.

It is obvious that the distance of CD from the horizontal plane of
projection is equal to the distance of its elevation from XY, also the
distance of EF from the vertical plane of projection is equal to the
distance of its plan from XY.

A knowledge of the foregoing principles will enable the student to
solve a number of problems.

For example. — Given the traces of a plane (Fig. 415) and the plan
p of a point contained by the plane, to find the elevation of the point.
Through p draw ab to cut the horizontal trace of the plane at a and
XY at h. Consider ah as the plan of a line lying in the plane. The
elevation a'h' of this line is found as shown, and a'h' must contain p'
the elevation required. Instead of taking the line ah, a'h' lying in the

212

PRACTICAL GEOMETRY

plane and containing the point P, the horizontal line cd, c'd' lying in
the plane and containing the point P may be taken, or the line ef, e'f

Fig. 414.

Fig. 415.

parallel to the vertical plane of projection may be used to find p'. If
p' is given |> may be found by working the foregoing construction
backwards.

Again, given the traces of a plane and the distances of a point in
the plane from the planes of projection to find its plan and elevation.
Place in the plane (Fig. 415) a horizontal line cd, dd' whose distance
from the horizontal plane of projection is equal to the given distance
of the point from the latter plane. Next place in the plane a line ef,
e'f whose distance from the vertical plane of projection is equal to the
given distance of the point from the latter plane, p the intersection
of cd and ef is the plan, and p the intersection of c'd' and e'f is the
elevation required.

In another set of problems it is required to find the traces of a
plane which contains given points or lines. Three points which are
not in the same straight line will fix a plane, also two parallel straight
lines or two intersecting straight lines, or one straight line and one
point not in that line. These problems are solved by making use of
the facts that if two points are in a plane the straight line joining
them is in that plane and the traces of this line are on the traces of
the plane.

173. To draw the Traces of a Plane which shall contain
a given Line and have a given Inclination to one of the
Planes of Projection. — Let ah, a'V (Fig. 417) be the given line, and
let the given inclination of the plane be to the horizontal plane.

Draw the projections of a cone having its vertex at the vertical

STRAIGHT LINE AND PLANE

213

trace of the given line, its base on the horizontal plane of projection,
and its slant side inclined to its base at the given angle {$).

A line through the horizontal trace of the given line, to touch the
base of the cone, will be the horizontal trace of the plane required,
and a line through the point where this horizontal trace meets the
ground line and through the vertical trace of the given line will be
the vertical trace of the plane required. The foregoing construction
will be readily understood by referring to the pictorial projection
shown in Fig. 416.

Fig. 416.

Fig. 417.

If the horizontal trace of the given line falls outside the base of
the cone two tangents from this point to the base of the cone can be
drawn and there will therefore be two planes which will satisfy the
given conditions. In Figs. 416 and 417 only one of the two possible
planes is shown. If the horizontal trace of the given line falls on the
circumference of the base of the cone only one tangent from this point
to the base of the cone can be drawn and there will then be only one
plane which will satisfy the given conditions, and this plane will have
the same inclination as the given line. If the horizontal trace of the
given line falls inside the base of the cone no tangent from this point
to the base of the cone can be drawn, which shows that the problem is
impossible when the inclination of the plane is less than the inclination
of the line.

In the construction just explained it has been assumed that the
horizontal and vertical traces of the given line have been accessible, it
now remains to be shown how the problem is to be dealt with when
these points are inaccessible.

Referring to Fig. 419, the horizontal and vertical traces of the
given line ah, a'h' are supposed to be inaccessible.

Draw the projections of two cones having their vertices at points
cd and dd' in the given line, their bases on the horizontal plane,

214

PRACTICAL GEOMETRY

and their slant sides inclined to their bases at the given angle (0).
The horizontal trace of the required plane is a common tangent to the
bases of the two cones. The vertical trace of the plane may be found
by taking a point ee' in the horizontal trace of the plane and joining
this point with two points (say cc and dd') in the given line. The line

Fig. 418.

Fig. 419.

joining the vertical traces of these two lines is the vertical trace of the
plane required. A reference to the pictorial projection (Fig. 418) will
make the construction clear.

It should be noticed that although four tangents may be drawn to
the bases of the cones shown in Figs. 418 and 419 only the outside
tangents, and not the cross tangents, can be the horizontal traces of
planes which are tangential to the two cones.

If the inclination of the required plane be given to the vertical
plane of projection instead of to the horizontal plane the bases of the
cones must be placed on the former plane instead of on the latter.

174. In a given Plane to place a Line having a given
Inclination to one of the Planes of Projection. — The inclina-
tion of the line must not exceed the inclination of the plane.

Let the given inclination be to the horizontal plane.

.Take a point C (Eig. 421) on the ground line and draw Ch' making
the given angle 6 with XY and meeting the vertical trace of the
given plane at h'. Draw h'h perpendicular to XY meeting the latter
at h. With centre h and radius 6C describe an arc of a circle to cut
the horizontal trace of the given plane at a. Draw aa' perpendicular
to XY meeting it at a', ah is the plan and ah' the elevation of the
line required.

The correctness of the construction is obvious, for AB (Fig. 420)
is evidently on the given plane and it is also on the surface of a cone
whose base is on the horizontal plane and whose base angle is 6.

The construction is applicable whatever be the positions of the
traces of the given plane.

STRAIGHT LINE AND PLANE

215

If the given inclination be to the vertical plane of projection
instead of to the horizontal plane, the construction will present no
difficulty to the student who has followed that just given.

Fig. 420.

Fig. 421.

175. To draw the Projections of a Line which shall pass
through a given Point, have a given Inclination, and be parallel
to a given Plane. — By the construction of the preceding Art.
place in the given plane a line having the given inclination. Through
the plan and elevation of the given point draw lines parallel to the
plan and elevation respectively of the line which has been placed in
the given plane. These will be the projections required.

That the line whose projections have thus been found fulfils the
given conditions is not difficult to see ; for it is parallel to a line
which has the given inclination, and must therefore have itself the
given inclination. Also it can never meet the given plane, for if
it did it could not be parallel to any line in that plane, therefore
it is parallel to the given plane.

176. To draw the Traces of a Plane which shall contain
a given Point and be parallel to a given Plane. — The horizontal
and vertical traces of the plane required must be parallel to the
horizontal and vertical traces respectively of the given plane.

Fig. 422,

Fig. 423.

pp' (Figs. 422 and 423) is the given point and H.T. and V.T. are
the traces of the given plane.

216 PRACTICAL GEOMETRY

The projections of a line passing through 'pp and parallel to the
given plane are first drawn. The traces of this line are on the traces
of the plane required.

In Fig. 422 a horizontal line pq, p(][ parallel to the given plane
is shown. The vertical trace of this line is a point on the vertical
trace of the plane required, and ;the point where the vertical trace of
the plane meets XY is a point on the horizontal trace of the plane.

In Fig. 423 a line ah, a'h' is placed in the given plane and the
projections rs, r's' of a line parallel to ah, a'h' are drawn through p
and p' respectively. The traces of rs, r's' are on the traces of the
plane required.

The construction shown in Fig. 423 is used when the vertical
trace of a horizontal line through pp' parallel to the given plane is
inaccessible, or when the horizontal trace of a line through pp' parallel
to the vertical plane of projection and parallel to the given plane is
inaccessible.

177. In a given Plane to place a Line which shall be
parallel to and at a given distance from another given
Plane not parallel to the first. — Draw the traces of a plane
parallel to the second given plane and at the given distance from it
(Art. 171, p. 208). The line of intersection of this third plane with
the first given plane is the line required.

178. To draw the Projections of a Line which shall pass
through a given Point and be parallel to two given inter-
secting Planes. — The required projections will pass through the
projections of the given point, and be parallel to the projections of
the line of intersection of the two given planes.

179. To draw the Traces of a Plane which shall contain
a given Line and be parallel to another given Line. — Draw
the projections of a line to intersect the first given line and be parallel
to the second. The plane containing these two intersecting lines is
the plane required.

180. To draw the Traces of a Plane which shall contain
a given Point and be parallel to two given Lines. — Draw the
projections of two lines to pass through the given point and be parallel
respectively to the two given lines. The plane containing the lines
passing through the given point is the plane required.

181. Intersection of a Line and Plane. — Let LMN (Fig. 425)
be a given plane and ah, a'h' a given line. It is required to show the
projections of the point of intersection of the line and plane.

Draw the traces LR and RN of a plane, preferably perpendicular
to one of the planes of projection, to contain the given line. Find the
line of intersection of this plane with the given plane. The points
p and p' where ah and a'h' meet the plan and elevation respectively
of the intersection of the planes are the projections required.

Referring to the pictorial projection (Fig. 424) it will be seen
that AB and LN being in the same plane LRN they must intersect
at some point P unless they happen to be parallel. Again, since LN
is the line of intersection ,of the planes LMN and LRN the line LN

STRAIGHT LINE AND PLANE

217

isin the plane LMN, therefore the point P is in the plane LMN. But
P is in AB, therefore the point P is the point of intersection of AB
and the plane LMN. If AB is parallel to LN then AB is parallel to
the plane LMN.

Fig. 424.

Fig. 425.

182. To draw the Projections of a Line which shall pass
through a given Point and be perpendicular to a given
Plane. — If a line is perpendicular to a plane the projections of the
line are perpendicular to the traces of the plane, the plan to the
horizontal and the elevation to the vertical trace.

The construction is therefore as follows. Through the plan of the
point draw a line at right angles to the horizontal trace of the plane,
and through the elevation of the point draw a line at right angles to
the vertical trace of the plane ; these will be the projections required.

183. To draw the Traces of a Plane which shall contain
a given Point and be perpendicular to a given Line. — Let
pp (Fig. 426) be the given point and a6, a'h' the given line.

Through p draw pq perpendicular to ah, meeting XY at q. Draw
q(J perpendicular to XY, meeting a parallel to XY through p' at q'.
Through (^ draw LM perpendicular to a'h', meeting XY at M. Through
M draw MN perpendicular to ah. LM and MN are the vertical and
horizontal traces respectively of the plane required.

In the foregoing construction it has been assumed that the line
through p perpendicular to ah meets XY within the paper, but as
this is not always the case another construction will now be given
which will apply to any case.

Let pp' (Fig. 427) be the given point and ah, a'h' the given line.

On ah as a ground line make another elevation, ^/, of the given
point P, and a^h^ of the given line AB. Through pi draw c^d per-
pendicular to a-lh-l, meeting ah at d. c^d is the trace on the vertical
plane, of which ah is the ground line, of the plane perpendicular to AB
and containing the point P. It is obvious that the point d is a. point

218

PRACTICAL GEOMETRY

in the horizontal trace of the plane required, and that this trace will
be determined by drawing a line through d perpendicular to ah.

Produce ah to meet XY at c and draw cc/ pei-pendicular to ah
jting c^d at c/. The point c/ is an auxiliary elevation of a point

meetmar

' y

Y X

Fig. 426.

which is in the required plane and also in the vertical plane of which
XY is the ground line. If therefore cd be drawn perpendicular to
XY and made equal to ccl^ c' will be a point in the vertical trace
required which will be perpendicular to a'h'.

An examination of the figures will show that the plane determined
satisfies the given conditions.

184. To draw the Traces of a Plane which shall contain
a given Point and be perpendicular to two given Planes.—'
The plane whose traces are required will be perpendicular to the line
of intersection of the two given planes. Hence one construction is —
find the line of intersection of the given planes, and then by the
construction of the preceding article determine the traces of the plane
which contains the given point and which is perpendicular to this line
of intersection.

Another construction is — determine the projections of two lines
which shall pass through the given point and be perpendicular one to
one plane and the other to the other. The plane containing these
two intersecting lines is the plane whose traces are required.

185. To draw the Projections of a Line which shall pass
through a given Point and meet a given Line at right angles.
— Determine (Art. 183, p. 217) the traces of a plane which shall
contain the given point and l3e perpendicular to the given line. Find
(Art. 181, p. 216) the projections of the intersection of this plane and
the given line. The line joining the plan of the given point with the
plan of the intersection of the line and plane will be the plan, and the line
joining the elevations of the same points will be the elevation required.

STRAIGHT LINE AND PLANE

219

186. To draw the Traces of a Plane which shall contain
a given Line and be perpendicular to a given Plane. — Deter-
miDe the projections of a line to intersect the given line and be
perpendicular to the given plane. Find the traces of a plane to
contain these two intersecting lines. These will be the traces required.

187. Given the Inclinations of two intersecting Lines and
the Angle between them, to draw their Projections and the
Traces of the Plane containing them. — Draw C^A and CiB
(Fig. 428) making the angle AC^B equal to the given angle between
the lines. From a point A in CjA draw AD making the angle CjAD
equal to a, the given inclination of one of the lines. Draw CjD per-
pendicular to AD. With centre C^ and radius C^D describe an arc
DEF and draw BF to touch this arc and make the angle C^BF equal
to (3, the given inclination of the other line. Join AB. Consider the
triangle AC^B to be on the horizontal plane and imagine this triangle
to rotate about the side AB until the point Cj is at a distance equal
to CjD or C^F from the horizontal plane. Denoting the new position
of the point C^ by C (see the pictorial projection, Fig. 429), the lines

Pig. 429.

CA and CB will be inclined to the horizontal plane at angles equal to
a and /3 respectively, and their plans will be equal in length to AD
and BF respectively. Hence, if with centre A and radius AD the arc
Dc be described, and if with centre B and radius BF the arc Be be
described, meeting the former arc at c, Ac and Be will be the plans of
the lines required.

AB will be the horizontal trace of the plane containing the lines
CA and CB.

An elevation of the lines on any vertical plane can easily be
obtained, since A and B are on the horizontal plane and the distance
o£ C from the horizontal plane is known. In Fig. 428 an elevation is
shown on a ground line perpendicular to AB. The distance of c from

220

PRACTICAL GEOMETRY

XY is equal to C^D or CiF. Nc is the vertical trace of the plane
containing the lines CA and CB.

Instead of using the arcs T>c and ¥c to find the point c, this point
maybe determined as follows: Draw XY perpendicular to AB and
produce AB, if necessary, to meet XY at N. Draw C^C/ perpendicular
to XY. With centre N and radius NC/ draw the arc C/c' to meet at
c a parallel to XY at a distance from XY equal to C^D or C^F. A
perpendicular to XY from c' to meet a parallel to XY from C^ deter-
mines the point c.

The student should after drawing the figure ABFC^DA on a piece
of paper cut it out and fold it along the lines ACj and BC^ until the
edges C^D and C^F coincide. Placing this model so that the edges
AB, BF, and DA are on the table, he should then have no difficulty
in fully understanding this important problem.

188. Given the Traces of a Plane, to determine the Trace
of this Plane on a new Vertical Plane.— Let MN (Fig. 431) be
the horizontal trace of a plane, and let LM be the trace of this plane
on a vertical plane of which XY is the ground line ; it is required to
find the trace of this plane on a vertical plane of which X^Y^ is the
ground line. Let MN meet XjY^ at S.

Fig. 430.

Take a point p in the new ground line and consider this as the plan
of a point lying in the given plane LMN. Find the elevation p of
this point on the ground line XY. Draw jpp^' perpendicular to XjYi
and make pp/ equal to the distance of p' from XY. Sp/ is the vertical
trace required. For, P is a point in the given plane and also a point
in the new vertical plane, therefore pi must be a point in the new
vertical trace. Also the point S, where the horizontal trace meets
XjYj, is evidently a point in the new vertical trace, therefore the line
Sp/ is the trace required.

In the case where the new ground line does not meet the horizontal
trace of the plane within a convenient distance, a second point in the
required vertical trace may be found in the same way as the first
point pi'.

STRAIGHT LINE AND PLANE 221

189. To rabat a given oblique Plane, with any Points
or Lines on it, into the horizontal Plane.— Let LMN (Fig. 432)
be the given oblique plane, and
let ahc, a'h'c' be a triangle lying
on this plane. Take a new
ground line XiYi at right angles
to MN, the horizontal trace of
the given oblique plane. Find
(Art. 188, p. 220) ST the verti-
cal trace of the given plane on
the new vertical plane of which
X^Yj is the ground line. This
line ST will be the edge view
of the given plane and a/c/fe/
the elevation of the triangle
ACB on the new vertical plane
will be on ST as shown.

Now imagine the plane TSN Fig. 432.

to turn about SN as an axis.

The point A will describe an arc of a circle whose plane is perpendicular
to SN, and will therefore be vertical, and its plan will be a straight
line aAj perpendicular to SN. The elevation (on XiYj) of this arc
will be an arc of a circle a/ A/ whose centre is S. The point A/ on
XjYi is the elevation of the point A when the plane with A on it has
been brought into the horizontal plane. Hence a perpendicular from
A/ to Xi Yj to meet aA^ determines Aj the plan of A in the new position.
In like manner the points B^ and Cj may be determined, and these
points Aj, Bi, and C^ being joined a triangle is formed which is the
triangle ABC brought into the horizontal plane by turning it
about SN.

It is evident that the triangle AiBjCi is the true form of the
triangle ABC.

The construction given above will be found of great use in solving
numerous problems.

If any line be drawn in the triangle AjB^Cj, or if any figure be
built up on that triangle, it is obvious how this line or figure may be
turned about SN into the plane TSN and their projections in the new
position found.

Exercises XVIa

In reproducing the figures which are shown on a squared ground take the small
squares as of half-inch side.

1. The traces of a plane are given in Fig. 433, also one projection of each of
three points A, B, and C contained by the plane. Find the true form of the
triangle ABC.

2. A plane is given by its traces in Fig. 434, the plan abc of a triangle lying
on the plane is also given. Draw the elevation of the triangle. Show also the

222

PRACTICAL GEOMETRY

plan and elevation of a point which is in the given plane and 1-25 inches from
each of the planes of projection.

Fig. 433.

.^•>

—

■*x

„>

JUt

c nu

c ^

's:

5*.

—

•^

Fig. 434.

Fig. 485.

Fig. 436.

3. a! and h' (Fig. 435) are the elevations of two of the angular points of a
triangle lying on the plane whose traces are given, and c is the plan of the other
angular point. Draw the plan and elevation of the triangle and find its true
form.

4. Draw the traces of the plane which contains the given triangle ahc^ a'h'c'
(Fig. 436). Show also th3 traces of the plane which contains the line a&, a'b' and
is parallel to the ground line.

5. The horizontal trace of a plane is given in Fig. 437, also the plan and eleva-
tion of a point P in the plane. Draw the vertical trace of the plane. Show also
the plan and elevation of a point Q which is in this plane, and which is 2 inches
from P and 1 inch above the horizontal plane. Assume that the point where the
given horizontal trace meets the ground line is inaccessible.

—

d
I'

—

c a'

h' -

' —

*N,

—

-
c

b *!

■

—

ih.

H.T.

Fig. 437.

Fig. 438.

Fig. 439.

Fig. 440.

Fig. 441.

6. The horizontal trace of a plane is given in Fig. 438, also the elevation a'h'
of a line AB contained by the plane. The length of AB is 3 inches. Draw the
plan of AB and the vertical trace of the plane. Assume that the point where the
horizontal trace of the plane meets the ground line is inaccessible.

7. Draw the traces of the plane which contains the line ah, a'h', and the point
cc' given in Fig. 439.

8. The plan and elevation of a line AB are given in Fig. 440. This line cuts
a plane whose horizontal trace is given at a point 1*5 inches from B. Draw the
vertical trace of the plane.

9. Draw the traces of the two planes which contain the line ah, a'h' (Fig. 441)
and which are inclined at 60^^ to the horizontal plane. Assume that points to the
left of aa' and points to the right of hh' are inaccessible.

10. The horizontal trace of a vertical plane makes an angle ofi 45° with XY.
Draw the elevation of two lines contained by this plane, one line being inclined
at 30° to the vertical plane of projection and the other inclined at 60° to the
horizontal plane. The first line intersects the ground line and the second inter-
sects the first at a point 1-5 inches from the vertical plane of projection.

11. The vertical and horizontal traces of a plane make angles of 60^ and 45°

STRAIGHT LINE AND PLANE

223

H.T.

Fig. 442.

-yw

\^^^

^.'<.

/ "

-H'

^x«

<-"'

-^h

Fig. 443.

Fig. 444.

respectively with XY. Draw the projections of a line lying in this plane, inclined
at 30^ to the horizontal plane and passing through a point 1 inch from each of
the planes of projection.

12. Taking the plane given in Fig. 435 place in this plane a line inclined at
30° to the horizontal plane, the portion of the line between the traces of the
plane to be 2-5 inches long.

13. Draw the projections of a line passing through the given point pp' (Fig.
442) having an inclination of 45° to the horizontal plane and parallel to the given
plane.

14. Draw the traces of a plane which shall contain the given point pp' (Fig.
442) and be parallel to the given plane.

15. In the given plane
LMN (Fig. 443) place a line
which shall be parallel to
the plane PQR and at a
distance of 0-5 inch from it.

16. Draw the projec-
tions of a line which shall
pass through the point aa'
(Fig. 443) and be parallel
to each of the given planes.

17. The vertical and
horizontal traces of a plane
make angles of 60° and
45"^ respectively with XY.
Draw the plan and eleva-
tion of a line perpendicular to this plane, meeting it at a point A, and intersecting
XY at B so that AB is 1-5 inches long.

18. A line inclined at 35° to XY is both horizontal and vertical trace of a
plane. A point P in XY is 2-5 inches from the point where the plane cuts XY.
Find the perpendicular distance of P from the plane.

19. The projections of two non-intersecting lines AB and CD are given in
Fig. 444. Draw the traces of the plane which contains the line CD and is
parallel to AB. Show also the plan and elevation of the projection of AB on this
plane.

20. Draw the traces of the plane which bisects the line,a6, a'b' (Fig. 440) at
right angles.

21. Draw the traces of the plane which shall contain the point aa' (Fig. 443)
and be perpendicular to each of the given planes.

22. Draw the projections of a line which shall pass through the point cc'
Fig. 439) and meet the line ab, a'b' at right angles.

23. Show by its traces a plane perpendicular to the plane of the triangle abc,
a'b'c' (Fig. 436) and bisecting the sides AC and BC.

24. Two intersecting straight lines containing an angle of 60° are placed so
that one of them is inclined at 35° while the other is inclined at 45° to the
horizontal plane. Find the inclination of the plane containing the two lines.

25. A person using a theodolite at a station point A observes the angular
elevations of two objects P and Q above the horizon to be 35° and 65° respectively.
The horizontal angle (or azimuth) between their directions, that is between
vertical planes containing them, measures 75°. Find the angle PAQ subtended
by the two objects at the place A, as would be measured by a sextant. [b.e.]

190. Inclination of a Line to a Plane. — The inclination of
a given line to a given plane is the angle which the line makes with its
projection on the plane. Let PQ be the given line and let it intersect
the plane at Q. Let PR be a normal to the plane, intersecting the
plane at R. If Q and R be joined, then the line RQ is the projec-
tion of PQ on the plane and the angle PQR is the inclination of the
line PQ to the plane. But since the angle PRQ is a right angle the

224

PRACTICAL GEOMETRY

angle QPR is the complement of the angle PQR. Hence, if from a
point P in the given line a normal PR be drawn to the given plane
(Art. 182, p. 217) the true angle between this normal and the given
line may be found as explained in Art. 147, p. 177, and the comple-
ment of this angle is the inclination required.

It is obviously unnecessary to find the points in which the given
line PQ and the normal PR intersect the given plane.

191. To find the Angle between two Planes. — Referring to
the pictorial projection, Fig. 445, AB and CD are two planes intersect-
ing in the line CE. FH is a third
plane which is at right angles to the
line CE. The plane FH intersects the
planes AB and CD in the lines KH
and KL respectively. Since CE is
at right angles to the plane FH it
follows that CE is at right angles to
each of the lines KH and KL. Hence
the angle between the planes AB and
CD is the angle between the lines KH
and KL. If from a point P in the
plane FH perpendiculars PR and PQ
be drawn to KH and KL, then, the angle QKR is the supplement of
the angle QPR, and if RP be produced to S the angle QKR is therefore
equal to the angle QPS. It may be proved that the lines PR and PQ
are perpendicular to the planes AB and CD respectively.

Hence, to find the angle between two given planes, draw, from any
point, two perpendiculars, one to each plane, and find the angle between
these perpendiculars. The acute angle between these perpendiculars
will be equal to the acute angle between the planes.

Another method is as follows. Let LMN and LRN (Fig. 447) be
the two given planes. Draw ZN, the plan of the line of intersection

Fig. 445.

Fig. 446.

Fig. 447.

of the planes. Draw ACB at right angles to IN intersecting MN, /N,
and RN at A, C, and B respectively. With centre / and radii ZC and

STRAIGHT LINE AND PLANE 225

LN describe arcs CCg and NNg cutting XY at Cg and Ng. Join LN^
and draw C2P2 perpendicular to LNg. On ZN make CV^ equal to CgPg.
Join AP, and BP^. The acute angle between the lines AI^ and EPj
is the acute angle between the planes LMN and LRN. In this con-
struction AB is the horizontal trace of a plane which is perpendicular
to the line of intersection of the given planes. This plane intersects
the given planes in two lines .AP and BP (see the pictorial projection,
Fig. 446 ) and the acute angle between these lines is the angle between
the given planes. The lines AP and BP together with the line AB
form a triangle of which APjB is the true form.

The points A and B where the line perpendicular to ZN meets the
horizontal traces of the given planes may be on the same side of ZN
instead of on opposite sides as in Fig. 447. In particular cases AB
may be parallel to the horizontal trace of one of the given planes and
the point A or the point B will then be at an infinite distance from C,
and APi or BP^ will be perpendicular to ZN.

192. To draw the Traces of a Plane which shall make a
given Angle with a given Plane and contain a given Line in
it. — Let LMN (Fig. 447) be the given plane, and ZN the plan of the
line in it which the required plane is to contain.

Draw AC perpendicular to ZN, meeting the latter at C and the
horizontal trace of the given plane at A. With centre Z and radii ZC
and ZN describe the arcs CCg and NNg cutting XY at Cg and Ng. Join
LN2 and draw C0P2 perpendicular to LNg. On ZN make CP^ equal to
C2P2. Join AP^. Draw P^B making the angle AP|B equal to the
given angle. Let P^B meet AC or AC produced at B. Join NB and
produce it to meet XY at R. Join LR. NR and RL are the traces
of the plane required.

If the student has understood the construction of the preceding
Art. referring to Fig. 447, he should have no difficulty in satisfying him-
self as to the correctness of the construction given above for finding
the plane LRN.

193. To draw the Traces of a Plane which shall make a
given angle with a given Plane, and contain a given Line not
contained by the given Plane. — Let LMN (Fig. 448) be the given
plane. This plane is shown as an inclined plane, that is a plane inclined
to the horizontal plane but perpendicular to the vertical plane of pro-
jection. If the plane is given as an oblique plane it should be converted
into an inclined plane by the construction of Art. 188, p. 220.

ah, a'h' is the given line.

Take two points in the given line and consider them to be the
vertices of two cones whose bases are on the given plane, and whose
slant sides are inclined to their bases at the given angle (/?).

It is evident that a plane which touches these two cones will fulfil
the conditions of the required plane.

Turn the given plane about its horizontal trace until the plane, with
the circles which are the bases of the cones, comes into the horizontal
plane. Draw the tangent CjN to these circles when in this position.
Now turn the plane with the tangent on it back to its former position.

226

PRACTICAL GEOMETRY

The plane which contains the lines AB and CN is the plane required ;
its traces are NR and ST. There are two planes which satisfy the
given conditions but only one
of them is shown in Fig. 4-1:8.

194. To draw the Traces
of a Plane which shall con-
tain a given Point, have a
given Inclination, and make
a given Angle with a given
Plane. — Conceive a cone hav-
ing its vertex at the given point,
its base on the horizontal plane,
and having a base angle equal
to the angle which the required
plane is to make with the hori-
zontal plane. Conceive also a
second cone having its vertex
at the given point, its base on
the given plane, and having a
base angle equal to the angle
which the required plane is to
make with the given plane. It
is evident that a plane which
touches these two cones will
fulfil the conditions of the pro-
blem.

Referring to Fig. 449, vv' is
the given point, and LMN is the given plane, which is assumed to be
perpendicular to the vertical plane of projection, v'a'h' is the elevation
of the cone whose base is on the horizontal plane and whose base angle a
is equal to the angle which the required plane is to make with the
horizontal plane, v'c'd! is the elevation of the cone whose base is on
the given plane and whose base angle ^ is equal to the angle which the
required plane is to make with the given plane.

Let the surface of the second cone be produced to cut the horizontal
plane of projection. The curve of section, or horizontal trace of the
cone, will be a " conic section " and it may be found as follows. Draw
t'r perpendicular to v'e', the elevation of the axis of the cone, meeting
v'e' produced at t', XY at s and v'd! produced at r. With t' as centre
and t'r' as radius describe an arc, and through s' draw s'H.^ perpendicu-
lar to t'r' to meet the arc at Ho. Draw s's perpendicular to XY to meet
a parallel to XY through v at 8. On ss' make sli equal to s'Hg. h is a
point on the horizontal trace of the cone. Another point will be at a
distance below 8 equal to sh. By taking different positions for the line
t'r' any number of points on the horizontal trace of the cone may be
found.

The base of the cone whose axis is vertical will be its horizontal
trace and will be a circle whose centre is v and diameter equal to a'h'.

The horizontal traces of all planes which touch the two cones will

Fig. 448.

STRAIGHT LfNE AND PLANE

22^

be tangents to the horizontal traces of these cones. OQ is one of these
tangents meeting XY at Q. Through v draw vp parallel to OQ to

/:\\

/l\\

V. .^

V\^k^

^^'

Y '

h\\ \i

)¥

I ^

1 \y

N /.

■ Y

-^f

k '

» V

Fig. 449.

leet XY at p. From p draw pp perpendicular to XY to meet a line
through v' parallel to XY at p'. Join p'Q. The plane p'QO will satisfy
the conditions of the problem.
If the horizontal trace OQ does
not meet XY within a convenient
distance, the vertical trace may-
be found by the construction shown
in Fig. 419, p, 214.

I'he drawing of the conic section
which is the horizontal trace of
the cone whose base is on the given
plane may be avoided by the fol-
lowing construction. Conceive a
sphere to be inscribed in this cone.
The elevation of this sphere will
be a circle touching the lines v'd
and v'd\ as shown in Fig. 450, its
centre being on v'e'. Next con-
ceive a cone to envelop this sphere
and have its axis vertical and its
base angle equal to a. u' is the
elevation of the vertex of such a
cone and the circle shown whose
centre is w is its horizontal traee, p ,k^

228

PRACTICAL GEOMETRY

A plane which touches the cones whose vertices are at vv' will also
touch the cone whose vertex is at uu', and the horizontal trace of the
plane will be a tangent to the two circles shown in the plan. A little
consideration will show that if the horizontal trace of the required
plane is be a cross tangent to the circles the vertices vv' and uv! must
be on opposite sides of the horizontal plane of projection as shown in
Fig. 450.

In the particular case where the required plane is to be perpen-
dicular to the given plane the cone whose base is on the given plane
will become a straight line perpendicular to the given plane, and
the horizontal trace of the required plane will pass through the hori-
zontal trace of this line and touch the base of the other cone.

195. To draw the Projections of a Line which shall pass
through a given Point and meet a given Line at a given Angle.
— Let jpp' (Fig. 451) be the given
point and aZ>, a'h' the given line.
Determine (Art. 172, p. 211) the
plane LMN which contains the
point P and the line AB. Rabat
this plane with the point P and
line AB into the horizontal plane
(Art. 189, p. 221). Pi and A^fc are
the rabatted positions of P and AB
respectively. Through Pi draw
PiCi making with A^ an angle
equal to the given angle. Through
Ci draw CjC perpendicular to MN
to meet ah at c. Draw cc perpen-
dicular to XY to meet ah' at c'.
The lines pc and p'c' are the pro-
jections of the line required.

If the required line is to be perpendicular or parallel to the given
line there is only one solution, in other cases there will be two
lines which will satisfy the given conditions. In Fig. 451 only one
of the two lines which may be drawn to satisfy the given conditions
is shown.

196. To find the Traces of a Plane which shall contain
one given Line and make a given Angle with another given
Line. — Let AB denote the first given line and CD the second.

Determine the projections of a line EF which shall be parallel to
CD and meet AB at a point E.

Take E as the vertex and EF as the axis of a cone whose semi-
vertical angle is equal to the given angle. Find as in Art. 194 the
horizontal trace of this cone. A tangent from the horizontal trace of
AB to the horizontal trace of the cone will be the horizontal trace of
the plane required. Having found the horizontal trace of the plane,
its vertical trace may be determined from the condition that the plane
is to contain the line AB.

If the horizontal trace of AB falls outside the curve which is thQ

Fig. 451.

STRAIGHT LINE AND PLANE

229

horizontal trace of the cone, there will be two solutions of the problem ;
if it falls on the curve there will be one solution only, and if it falls
inside the curve there can be no solution.

197. To draw the Traces of two Planes which shall be
perpendicular to one another and have given Inclinations.
— Let the given inclinations be o\vi'
to the horizontal plane and let n\«<^L
them be a and ^. Draw MN
(Fig. 452), the horizontal trace
of one of the planes, at any angle
to XY (preferably a right angle),
and determine its vertical trace
LM from the condition that the
inclination of the plane is the
given angle a. Take any point
pj} in this plane (preferably in
its vertical trace) and determine
a line pq^ p'q' perpendicular to
the plane. Any plane which
contains this perpendicular will
be perpendicular to the plane
LMN. Determine also a cone
having its vertex at pp', its base
on the horizontal plane and
having a base angle y8 the given inclination of the second plane. A
line through the horizontal trace of pq, p'q to touch the base of this
cone will be the horizontal trace ST of the second plane. The vertical
trace RS of the second plane is easily found from the condition that
the plane is to contain the line pq, p'q'.

198. To draw the Traces of three Planes which are
each perpendicular to the other two, having given the
Inclinations of two of the Planes. — Determine by the preceding
Article the traces of the two planes whose inclinations are given.
Next determine the line of intersection of these two planes. The
third plane will be perpendicular to this line and its horizontal and
vertical traces will be perpendicular to the plan and elevation of the
line respectively.

199. Projections of a Solid Right Angle. — A solid right
angle is formed by three planes which are mutually perpendicular.
These three planes will intersect in three straight lines which are each
perpendicular to the other two, and the solid angle is represented on
paper by the projections of these three mutually perpendicular straight
lines.

Any three intersecting straight lines on the paper may be taken
as a projection of three lines which are mutually perpendicular. Let
oa, ob, and oc (Fig. 453) be the plans of three lines which are mutually
perpendicular.

Since each line is perpendicular to the other two it will be per-
pendicular to the plane containing the other two, and therefore the

230

PRACTICAL GEOMETRY

plan of each will be perpendicular to the horizontal trace of the plane
of the other two. Also the horizontal trace of the plane containing
any two of the lines will pass through the horizontal traces of these
lines.

It is evident that the form of any elevation of the lines will not be
affected by the position of the horizontal plane of projection ; the
horizontal plane of projection may therefore be taken at any level.

Take any line ab, perpendicular to oc, as the horizontal trace of the
plane containing the lines OA and OB. A perpendicular from h to oa
to meet oc at c will be the horizontal
trace of the plane containing OB
and OC, and ac will be the hori-
zontal trace of the plane of OA
and OC.

Produce co to meet ah at d, and
consider od as the plan of a line
lying in the plane AOB. Since OC
is perpendicular to the plane AOB,
the angle COD will be a right
angle.

Conceive the triangle COD to
revolve about CD as an axis until it
comes int^ the horizontal plane.
The point will describe an arc of
a circle whose plan will be a line
through perpendicular to cd, and
the point O will, when brought into
the horizontal plane, occupy such
a position that the lines drawn from
it to c and d will contain a right angle. Hence on cd describe a
semicircle to cut the line oO^ perpendicular to cd at O^. oO, will be
the distance of the point O from the horizontal plane containing the
points A, B, and C. Having found this distance, an elevation of the
lines on any vertical plane can be drawn. In the figure an elevation
is shown on XY.

This problem is one of great importance in connection with iso-
metric and axometric projections and it is further considered in
Chap. XXI.

200. To find a Point on a given Plane such that the sum
of its Distances from two given Points shall be the least
possible. — Let LMN (Fig: 454) be the given plane and aa', hh' the
given points. These given points are on the same side of the given
plane. Determine the projections of a line to pass through aa' and
be perpendicular to the plane LMN. Pind the point cc' where this
perpendicular meets the plane LMJST. Determine a point dd' in
ac, a'c' produced such that CD is equal to AC. The projections d
and d' of D are such that cd = ac, and cd' = a'c. Join hd and b'd'.
Find the point ee where the line hd, h'd' intersects the plane LMN.
ee' is the point required.

Fig. 453.

STRAIGHT LINE AND PLANE

231

LMN (Fig. 455) be the given plane

of the line AB on the plan ah as a
construction of Art. 16, p. 20, the

The student should have no difficulty in proving the correctness of
the above construction after referring to the corresponding problem in
plane geometry discussed in
Art. 9, p. 8.

It is easy to show that the
lines ae, a'e' and he, h'e' are
equally inclined to the plane
LMN, and hence if a ray of
light from aa' impinges upon
the plane LMN and on re-
flection from the plane passes
through hh', the incident ray
must strike the plane at ee'.

201. To find the Locus
of a Point which moves on
a given Plane so that the
Ratio of its Distances from
two given Points shall be
equal to a given Ratio. — Let
and aa' and hb' the given points.

Draw a new elevation a/?>/
ground line. Determine by the
circle whose centre is o^' and
radius o/c?/ which is the locus of
a point which moves in the plane
of the new elevation and whose
distances from a/ and />/ are to

ratio.
^. ... ..... ..... .. 2:1.

The point o/ is in ai'6/ produced.
Find o the plan and o' the eleva-
tion on XY of the point of which
o/ is the elevation on ah as a
ground line. The surface of a
sphere whose centre is at oo and
whose radius is o/(Z/ will contain
all points in space whose distances
from aa' and hh' are to one an-
other in the given ratio. The
circle which is the section of this
sphere by the plane LMN is the
locus required. In Fig. 455 the
plane LMN being perpendicular
to the vertical plane of pro-
jection, the circle which is the
section of the sphere by the plane

LMN will have the straight line c'd' for its elevation and the ellipse cd
for its plan.

20.2. To determine a Line which shall pass through a given

one another
In Fig. 455

in the given
this ratio is

Fig. 455.

232

PRACTICAL GEOMETRY

Point and intersect two given Lines.^ — Let P be the given point
and AB and CD the given lines.

First method. Determine the traces of the plane containing the
point P and the line AB. Determine also the traces of the plane
containing the point P and
the line CD. The line of
intersection of these two
planes is the line required.

Second method. Fig. 456.
Determine the projections of
two lines, PKL and PMN,
intersecting AB at points K
and M. Let these lines meet
the vertical plane containing
CD at the points L and N.
The line LN will be the in-
tersection of the plane con-
taining P and AB with the
vertical plane containing CD.
Let LN meet CD at Q. PQ
is the line required.

203. To determine the common Perpendicular to two
given Lines which are not in the same Plane. — Let AB and CD
(Figs. 457 and 458) be the two given lines. Determine a line FEH

Fig. 456.

Fig. 457

Fig. 458.

to intersect AB at E and be parallel to CD. The plane containing
AB and FEH will be parallel to CD. The traces of this plane are
KL and MN.

STRAIGHT LINE AND PLANE 233

Take any point P in CD and determine the normal PQ to the
plane KLMN, meeting the plane at Q. The length of PQ is the
shortest distance between the lines AB and CD. Determine a line
through Q parallel to CD to meet AB at R. This line will lie in the
plane KLMN. A line RS parallel to QP and meeting CD at S will
be the common perpendicular to the lines AB and CD.

It is evident that the line QR, produced both ways, is the projection
of CD on the plane KLMN.

204. To determine the Centre and Radius of a Sphere the
Surface of which shall contain four given Points. — It should
first be observed that no three of the given points must be in the same
straight line, and if all four lie in the same plane they must lie on the
circumference of a circle. In the latter case the problem is indeter-
minate, as any number of spheres can be found on the surface of which
the points will lie.

Let A, B, C, and D be the given points. Determine by the con-
struction of Art. 183 a plane to bisect the line AB at right angles.
Since every point in this plane is equidistant from the points A and B
it must contain the centre of the required sphere. In like manner
the centre of the sphere must lie on the planes bisecting BC and CD
at right angles. Therefore the point of intersection of these three
planes must be the centre required, and the distance of this point from
any one of the given points will be the radius of the sphere.

As there are four given points, six lines can be got by joining them,
and the point of intersection of three planes bisecting at right angles
any three of these six lines not lying in the same plane will be the centre
of the sphere.

Another metlwd of finding the centre of the sphere is to determine
the centre of the circle containing three of the given points and then
determine the normal to the plane of this circle through its centre.
This normal will contain the centre of the required sphere. Taking
another three of the points and proceeding in the same way another
normal is found which contains the centre of the sphere, and the point
of intersection of these two normals is the centre required.

205. Trihedral Angles — Spherical Triangles. — The angle
formed by three planes meeting at a point is called a trihedral angle.
The angles between the lines in which the three planes intersect are
the sides or faces, and the dihedral angles between the planes are the
angles of the trihedral angle. Referring to Pig. 459, the three planes
OAB, OBC, and OCA form at O a trihedral angle. The plane angles
AOB, BOC, and COA are the sides or faces of the trihedral angle
referred to, and the dihedral angles between these faces are its
angles.

If a sphere be placed with its centre at 0, the three planes which
form the trihedral angle will intersect the surface of the sphere in
three great circles, which will intersect one another and form what
is called a spherical triangle. In Fig. 459, ABC is a projection of a
spherical triangle.

The sides and angles of a spherical triangle are the sides and

234

PRACTICAL GEOMETRY

Fig, 459.

angles of the trihedral angle which it subtends at the centre of the
sphere on the surface of which it is drawn. Hence the solution of a
spherical triangle is the solution of a trihedral angle.

If from a point S within
the trihedral angle (Fig.
459), perpendiculars SP,
SQj and SR, be let fall on
the faces OBC, OCA, and
OAB respectively of the
trihedral angle, these per-
pendiculars will form the
edges of another trihedral
angle S, which has the
remarkable properties that
its angles are the supple-
ments of the sides of the
trihedral angle O, and its
sides are the supplements
of the angles of the trihe-
dral angle O. The trihedral
angles and S are there-
fore called suj)plementary
trihedral angles, either be-
ing the supplementary trihedral angle of the other.

If from the point O (Fig. 459) perpendiculars OA', OB', and OC
be drawn outwards from and to the faces OBC, OCA, and OAB
respectively to meet the surface of the sphere at A', B', and C, then
A', B', and C are poles of the great circles BC, C A, and AB respectively,
and great circles of the sphere through A' and B', B' and C', and C
and A' form another spherical triangle A'B'C which is called the polar
triangle of the triangle ABC. It is obvious that the trihedral angle
subtended by A'B'C at the centre of the sphere is equal in every
respect to the trihedral angle at S. Hence the polar triangle A'B'C'
and the triangle ABC are supplementary.

The following notation is used to denote the elements of a trihedral
angle or spherical triangle. The three angles are denoted by the
capital letters A, B, and C, and the three sides by the italic letters a,
h, and c, the sides a, b, and c being opposite to the angles A, B, and C
respectively. The elements of the supplementary trihedral angle or
the polar triangle are denoted by the same letters accented ; thus the
angles are A', B', and C, and the sides a', V, and c'.

The relations between the elements of two trihedral angles which
are supplementary or between a spherical triangle and its polar triangle
are expressed by the following equations, all angles being measured in
degrees —

A' = 180 - a a' = 180 - A

B' = 180-6 6' = 180 - B

C = 180 - c c' = 180 - C

From the above equations the following are obtained —

STRAIGHT LI:N^E AND PLANE

235

a = 180 - A' A = 180 - a'

6 = 180 - B' B = 180 - V

= 180 - C C = 180 - c'

Any three of the six elements of a trihedral angle or spherical
triangle being given, the other three can be found. There are in all
six cases to consider. — I. Given the three sides. II. Given two sides
and the angle between them. III. Given two sides and the angle
opposite to one of them. IV. Given the three angles. V. Given one
side and the two adjacent angles. VI. Given one side, an adjacent
angle, and an opposite angle.

By means of the properties of the supplementary trihedral angle,
cases IV, V, and VI may be reduced to cases I, II, and III respectively.
Thus, in solving case IV, find by the solution of case I the angles of
the trihedral angle whose sides are the supplements of the given angles,
then the supplements of the angles found will be the sides required.
Or, using symbols, given A, B, and C, then a' = 180 — A,
6' = 180 — B, and c' = 180 — C; now find, by the construction for
case I, A', B', and C', then a = 180 - A', & = 180 - B', and
c = 180 - C

In case V suppose a, B, and C given, then A' = 180 — a,
6' = 180 — B, and c' = 180 — C. Now, by the construction for case
II, find a', B', and C', then A = 180 - a', 6 = 180 - B', and
c = 180 — C. In like manner case VI is reduced to case III.

The solutions of cases I, II, and III will now be given.

Case I. Given the three sides a, h, and c of a trihedral angle to
determine the three angles A, B, and C. Construct the three sides a, h,
and c, as shown in Fjg. 460, forming the development of the faces of

Fig. 460.

the trihedral angle. Consider this development to be lying on the
horizontal plane. Mark off equal lengths ONj and ON.^ on the outer
lines of the development. Draw IS^n perpendicular to OL, and N.^w

236 PRACTICAL GEOMETRY

perpendicular to OM to meet N^n at n. Join On. The three lines
OL, OM, and 0/i will form the plan of the trihedral angle, the face h
being on the horizontal plane. In finding the point n the faces a and
c have been supposed to turn about the edges OL, and OM respectively
until the lines ONi and ON^ coincide. It is evident that the point N^
describes an arc of a circle whose plan is the straight line N^n ; also,
the point N2 describes an arc of a circle whose plan is the straight line
N2W ; and since N^ and Ng are equidistant from O, n must be the plan
of the point where they meet, and On is the plan of the edge between
the faces a and c.

To determine the angles A and C, draw ?iw/ perpendicular to NjW,
and nn.J perpendicular to N3W. With Q as centre, and QNi as radius,
describe an arc to cut w?^/ at ??/, and with R as centre, and RNo as
radius, describe an arc to cut nnj at nj. Join w/Q and n^R. w/Qn is
the angle C, and w^'Rw is the angle A. It is evident that when the
face a is brought into its natural position, the line N^Q on that face
will remain perpendicular to OL ; also the line wQ on the face h is
perpendicular to OL ; therefore the angle w/Qw, which is the angle
between these lines, will be equal to the angle between the faces a
and b, that is, the angle C. In like manner the angle n-^Hn is equal
to the angle A.

To find the angle B, draw N^L perpendicular to ON^ to meet OL
at L, and draw N2M perpendicular to ON^ to meet OM at M. With
centre L, and radius LNj, describe an arc to cut On at N. Join N to
L and M. LNM is equal to the angle B. It is evident that when
the faces a and c are brought into their natural positions, the lines
LNi and MNg lying on these faces will be perpendicular to their line
of intersection, and therefore the angle between them, which is
obviously equal to the angle LNM, will be equal to the angle B.

The student will more clearly understand the foregoing constructions
if, after drawing the figure shown, he cuts it out along the lines LN^,
NjO, ON2, and N.^M, and then folds the triangles LN^O and MN.p
about OL and OM respectively until the lines ON^ and ON^ meet one
another.

Case II. Given two sides
of a trihedral angle, and the
angle betioeenthem, to determine
the third side and the other
angles. Let a and h be the
given sides, and C the angle
between them. Construct the
two sides a and h on the hori-
zontal plane as shown in Fig.
461. From a point Nj in ONj
draw NjQw perpendicular to
OL and intersecting it at Q.
Draw Qw/ making the angle
w/Qn equal to C. Make Q^i'
equal to QNj, and draw w/w perpendicular to Nin to meet the latter at ».

STRAIGHT LINE AND PLANE

237

Join On. The lines OL, OM, and On will form the plan of the trihedral
angle, the face b being on the horizontal plane.

To determine the side c, draw nNa perpendicular to OM, and with
centre O and radius ONj describe an arc to cut wN.2 at N^. Join ON2.
N2OM is the side c brought into the horizontal plane.

The angles A and B may now be determined exactly as in case I.
If l^e student has understood the constructions in case I, he should
have no difficulty in demonstrating the above constructions for
case II.

Case III. Given two sides of a trihedral angle, and the angle opposite
to one of them, to determine the third side and the other angles. Let a an^
h be the given sides

and A the given angle. ^^

As before, construct
the given sides on the
horizontal plane as
shown in Fig. 462.
Take a ground line
XY at right angles to
OL, cutting OL at Q,
ONi at Ni and OM at
M. From Q draw QT
at right angles to OM
meeting it at T. With
Q as centre, and QT as
radius, describe the arc
TT2 to meet XY at T2.
Draw T2V making the
angle QT2V equal to the angle A and meeting OL at V. Join MV.
MV is the vertical trace of the face c, OM being its horizontal trace.
Now let the face a revolve about OL. Since OL is perpendicular to
XY the elevation of the circle described by Nj will be a circle having
a radius equal to QNj, with Q as centre. Let this circle cut MV at n'.
Join n'Q. The line n'Q, is the vertical trace of the face a when it is
in its natural position, and OQ is its horizontal trace ; therefore, if
a perpendicular be let fall from n' to XY to meet XY at w, the line Ow
will be the plan of the intersection of the faces a and c. Hence the
lines OL, OM, and Ow form the plan of the trihedral angle, the face h
being on the horizontal plane. The third face c may now be deter-
mined as in case II and the angles B and C may be found as in
case I.

The arc Njw' may meet the line MV at two points (it does so in
Fig. 462), and there are then two solutions, that is, there are two
trihedral angles which satisfy the conditions of the problem. This is
then called the amhiguous case.

206. Circumscribed and Inscribed Circles of Spherical
Triangles. — A plane containing the three angular points A, B, and
C of a spherical triangle (Fig. 463) will intersect the surface of the
sphere in a circle which is the circumscribed circle of the spherical

Fig. 462.

238

PRACTICAL GEOMETRY

triangle. A right circular cone, having its vertex at O, the centre of

the sphere, and having its curved surface touching the three faces

of the trihedral angle subtending _~-..

the spherical triangle at O, will

intersect the surface of the

sphere in a circle which is the

inscribed circle of the spherical

triangle. The inscribed cone is

determined by first drawing a

sphere inscribed in the trihedral

angle. The cone envelops this

sphere.

Exercises XVIb

1. Find the angle between the
pairs of planes shown at (a), (&), (c),
and (d), Fig. 464. Show also in each
case the traces of the plane which
bisects the angle between the given
planes.

Fig. 464.

2. The horizontal and vertical traces of a plane (1) make angles of 90^ and
45'^ respectively with XY. A line inclined at 40" to the vertical plane of projec-
tion lies in this plane and the distance between its horizontal and vertical traces
is 3 inches.- This line is the intersection of a plane (2) with the plane (1), the
inclination of the plane (2) to the horizontal plane being 60°. Draw the traces of
the planes (1) and ^2) and find the angle between the planes.

3. OA and OB (Fig. 465) are two straight lines lying on the horizontal plane.
OC is another line of which Oc is
the plan. The inclination of 00
to the horizontal plane is 50-. De-
termine the angle between the
planes AOC and BOG, also the angle
between the planes AOC and AOB,
and the angle between the lines
OA and 00.

4. ab (Fig. 466) is the plan of a
line lying in a plane of which H.T. is the horizontal trace. The inclination of
AB to the horizontal plane is 50°. Draw the traces of second plane to contain
AB and make an angle of 60" with the first plane.

5. The vertical and horizontal traces of a plane (1) make angles of 40° and 45°
respectively with XY. A point in the vertical trace of this plane and 1-5 inches

Fig. 465.

Fig. 466.

* STRAIGHT LINE AND PLANE

239

above XY is the elevation of a line AB. Draw the tra-ces of a plane (2) to contain
the line AB and make an angle of 75*^ with the plane (1).

6. Draw the traces of a plane (1) which is inclined at 45^ to the horizontal
plane and is perpendicular to the vertical plane of projection. Then draw the
traces of a plane (2) which is inclined at 65° to plane (1) and inclined at 75° to
the horizontal plane.

7. The traces of a plane LMN are both inclined at 45° to XY. A point P in
this plane is 2 inches above the horizontal plane and 1 inch in front of the vertical
plane. Draw the projections of a line passing through the point P, inclined at
30° to the horizontal plane and inclined to the plane LMN at 60°.

8. (the is the plan of a triangle, ah = 2-1 inches, be = 2*9 inches, and
ca = 1-4 inches. The heights of the points A, B, and C above the horizontal
plane are, 0, 2*5, and 1-25 inches respectively. Find the plan of a point D in AB
such that the angle ADC is 50°.

9. A line AB of indefinite length passes through a point G which is in the
vertical plane of projection and 2 inches above XY. AB is inclined at 40° to the
horizontal plane and its plan makes 30° with XY. A point P is 1*5 inches in
front of the vertical plane and 1 inch above the horizontal plane, and its plan is
at a perpendicular distance of 1-8 inches from the plan of AB. Draw the plan
and elevation of an equilateral triangle which has one angular point at P and the
opposite side on AB.

10. aMN6 is a quadrilateral, the angles at M and N being right angles.
MN = 1-5 inches, aM = 1 inch, and 6N = 2*5 inches, ab is the plan of a horizontal
line which is 0*8 inch above the horizontal plane. MN is the horizontal trace
of a plane which contains the point A. Draw the plan of a line which lies in the
plane AMN and makes an angle of 75° with AB.

11. Draw the traces of a plane which shall contain the line AB (Exercise 17)
and be inclined to the line CD at an angle of 15°.

12. One plane is inclined at 60° to the horizontal plane and 45° to the vertical
plane. Another plane is perpendicular to the first and inclined at 70° to the
horizontal plane. Draw the traces of the two planes.

13. The horizontal and vertical traces of a plane are in one straight line
inclined at 40° to XY. A second plane is perpendicular to the first and inclined
at 60° to the horizontal plane. Draw the traces of the two planes.

14. Three planes are mutually perpendicular. One is inclined at 50° and
another at 45°. Find the inclination of the third plane.

15. H.T. (Fig. 467) is the horizontal trace of a plane which contains the point
whose plan is c. The height of C above the horizontal plane is 1 inch, a and b
are the plans of two points whose heights above the horizontal plane are 1*5 inches
and 2 inches respectively. Find the projections of a point P on the given plane
such that AP + BP shall be a minimum.

-Y

ftf

Fig. 467.

Fig. 469.

16. Taking the same plane and the same points as in the preceding exercise,
draw the projections of the locus of a point P which moves on the plane in suck
a manner that the ratio of AP to BP shall be equal to the ratio of 8 to 2.

240

PRACTICAL GEOMETRY

17. ah and cd (Fig. 468) are the plans of two straight lines. The heights of
the points A, B, C, and D above the horizontal plane are, 1'5, 0-5, 1-0, and 2-5
inches respectively. Draw the plan and elevation of the common perpendicular
to the lines AB and CD and find its true length.

18. a, b, c, and d (Fig. 469) are the plans of four points. The heights of
A, B, C, and D above the horizontal plane are, 2 5, 0'25, 2-6, and 0-6 inches
respectively. Determine the plan and elevation of the sphere whose surface
contains the points A, B, C, and D.

19. Taking the lines given in exercise 17 and Fig. 468, determine the locus of
a point which moves on the horizontal plane so as always to be equally distant
from these lines.

20. The common perpendicular EF to two straight lines AB and CD is 1 inch
long and its inclination is 40^. The inclination of AB is 35^ and the angle
between the plans of AB and CD is 90°. Draw the plans of AB, CD, and EF, and
find the inclination of CD.

21. A person standing in the corner of a quarry notices that on one vertical
face the line of a particular bed slopes downwards from the corner at an angle of
30^ with the horizontal, whilst on an adjacent vertical face the corresponding
angle is 48°. The angle between the two facps is 110°. What is the true dip or
inclination of the strata ? [b.e.]

22. Two inclined faces of rock A, B in a quarry intersect in L. The lines of
steepest ascent on A make 10° with the vertical and go due North. Those on B
make 20° and have a direction 12° North of East. A seam of coal intersects these
faces in lines which both go upwards from L, making 70° with L on the face A
and 65° with L on the face B. Find the direction, and inclination to the
horizontal, of the lines of st'eepest descent of the seam. [b.e.j

23. The plans of four points O, A, B, and C are given in Fig. 470. O is the
centre of a sphere of 2*5 inches radius. A, B,
and C lie on the surface of the sphere, A and C
being above, and B below the level of O. Draw
the plan of the spherical triangle ABC,* and an
elevation on a ground line parallel to ox. Add
the plan and elevation of the circumscribed
circles of the spherical triangle. Lastly, determine
the plan and elevation of the polar triangle.

24. Solve, by construction, the following
trihedral angles or spherical triangles. The
results, as obtained by calculation, correct to the
nearest tenth of a degree, are added.

(1) Given, a = 40°, 6 = 50°, c = 60°.

Results. A = 47-9°, B = 62-2°, C = 89-1°.

K-r

o-2L^^

.2-3- -

0-3

I I
I I

Fig. 470.

(2)
(3)

Given,
Results.
Given,
Results.

(4) Given,
Results.

(5) Given,
Results.

or,

(6) Given,
Results.

or,

(7) Given,

a = 70°, b = 80°, c =
A = 69-7°, B = 79-4°,
a = 45°, b = 55°, C =
A = 58-4°, B = 80-7°
a = 60°, 6 = 70°, C •=
A = 50-5°, B = 56 9°,
a = 45°, b = 56°. A =

90°.

C = 93-7°.
60°.

c = 46°.
120^
c = 103-6".

:45°.

c = 75-1°, B = 55°, C = 104-9°,
c = 15-5°, B = 125°, C = 15-5°.
a = 60°, A = 70°, C = 90°.
b = 39-1°, c =-- 67-2°, B = 43-2°,
b = 140-9°, c = 112-8°, B = 136-8°.
A = 120°, B = 45°, C = 35°.

Results, a = 78-7°, b = 53-2°, c = 40*5°.
(8) Given, A = 130°, B = 120°, c = 90°.

Results, a = 126°, b = 113-9°, C = 108-7°,

CHAPTER XVII

SECTIONS OF SOLIDS

207. Sections of Solids. — Many objects of which mechanical
drawings have to be made are of such a form that their construction
is not completely shown by outside views only. The construction of
the interior of a house, for instance, cannot be seen from the outside.
In order to exhibit the interior of such an object it is imagined to be
cut in pieces by planes and these pieces are then represented separately.
But in representing an object of comparatively simple form, the
addition of a sectional drawing of it often adds very much to the illus-
tration of it, although such a sectional drawing may not be absolutely
necessary for the complete representation of the object.

That surface which is produced when a plane cuts a solid is called
a section, and if that part of the solid which is below or behind the
cutting plane is also shown on the projection of a section the projection
is called a sectional plan or a sectional elevation. But in the application
of these terms to architectural and engineering drawings the term
section is often used in the same sense as sectional plan or sectional
elevation.

The projection of a section is distinguished in various ways, one
way, used in this work, is by drawing across it parallel diagonal lines
at equal distances apart. These lines are called section lines.

If the true form of a section is required it must be projected on a
plane parallel to that of the section.

In this chapter sections of solids having plane faces are considered.
Sections of the sphere, cylinder and cone are considered in chapter
XVIII.

208. Section of a Prism by a Plane perpendicular to one
of the Planes of Projection. — A plane section of a prism will be a
rectilineal figure whose angular points are at the points where the
plane of section cuts the edges of the prism. Since the plane of section
is perpendicular to one of the planes of projection one of the projections
of the section will be a straight line coinciding with the trace of the
plane of section on the plane of projection to which the plane of section
is perpendicular.

The determination of sections of a right square prism is illustrated
by Fig. 471. The plan (1) and the elevation (2) represent the prism

242

PRACTICAL GEOMETRY

when standing with one end on the horizontal plane, PQ is the
horizontal trace of a vertical plane of section. This plane of section
cuts two of the vertical faces
of the prism in vertical lines of
which the points a and b are
the plans, and a'a' and b'h' per-
pendiculars to XY, the eleva-
tions. The same plane inter-
sects the ends of the prism in
horizontal straight lines of
which ah is the plan and a'h'
and a'h' the elevations. The
rectangle a'b'h'a' is the com-
plete elevation of the section.

RS is the vertical trace of
another plane of section. This

second plane of section is per- Fig. 471.

pendicular to the vertical plane

of projection. On the plan (1) this second section appears as the figure
cdec. Another plan (3) of this second section is shown on a ground
line XiYj parallel to RS. The plan (3) shows the true form of the
section of the prism by the plane RS. The whole solid is also shown
in plan (3), the part above the section by the plane RS being repre-
sented by thin dotted lines

209. Section of a Pyramid by a Plane perpendicular to
one of the Planes of Projection. — The first paragraph of the pre-
ceding Article on the section
of a prism applies also to the
section of a pyramid, and all
that need be done here is to
give an example which the
student should work out.

In Fig. 472, (1) is the plan
and (2) the elevation of a right
hexagonal pyramid when stand-
ing on the horizontal plane
with two sides at right angles
to XY. (3) is a plan of the
pyramid when one triangular
face is horizontal. The plan
(3) is projected from the eleva-
tion (2) as shown. (4) is an
elevation projected from (3) on
a ground line X0Y2 parallel to
XY. The straight" line PQ is
the horizontal trace of a plane

Fig. 472.

of section which is perpendicular to the plane of the plan (3). PQ
is parallel to XY and passes through c, the plan (3) of one angular
point of the base of the pyramid. The section of the pyramid by the

SECTIONS OF SOLIDS

243

plane PQ is shown projected on to each of the other views of the
pyramid, and in each view the part of the solid between the vertex
and the section is shown by thin dotted lines. The side of the hexa-
gonal base may be taken 125 inches long.

210. Sections of Mouldings. — A moulding is an ornament of
uniform cross section which may be formed on a piece of a structure
or it may be a separate piece attached to the structure for ornament
only, or it may be the main part of the structure as in a picture frame.
Mouldings are of frequent occurrence in cabinet making, joinery, and
masonry.

The principal problem in the geometry of mouldings is the determina-
tion of the cross section of one moulding which will mitre correctly

Fig. 473.

Fig. 474.

with another when faces of the one are in different planes from corre-
sponding faces of the other.

Given the cross section of a moulding the determination of any other
section is a simple problem and is worked exactly as for a prism. In
Fig. 473, (1) is the true form of the cross section of a straight piece of
moulding. (2) is a plan of the moulding, the long edges being horizontal.
PQ is the horizontal trace of a vertical plane of section, and (3) is the
elevation of the section PQ on a ground line X, Yj parallel to PQ.

The same moulding is shown in Fig. 474. (1) is the cross section,
(2) is a side elevation, and (3) is a plan when the long edges are parallel
to the vertical plane of projection but inclined to the horizontal plane.
RS is the horizontal trace of a vertical plane of section, a'h' is the
projection of the section RS on the plane of the elevation (2), and a/fc/
is a projection of the same section on a plane parallel to it and which
therefore shows the true form of the section.

244

PRACTICAL GEOMETRY

In Fig. 474 the projections (1), (2), (3), and (4) have been drawn
in the order in which they are numbered, but a study of the figure will
show that if the section a/6/ by the plane RS be given instead of the
cross section (1) the latter may be found by working backwards.

In Fig. 475 the section a! is the cross section and a is the plan of a
straight piece of moulding whose long edges are horizontal and which
is fixed to the face of a vertical wall whose plan is rs. h is the plan
and h' the elevation of another straight piece of moulding whose long
edges are inclined to the horizontal plane and which is fixed to the
face of a vertical wall whose plan is st. The faces of the two walls are
at right angles to one another, and the two mouldings intersect in a
vertical plane, whose horizontal trace, or plan, is the straight line ns
which bisects the angle between sr and st. A moulding such as hh'

///////////////A

"T'^y.*-- -/»

Fig. 475.

Fig. 476.

whose long edges are inclined to the horizontal is called a raking
moulding.

The problem is now to find the true form of the cross section of the
raking moulding. The condition which the raking moulding must
satisfy is that its section by the plane ns of the joint with the other
moulding must be the same as the section of the other moulding by
that plane. From this condition it follows that the corresponding
longitudinal edges or lines on the two mouldings must intersect in the
plane of the joint. The projections of the longitudinal lines on the
raking moulding may therefore now be drawn and the cross section
c' determined as shown.

The case illustrated by Fig. 470 differs from that illustrated by
Fig. 475 in that the angle rat between the faces of the walls in Fig. 47G
is greater than a right angle. «,', the cross section of the horizontal

SECTIONS OF SOLIDS

245

Fig. 477.

Fig. 477 shows a curved

moulding, is shown rabatted into the vertical plane of projection. It
will be seen that in order that the elevations of the longitudinal lines
on the raking moulding may be drawn the elevation n's' of the joint
must first be determined.

Two straight 'mouldings, of the same cross section, may be mitred
correctly together if they are
attached to a plane surface, or if
they are attached to plane sur-
faces inclined to one another,
provided that the longitudinal
lines of the two mouldings are
perpendicular to the line of in-
tersection of the plane surfaces
to which they are attached.

Two curved mouldings of the
same cross section but of different
curvatures and attached to a
plane surface may be jointed to-
gether but the joint is a curved surface

moulding A jointed to a straight moulding B both having the same
cross section C. The two mouldings are supposed to be attached to
the plane of the paper. The projection of the joint on the plane of
the paper is the curved line DEF determined as shown.

211. Geometry of Rafters. — Practical problems on the sections
of prisms occur in timber construction, and illustrations will now be
given of these problems as applied to a timber roof frame.

Fig. 478 shows plan and elevation line diagrams of part of what is
called a hipped roof. The various members of the frame of this roof
are mainly of rectangular cross section. vUj v'u' is the ridge piece, vr, v'r'
is the hip or hip rafter which passes from one end of the ridge piece to
the top corner of two intersecting walls, cc?, c'd' is one of the jacic
rafters which lie between the hip rafter and the wall plate on the top
of one of the walls, ef ef is one of the common rafters which lie
between the ridge piece and a wall plate.

The dihedral angle between the two roof surfaces intersecting in
the line VR is determined by the construction of Art. 191, p. 224, or
by that construction slightly modified as shown here, vv^' is drawn
at right angles to vr and is made equal to the height of v' above /.
Joining 7;/r determines the true length of VR. mn is drawn at right
angles to vr to meet the horizontal traces of the roof surfaces at the
level of R at m and n. mn is the horizontal trace of a plane at right
angles to VR. This plane intersects the roof surfaces whose line of
intersection is VR in two straight lines the angle between which is
the angle required, os' is drawn at right angles to v^'r and oS is made
equal to os'-. Joining S to m and n determines mSw the dihedral angle
required.

At h and k are shown two forms of the cross-section of the hip
rafter to an enlarged scale. TP and TQ are parallel to Sm and Sr?.
respectively. This determines the angle ^ which the carpenter requires

246

PRACTICAL GEOMETRY

in backing off* the top face of the rafter to bring it into the planes or
plane of the roof surface.

Fig. 479 shows in detail a plan (1), a side elevation (2) and a front
elevation (3) of the upper part of a jack rafter where it joins the hip
rafter. All the rafters have their side faces vertical, and XoY., is the
plan of the vertical face of the hip rafter to which the jack rafter is

Fig. 478.

Fig. 479.

attached. This end of the jack rafter requires to be bevelled in two
directions in order to lit against the face of the hip rafter. The angles
for these bevels are marked a and /?. The angle y8 is shown on the side
elevation (2). To determine a the upper surface of the rafter is brought
into a horizontal position by turning it about a horizontal axis through
B at right angles to XY, as shown in the plan (1) and elevation (2).
The true form of the end of the rafter is shown at (4).

Exercises XVII

1. The solid whose projections are given in Fig. 480 is cut in two by a vertical
plane whose horizontal trace is HT. Draw an elevation of the part to the left
of the plane of section on a ground line parallel to HT.

2. The solid whose projections are given in Fig. 481 is cut in two by a vertical
plane whose horizontal trace is PQ. Draw an elevation of the larger portion on
a plane parallel to the plane of section,

3. The solids whose projections are given in Fig. 482 are cut by a vertical plane

SECTIONS OF SOLIDS^

247

whose horizontal trace is RS. Draw an elevation of the portions whose plans
lie between RS and XY on a ground line parallel to RS.

4. The plan of a right square prism with a rectangular hole through it is given
in Fig. 483. A side of the square base is on the horizontal plane. Draw the

Jl\

^>.

y. "^

■J

JT"

^H,^

LLI

Fig. 480. Fig. 481. Fig. 482. Fig. 483.

In rep7oducing the above diagrams take tlie small squares as of 0-3 inch side.

elevation on XY and show on this elevation the section of the solid by the vertical
plane of which HT is the horizontal trace. Determine also the true form of the
section.

5. A right square prism, side of base 1 inch, altitude 3 inches, is cut into two
equal parts by a plane. The true form of the section is a rhombus of 1-5 inches
side. Draw a plan of one part when it stands with its section face on the
horizontal plane.

6. A pyramid has for its base a square of 2 inches side which is on the
^horizontal plane. The altitude of the pyramid is 2*5 inches and the plan of the

vertex is at the middle point of one side of the plan of the base. This pyramid
is cut into two portions by a horizontal plane which is 1 inch above the base.
Draw the plan of the lower portion.

7. A right hexagonal pyramid, side of base 1 inch, altitude 2 inches is cut by
a plane which contains one edge of the
base and is perpendicular to the opposite
face. Determine the true form of the
section. ,

8. The plan and elevation of one of the
rectangular faces ABCD of a right hexa-
gonal prism are given in Fig. 484, AB and
CD being on the ends of the prism. Com-
plete the plan and elevation of the prism
and show on the elevation the section by
the vertical plane whose horizontal trace
isPQ.

9. The plan and elevation of one of the
triangular faces VAB of a right square
pyramid are given in Fig. 485, V being the
vertex of the pyramid. Complete the plan
and elevation of the pyramid and show on
the elevation the section by the vertical
plane whose horizontal trace is RS.

10. A horizontal moulding has (e), Fig. 486, for its cross-section. Determine
the true form of a vertical section which is inclined at 45° to the longitudinal
lines of the moulding.

11. A moulding whose cross-section is given at (/) in Fig. 486 is placed with
its back on the vertical plane of projection and its longitudinal lines inclined at
80° to the horizontal plane. Draw the plan and elevation and show the true
form of a section by a vertical plane whose horizontal trace is inclined at 60°
to XY.

12. Two mouldings of the same cross-section (c), Fig. 486, are fixed to the face

Fig. 484.

Fig. 485.

Tahe small squares 0'4 inch side.

I^RACTICAL GEOMETRY

of a vertical wall. One moulding is horizontal and the other rakes at 30° to the
horizontal. The two mouldings join correctly together. Determine the true
form of the joint section.

(a) (b) (c) (d) (c) if) (g)

Fig. 486.
In reproducing the above sections take the small squares as of \ inch side.

13. A horizontal moulding mitres correctly with a raking moulding. Inclina-
tion of raking moulding to the horizontal, 25°. The backs of the mouldings are
in vertical planes which are at right angles to one another as

shown in plan in Fig. 487. Select from Fig. 486 a cross- ^
section for the horizontal moulding and then determine the"
cross-section of the raking moulding.

14. Same as exercise 13 except that the angle between
the vertical planes is 120° instead of 90°.

15. Two raking mouldings on vertical walls which are at
right angles to one another as in Fig. 475, p. 244, mitre cor-
rectly. Both mouldings are inclined at 20^ to the horizontal.
Select a cross-section for one of the mouldings from Fig. 486
and then determine the cross-section of the other.

16. Same as exercise 15 except that the angle between the walls is 120°
instead of 90°.

17. A, C, and E (Fig. 488) are three pieces forming part of a timber frame,
each piece being of rectangular cross-section. The projections a' and c are
incomplete. Complete the projections indicated and draw an elevation on a
ground line inclined at 45° to XY.

18. A piece of timber having plane faces is shown in plan and elevation in
Fig. 489. Draw two other elevations, one on a ground line perpendicular to XY,
and the other on a ground line parallel to the longer sides of the plan. Determine
the true form of the cross-section of the piece.

Fig. 487.

\e, 1

',m^

^\\^\v\\\\\\\\\\\\\\\\\\\\''

^<;

Fig. 488. Fig. 489. Fig. 490.

In reproducing the above diagrams take the small squares as of half-inch side,

19. Fig. 490 shows the plan of a piece of timber A, of uniform square cross-
section, which lies between the faces RT and ST of two vertical walls. The
longitudinal edges of A are inclined at 20° to the horizontal, rising from R to S.
B is another piece of timber of uniform square cross-section. The longitudinal
edges of B are horizontal. The piece B fits into a notch on A, the greatest vertical
depth of this notch being equal to half the thickness of A. Draw two elevations
of A and B, one on a ground line parallel to rs, and the other on a ground line
parallel to rt. Determine all the bevels for the ends and the notch of the piece A.

CHAPTER XVIII

THE SPHERE, CYLINDER, AND CONE

212. Surfaces and Solids. — When a surface encloses a space
and that space is filled with solid matter the whole is called a soli<^,
and when the surfaces and solids have certain definite shapes they
Imve certain definite names. But in speaking of certain solids and
their surfaces the same name is frequently used to denote the solid
and its surface. For example, in speaking of the sphere, the cylinder,
and the cone the solids bearing these names may be referred to or it
may be that it is their surfaces that are referred to, although the
words " solid " or " surface " may not be used. In general the omission
of the words " surface of" when a solid is mentioned by name will not
load to any confusion when it is the surface that is referred to. For
example, by " the curve of intersection of two cylinders" is obviously
meant " the curve of intersection of the surfaces of two cylinders."

213. The Sphere. — The sphere may be generated by the revolution
of a semicircle about its diameter which remains fixed. The surface
of a sphere is also the locus of a point in space which is at a constant
distance from a fixed point called the centre of the sphere. The constant
distance of the surface of the sphere from its centre is the radius of
the sphere. A straight line through the centre of the sphere and
terminated by the surface is a diameter of the sphere.

The orthographic projection of a sphere is always a circle of the
same diameter as the sphere.

214. Plane Sections of a Sphere. — All plane sections of a
sphere are circles. When the plane of section contains the centre of
the sphere the section is a great circle of the sphere. All sections of
the sphere by planes which do not contain its centre are called small
circles.

A portion of a sphere lying between two parallel planes is called a
zone^ and a portion lying between two planes containing the same
diameter is Called a lune. See Fig. 491.

The determination of a plane section of a sphere is illustrated by
Figs. 492 and 493. In Fig. 492 the plane of section is perpendicular
to the vertical plane of projection and has V.T. for its vertical trace.
In Fig, 493 the plane of section is inclined to both planes of projection
and has V.T. for its vertical trace and H.T. for its horizontal trace.
In both cases points on the projections of the section may be found

250

PRACTICAL GEOMETRY

as follows. PQ parallel to XY is taken as the vertical trace of a
horizontal plane. This plane cuts the sphere"" in a circle whose plan is
a circle concentric with the plan of the sphere and whose diameter
is equal to the portion of PQ within the circle which is the elevation
of the sphere. This same plane cuts the given plane of section in
a straight line. The points rr in which the plan of this line intersects
the plan of the circle are the plans of points on the required inter-
section, and r'r' the elevations of these points are on PQ. In like manner
by taking other positions for PQ any number of points on the projections
of the required intersection may be found.

Since the required intersection is a circle and the projections of a
circle are ellipses the axes of these ellipses may be found and the
ellipses then be drawn by the trammel method (Art. 45, p. 41).

Referring to Fig. 492 a'a' the intercept of V.T. on the elevation of

; LUNE

Fig. 491.

Fig. 492.

the sphere is the elevation of the circle which is the section of the
sphere by the given plane. Projectors from a' and a! to meet a parallel
to XY through o the plan of the centre of the sphere determines aa
the minor axis of the ellipse which is the plan of the required section.
The major axis bisects the minor axis at right angles and has a length
equal to a!a'.

Referring to Fig. 493, draw aah through o at right angles to H.T.
aah is the horizontal trace of a vertical plane which intersects the
sphere in a great circle and the given plane of section in a straight
line, and A A the portion of this line within the great circle has for
its plan the minor axis of the ellipse which is the plan of the required
section. Let this vertical plane with the line AAB and the great
circle in it be turned round about a vertical axis through O until it is
parallel to the vertical plane of projection. AjAjBi is the elevation of
the line AAB in its new position and the elevation of the sphere is the

THE SPHERE, CYLINDER, AND CONE

251

elevatioQ of the new position of the great circle. This determines
the levels of the points A and A on the given plane of section and the
plans a and a are then readily found, aa is the minor axis of the
ellipse, the major axis of which is equal to AjAj. By a similar
construction, using a plane perpendicular to the vertical plane of
projection with its vertical trace through o' at right angles to V.T., the
axes of the ellipse in the elevation may be found. It should be
remembered that the major axes of these ellipses are parallel to the
horizontal and vertical traces respectively of the given plane of section
and the minor axes are perpendicular to these traces.

By taking a horizontal section of the sphere and given plane of
section through the centre of the sphere the points 8 and s where the
ellipse in the pla-n touches the circle which is the plan of the sphere
are determined (Figs. 492 and 493). These points determine the
limits of the portions of the required section which are on the upper
and lower halves of the sphere respectively. In like manner (Fig. 493)
a section of the sphere and given plane by a plane parallel to the
vertical plane of projection and through the centre of the sphere
determines n' and w' where the ellipse in the elevation touches the
circle which is the elevation of the sphere. And these points determine
the limits of the portions of the required section which are on the front
and back halves of the sphere respectively.

Fig. 494.

Fig. 495.

Useful exercises on the projection of plane sections of the sphere
are illustrated by Figs. 494 and 495. In Fig. 494 the sections, except
the middle one, are small circles of the sphere, while in Fig. 495 they are
all great circles. The different views in each Fig. should be drawn in
the order in which they are numbered. The sphere may be taken,
say, 2-5 inches in diameter.

Another useful exercise is illustrated by Fig. 496. This is a
hexagonal nut with a spherical chamfer. The various curves, other
than those representing the hole in the nut, are plane sections of a
sphere whose centre is O and radius R. In this example the various

252

PRACTICAL GEOMETRY

Fig

views are arranged according to the American system (p. 169). In
the elevations (2 ) and (3) the elliptic arcs a' and h' are usually drawn
as arcs of circles. The
arc c in the elevation
(3) is a true arc of a
circle whose centre is d.
In working this example
the following dimen-
sions may be taken. —
Diameter of screw, 1*75
inches; width of nut
across the flats, 2*75
inches; height of nut,
1'75 inches; radius of
spherical chamfer, 2*1
inches.

215. Projections

of Points on the Surface of a Sphere. — Suppose that one pro-
jection of a point on the surface of a sphere to be given and that it
is required to find the other. Take a plane parallel to one of the
planes of projection and containing the point. The section of the
sphere by this plane will have for one projection a circle and for the
other a straight line, and the given projection of a point will lie on
one of these and the other on the other.

216. Intersection of a Straight Line and a Sphere.— Take a
plane perpendicular to one of the planes of projection and containing the
given line. Draw a projection (A) of the section of the sphere by this
plane on a plane (B) parallel to it. Draw also on the plane (B)
another projection (C) of the given line. (A) is a circle which
intersects (C) at points which are projections of the points of inter-
section of the line and sphere. From these the projections of the
points of intersection on the original planes of projection may be
found.

217. The Cylinder. — The cylinder may be generated by a straight
line which moves in contact with a fixed curve and remains parallel
to a fixed straight line. If the fixed curve is a circle whose plane
is at right angles to the fixed straight line, the cylinder is a right
circular cylinder, and a straight line through the centre of the fixed
circle at right angles to its plane is the axis of the cylinder. The fixed
circle referred to above is a normal section of the right circular cylinder,
and its diameter is the diameter of the cylinder.

When a cylinder is limited in length by two planes which intersect
all positions of the generating line, the sections of the cylinder by these
planes become the ends of the cylinder, and when these ends are at right
angles to the generating line the cylinder is a right cylinder. In
speaking of a right circular cylinder, when its ends are considered, the
ends are generally understood to be at right angles to the axis. But
a right circular cylinder may have its ends inclined to its axis at an
angle which is not a right angle.

THE SPHERE, CYLINDER AND CONE

253

218. Projections of a Right Circular Cylinder. — Starting
with the cylinder in the position in which its axis is vertical, its plan
is a circle and its elevation a rectangle as shown at (1) and (2) in
Fig. 497.

Taking a second ground line XiY, inclined to rV the first elevation
of the axis of the cylinder, and projecting from (2), a new plan (3) is
obtained. In this new plan (3), rs, the plan of the axis of the cylinder,
is parallel to XiYi and at a distance from it equal to the distance of
y,s in (1) from XY. Also the plans of the ends of the cylinder are equal
ellipses whose major axes are at right angles to rs and equal in length
to the diameter of the cylinder. The minor axes bisect the major axes
at right angles and their lengths are found by projectors from (2) as
shown.

Taking a third ground line X2Y2 inclined to rs in the plan (3), r's',
a new elevation of the axis of the cylinder is first obtained, the
distances of r' and s' in (4) from XgYg being equal to the distances of

Fig. 497.

r' and s' respectively in (2) from XjY,. Before going further with the
elevation (4), take another ground line X^Y., parallel to rs in (4) and
draw rs a new plan of the axis of the cylinder as shown in (5). The
distances of r and s in (5) from X.^Y.^ are equal to the distances of r and
8 respectively in (3) from XgYg. The new plan (5) being on a plane
parallel to the axis of the cylinder will be a rectangle as in (2) and
may therefore be now drawn. The axes of the ellipses in (4) are
next determined by projectors from (5) as shown, just as the axes of
the ellipses in (3) were obtained by projectors from (2).

Instead of finding the axes of the various ellipses which are the
projections of the ends of the cylinder and then constructing the
ellipses in the usual way, a number of points on the ends may be pro-
jected as is shown for one point N.

If rs in (3) and r's in (4) be given and also the diameter of the
cylinder, it is obvious that by drawing the projections (2) and (5) the
projections (3) and (4) may be completed.

219. Plane Sections of a Right Circular Cylinder. — All

254

PRACTICAL CxEOMETRY

sections of a right circular cylinder by planes at right angles to its
axis are circles of the same diameter, and all sections of the curved
surface by planes parallel to the axis are parallel straight lines. All
sections of the curved surface by planes inclined to the axis are ellipses
whose centres are on the axis of the cylinder.

Referring to Fig. 498, PQ is the vertical trace of a horizontal plane
cutting the inclined cylinder whose axis SSj is parallel to the vertical
plane of projection. S and Sj are the centres of two spheres inscribed
in the cylinder and touching the plane of section at F and.Fj respec-
tively. The points F and Fi are the foci of the ellipse which is the
section of the cylinder by the given plane of section, a'a' the portion
of PQ lying within the elevation of the cylinder is the elevation of the
major axis of the ellipse and from this the plan aa, is projected. The

Fig. 498.

Fig. 499.

minor axis libx of the plan of the ellipse is equal to the diameter of the
cylinder.

Planes through S and Sj at right angles to the axis of the cylinder
intersect the given plane of section in straight lines XM and XiMj
which are the directrices of the ellipse (Art. 35, p. 32).

Fig. 499 differs from Fig. 498 in that the axis of the cylinder is
inclined to both planes of projection instead of being inclined to the
horizontal plane only. The intercept of J*Q within the elevation of
the cylinder does not now give the major axis of the ellipse, but the foci
and the minor axis are determined as before, but by making ca and ca^
each equal to hf, the major axis is determined.

The determination of the section of a cylinder, whose axis is

THE SPHERE, CYLINDER, AND CONE

255

inclined to both planes of projection, by an oblique plane is illustrated
by Fig. 500. AB is the axis of the cylinder and H.T. and V.T. are
the traces of the plane of section.

The ellipse uv, which is the
horizontal trace of the cylinder, is
the section of the cylinder by the
horizontal plane of projection and
it may be determined by the con-
struction shown in Fig. 499, or by
the construction shown in Fig.
503.

Taking mw, parallel to ah, as
the horizontal trace of a vertical
plane, this plane intersects the
given plane of section in a line of
which m'n' is the elevation. This
same vertical plane intersects the
cylinder in two straight lines
whose elevations c'd' and c/dfj' are
parallel to a'b', the points c' and c/
being projected from c and Cj, the
points where mn cuts the ellipse uv.
The points r' and s' where m'n
intersects c'd' and f/c?/ are the
elevations of points on the required
section. Projectors from r' and s

to meet mn determine points r and s on the plan of the required
section.

By taking other vertical planes parallel to the axis of .the cylinder
other points on the required
section may be found.

It should be noted that
all the lines of intersection of
the assumed vertical planes
with the given plane of section
will be parallel.

The student should work
out this example to the dimen-
sions marked on the figure.

220. Circular Sections
of a Right Elliptical
Cylinder. — A right elliptical
cylinder is one in which a
section by a plane at right
angles to the axis of the
cylinder is an ellipse. In Fig.

501, (?«) is the plan and (y) the elevation of such a cylinder, the axis
of the cylinder being perpendicular to the vertical plane of projection,
and the major axis of a section at right angles to the axis is vertical.

Fig. 500.

Fig. 601.

Fig. 502.

256

PRACTICAL GEOMETRY

The elevation (v) shows the true form of the normal section and is an
ellipse whose semi-major axis is o'a'.

Taking a point o on the plan of the axis as centre and a radius
equal to o'a' aji arc of a circle is described cutting the outlin-e of the
plan which is parallel to the plan of the axis at r and s. The straight
line rs is the horizontal trace of a vertical plane which will cut the
cylinder in a circle whose diameter is equal to the major axis of the
normal section. An elevation (w) of the circular section on a ground
line XiYi parallel to rs is shown. All sections of the curved surface
of the cylinder by planes parallel to one circular section are equal
circles.

The property of a right elliptical cylinder of having circular
sections makes it possible to bore such a cylinder with an ordinary
boring bar as shown in Fig. 502. The axis of the boring bar is set at
an angle to the axis of the cylinder to be bored, being such that

cos = y., where d is equal to the minor axis and D is equal to the

major axis of the normal elliptic section of the cylinder. While

the boring bar is rotated the cylinder is moved in the direction of its

axis, or if the cylinder is stationary the boring bar is moved bodily in

the direction of the axis of the cylinder. It is evident that with the

arrangement sketched in

Fig. 502 the length of

cylinder which may be

bored is limited, the limit

depending on the angle

and the diameter of the

boring bar.

221. Cylinder En-
veloping a Sphere. —
A cylinder which en-
velops a sphere will have
its axis passing through
the centre of the sphere
and the curve of contact
will be a "great circle"
of the sphere whose plane
is perpendicular to the
axis of the cylinder.

Referring to Fig. 503,
oo' is the centre of the
sphere and mn, m'n' is the
axis of the cylinder.
The sphere and the direc-
tion of the axis of the
cylinder are supposed to
be given. In problems
requiring a cylinder to envelop a sphere the cylinder is generally of
indefinite length and the ends are not required to be shown but

X i

\'^

Fig. 503.

THE SPHERE, CYLINDER, AND CONE

257

usually the trace of the surface on one of the planes of projection is
necessary. Neglecting the ends, the projections of the cylinder will
consist of the tangents to the projections of the sphere parallel to the
projections of the axis of the cylinder. Fig. 503 shows the construc-
tion for finding the horizontal trace of the cylinder and the plan of the
circle of contact. An elevation of the cylinder and sphere is drawn
on mn as a ground line, w/w is the new elevation of the axis of the
cylinder and 0/ that of the centre of the sphere. The points a and 6,
where the tangents to the circle which is the new elevation of the
sphere, parallel to wi/w, meet mn, are the extremities of the major axis
of the ellipse which is the horizontal trace of the cylinder. The
minor axis of this ellipse passes through n, the horizontal trace of the
axis of the cylinder, and is equal to the diameter of the cylinder or
sphere.

The line c/d/, passing through 0/, and perpendicular to m/»i is the
new elevation of the circle of contact. Perpendiculars from c/ and d^'
to mn determine cd, the minor axis of the ellipse which is the plan of
the circle of contact. The major axis of this ellipse is a diameter of
the circle which is the plan of the sphere and is at right angles to mn.

The elevation of the circle of contact is obtained by making an
auxiliary plan of the cylinder and sphere on a plane parallel to the
axis of the cylinder, say on m'n' as a ground line. The construction
is similar to that already described for the plan. The vertical trace
of the cylinder may also be obtained from this same auxiliary plan.

222. Projections of Points on the Surface of a Right
Circular Cylinder. — Suppose that one projection, say the plan r of
a point R, on the surface
of a right circular cylinder
to be given and that it is
required to find the other.
Take an elevation of the
cylinder on a vertical
plane parallel to its axis.
Take any point 00' (Fig.
504) on the axis of the
cylinder as the centre of
a sphere inscribed in the
cylinder. Through r draw
a straight line rs parallel
to the plan of the axis of
the cylinder and assume
this to be the horizontal
trace of a vertical plane
which is parallel to the axis of the cylinder. This vertical plane
cuts the sphere in a circle whose diameter is equal to ac the intercept
of rs on the plan of the sphere. The elevation of this circle is a
circle with its centre at o'. This same vertical plane cuts the curved
surface of the cylinder in, two straight lines whose elevations are
tangents to the circle just mentioned and are parallel to the elevation

Fig. 504.

Fig. 505.

258 PRACTICAL GEOMETRY

of the axis of the cylinder. There are two points on the surface of
the cyUnder having the point r for their plan. The elevations of
these points are on the projectors from r and on the lines which are
the elevations of the lines of intersection of the assumed vertical
plane and the cylinder.

Another solution of the problem is shown in Fig. 505. A sphere
is inscribed in the cylinder as before and a second cylinder, with its
axis vertical, is made to envelop this sphere. The two cylinders inter-
sect in two ellipses whose elevations on the vertical plane parallel to
the axes of the cylinders are the straight lines mV and s't'. The points
whose plan is r are evidently on the intersection of the two cylinders
and their elevations are found as shown.

223. Intersection of a Straight Line and a Right Circular
Cylinder. — First obtain from the given projections of the line and
cylinder projections of them, on planes parallel and perpendicular
respectively to the axis of the cylinder. Next take a plane parallel
to the axis of the cylinder and containing the given line. This plane
will intersect the curved surface of the cylinder in two straight lines
which will intersect the given line at the points of intersection required.
Working backwards to the original projections the required projections
of the points of intersection of the line and cylinder are found.

224. The Cone. — The cone may be generated by a straight line
which moves in contact with a fixed curve and passes through a fixed
point. The fixed point is the vertex of the cone. If the fixed curve is
a circle and the fixed point is on the straight line which passes through
the centre of the circle and is at right angles to the plane of the circle,
the cone is a right circular cone. The fixed circle is a circular section
of the right circular cone. The straight line joining the vertex of a
right circular cone to the centre of a circular section is the axis of the
cone. The right circular cone may also be defined as the surface
described by a straight line which intersects a fixed line at a fixed
point and moves so that its inclination to the fixed line is constant.
The fixed line in this case is the axis of the cone.

If a right circular cone terminates at one end at the vertex and
at the other end at a circular section, that circular section is called
the hase of the cone.

225. Projections of a Right Circular Cone. — Starting with
the cone in the position in which its axis is vertical, its plan is a
circle and its elevation is an isosceles triangle as shown at (1) and (2)
in Fig. 506.

Taking a second ground line XjYj inclined to v'r' the first elevation
of the axis of the cone, and projecting from (2), a new plan (3) is
obtained. In this new plan (3), vr the plan of the axis of the cone,
is parallel to X,Yi and at a distance from it equal to the distance of
vr in (1) from XY. Also the plan of the base of the cone is an ellipse
whose major axis is at right angles to vr and equal in length to the
diameter of the base of the cone. The minor axis bisects the major
axis at right angles and its length is found by projectors from (2) as
shown.

THE SPHERE, CYLINDER, AND CONE 259

Taking a third ground line XoYg inclined to vr in the plan (3),
rV, a new elevation of the axis of the cone is first obtained, the distances
of v' and r' in (4) from XgYg being equal to the distances of v' and /
respectively in (2) from XiYj. Before going further with the eleva-
tion (4), take another ground line X3Y3 parallel to v'r' in (4) and
draw vr a new plan of the axis of the cone as shown in (5). The
distances of v and r in (5) from X3Y3 are equal to the distances of v
and r respectively in (3) from X^Yg. The new plan (5) being on a
plane parallel to the axis of the cone will be an isosceles triangle as

Fig. 506.

in (2) and may therefore now be drawn. The axes of the ellipse in
(4) are next determined by projectors from (5) as shown, just as the
axes of the ellipse in (3) were obtained by projectors from (2).

Instead of finding the axes of the various ellipses which are the
projections of the base of the cone and then constructing the ellipses
in the usual way, a number of points on the base may be projected as
is shown for one point N.

If vr in (3) and v'r' in (4) be given and also the diameter of the base
of the cone it is obvious that by drawing the projections (2) and (5)
the projections (3) and (4) may be completed.

226. Plane Sections of a Right Circular Cone. — In what
follows only the curved surface of the cone is considered, the base
being supposed to be beyond the limits of the section or the part of
the section represented. When the term '* cone " is used " right
circular cone " is to be understood.

In studying the sections of the cone it is necessary to consider that
the straight line which generates the surface is of unlimited length
and consequently that the vertex of the cone is not at one end of the
generating line. The cone then consists of two sheets generated by the
parts of the generating line which are on opposite sides of the vertex.
The two sheets of a cone are exactly alike and the axis of one is the
axis of the other produced. The two sheets of a cone are shown in
Fig. 510.

The following are the various sections of the cone. (1) Two

260

PRACTICAL GEOMETRY

straight lines when the plane of section passes through the vertex of
the cone. (2) The circle when the plane of section is perpendicular to
the axis of the cone. (3) The ellipse when the plane of section cuts
all the generating lines on the same side of the vertex. (4) The
hyperbola when the plane of section cuts both sheets of the cone and
does not pass through the vertex. (5) The parabola when the plane
of section is parallel to a tangent plane to the cone, or, to give a
common but less exact definition, when the plane of section is parallel
to the slant side of the cone.

The circle (2) is a particular case of the ellipse (3), and two straight
lines (l)area particular case of the hyperbola (4). As the plane of

Fig. 507.

Fig. 508.

section turns round from a position which gives an ellipse to a position
which gives an hyperbola it passes through the position which gives
a parabola. Sections (1) and (4) are those which lie on both sheets of
the cone.

The sections which will now be considered are, the ellipse, the
parabola, and the h)^perbola.

Taking the elliptic section first and referring to Fig. 507, a cone,
whose axis VS is inclined to the horizontal plane but is parallel to the
vertical plane of projection, hi cut by a horizontal plane of which PQ
is the vertical trace. S anvd Sj are the centres of two spheres
inscribed in the cone and touching the plane of section at F and F,

THE SPHERE, CYLINDER, AND CONE 261

respectively. The points F and Fj are the foci of the ellipse which is
the section of the cone by the given plane of section, a'a^y the portion
of PQ lying within the elevation of the cone, is the elevation of the
major axis of the ellipse and from this the plan aa^ is projected. The
minor axis h\ of the plan of the elUpse bisects aa^ at right angles at c.
The length of the minor axis may bo determined by taking / as centre
and a radius equal to ac and describing arcs to cut hh^ at h and h^.

Planes containing the circles of contact between the cone and
the inscribed spheres intersect the given plane of section in straight
lines XM and XjMi which are the directrices of the ellipse (Art. 35,
p. 32).

Fig. 508 differs from Fig. 507 in that the axis of the cone is
inclined to both planes of projection instead of being inclined to the
horizontal plane only, w'w/ the intercept of PQ within the elevation of
the cone does not now give the major axis of the ellipse, but the foci
are determined exactly as before. A projector from n' to the plan
will however be a tangent to the plan of the ellipse. From / draw fe
at right angles to this tangent, meeting it at e. With centre c, the
middle point of ffy, and with a radius equal to ce describe arcs to cut ff^
produced at a and a^. Then aa^ is the major axis of the plan of the
ellipse. This follows from the fact that the foot of a perpendicular
from a focus of an ellipse on to a tangent lies on the auxiliary circle
(Art. 42, p. 39). The minor axis hh^ may now be determined as
before.

Consider next a parabolic section, which is illustrated by Fig. 509.
The cone is placed so that it is tangential to a horizontal plane and
the axis is parallel to the vertical plane of projection. PQ is the
vertical trace of a horizontal plane of section. S is the centre of the
sphere which is inscribed in the cone and which touches the plane of
section. Only one such sphere can be drawn in this case. The
point F at which the inscribed sphere touches the plane of section is
the focus of the parabola and A is the vertex.

The plane of the circle of contact between the cone and the
inscribed sphere intersects the plane of section in the straight line XM
which is the directrix of the parabola. It will be found that ax is
equal to af. The parabola may now be constructed as described in
Art. 34, p. 30, or as follows. Take a point o' on the elevation of the
axis of the cone. Draw o'd! at right angles to v'o' to meet the outline
of the elevation of the cone at d' and PQ at /. With o' as centre
and o'c?' as radius describe an arc to meet a straight line r'R^ parallel
to v'o' at Rp Through r' draw a projector to the plan cutting the
plan of the axis of the cone at t. On this projector make <r, above
and below t, equal to r'Rj, then r and r are two points on the plan of
the parabola. The theory of this construction is that r'o'd' is the
part elevation of a circular section of the cone and the arc d'M^ is a
rabatment of part of this circle into the plane of the elevation. It
will be seen that r'Rj is the half width of the cone measured at right
angles to the vertical plane of projection through the point R. In like
manner any number of points on the parabola may be found. This

262

PRACTICAL GEOMETRY

construction may also be used to find points on the ellipse of an elliptic
section when the cone is placed as in Fig. 507.

An hyperbolic section is illustrated by Fig. 510. The axis of the
cone is parallel to the vertical plane of projection but is inclined to
the horizontal plane. A horizontal plane whose vertical trace is
PQ cuts both sheets of the cone. The foci, the transverse axis, and
the directrices of the hyperbola are determined as in the case of the
elliptic section Fig. 507.

The hyperbola may be constructed from its foci and transverse
axis as described in Art. 43, p. 40, or points such as r and r may be

P afa/

Fig. 510.

found by the construction shown and already explained in connection
with the parabolic section.

The asymptotes pass through C the middle point of the transverse
axis and are parallel to the straight lines which form the section of
the cone by the plane UW through the vertex and parallel to the
given plane of section PQ. vn and vn are the plans of the lines which
form the section of the cone by the plane UW, the points n and n
being determined in a manner similar to that for the points r and r as
shown. The lines vn and vn may however be found by drawing
them as tangents through v to the circle which is the plan of the
section of one of the inscribed spheres by the plane UW as shown.

THE SPHERE, CYLINDER, AND CONE

263

The asymptotes Id and Id may now be drawn through c parallel
to vn and vn.

227. Cone Enveloping a Sphere. — A cone which envelops a
sphere will have its axis passing through the centre of the sphere and
the curve of contact will be a "small circle" of the sphere whose plane
is perpendicular to the axis of the cone.

Referring to Fig. 511, oo' is the centre of the sphere and vv' is the
vertex of the cone. The sphere and the vertex of the cone are supposed
to be given. The constructions for finding the horizontal trace of the
tone and the plan of the circle of contact are shown. An auxiliary

\ ^ ^"^"^^

\ ^ ^""^^N^

\ "^ .^"^ ^*5)^^

\ / "- ' ' ' '^ / ^\Vw

\ / >>' ' / \ ^^

\/ / ^yi \ ^^

\/^'Ao ^-s^

v ^ .^^ ^ / ^^w.

v--"'^ ' ^ / ^

V ^ ^ /

^\^

^"v.

^*"\^

\ ' ^ '

\ ' \n

>— , ^^

— -^

/^^ "y^-r^-—.^

T^i— -^

Jr-ky] \ \ 1

^-V

N. \ / \ \ yXiV^/ 1 ^ /

^ ■' \

^^k/yP^ \ "- ^ - / X

"■^v^..^^^ ',

\
\

/IHvy ) ^ ^

\
\

\/-'\K)J

Fig. 511.

elevation of the cone and sphere is drawn on the plan vn of the axis of the
cone as a ground line, v^n is the new elevation of the axis of the cone
and o/ that of the centre of the sphere. The points a and I where the
tangents from v/ to the circle which is the new elevation of the sphere
meet vn are the extremities of the major axis of the ellipse which is the
horizontal trace of the cone. The minor axis of this ellipse must of course
bisect ah at right angles, but it should be noticed that the middle point
of ah is not at n the horizontal trace of the axis of the cone. To deter-
mine the length of the minor axis of the ellipse which is the horizontal
trace of the cone the construction is as follows. Through e the middle
point of ah draw ,^/e/i/ at right angles to v^n meeting v^n at gx and

264-

PRACTICAL GEOMETRY

Vyh at 111. With centre gl and radius glh^ describe the arc A// and
draw ef at right angles to eA/ to meet this arc at /. ef is equal in
length to the semi-minor axis. The theory of this construction is that
a section of the cone perpendicular to its axis has been taken whose
trace on the plane of the auxiliary elevation is glh^. The true form of
this section is a circle a portion of which is shown turned round into
the plane of projection. The chord of this circle drawn through e per-
pendicular to eA/ gives the greatest width of the cone at the level of the
horizontal plane.

The line c/c?/ joining the points of contact of the tangents from v^
to the circle which is the auxiliary elevation of the sphere is the
auxiliary elevation of the circle of contact. Perpendiculars from c/
and d^ to vn determine c and d the extremities of the minor axis of the
ellipse which is the plan of the circle of contact. The major axis of this
ellipse is equal to the diameter of the circle of contact and is therefore
equal to c/<i/.

The axes of the ellipse which is the elevation of the circle of contact
are found by making an auxiliary plan of the cone and sphere on a plane
parallel to the axis of the cone, say on v'ri! as a ground line. The con-
struction is similar to that already described for the plan. The
vertical trace of the cone may also be found from this same auxiliary
plan.

228. Projections of Points on the Surface of a Right
Circular Cone. — Suppose that one projection, say the plan r, of a
point R on the surface of a given right circular cone to be given and
that it is required to find the other. Take an elevation of the axis of
the cone on a ground line parallel to vr (Fig. 512). Take a point oo' on

Fig. 513.

the axis of the cone as the centre of a sphere inscribed in the cone.
Assume vr to be the horizontal trace of a vertical plane. This vertical
plane cuts the sphere in a circle whose diameter is equal to ac the
intercept of vr on the plan of the sphere. The elevation of this circle
is a circle with its centre at o'. This same vertical plane cuts the

THE SPHERE, CYLINDER, AND CONE

265

curved surface of the cone in two straight lines whose elevations pass
through v' and are tangents to the circle just mentioned. There are
two points on the surface of the cone having the point r for their plan.
The elevations of these points are on the projector from r and on tho
elevations of the lines of intersection of the assumed vertical plane and
the cone.

Another solution of the problem is shown in Fig. 513. A sphere is
inscribed in the cone so that the plan of the sphere passes through r,
and a cylinder with its axis vertical is made to envelop this sphere.
The cone and cylinder intersect in two ellipses whose elevations on the
vertical plane parallel to the axes of the cone and cylinder are the
straight lines m'ln! and s't'. The points whose plan is r are evidently on
the intersection of the cone and cylinder and their elevations are there-
fore found as shown.

229. Intersection of a Straight Line and a Right Circular
Cone. — First obtain from the given projections of the line and cone
projections of them on planes parallel and perpendicular respectively
to the axis of the cone. Next take a plane to contain the vertex of
the cone and the given line. This plane will intersect the curved
surface of the cone in two straight lines which will intersect the given
line at the points of intersection required. Working backwards to the
original projections the required projections of the points of intersection
of the line and cone are found.

230. The Oblique Cylinder. — The oblique cylinder may be
generated by a straight line which moves in contact with a fixed circle
and remains parallel to a fixed
straight line which is not per-
pendicular to the plane of the
circle. The line parallel to the
fixed line and passing through
the centre of the fixed circle is
the axis of the cylinder. A
section of the oblique cylinder by
a plane perpendicular to the
plane of the fixed circle and con-
taining the axis of the cylinder
is called the principal section.

An oblique cylinder is shown
in Fig. 514 by a plan (1) and
elevation (2). The fixed circle is
on the horizontal plane of pro-
jection and may be called the
base of the cylinder. The axis
of the cylinder is parallel to the
vertical plane of projection.
The principal section is therefore
also parallel to the vertical plane
of projection.

PQ is the vertical trace of a plane of section which is perpendicular

Fig. 514.

266 PRACTICAL GEOMETRY

to the vertical plane of projection. The section by this plane PQ is an
ellipse whose major axis is equal to a'a/ and whose minor axis is equal
to the diameter of the base circle. Two plans of this section, which
are ellipses, are shown, the lower one (1) on XY as a ground line, and
the upper one (3) on PQ as a ground line. The latter plan shows the
true form of the section.

To find points on the lower plan (1) of the section, take a horizontal
plane whose vertical trace is ST and which cuts the axis of the cylinder
at o€> . The section of the cylinder by this plane is a circle equal to
the base circle. The plan of this circle is a circle whose centre is o.
Draw this circle, and from /, the point of intersection of PQ and ST,
draw the projector r^er cutting the circle at r and r which are points
on the plan of the section. In like manner other points may be found.
The plan of the section may however be constructed from its axes aa^
and hhy by the trammel method.

Referring next to the upper plan (3) which shows the true form of
the section, aa-^^ the major axis, is parallel and equal to a'a/. The
semi-minor axis ch is equal to the radius of the base circle and er in (3)
is equal to er in (1).

A section by a plane LM which is perpendicular to the principal
section and makes with the axis an angle Q equal to the angle between
the axis and the base is called a sub-contrary section and is a circle equal
to the base circle.

Planes of section parallel to the base or to a sub-contrary section
are called cyclic planes because the sections by these planes are
circles.

Since a section of an oblique cylinder by a plane at right angles to
its axis is an ellipse the oblique cylinder is evidently also a rigid
elliptical cylinder (Art. 220).

231. The Oblique Cone. — The oblique cone may be generated by
a straight line which moves in contact with a fixed circle and passes
through a fixed point which is not on the straight line through the
centre of the circle at right angles to its plane. The fixed point is the
vertex of the cone and the straight line joining the vertex to the centre
of the fixed circle is the axis of the cone. A section of the oblique
cone by a plane perpendicular to the plane of the fixed circle and
containing the axis of the cone is called the principal section.

An oblique cone is shown in Fig. 515 by a plan (1) and elevation
(2). The fixed circle is on the horizontal plane of projection and
may be called the base of the cone. The axis of the cone is parallel
to the vertical plane of projection. The principal section is therefore
also parallel to the vertical plane of projection.

PQ is the vertical trace of a plane of section which is perpen-
dicular to the vertical plane of projection. The section by this plane
PQ is an ellipse whose major axis is equal to a'a/. Two plans of this
section, which are ellipses, are shown, the lower one (1) on XY as a
ground line, and the upper ono (3) on PQ as a ground line. The latter
plan shows the true form of the section.

To find points on the lower plan (1) of the section, take a horizontal

THE SPHERE, CYLINDER, AND CONE

267

plane whose vertical trace is ST and which cuts the axis of the

cone at od . The section of the cone by this plane is a circle whose

diameter is the intercept of ST on

the elevation of the cone. The

plan of this circle is a circle whose

centre is o. Draw this circle and

from r', the point of intersection

of PQ and ST, draw the projector

r'er cutting the circle at r and r,

which are points on the plan of

the section. In like manner other

points may be found. The above

construction when applied to the

point c' which is the middle point

of dal gives the minor axis hh^ of

the ellipse which is the plan of the

section.

Referring next to the plan (3)
which shows the true form of the
section, aa^, the major axis, is
parallel and equal to dal. The
semi-minor axis cb in (3) is equal

to cb in (1), also er in (3) is equal

to er in (1). Fig. 515.

A section by a plane LM
which is perpendicular to the principal section and makes with the
axis an angle B equal to the angle between the axis and the base is
called a suh-contrary section and is a circle.

Planes of section parallel to the base or to a sub-contrary section
are called cyclic planes because sections by these planes are circles.

As in the case of a right circular cone, a section of an oblique cone
by a plane which is parallel to a tangent plane to the cone is a para-
bola. Also a section of an oblique cone by a plane which cuts both
sheets of the cone is an hyperbola.

Exercises XVIII

1. The circle (Fig. 516) is the plan of a sphere resting on the horizontal
plane. H\Ti and HoTg, parallel to XY, are the horizontal traces of two parallel
planes. The plane of which H,Ti is the horizontal trace contains the centre of
the sphere. Draw the plan and elevation of the zone of the sphere which lies
between these two planes.

2. The parallel lines HiTi and HgTg (Fig. 517) are the horizontal traces of
two planes which pass through the centre of a sphere resting on the horizontal
plane, the given circle being the plan of the sphere. Draw the plan and elevation
of the lower lune of the sphere lying between the given planes.

3. The circle (Fig. 518) is the plan of a sphere resting on the horizontal plane.
HiTi inclined at 45° to XY and HjTj perpendicular to XY are the horizontal
traces of two planes which contain the centre of the sphere. Draw the plan and
elevation of the lower portion of the sphere which lies between these two planes.

4. The given circle (Fig. 519) is the plan of a sphere, centre C. The points
a, a, b, are the plans of points A, A, B on its upper surface. N and S are the
upper and lower ends of a vertical diameter.

268

PRACTICAL GEOMETRY

Draw the plan of a figure on the surface made up of three arcs of great circles
joining NA, NA, AA, and the arc of a small circle ABA.

Draw the stereographic projection of this figure on the horizontal plane

X >- - v^- Y X , ^ ^\ Y X-^

Fig. 516.

Fig. 517.

through G. That is, find the plane section of a cone, vertex S, of which the
figure is a spherical section. [b.e.]

5. The horizontal and vertical traces of a plane make angles of 30^ and 45*^
respectively with XY. A hemisphere of 1*25 inches radius has its base on this
plane and touching the planes of projection. Draw the plan and elevation of the
hemisphere.

6. A square of 1*5 inches side and a circle of 1-25 inches radius, the centre of
the circle being at the centre of the square, form the plan of a sphere with a
square hole through it. From this plan project two elevations, one on a ground
line XY parallel to a side of the square and the other on a ground line XjY,
parallel to a diagonal of the square. Also from the second elevation project a
plan on a ground line XgYg making an angle of 45° with XjYi.

7. A cylinder 2 inches in diameter and 3 inches long has its axis horizontal
and inclined at 45° to the vertical plane of projection. The cylinder is cut in
halves by a vertical plane which is inclined at 60° to the axis of the cylinder and
15° to the vertical plane of projection. Draw the elevation of one of the halves
of the cylinder.

8. rs, r's' (Fig. 520) is the axis of a hollow cj'linder, whose external diameter
is 2*1 inches and whose internal diameter is 1-4 inches. Draw the plan and
elevation.

9. A plan and an elevation of a hollow half cylinder, with one end closed, are

—

♦;

•

!::

Fig. 520.

Fig. 521.

Fig. 622.

Fig. 523.

given in Fig. 521. Draw these projections and from the plan project an elevation
on a ground line inclined at 60° to XY.

10. A plan {u) and elevations {v) and [w) of a solid are given in Fig. 522.
Draw these projections and from the plan project an elevation on a ground line
parallel to rs. Also from the elevation {v) project a plan on a groimd line perpen-
dicular to rs.

11. A plan (1) and an elevation (2) of a right circular cylinder with two longi-
tudinal grooves in it are given in Fig. 523. Draw these projections and add others

THE SPHERE, CYLINDER, AND CONE

269

corresponding to (3), (4), and (5) in Fig. 497, p. 253. The angle between X^Y^
and XY to be 60°, and the angle between X^Y^ and X,Yi to be 45°.

12. A right circular cylinder of indefinite length has its axis vertical. A
plane whose horizontal and vertical traces make angles of 30° and 46° with the
ground line cuts the cylinder. Draw the elevation of the section and determine
its true form.

13. A split wrought-iron collar is shown in Fig. 524. SS is a vertical section
plane. Draw a sectional elevation on a ground line parallel to SS, the portion
A in front of the section plane being supposed removed. [b.e.]

Note. Fig. 524 is to be reproduced double size.

Fig. 525.

14. Referring to Fig. 525, (u) is a plan and (v) a side elevation of a quarter of
a ratchet wheel having 24 teeth of the form shown, (w) is an elevation on a
vertical plane inclined at 30° to the side of the wheel. Draw these projections
for the whole wheel to the dimensions given.

15. A sphere 2-2 inches in diameter rests on the horizontal plane and touches
the vertical plane of projection. The plan of the axis of a cylinder which
envelops the sphere makes 45° with XY and the elevation of the axis makes 60°
with XY. Draw the horizontal and vertical traces of the surface of the cylinder,

Fig. 527.

Fig. 528.

Fig. 529.

also the plan and elevation of the circle of contact between the sphere and
cylinder.

It. WW, m'n' (Fig. 526) is the axis of a right circular cylinder 1-5 inches in

270

PRACTICAL GEOMETRY

diameter, r is the plan of a point on the upper surface of the cylinder and s' is
the elevation of a point on the front surface. Find r' and s. Find also the plans
and elevations of the points of intersection of the line ah, a'h' with the surface of
the cylinder.

17. The elevation of a solid made up of two truncated right circular cones is
given in Fig. 527. Draw the plan of the solid. Show also the plan of the section
of the solid by a plane which is perpendicular to the vertical plane of projection
and which has VT for its vertical trace.

18. A right circular cone, base 2*5 inches in diameter and altitude 2 inches,
lies with its slant side on the ground. Draw the plan of the cone and show the
plan of a straight line which lies on the surface of the cone and is inclined at 45"^
to the ground.

19. The plan and an elevation of a right circular cone, whose axis is vertical
and which has a triangular hole through it, are given in Fig. 528. Draw these
projections and add an elevation on a ground line inclined at 45° to XY.

20. A plan and an elevation of a square nut with a conical chamfer are given
in Fig. 529. Draw these projections and add an elevation on a gromid line
parallel to one diagonal of the plan. Also, project from the given elevation a new
plan on a ground line inclined at 60° to r's'.

21. An equilateral triangle v'a'h', of 2 inches side, is the elevation of a right
circular cone, a'h' the elevation of the base being parallel to XY. A circle
inscribed in the triangle is the elevation of a plane section of the cone. Draw
the plan of the section.

22. A right circular cone, base 2 inches diameter, altitude 2*5 inches, lies
with its slant side on the horizontal plane. The cone is divided by a vertical
plane which passes through the centre of the base and whose horizontal trace is
inclined at 30° to the plan of the axis. Draw the elevation of the larger part of
the cone on a plane parallel to the plane of section.

23. vr, v'r' (Fig. 530) is the axis of a right circular cone, vv' being the vertex
and rr' the centre of the base. The cone rests with its slant side on the hori-

FiG. 532.

Fig. 533.

zontal plane. Draw the plan and elevation of the cone. Show the projections of
a point on the surface of the cone which is 0*75 inch above the horizontal plane
and 1-75 inches distant from the vertex of the cone.

24. The given circles a, h, and c (Fig. 531) are each 2 inches in diameter.
These circles touch one another, a and h touch XY and the straight line HT is
tangential to a and c. a is the plan of a cone, altitude 2'5 inches ; h is the plan
of a cylinder, length 4 inches, and c is the plan of a sphere. All these solids
stand on the horizontal plane. HT is the horizontal trace of a plane which cuts
the solids and is inclined at 45° to the horizontal plane. Show in plan and
elevation the sections of the given solids by the given plane.

25. A solid cut from a hollow cylinder is given in Fig. 532 by a plan and an
elevation. Draw an elevation on a ground line which is perpendicular to XY,
also an elevation on a ground line which is inclined at 45^ to XY.

26. Show the points of intersection of the given line ah, a'h' (Fig. 533) with
the surface of the given cone.

THE SPHERE, CYLINDER, AND CONE

271

27. A sphere 1-5 inches in diameter rests on the horizontal plane and its
centre is 1*5 inches in front of the vertical plane of projection. A point on the
horizontal plane, 3'75 inches in front of the vertical plane of projection and 2-25
inches from the plan of the centre of the sphere is the vertex of a cone which
envelops the sphere. Draw the vertical trace of the cone, also the plan and
elevation of the circle of contact between the cone and sphere.

28. A circular section of an oblique cylinder is 2 inches in diameter and the
axis of the cylinder is inclined at 40° to this section. Draw the true form of the
section of the cylinder by a plane perpendicular to its axis,

29. An oblique cylinder is given in Fig. 534. Draw the vertical trace of the

Pig. 534.

Fig. 535.

Fig. 536.

cylinder, that is, determine the section of the cylinder by the vertical plane of
projection. Draw also the plan and elevation of a sub-contrary section.

30. An oblique cone is given in Fig. 535. Draw the plan and also the true
form of the parabolic section of this cone by a plane whose vertical trace is PQ
and which is perpendicular to the vertical plane of projection. Draw also the
elevation of the hyperbolic section by the vertical plane whose horizontal trace
isRS.

31. Draw the plan and elevation of the section of the oblique cone (Fig. 536)
by the plane whose traces are H.T. and V.T.

CHAPTEK XIX

SPECIAL PKOJECTIONS OF PLANE FIGURES AND SOLIDS

232. Projections of a Figure whose Plaiie and a Line
in it have Given Inclinations. — First determine the traces of a
plane containing the figure. The working of the problem is much
simplified by first assuming this plane perpendicular to the vertical
plane of projection. In this position the elevation of the figure will
be a straight line coinciding with the vertical trace of the plane.
Afterwards any other elevation may be obtained by Art. 157, p. 191.

Draw L'M (Fig. 537) making with XY an angle equal to the
given inclination of the plane of the figure. Draw MN the horizontal
trace at right angles to XY. In the
plane L'MN place a line RS having
the inclination a of the line of the
figure given (Art. 174, p. 214).

Now imagine the plane L'MN,
with the line RS upon it, to rotate
about its horizontal trace MN until
it comes into the horizontal plane.
The point S being in MN will remain
stationary while the point R will
describe an arc of a circle in the
vertical plane of projection with M
as centre. Hence, if with M as
centre and Mr' as radius the arc
r'Rj be described, meeting XY at
Rj, RjS will be the position of the
line RS when that line is brought
into the horizontal plane.

Mark off on R^s a length AjEj equal to the length of that line of
the figure whose inclination a is given, and on AjBi construct the
given figure. Note that the line AB of the figure is not necessarily
one of its sides, but may be any line whatever in the plane of the
figure and occupying a definite position in relation to the figure.
Next imagine the figure thus drawn on the horizontal plane to rotate
about MN until its inclination is 6. All points in the figure will
describe arcs of circles whose plans will be straight lines perpendicular
to MN and whose elevations will be arcs of circles having their
centres at M and having radii equal to the distances of the points

SPECIAL PROJECTIONS

273

from MN. Thus the point Cj will describe an arc of a circle whose
plan is CiC perpendicular to MN and whose elevation is the arc c-ld of
a circle whose centre is M. c is of course in a straight line through
d at right angles to XY.

As a verification of the construction it should be noticed that if
any line, say C^B,, of the figure constructed on the horizontal plane
be produced, if necessary, to meet MN it will do so at the point t
where the plan he of the same line in its inclined position meets it.

It must be borne in mind that no line of the figure can have a
greater inclination than that of its plane, that is, a must not be
greater than 6.

233. Projections of a Plane Figure, the Inclinations of
two Intersecting Lines in it being given. — Determine by Art.
187, p. 219, the traces of the plane containing the lines whose inclina-
tions are given, taking the horizontal trace at right angles to the
ground line. Eotate this plane, with the lines upon it, about its hori-
zontal trace, as in the preceding Article, until it comes into the hori-
zontal plane. On the lines thus brought into the horizontal plane
construct the given figure and proceed to determine the plan and ele-
vation of the figure in its inclined position exactly as in the preceding
Article.

234. Projections of a Plane Figure having given the
Heights of three Points in it. — Note that the difference between
the heights of any two points must not exceed the true distance
between the points.

Let A, B, and C be the points whose heights are given. '

Construct on the horizontal plane a triangle A^BiCj (Fig. 538)
equal to the triangle ABC. With centre Aj and radius equal to the
height of the point A describe the circle
EFH. With centre Bj and radius equal to
the height of the point B describe the circle
KLM. With centre Cj and radius equal to
the height of the point C describe the circle
NPQ. Draw HL to touch the circles EFH
and KLM, and produce it to meet A^Bj
produced at R, Also draw EP to touch
the circles EFH and NPQ, and produce it
to meet A^Ci produced at S. RS is the
horizontal trace of the plane which will
contain the points A, B, and C.

Draw XY perpendicular to RS meeting
it at O. Draw A^a/ perpendicular to XY
to meet it at a/. With centre O and radius
Oa/ describe an arc of a circle to meet at
a' a parallel to XY which is at a distance
from XY equal to the height of the point
A. Oa! is the vertical trace of the plane
which will contain the points A, B, and C.

The theory of the above construction for finding the traces of the

CCy

538.

274

PRACTICAL GEOMETRY

plane containing the points A, B, and C is similar to that of the con-
struction given in Art. 187, p. 219, which should be again referred to.
It is evident that the angles A^RH and AjSE are the inclinations of
the straight lines A^ and AC respectively.

The plan abc and the elevation a'h'c of the triangle ABC are
next determined as shown, the construction being the same as in
Art. 232.

If ABC is not the complete figure given, a figure equal to it
must be built up on the triangle A^BiCp Then the projections of the
remainder of the figure are obtained in the same way as the projections
of the part ABC.

235. Projections of a Plane Figure when it has been
turned about a Horizontal Line in it till the Plan of an
Opposite Angle is equal to a given Angle. — Let ABCD denote
the given figure, AC the horizontal line about which it is turned, and
B the angle whose plan is to be equal to a given angle.

Construct the figure aB^cDi (Fig. 539) equal to the given figure
ABCD. On ac describe a segment of a circle (Art. 14, p. 18) to
contain an angle equal to that into which
the angle B is to be projected.

As the figure revolves about AC the
point B will describe an arc of a circle
whose plan is a straight line through B^
perpendicular to ac. Let this perpendicular
meet the arc of the segment of a circle
which has been described on ac at the
point b. Join ah and he. Then ahc is the
plan of the angle B when the figure has
been turned as required.

Draw XY at right angles to ac, meeting
ac produced at a'. Draw B^?)/ at right
angles to XY to meet it at 6/. With
centre a' and radius a'6/ describe the arc
6/6', and through 6 draw a perpendicular
to XY to meet this arc at 6'. a'b' is the
vertical trace and aa' is the horizontal trace
of the plane containing the figure ABCD
when it occupies the position required.

From Di draw DjC?/ perpendicular to XY
to meet it at d^'. With centre a' and radius a'd^' describe the arc d^^d'
to meet h'a' produced at d'. Draw d'd perpendicular to XY to meet
a line through D^ parallel to XY at d. Join ad and cd. abed is the
plan and d'h is an elevation of the figure ABCD as required.

236. Projections of a Solid when a Plane Figure on it
and a Line in that Figure have given Inclinations. — The
plane figure may be a face of the solid or it may be a section of it.
Considering the most general case, the first step is to determine the
projections of the plane figure as in Art. 232, p. 272. On the rabat-
jnent of the plane figure on the horizontal plane the feet of the

SPECIAL PROJECTIONS

275

perpendiculars from the angular points of the solid on the plane of
the figure must next be located. The projections of the feet of these
perpendiculars are next found, and the projections of the perpen-
diculars can then be drawn. The projections of the angular points of
the solid are therefore determined.

Two examples are illustrated by Figs. 540 and 541.

In the first example (Fig, 540) a right square prism, side of base
1*5 inches, altitude 1"8 inches, is shown when its base is inclined at 50*^
and one side of that base is inclined at 20°. The plan ahcd and eleva-
tion a'b'c'd' of the square base are first determined as in Art. 232, p. 272.
In this case the feet of the perpendiculars from the angular points of
the solid on the plane of the base are at the angular points of that base.
Hence it is now only necessary to draw from a', h\ c\ and d! perpen-
diculars to the vertical trace of the plane of the base and make them
1-8 inches long in order to obtain the elevations e', /', g', and h' of the
other angular points of the solid. The plans of these perpendiculars
are at right angles to the horizontal trace of the plane of the base and
are therefore parallel to XY.

Fig. 540.

Fig. 541.

In the second example (Fig. 541) a cube of 1-5 inches edge is
shown when the plane of two diagonals of the solid is inclined at 60°
and one of these diagonals is inclined at 40°. The plane of the two
diagonals divides the cube into two right triangular prisms, the
triangular ends of which are each equal to one half of a face of the
cube. The common face of these two triangular prisms is a rectangle
of which two opposite sides are edges of the cube while the other two

276 PRACTICAL GEOMETRY

sides are diagonals of opposite faces of the cube and the diagonals are
diagonals of the solid.

Tlie plan ahde and elevation a'h'd'e' of the rectangle which is the
section of the cube by the plane of two of its diagonals are determined
by Art. 232, p. 272. In this case the feet of the perpendiculars from
the other angular points of the cube on to the plane of two diagonals
are at the middle points of the longer sides of the rectangle ABDE
and the lengths of £hese perpendiculars are each equal to the half of a
diagonal of a face of the cube. Hence, perpendiculars to the vertical
trace of the plane of the above mentioned rectangle from the middle
points of a'h' and d!e' and equal to half of a diagonal of a face of the
cube determines the elevations c', c', /', and /' of the other corners of
the cube. The plans of these perpendiculars are parallel to XY.

237. Projections of a Solid when two Intersecting Lines
connected with it have given Inclinations. — Determine by
Art. 187, the plane containing the intersecting lines whose in-
clinations are given. Rabat this plane with the lines upon it into
the horizontal plane by turning it about its horizontal trace. On
these lines thus brought into the horizontal plane complete the
plan of the solid and then proceed exactly as in Art. 236.

238. Projections of a Solid when the Heights of three
Points in it are given. — Determine by Art. 234, the plane con-
taining the three points whose heights are given. Rabat this plane
with the points upon it into the horizontal plane by turning it about
its horizontal trace. About these points thus brought into the hori-
zontal plane complete the plan of the solid and then proceed exactly as
in Art. 236.

239. Projections of a Solid when two Faces A and B,
which are at Right Angles to one another, have given
Inclinations. — If one of the two faces referred to is a base denote it
by A. First determine L'MN, the plane of the face A, the horizontal
trace MN being perpendicular to XY. Next determine by Art. 194,
p. 226, a plane L'PN perpendicular to the plane L'MN and inclined at
an angle equal to the given inclination of the face B.

Find LN the line of intersection of these two planes. Now rabat
the plane L'MN with the line LN upon it into the horizontal plane by
turning it about MN. On this line thus brought into the horizontal
plane construct the face A, that side of the face A which is adjacent
to the face B being made to coincide with the line, and then proceed
exactly as in Art. 236.

240. Projections of a Solid when two Faces A and B,
which are not at Right Angles to one another, have given
Inclinations. — If one of the two faces is a base denote it by A.
First determine L'MN, the plane of the face A, the horizontal trace
MN being perpendicular to XY. Next determine by Art. 194, p. 226,
a plane L'PN having an inclination equal to that of the face B and
making with the plane L'MN an angle equal to the angle between the
faces A and B. The procedure is now the same as in the second
paragraph of the preceding Article.

SPECIAL PROJECTIONS 277

Exercises XIX

1. Draw the plan of an equilateral triangle of 2*5 inches side when its plane
is inclined at 45°, one side inclined at 30°, and one angular point on the hori-
zontal plane. From this plan project an elevation on a ground line parallel to
the plan of that side qi the triangle which is inclined at 30°.

2. Draw the plan of a square of 2*5 inches side when its plane is inclined at
60° and one diagonal is inclined at 45°. What is the inclination of the other
diagonal ?

3. A regular hexagon ABGDEF of 1-25 inches side has the side AB horizontal
and the plan of the diagonal BD is 2 inches long. Draw the plan and from it
project an elevation on a ground line parallel to ab. What is the inclination of
the plane of the hexagon ?

4. The plane of a square of 2 inches side is inclined at 60° and the plan of one
diagonal is 2 inches long. Draw the plan of the square.

5. The horizontal trace of a plane makes an angle of 45° and the vertical trace
makes an angle of 50° with XY. A regular pentagon of 1-5 inches side lies on
this plane with one side inclined at 30° to the horizontal plane. Draw the plan
and elevation of the pentagon.

6. A square of 2 inches side has one side inclined at 45° and an adjacent side
inclined at 30°. Draw the plan of the square and an elevation on a ground line
parallel to the plan of the diagonal which is inclined at 45°.

7. ABC is a triangle. AB = 2 inches, BC = 2*75 inches, and CA = 3 inches.
Draw a plan of this triangle when the sides AB and BC are inclined at 40° to the
horizontal plane.

8. A regular pentagon of 2 inches side has one side inclined at 30° and a
diagonal through one end of that side inclined at 40°. Draw the plan.

9. ABC is an equilateral triangle of 2*5 inches side. A is on the horizontal
plane, B is 1 inch and C is 1*5 inches above the horizontal plane. Draw the plan
and an elevation on a ground line parallel to ab.

10. Draw the plan of a square of 2 inches side when the heights of its centre-
and two angular points above the horizontal plane are 2 inches, 1-5 inches, and
0'75 inch respectively.

11. Draw the plan of a regular pentagon of 1*5 inches side when one side is on
the horizontal plane and the plan of the opposite angle is 120°.

12. A 60° set-square revolves about its shortest side,, which is horizontal,
until the plan of the opposite angle is 60°. Find what is then the inclination of
the plane of the set-square.

13. ABC is a triangle, AB = 1*5 inches, BC = 3 inches, and CA = 2 inches.
Draw the plan of this triangle when the side AB is horizontal and the plan of the
angle C is 20°.

14. Draw a triangle oab, oa = 1 inch, ob ^ 1*25 inches, and a6 = 1 5 inches.
o is the plan of the centre and ab is the plan of one side of a regular hexagon.
Complete the plan of the hexagon.

15. Draw a triangle abc. ab = 1 inch, be = I'd inches, and ca = 2 inches,
ab and be are the plans of adjacent sides of a regular octagon. Complete the plan
of the octagon.

16. Draw ab 2 inches long. From the middle point c of o6 draw cd 3 inches
long and making the angle acd = 60°. These are the plans of two intersecting
straight lines which are at right angles to one another. The points A and D are
each 0-5 inch above the horizontal plane. Find the height of the point B.

17. a is a point on a straight line HT. ab is a straight line 2 inches long
making an angle of 30° with HT. ab is the plan of one edge of a regular tetrahe-
dron of 2-5 inches edge. HT is the horizontal trace of the plane of a face of the
tetrahedron containing the edge AB. Complete the plan of the tetrahedron and
draw an elevation on a ground line parallel to HT.

18. Draw the plan of the solid given in Fig. 373, p. 196, when the base is in-
clined at 45° and one side of that base is inclined at 30°. From this plan project
an elevation on a ground line parallel to the horizontal trace of the plane of the
base.

278 PRACTICAL GEOMETRY

19. Draw the plan of the solid given in Fig, 372, p. 196, when the base is in-
clined at 50° and one diagonal of the base is inclined at 40°, From this plan
project an elevation on a ground line parallel to the plan of the diagonal which
is inclined at 30°.

20. A right pyramid has for base a regular pentagon of which the diagonals
measure 2-5 inches. The vertex is 2 inches above the base. Draw the plan and
elevation of the pyramid, with its base in a plane inclined at 55° to the vertical
plane and at 60° to the horizontal plane ; one diagonal inclined at 30°, and one
end of that diagonal in the vertical plane. [b.e.]

21. Two diagonals of a cube of 2 inches edge are inclined at 35° to the hori-
zontal plane. Draw the plan of the cube.

22. A right prism 3 inches long has for its ends regular hexagons of 1*25
inches side, AB is an edge of one end and BH is one of the long edges. Draw the
plan of this prism when the heights of the points A, B, and H above the hori-
zontal plane are 0*2 inch, 1 inch, and 2-5 inches respectively.

23. ABC is the base of a pyramid and V its vertex. AB = 2*5 inches, AC =
2 inches, BG = 2-75 inches, AV := CV = 3 inches, and BV = 2-5 inches. Draw
the plan of the pyramid when A is 1-5 inches, B 2*5 inches and C 1 inch above
the horizontal plane. [b.e.]

24. Draw the plan of a cube of 2 inches edge when one face is inclined at 55°
and another face is inclined at 75°.

25. One face of a regular tetrahedron of 2*5 inches edge is inclined at 45° and
another face is inclined at 70°. Draw the plan of the tetrahedron.

CHAPTER XX

HORIZONTAL PROJECTION

241. Figured Plans. — If the plan of a point is given, and
also its distance from the horizontal plane, the position of the
point is fixed without showing an elevation of it. The distance of
the point from the horizontal plane is shown by placing a number
or index adjacent to its plan. For example, if the point is 6
units above the horizontal plane the figure 6 is placed adjacent to its
plan, and if the plan is also lettered the figure is placed to the right of
the letter and at a slightly lower level thus, a^. If the negative or
minus sign be placed in front of the index this denotes that the given
distance is below the horizontal plane. Thus, a point whose plan is
marked — 6 or a.^ is 6 units below the horizontal plane. A plan
such' as has been described is called a figured plan or indexed plan.

If the plan of a straight line is given and also the figured
plans of two points in it, it is obvious that the line is completely
fixed.

In horizontal projection points and lines are shown by their figured
plans. The form and position of a curved line can only be shown in
horizontal projection by giving the figured plans of a sufficient number
of points in it.

242. Scales of Slope. — It has already been shown that planes
may be represented by their traces on the co-ordinate planes. In
horizontal projection, planes are represented by their scales of slope.
The scale of slope of a plane is really the figured plan of a straight
line lying in the plane and perpendicular to the horizontal trace of the
plane. If a straight line be given and it is understood that a piano
whose horizontal trace is perpendicular to the line contains the line
then it is obvious that the plane is completely
fixed. To show that the given line represents a
plane and not simply a line a second and thicker
line is ruled near to and parallel to it. It is
usual to place the thicker of the two lines to the
left of the other when looked at by a person
ascending the plane.

Fig. 542 shows the connection between the
scale of slope ah and the traces L'M, MN of a
plane. The numbers at the difierent points of
the scale denote the distances of these points
from the horizontal plane.

280

PRACTICAL GEOMETRY

Fig. 643.

243. Applications of Horizontal Projection.^ — Most of the
problems on points, lines and planes can be worked as conveniently by
horizontal projection as by ordinary plan and elevation, but it may be
observed that although the solution of a problem may be given by a
figured plan only the working may involve constructions which are
equivalent to the drawing of one or more elevations.

As the principles involved in the solution of problems on points,
lines, and planes are generally the same whether they are solved by
plan and elevation or by horizontal projection, a selection of a few
problems only will be taken in this chapter to illustrate the method of
horizontal projection.

244. Simple Problems on the Straight Line.— Let aAs
(Fig. 543) be the figured plan of a straight line AB. From a and b
draw perpendiculars to ah, and make them
respectively equal to the indices of a and b.
The line AjBj joining the tops of these per-
pendiculars will have a length equal to the
true length of AB. The angle between A^Bi
and ah will measure the inclination of AB,
and the point where AjB^ meets ah will be
the horizontal trace of AB. It is evident that
the index of the horizontal trace of a line is
O. If one of the indices is positive and the other negative the per-
pendiculars aAi and feB^ must be drawn on opposite sides of ah.

In the above construction A^Bj maybe looked upon as an elevation
of AB on ab as a ground line, or it may be taken as the line AB
rebatted on to the horizontal plane about ab as an axis.

The true length and inclination of AB may also be found by
drawing 5P perpendicular to ab and equal to the difference between
the indices of a and b. The line aP will then be the true length of
AB, and the angle Pah will be the inclination of AB.

To determine a point c in ab which shall have a given index, say
6*5, make 6Q equal to 6*5, draw QC^ parallel to ab to meet AjB^ at
Cp A perpendicular C^c to ab determines the point required.

To determine the index of a given point c in ab, draw cC^ per-
pendicular to ah to meet A^Bj at C^. The length of cCi is the index
required.

To draw a line through a given point c^ (Fig. 544) parallel to a
given line «2^9 > draw through c a line cd parallel to aS, and make
cd = ab. The index of d will be 7 greater than
5 the index of c because the index of & is 7
greater than the index of a. If the line cd be
produced in the opposite direction, and ce be
made equal to ba, then the index of e will be 7
less than 5 the index of c because the index
of a is 7 less than the index of b. The index
of e will therefore be — 2, or the point e is 2
units below the horizontal plane.

245. Inclination of a Given Plane.— Let ab (Fig. 545) be

HORIZONTAL PROJECTION

281

the scale of slope of the plane. Since the long lines of the scale of

slope are at right angles to the horizontal trace of the plane, the

inclination of these lines must be the same as

the inclination of the plane. Hence if 6Bi be

drawn perpendicular to ah, and made equal to

the difference between the indices of h and a,

the angle B^afe will measure the inclination of the

plane.

246. Two Problems on Parallel Planes.
— (1) To determine a plane to contain a given point p^ (Fig. 646),
and be parallel to a given plane ah. Since the horizontal traces of
parallel planes are parallel it is clear n ,

that their scales of slope which are at ^ | i | |~"^

right angles to these traces, must also 5 | 10 15

be parallel. Draw, therefore, the long Pj

lines cd of the required scale of slope ^

parallel to ah^ and in any convenient ^ j'q jj

position. Through p draw a line pq^ -^^^ -^g

at right angles to cd. pq will be

the plan of a horizontal line lying in the required plane, and q will

therefore have the same index as p. The scale of slope cd must be

graduated in the same way as ab ; that is to say the difference

between the indices of a given length on cd

must be equal to the difference between

the indices on an equal length of ab.

(2) To determine the distance between
two parallel planes ah and cd (Fig. 547).
If the traces of these planes on a vertical
plane perpendicular to their horizontal traces,
and therefore parallel to their scales of slope,
be drawn, the distance between these traces
will be the distance between the planes.

ah is taken as the ground line, and ac'
and hh' are drawn perpendicular to ah. hV p^^ ^^rj

is made equal to the index of h, and ad equal

to the index of c. As in the example shown in the figure, the index
of a is 0, ah' is the vertical trace of the plane whose scale of slope is
ah, and as the other plane is parallel to this one, a line c'd! parallel to
ah' will be its vertical trace on the assumed
vertical plane. The distance between ah' and
c'd' is the distance between the given planes.

247. The Plane containing three

given Points. — Let a^, , hy^ , and

(Fig.

648) be the given points. Find by Art. 244
a point d in ah having the same index as c.
Join cd. cd is the plan of a horizontal line
lying in the plane containing the three given
points; therefore the scale of slope of the
plane containing these three points must be perpendicular to cd.

282 PRACTICAL GEOMETRY

Where cd cuts the scale of slope determines a point on it having
the same index as c or d. If through a a line be drawn parallel to cd
to meet the scale of slope, a point is determined on the latter having
the same index as a. These two points on the scale of slope having
been found, the scale may be graduated if required.

248. In a Given Plane to place a Line having a Given
Inclination so that it shall contain a Given Point in the
Plane. — Let ah (Fig. 549) be the given plane
and pao ttie given point. Through r, any point
in the scale of slope not having the same index
as jp, draw rq perpendicular to ah. Draw pQ^,
in any convenient direction, say parallel to rq,
and a linepPi at right angles to jpQ^. Make
2?Qi equal to the difference between the indices
of p and r, and draw PiQi making the angle
PiQiP equal to the given inclination. With Fig. 549.
centre p and radius pQ^ describe an arc cutting

rq at q. po^qi^ is the line required. Except when the given inclination
is the same as the inclination of the given plane there are obviously
two straight lines which will satisfy the given conditions.

249. Intersection of Two Given Planes. — Let ah and cd
(Fig. 550) be the given planes. Through a and 6, any two points in
the scale of slope ah, draw the lines ap and
hq at right angles to ah. Through points
c and d in cd, having the same indices re-
spectively as a and h, draw lines perpen-
dicular to cd. The line op is the plan of a
horizontal line lying in the plane ah. cp is
the plan of a horizontal line lying in the
plane cd. Since these lines have the same
indices, they are in the same horizontal
plane, therefore they will intersect at a

point of which p is the plan. Therefore ^20 is a point in the inter-
section of the two planes. In like manner, q^^Q is a point in the
intersection, therefore the line p2o(Zio is the intersection of the given
planes.

If the given scales of slope are parallel, the above construction will
evidently fail, because the horizontals of both planes will be parallel,
and therefore never meet. In this case a third plane may be taken,
not parallel to either of the given planes, and its intersection with
each of them found by the method just given. This determines two
lines whose intersection with one another will be a point in the
intersection required. It is evident that the intersection of the given
planes in this case will be a horizontal line, therefore a line perpen-
dicular to the given scales of slope through the point found determines
the intersection of the given planes.

If the third plane mentioned above be taken perpendicular to each
of the given planes, it will be a vertical plane, and the plans of its
intersection with the given planes will coincide, and an elevation of

HORIZONTAL PROJECTION' 283

them must be drawn to determine the point where they meet. This
elevation is best taken on the third plane itself.

If the given scales of slope are nearly parallel, the horizontals of
the planes will meet at a very acute angle, and it is more accurate to
find two points in the required intersection by the method explained
for planes whose scales of slope are parallel, that is, by cutting each of
the given planes by two other planes.

250. Intersection of a Line and Plane. — Let aj)^^ (Fig. 551)
be a given line and cd a given plane. Through any two points a and
6 in the given line, draw the parallel lines ar and
hq in any convenient direction. Through c and
d, points on the scale of slope having the same
indices as a and h respectively, draw the hori-
zontals cr and dq^ meeting the parallels ar and hq
respectively at r and q. Join rq. The point P
where the line rq or rq produced meets ah or ah
produced is the plan of the point where the given
line meets the given plane.

The theory of the above construction is as
follows — «28*'28 ^^d ?>ia5i3 are horizontals of a plane
containing the given line. The line r^^ia ^^
evidently the intersection of this plane with the

given plane. The point V is therefore a point in both planes and also
in AB, therefore it is the intersection required.

The intersection may also be found by taking an elevation of the
line and plane on a ground line parallel to the scale of slope.

251. The Normal to a given Plane through a given Point.
— The plan of a line which is perpendicular to a plane is at right angles
to the horizontal trace of the plane, and will therefore be parallel to its
scale of slope.

To figure the plan of the line (which of course passes through the
figured plan of the point), determine the trace of the plane and the
elevation of the point on a vertical plane parallel to the scale of
slope. Through the elevation of the point draw a perpendicular to the
trace of the plane, this perpendicular will be the elevation of the
normal and from it the plan may be figured.

252. The Plane through a given Point perpendicular to a
given Line. — Taking the figured plan of the line as a ground line,
determine the elevations of the point and line. Through the elevation
of the point draw a line perpendicular to the elevation of the given
line. This perpendicular will be the vertical trace of the required
plane. The scale of slope of the plane will be parallel to the given
figured plan of the line, and it may be graduated from the vertical trace
found as above.

253. Contour Lines. — The plan of a portion of the earth's
surface is made to show the form of that surface very clearly, by
drawing on it the sections of the surface by a series of horizontal
planes, at equal distances from one another. These sections are called
contours or contour linen. It is evident that the relative closeness of the

284 PRACTICAL GEOMETRY

contour lines shows the relative steepness of the different parts of the
surface, the surface being steepest where the contour lines are closest
together. It is usual to affix to the contour lines their heights above
some fixed horizontal plane.

254. Intersection of a Plane and a Contoured Surface.—
Fig. 552 shows the scale of slope of a plane and the contoured plan of

Fig. 652.

a surface. To determine the intersection of the plane and the surface,
draw the plans of a number of horizontal lines lying in the plane, and
having the same indices as the contour lines. The points where these
lines meet the contour lines having the same indices, are points in the
intersections required. The complete intersection is obtained by drawing
a fair curve through the points thus found.

255. Contouring a Surface from its Equation. — A formula
or equation which contains three variables is represented graphically
by a surface and in representing this surface on paper a system
of contour lines may be necessary. As an example take the

, ^^ (62800 - Si^)d:'v , . , . , . . .

formula, H = 9^0000 which gives the relation between the

horse-power H transmitted by a cotton rope and the diameter d of
the rope in inches and the velocity v of the rope in feet per second,
allowing for the stress in the rope due to centrifugal force.

For each diameter of rope the relation between H and v is shown
by a plane curve, and if the curves for ropes of different diameters be
constructed these curves will be contours of the surface which
represents the original formula. The contours of the surface correspond-
ing to diameters of J inch, 1 inch, 1^ inches, 1^ inches, 1| inches, and
2 inches are shown at (a) Fig. 553 between the limits v = 0, and v = 1 20.
To construct these curves it is first necessary to construct, by calculation,
the table given on the opposite page.

HORIZONTAL PROJECTION

285

Horse-power H, for different values of v and d.

Diameter d.

1* 1 U

1-5

2-7

4-2

8-3

10-9

5-4

8-4

121

16-4

21-4

4-4

7-8

12-2

17-6

24-0

31-4

5-7

10-1

15-8

22-7

30-9

40-3

6-8

12-0

18-8

27-0

36-8

48-1

7-6

13-6

21-2

30-5

41-5

54-3

8-2

14-6

22-9

32-9

44-8

58-6

8-5

15-2

23-7

34-1

46-4

60-7

83-5

8-6

15-2

23-8

34-2

.46-6

60-8

8-5

151

23-5

33-9

46-1

60-3

100

8-0

14-3

22-3

32-1 .

43-7

57-0

110

7-1

12-7

'19-8

28-5

38-8

50-7

120

5-8

10-2

16-0

230

31-3

40-9

In addition to the velocities at intervals of 10 feet per second
given in the first column of the table it will be noticed that the
velocity 83-5 feet per second is given. This velocity 83 5 is the
velocity at which the horse-power for any given rope is a maximum.
The velocity 83-5 is obtained by calculation as follows. The horse-
power for any rope will evidently be a maximum when 62800 v — 3v'^

is a maximum. Let y = 62800?; — 3v\ Differentiating this, -=^

= 62800 — %^ and yisa, maximum when 62800 — 9v^ = o, that is when
3v = V62800; or V = 83-5.

Assuming that the curves at (a) Fig. 553 are in vertical planes,
horizontal sections of the surface at levels whose intervals are 10
horse-power are shown at (6) Fig. 553 between the limits d = ^
and d = 2,

The student should work out this example to the following scales.
H, 1 inch to 10 horse-power, d, 1 inch to J inch diameter, v, 1 inch
to 20 feet per second. In addition to the contours shown at (a)
Fig. 553 the student should add the curves for ropes of | inch,
IJ inches, If inches. If inches, and 1~ inches diameter. Also in
addition to the contours shown at (p) he should add the curves for
5, 15, 25, 35, 45, and 55 horse-power. Lastly, on a base line parallel
to MN he should construct the contours of the surface showing the
relation between H and d for velocities at intervals of 10 feet per
second. These contours are vertical sections of the surface parallel
to MN just as the curves at (a) are vertical sections by planes parallel
to LM.

280

PRACTICAL GEOMETRY

10 20 30 40 50 60 70 80

Velocity V 83-5

90 100 ilO 120

10 20 30 40 SO 60 70

WjfcUy V «3-5

Fig. 553.

90 too 110 120

10 20 30 40 50 60 70
Velocity V.
Fig. 564.

90 100 110 120

HORIZONTAL PROJECTION

287

An oblique parallel projection of the surface which has just been
considered is shown in Fig. 554. The student should have no difficulty
in making such a projection after he has studied Chapter XXI. This
projection may be very readily drawn on squared paper, taking the axis
for (Z at 45° to the axis for v.

When the student draws the oblique parallel projection shown in
Fig. 554 he should put in the additional contours suggested with
reference to Fig. 553.

Exercises XX

Note. Unless otherwise stated the unit for the indices is Vg or 0*1 of an inch, so
that an index 25 denotes a height of 2-5 inches above the horizontal plane.

1. Two points a and b in the plan of a straight line are -2 inches apart. The
index of a is 3 and the index of b is 11. Determine, (1) the index of a point c in
ab which is J inch from a ; (2) a point d..^ in the plan of the line ; (3) a point e^
in the plan of the line ; (4) a point /-j in the plan of the line ; (5) the true length
of EF.

2. Draw a triangle a^ 6-4 Cig {ab = 2-7 inches, 6c = 2*2 inches, ca = 1*7 inches).
Find a point d in be whose index is 8, and show the figured plans of two
straight lines passing through B and C, and parallel to AD.

3. A straight line making an angle of 30"^ with the ground line is both hori-
zontal and vertical trace of a plane. Show the scale of slope of this plane.

4. Show the scales of slope of two parallel planes inclined at 50°, the distance
between the planes being 0*7 inch.

5. Determine the scale of slope of the plane of the triangle given in exercise 2,
also the true shape of the triangle.

6. An equilateral triangle of 2*5 inches side has
its angular points indexed 3, 12, and 15. Show the
figured plans of the bisectors of the angles of the
triangle of which the given triangle is the plan.
Draw also the ellipse which is the plan of the cir-
cumscribing circle of the triangle.

7. Draw the scale of slope of a plane whose
inclination is 60° and show the figured plan of a
triangle ABC which lies in this plane. The inclina-
tions of AB and BC are 30"^ and 45° respectively,
the indices of a, b and c are 3, 24, and 8 respec-
tively.

8. A plane is inclined at 35° to the horizontal,
the lines of steepest upward slope going due East.
A second plane has an inclination of 52°, the direc-
tion of the lines of upward slope being due North.
Represent the planes by scales of slope, unit for
heights 0-1 inch. Show the figured plan of the inter-
section of the planes. Find and measure the angle
between the planes. [b.e.]

9. Fig. 555 is a plan, drawn to a scale of 1 inch
to 200 feet, showing at a, b two places A, B on a
hill-side, the surface of the ground being an inclined
plane represented by a scale of slope, heights being indexed in feet. Draw Fig.
555 to a scale of 1 inch to 100 feet and add the plan of a zigzag path connecting
A and B, made up of three straight lengths of a constant inclination equal to
half that of the hill, and deviating equally on each side of a straight line through
A and B. Ascertain and measure in feet the total length of this path. [b.e.]

^^ ._

140

120

iOO

-80

-40

Fig. 555.

288

PRACTICAL GEOMETRY

10. Two lines AB and CD are given by their figured plans in Fig. 556. Draw
the scale of slope of a plane which will
contain the line CD and make an angle of
15° with the line AB.

11. The plans of six points a^ h^, Cj^, d^^,
C20, and /as, taken in order, are situated at
the angular points of a regular hexagon of
IJ inches side. Find the intersection of the
plane containing A, C, and F, with the plane
containing B, D, and E, and state its in-
clination.

12. Draw a square a^^ b^^ c^^ d^ of 2 inches
side. Determine the plan of the sphere on
whose surface the points A, B, C, and D are
situated.

13. Draw the figured plan of the common perpendicular to the straight lines
AC and BD of the preceding exercise.

14. ^50 6eo Cgo ^50 (Fig. 557) is the figured plan of a straight roadway which
is to be made, partly by cutting, and partly by embankment, on the ground

—

■yy

—

_•

*■

«r

'C3

Fig. 556.

Fig. 557.

whose surface is given by its contoured plan. The sloping faces of the cutting
and embankment are to be inclined at 40°. Draw the plan of the intersections of
the faces of the cutting and embankment with the surface of the ground.

Show also, on EF as ground line, vertical sections at LL, MM, and NN.

N.B. — Fig. 557 is to be enlarged four times. The unit in this exercise is 1 foot.

Fig. 658.

HORIZONTAL PROJECTION

289

15. The surface of a piece of ground is given by contours at vertical intervals
of 5 feet (Fig. 558), the linear scale of the plan being 1 inch to 100 feet. A road
is to be cut at the given heights, the face of the cutting on each side having a
slope of 38° to the horizontal. Draw Fig. 558 to a scale of 1 inch to 50 feet and
complete the plan of the finished earthwork. [b.e.]

1 1 h ([

1 1 1 /Vti

t J- ^ -r tttU

t r ^11 ^ , TZutif

/„7 l.*^^ Vi t-tttW

Ml t:l u t tl ttu-

"t^^'U.' L o / , -'f

±1 z -/:--:#/-,- i--i- / ;/ s

113 ^jLJ Lr. J*.._j4ia_

4LI 4 1 ^( 1 ^ s

jzti 1 '^^4^1 ' t-?-4-

"Vr-g-ii: iiirsirs « "

(a)

Fig. 559.

16. You are given, in Fig. 559, the plan of the surface of a piece of , ground,
contoured in feet, the linear scale being 1 inch to 100
feet. The curve pp is the centre line of a road, 20 feet
wide, which is to be made partly by cutting and partly
by embankment, the former having a slope of 45^ and
the latter one of 38° to the horizontal. Draw Fig. 559
to a scale of 1 inch to 50 feet and complete the plan of the
finished earthwork within the limits of the data. [b.e.]

17. Represent by contour lines, as in the example
worked out in Art. 255, the surface whose equation is

M =- — ^. The axes to be arranged as shown at' (a), (h),

t "~ oU
and (c), Fig. 560. v denotes velocity in feet per minute,
and t denotes temperature in degrees Fahrenheit. Scales.
— For V, 1 inch to 50 feet per minute. For <, 1 inch to
20°. For M, 2 inches to 1. The limits of v and t are
shown at (a) and (6). j; —

Fig. 560.

CHAPTER XXI

PICTORIAL PROJECTIONS

256. Pictorial Projections. — Since the orthographic projection
of a line only shows its true length when the line is parallel to the
plane of projection, it is usual, in making working drawings of an object,
to arrange it so that as many of its lines as possible are parallel to at
least one of the co-ordinate planes. The object is then in a simple
position. Fig. 561 shows a plan and two elevations of a rectangular
solid when the solid is in the simplest possible position in relation to
the planes of projection. These projections are very easily drawn, but,
although they represent the solid completely, they are not at all
" pictorial," and a special training is necessary before the observer can
form from them a correct mental picture of the object.

FRONT ELEVATION. END ELEVATION

Fig. 561.

Fig. 562.

Now let the solid be tilted over as shown by the plan a and eleva-
tion a'. Fig. 562, and let a new elevation a/ be drawn in the manner
explained in Chap. XIV. It will now be observed that the elevation
a/ by itself gives a much better idea of the form of the object than
any of the other projections shown in Figs. 561 and 562, but unless a
much simpler method of drawing such a pictorial projection than that
shown in Fig. 562 can be devised, and unless some simple method of

PICTORIAL PROJECTIONS

291

measuring the various dimensions of the object direct from the drawing
can be given, such a pictorial projection would be of little value as a
substitute for ordinary working drawings.

In the subsequent articles of this chapter, methods are described
which enable the draughtsman to make pictorial projections resembling
that shown at a/ in Fig. 562 with almost as little trouble as is required
for the simple projections of the kind shown in Fig. 561, and it will be
found that the dimensions of the object represented may be determined
from these pictorial projections as easily as from the ordinary
projections.

Fig. 563 shows an isometric projection of the solid represented in
Figs. 561 and 562, and Fig. 564 shows an oblique parallel projection of

Fig. 663.

Pig. 564.

the same solid, and a glance at either of these pictorial projections
conveys to the mind at once a clear picture of the object. Since no
lines other than those shown require to be drawn, and since the
dimensions of length, breadth and thickness may be measured directly
on these projections, it will be seen that such drawings may often be
most useful.

The theorems upon which the methods of this chapter depend
are : —

(1) Parallel lines have parallel projections.

(2) Parallel lines have equal inclinations to the same plane.

(3) Parallel lines and lines which are equally inclined to the plane of
projection have the lengths of their projections to the same scale.

In stating the above theorems it is assumed that the projectors
are parallel to one another.

257. Isometric Projection. — When a solid is rectangular each
edge is parallel to one or other of three lines or axes which are
mutually perpendicular, and when the solid is placed so that these
axes are equally inclined to the plane of projection their projections
make angles of 120"^ with one another, and the projections of the
edges of the solid may be measured with the same scale since the edges
are parallel to lines which are equally inclined to the plane of projec-
tion. Hence follows the simple construction shown in Fig. 565 for
making an isometric projection of a rectangular solid. The edges of
the solid meeting at one angular point are taken as axes and their

292

PRACTICAL GEOMETRY

plans oa, oh, and oc are drawn by the aid of the T-square and 30° set
square as shown. The lengths of the plans of the edges are marked
off with the isometric scale, the
construction of which is explained
in the next article.

The axes referred to in this
article are called isometric axes
and a plane containing any two
of them or a plane parallel to
this plane is called an isometric
plane. The word isometric means
equal measure. It should be
noted that only those lines which
are parallel to one or other of
the isometric axes are isometric
lines.

258. The Isometric Scale.
— Let oa, oh, and oc (Fig. 566) be the projections of three lines OA,
OB, and OC which meet at O and are mutually perpendicular and
which are equally inclined to the plane of projection ; it is required to
find the ratio of say 06 to OB.

Draw he at right angles to oa. On he as diameter describe a
semicircle cutting oa at O^. 0^ is the true length of the line OB of
which oh is the projection. For the proof of this construction see Art.
199, p. 229, on the projection of a solid right angle.

Fig. 565.

Fig. 566,

ISOMETRIC SCALE.

ORDINARY SCALE.

Fig. 567.

Fig. 568.

If ao be produced to cut he at d, then it is easy to show that,
angle dhO^ = angle dO^b = 45° ; also that angle dho = 30°, and
angle dob = 60° ; hence angle ohO^^ = 15°. Then it follows that
oh : O16 : : a/2 : ^3, that is, the length of the isometric projection of a
line is to the true length of the line as ^2 is to a/^. Hence to draw
an isometric scale corresponding to a given ordinary scale it is necessary
to get two intersecting lines whose lengths are to one another as;
V2 to V3.

Fig. 567 shows one construction for the isometric scale. EF is the
ordinary scale. Angle FEH = 15°. Angle EFH s= 45°. EH is the
isometric scale, the sub-divisions of which are obtained by drawing lines
through the sub-divisions on EF parallel to FH a§ ibown. \ti w^l be.

PICTORIAL PROJECTIONS

293

seen that the triangle EFH in Fig. 567 is similar to the triangle hOiO
in Fig. 566.

Another construction for the isometric scale is shown in Fig. 568.
Angle FHK = 90°. FH = HK. HE = FK. The ordinary scale is
made on EF and the isometric scale on EH as shown. If FH = 1,
HK = 1, and FK = V^ = EH. Hence EF = ^3 and EH : EF

::V2:V3.

In actual practice in making drawings in isometric projection, the
isometric scale is seldom used. The lengths of the isometric lines are
marked off directly from the ordinary scale. The result of this is that
the isometric drawing is larger than it would be if an isometric scale
were used in the ratio of a/ 3 to a/ 2.

259. Projection of any Given Figure lying in an Iso-
metric Plane.— Let ABCDE (Fig. 569) be the given figure, the
plane of the paper being the plane of the figure, and let OX and OZ be
two lines at right angles to one another in that plane. Let the plane
XOZ be tilted up until the axes OX and OZ and the axis OY perpen-
dicular to OX and OZ are equally inclined to the plane of projection.
The projections ox, oz, and oy (Fig. 570) will be isometric axes. It is
required to add to the projections of the axes the projection of the
figure ABCDE.

From the extremities of the straight sides of ABCDE draw parallels
to OX to meet OZ at Ai, B,, C^ and Dj. Also, from a sufficient
number of points on the curved
side AED, draw parallels to
OX to meet OZ as shown in
Fig. 569.

The isometric projections
aa^, bbij cci, etc. of the lines
AAi, BBi, CCj, etc. can now be
determined and these determine
the projections a, b, c, etc. of
the points A, B, C, etc. Hence
the required projection of the
given figure can be completed.

260. Projection of a
Circle lying in an Isometric

Plane. — The method of the preceding article may be applied to find
the projections of a sufficient number of points on a circle lying in an
isometric plane, and a fair curve drawn through the points thus deter-
mined is the projection required. This is shown in Fig. 571, where two
diameters of the circle at right angles to one another are taken as two
of the three axes. Since, however, the orthographic projection of a
circle is an ellipse, a better construction is to find the axes of the
ellipse and then determine a sufficient number of points on it by means
of a trammel as explained in Art. 45, p, 41.

Referring to Fig. 572, take two diameters of the circle at right
angles to one another as the axes OX and OZ. Draw the inscribed
square ABCD having two sides parallel to OX and two sides parallel

Fig. 569.

294

PRACTICAL GEOMETRY

^A^^^' ^^ ^^® isometric projection, aocy the projection of the diameter
AOC will evidently be perpendicular to oy the third isometric axis.

Fig. 571.

Fig. 572.

aoc is therefore the projection of the horizontal diameter of the circle
and will therefore have a length equal to the true diameter. Con-
sequently aoc must be the major axis of the ellipse which is the
projection of the circle. Also hod must be the minor axis of the ellipse.
It is easy to show that the angle hao is 30°.

Hence, having found o the isometric projection of the centre of the
circle, the major and minor axes of the ellipse which is the isometric
projection of the circle are found as follows. — Draw aoc perpendicular
to the third isometric axis, and make oa equal to oc equal to the true
radius of the circle. Draw ah inclined at 30° to oa, or draw ah parallel
to the isometric axis ox. Draw hod perpendicular to aoc to meet ah at h.
Make od equal to oh. aoc and hod are the required axes of the ellipse.

If the ellipse is the projection of a circle connected with an object
which is to be projected isometrically without using an isometric scale,
then oa must be made greater than the true radius of the circle in the
ratio of the ordinary scale to the isometric scale, that is, in the ratio of
V3 : V2. .

ELEVATION.

Fig. 573.

261. Isometric Projection of an Object which is not
Rectangular. — If three axes, mutually perpendicular, be taken in

PICTORIAL PROJECTIONS 295

relation to any object, the object is said to be projected isometrically
when the three selected axes are projected isometrically ; but only
those Jines on the object which are parallel to the selected axes will
be projected truly isometrically.

The general construction is to locate a sufficient number of points
on the outline of the object by perpendiculars from them on to the
planes containing the axes referred to above, and then find the isometric
projections of these perpendiculars. For example, take the case of the
pyramid shown in Fig. 574 by ordinary plan and elevation. Take a
vertical axis OZ through the vertex V and two axes OX and OY at
right angles to one another and in the horizontal plane of projection.
Fig. 573 shows these three axes projected isometrically and the projec-
tion of the pyramid built up on these axes.

262. Axortietric Projection.^ — In axomeinc projeciion the three
principal axes of a rectangular object are not all equally inclined to
the plane of projection, and although all those lines which are parallel
to one axis have their projections drawn to the same scale, those which
are parallel to another axis will, in general, have their projections
drawn to a difierent scale. Thus in the axometric projection of a
rectangular box, if no two axes have the same inclination to the plane
of projection, one scale is required for measuring its length, another
for its breadth, and a third for its depth. These three scales are
determined in a manner similar to that for the isometric scale shown
in Figs. 566 and 567. The angles at E and F, Fig. 567, must of course
be found from the figure corresponding Ao Fig. 566. The angle at E
(Fig. 567) must be made equal to the angle O^to, Fig. 566, and the
angle at F equal to the angle hO^o, The scale thus found will be the
scale for measuring lines parallel to oh.

It will be observed that isometric projection is a particular case of
axometric projection.

263. Oblique Parallel Projection. — In parallel projection the
projectors from the different points of an object to the plane of
projection are parallel to one another. In orthographic projection the
projectors are perpendicular to the plane of projection and are therefore
parallel to one another ; hence, orthographic projection is a particular
case of parallel projection. It is usual, however, to restrict the term
parallel projection to the case where the parallel projectors are inclined
or oblique to the plane of projection. By parallel projection is therefore
meant oblique parallel projection.

A special and useful case of parallel projection is that in which
the plane of projection is parallel to one of the principal faces of the
object to be projected, and the projectors are inclined at 45° to the
plane of projection. It follows that all faces of the solid which are
parallel to the plane of projection will have for their projections
figures which are of the same size and shape as the faces themselves,
also all lines on the object which are perpendicular to the faces just
mentioned will have their projections of the same lengths as the lines
themselves.

In Fig. 575, ab is the ordinary or orthographic plan, and a'b' the

296

PRACTICAL GEOMETRY

corresponding elevation of a rectangular solid placed with two parallel
faces parallel to the vertical plane of projection and with two other
parallel faces horizontal. The solid being in this simple position in
relation to the planes of projection, another projection of it, A'B', is
made on the vertical plane of projection, the projectors being parallel
to one another but inclined at 45° to that plane of projection. The
elevations of these projectors are shown inclined to XY at 30°, but
they may be inclined at any other angle. Having fixed the inclinations
of the elevations of the projectors to XY the direction of their plans is
found from the further condition that the true inclination of the
projectors to the plane of projection is 45°.

X-\ 1^

Fig. 575.

Fig. 576.

It will now be seen that the projection A'B' may be drawn directly
without any preliminary plan and elevation. The rectangle A'D' is
first drawn, being of the same size and shape as the front face of the
solid. A'C is then drawn at 30° (or any other convenient angle) to
A'E', and its length is made equal to the true length of the edge of
which it is the projection. The projection may now be completed by
drawing parallel lines as shown.

Fig. 576 shows parallel projections of solids drawn as described
above, and the student should draw these, full size, to the dimensions,
in inches, marked on them.

If desired, the projections of those lines on the object which are
perpendicular to the front face, which is parallel to the plane of
projection, may be drawn to a scale which is either
smaller or larger than that used for the projection
of the front face, and the resulting projection will
be a correct parallel projection of the object. This
simply means that the inclination of the parallel
projectors to the plane of projection is taken either Fig. 677.

greater or less than 45°. For example, referring
to Fig. 577, if the face A'D' is drawn full size the length A'C may be
drawn half full size and the complete parallel projection of the solid

IPn

b^B

PICTORIAL PROJECTIONS

297

will then be as shown. When different scales are used the scales
;jhould be stated on the drawing.

When a face of the object perpendicular to the front face contains
a figure made up of curved lines, or lines which are not parallel to the
edges of that face, the projection of such a figure is determined by
using co-ordinates in the manner described in Art. 259 for isometric
projection. Also a parallel projection of an object of any form not
rectangular may be made by using co-ordinates as described in Art. 261
for isometric projection.

Exercises XXI

Note, The pictorial projections asked for in the following exercises are to be
drawn, as far as possible, directly without first drawing the object in ordinary plan
and elevatimi.

1-12. Ordinary orthographic projections of various objects are given in Figs.
578 to 589. Represent these objects in isometric projection.

For the dimensions of the objects shown in Figs. 578 to 589 take the sides of the
small squares as half an inch.

Fig. 578.

^' 1 1 -^ K k' "^-1 fr \a^ AT].

J i^E ij ^>o .i- i 4lvt

z ^ ^ r z\ \ //\\

1 1 r'^ ' > 1

j" 1 '11]

r\\/y

i--'-lt -~ X-^k^iiS^

oF.. X loK x] y oi.i 1 X VA^

Fig. 579. Fig. 580. Fig. 581. Fig. 582.

oyi\ — X -I — p — f-x

\o\ X _J__^_L_ac

^T^^

rEi-r::

S /

^^-^ _

J>^

-3

T^?^

-r

1 ■ ^

^'-

/ \

-O-

v-j

±:

u^±

Fig. 583. Fig. 584.

Fig. 585.

Fig. 586.

298

PRACTICAL GEOMETRY

^-"i "W

S '^

c-i ,.:S..

\ f

Mt\

SB^/\i

^-i

\y /-^ / /

_ 1

g^gg

^ /\

c-l

,,*^

^rf**

Fig. 687.

Fig. 588.

Fig. 589.

13-24. Kepresent the objects shown in Figs. 578 to 589 in axometric pro-
jection, the axes to be arranged as shown in Fig. 590.

25-36. Represent the objects shown in Figs. 578 to 589 in oblique parallel
projection, the axes to be arranged as shown
in Fig. 591. The scale is to be the same
for all the axes of projection.

37. The axes of two equal cylinders (2
inches in diameter) and their common per-
pendicular are in isometric projection.
Show the intersection of the cylinders with-
out using an auxiliary plan or elevation,
(a) when their axes intersect, (&) when
their axes are \ inch apart where they are nearest together.

38. The same as exercise 37 except that the projection of the axis of one
cylinder is parallel to ox and the projection of the axis of the other is parallel to
07/, Fig. 590.

39. Represent the cylinders referred to in exercise 37 in oblique parallel pro-
jection. The projection of the axis of one cylinder to be parallel to ox and the
projection of the axis of the other to be parallel to oy, Fig. 591. Show the inter-
section of the cylinders. The scale is to be the same for all the axes of projection.

40. A solid made up of a hemisphere and part of a circular cylinder is
shown in Fig. 592. The centre of the base of the hemisphere is on the axis of the
cylinder. Draw an isometric projection of this solid.

41. Fig. 593 shows a locomotive crank axle with the
fillets at the junctions of the various cylindrical parts with
one another and with the crank arms left out. Draw an
isometric projection of this axle to a scale of 1^ inches to
a foot.

42. Draw an oblique parallel projection of the crank
axle shown in Fig. 593 to a scale of IJ inches to a foot.
The axes of projection to be arranged as shown in Fig.
591, ox being the projection of the axis of the axle.

43. The relation between the pressure, P (in lb. per
square inch), the volume, V (in cubic feet), and the abso-
lute temperature, T (in degrees Fahrenheit) of 1 lb. of air
is given by the equation PV = 0*37 T. Fig. 594 shows,
in oblique parallel projection, contours of this surface,
between pressures of 10 and 100 lb. per square inch, at intervals of 500° from
500° to 2500°, the planes of these contours being parallel to the front pressure-

FiG. 592.

PICTORIAL PROJECTIONS

299

volume plane. Draw this Fig. to the following scales. — Pressure, 1 inch to 20
lb. per square inch. Volume, 1 inch to 4 cubic feet. Temperature, 1 inch
to 600^.

"~~"

_.-

K-7-^

K- 9 —

*A*

Jl
.4.

^-l!--*^

K4^.

k4^

k4*

*- 9 -*h-7-H

-00-

* ■ ■

' — '

Fig. 593.

Add the contours whose planes are parallel to the bottom volume-temperature
plane at intervals of 20 lb. per square inch. Also, add the contours whose planes

r"A -y'2S00

VOLUME. 18-5

Fig. 594.

are parallel to the left hand pressure-temperature plane at intervals of 4 cubic
feet.

Note that the curves in Fig. 594 are rectangular hyperbolas points on which
may be found by the construction given in Art. 50, p. 47.

CHAPTER XXII

PERSPECTIVE PROJECTION

264. Difference between Perspective and Orthographic
Projection. — In perspective, conical, or radial projection all the projec-
tors converge to a point, while in orthographic projection they are per-
pendicular to the plane of projection. The perspective projection of
an object represents it as it would actually appear to the eye of an
observer placed at the point to which all the projectors converge.

In perspective projection, the plane of projection, called the picture
plane, may be situated anywhere but it should meet all the projectors,
otherwise the projection or picture as it is called would be incomplete.
Generally the picture plane is assumed to be placed in a vertical
position between the eye of the observer and the object, and perpen-
dicular to the direction in which the observer is looking.

The point to which all the projectors converge is called the station
point or point of sight.

265. Direct Method of drawing a Perspective Projection.
— The following method of drawing a perspective projection of an
object follows directly from the definition of perspective projection,
and it will be seen that it is simply the determination of the plane
section of a pyramid or a combination of pyramids which have a
common vertex.

First draw the orthographic projections of the object, station point,
and picture plane so that the latter is perpendicular to the ground
line XY, as shown in Fig. 595, which illustrates the method applied
to a square prism standing on the ground, oc, o'c' is the picture plane
and 8s' is the station point.

Consider the edge ab, dV of the prism. Draw the plans sa, si and
the elevations s'a', s'b' of the conical or radial projectors of the extre-
mities of the edge under consideration. These radial projectors meet
the picture plane at points of which a and f3 are the plans and a and /3'
the elevations. Through a point o in oc draw on parallel to XY. With
centre o and radii oa and of3 draw arcs of circles to cut on at a^ and p^
respectively. From a^ and /?i draw perpendiculars to XY to meet
lines through a and ^' parallel to XY at A' and B' respectively. A
straight line A'B' will be the perspective projection of the edge ab, a'b'
of the prism. In like manner the perspectives of the other edges may
be obtained as shown.

It is evident that by the first part of the foregoing construction

PERSPECTIVE PROJECTION

301

edge views of the required picture are obtained, one being a plan and
the other an elevation, and by the second part of the construction the
true shape of this picture is determined.

The points c and c' are the feet of the perpendiculars from 8 and s'
respectively to the edge view of the picture plane. If with centre o
and radius oc an arc of a circle be drawn to cut on at fi, and if c^G be
drawn perpendicular to XY to meet the horizontal through a' at 0,
then C will be the position, in relation to the picture A'B', of the foot
of the perpendicular from the station point to the picture plane.

The foot of the perpendicular from the station point to the picture
plane is called the centre of vision. It will be observed that the centre
of vision is the orthographic projection of the station point on the
picture plane.

The horizontal line drawn on the picture plane through the centre
of vision is called the horizontal line.

Fig. 595.

The intersection of the picture plane with the ground plane is
called the ground line or base line.

The plane perpendicular to the ground, and containing the station
point and centre of vision is caUed the vertical plane.

The line in which the vertical plane intersects the picture plane is
called the vertical line.

266. The " School of Art " Method of drawing a Perspec-
tive Projection. — The direct method of drawing a perspective

302

PRACTICAL GEOMETRY

projection, described in the preceding Article, is exceedingly simple in
principle, and may be quickly learned, and is easily understood by a
student possessing a little knowledge of elementary solid geometry.
The direct method may be applied to draw the perspective of any solid
in any position, and in some cases it is the best method to use, but if
the solid has many parallel lines on it the construction can be much
simplified and greater accuracy obtained by making use of a number
of principles which are explained in succeeding articles. These prin-
ciples are applied in what may be called the " School of Art " method
of making a perspective drawing. Afterwards a combination of the
two methods very commonly used in practice will be described.

267. To determine the Perspective of a Given Point. — The
point is given by stating its distances from the ground and from the
vertical and picture planes. The first and second of these distances
determine the position of the orthographic projection of the point on
the picture plane.

In Fig. 596 the plane of the paper is the picture plane, C is the
centre of vision, CH is the horizontal line, CL is the vertical line, and
f' is the orthographic projection of the given point on the picture
plane. On CH make CD equal to the distance of the station point
from the picture plane. Draw p'K parallel to CH, and make p'K
equal to the distance of the given point P from the picture plane.
Join DK and jp'C. The point F where these lines intersect is the
perspectiTe of the point P. It is evident that CP' : Yp' ; : CD : p'K.

Fig. 696.

Fig. 597.

Referring now to Fig. 597 which is a distorted perspective projec-
tion of the lines of Pig. 596 together with the point P and the station
point S. Since SC and Pp' are perpendicular to the picture plane they
are therefore parallel to one another, and the triangles SCP' and Pp'P'
are similar. Hence CP' : Fp' : : SC : Pp' ; but CD is equal to SC and
p'K is equal to P^, therefore CP' : Pp' : : CD : p'K. But F is the point
where SP cuts the ^picture plane and is therefore the perspective of
the point P.

The point D, Fig. 596, is called the 'point of distance.

268. The Perspectives of Parallel Lines converge to a
Point.— Let A'B' and CD' (Fig. 598) be the perspectives of the
parallel lines AB and CD respectively. Let SV be a line parallel to

PERSPECTIVE PROJECTION

303

A', B', and V are in
same straight line.

Fig. 598.

AB and CD, and if it meet the picture plane at all, let it meet it at Y.
SV will be the line of intersection of the planes ABS and CDS.

Since each of the points A', B', and V are in the picture plane,
and also in the plane ABSV they must all lie on the intersection of
these two planes ; there-
fore
the

Similarly C, D', and V
are in a straight line ;
therefore A'B^ and CD',
the perspectives of the
parallel lines AB and CD,
converge to the point V.

The point towards
which the perspectives of
parallel lines converge is
called the vanishing point
of these lines.

The following particu-
lar cases are worthy of notice : —

(a) When the lines are perpendicular to the picture plane, SV will
also be perpendicular to that plane, and V will coincide with the centre
of vision. Hence, the perspectives of lines which are perpendicular to the
picture plane, converge towards the centre of vision.

(h) When the parallel lines are parallel to the picture plane, SV
will also be parallel to that plane ; therefore V will be at an infinite
distance from S, and A'B' and CD' will be parallel to one another.
Hence, the perspectives of parallel lines ichich are also parallel to the
picture plane, are parallel to one another.

(c) When the parallel lines are horizontal, SV will also be hori-
zontal ; therefore V must be on the horizontal line. Hence, the per-
spectives of parallel horizontal lines converge towards a point on the
horizontal line.

{d) When the parallel lines are parallel to the vertical plane, S V
will be in the vertical plane ; therefore V will be on the vertical line.
Hence, the perspectives of parallel lines ivhich are also parallel to the
vertical plane, converge towards a point on the vertical line.

e) The perspectives of vertical lines are parallel to the vertical line.
/) The perspectives of horizontal lines which are parallel to the
picture plane are parallel to the horizontal line.

(cj) The perspectives of parallel horizontal lines which are inclined at
45° to the picture plane converge towards the point of distance.

269. To Determine the Vanishing Point of a Line. —
Take the picture plane and the horizontal planfe containing the
station point for co-ordinate planes, and on these draw a plan cd) and
elevation a'V of the line whose vanishing point is required (Fig. 599).
HC, the horizontal line, will be the ground line for this plan and
elevation.

Draw Cs perpendicular to HC and equal to the distance of the

304

PRACTICAL GEOMETRY

Fig. 599.

station point from the picture plane, s will be the plan of the station

point and C its elevation. Through s draw sv parallel to ah, meeting

HC at V. Through C draw CV parallel

to a'h' to meet the perpendicular from v

to HC at V. The point V is the required

vanishing point, because V is the point in

which a line through the station point

parallel to the original line AB meets the

picture plane.

If the original line is horizontal CV
will coincide with HC, the horizontal line ;
therefore Y is on the horizontal line, and
the plan sv (or sV) makes an angle with
the horizontal line equal to the inclination of the original line to the
picture plane.

270. To mark off on the Perspective of a Line a Part
whose True Length is given. — Let TV (Fig. 600) be the per-
spective of a line whose trace on the
picture plane is T, and whose vanishing
point is V. As in Fig. 599, s is the
plan of the station point, and sv the
plan of the line which passes through
the station point, arid is parallel to the
line of which TV is the perspective.

With centre v and radius vs de-
scribe the arc sSg meeting the hori-
zontal line HC at Sg. VS2 is the real
distance of the vanishing point V from -p^Q^ 500.

the station point. Through V draw

VM parallel to HC, and with centre V and radius VS2 describe the
arc SgM.

Draw TQ parallel to HC. Let A' be a given point in TV. Draw
MA' and produce it to meet TQ at P. Make PQ equal to the given
length. Draw QM meeting TV at B'. The true length of the part
whose perspective is A'B' will be equal to the given length.

The construction is much simplified when the original line is hori-
zontal, for then V is on the horizontal line, and sv is equal to the true
distance of V from the station point
The construction for this case is shown
in Fig. 601.

The theory of the construction
shown in Figs. 600 and 601 will be
understood by reference to Fig. 602,
which is a perspective view of the
ground and picture planes, and the
various lines in their natural positions.

In Figs. 600, 601, and 602, the same
letters denote the same points. S is the station point, TAB the line
whose perspective is TV and V is its vanishing point. T is the trace

y_ / \V-^^

-•^

T \

M J

"/'

'^> \ \

Fig. 601.

!►

PERSPECTIVE PROJECTION

305

Fig. 602.

of AB on the picture plane. SV is parallel to TAB, and the per-
spectives of all lines which are parallel to TAB or SV will pass through
V as already explained

in Art. 268, o.nd il- ^"^^

lustrated by Fig. 598. ""

Similarly, the per-
spectives of all lines
which are parallel to
SM will pass through
M, and conversely,
lines whose perspec-
tives pass through M,
and which are in the
same plane, will be
parallel to SM. Now
AP and BQ are lines
whose perspectives
A'P and B'Q pass
through M, and they are in the same plane, therefore AP and BQ
are parallel to SM.

Comparing the triangles SVM and BTQ, TB is parallel to VS, TQ
is parallel to VM, and BQ is parallel to SM, therefore the triangles
are similar. But VM is equal to VS, therefore TB is equal to TQ,
and it follows that TA is equal to TP since AP is parallel to BQ,
therefore AB is equal to PQ. But PQ is a line in the picture plane,
therefore if PQ is made equal to the given length, the construction
given will determine the perspective of the part required.

The point M in Figs. 600, 601, and 602 is called a measuring point
for the vanishing point V. It is evident that each vanishing point
can have two measuring points, one to the right and another an equal
distance to the left of the vanishing point.

When the original line AB is parallel to the picture plane the
foregoing construction fails because the points T and V are "at
infinity." A simple and con-
venient construction in this case
is shown in Fig. 603 where A'E'
is the direction of the perspective,
and A' is the point in it from
which the part is to be marked
off. Take any point M on the
horizontal line. Join MA' and
produce it to meet at P a horizontal line PR on the picture, plane
which is at a height above the ground line equal to the height of
A above the ground. Draw PQ parallel to A'E', and make PQ equal
to the given length. Join MQ meeting A'E' at B'. A'B' is the part
required. The theory of this construction is as follows. PM and QM
are the perspectives of parallel horizontal lines passing through the
points A and B and meeting the picture plane at P and Q. The
figure ABQP in space is a parallelogram and AB is equal to PQ.

Fig. 603.

306

PRACTICAL GEOMETRY

But PQ is in the picture plane and may therefore be measured
directly.

271. Examples. — (1)^ square prism, 2 inclies X 2 inches x 4
incites, rests loith one rectangular face on the ground. Nearest corner
J inch to left of vertical plane and 1 inch hehind picture plane. Long
edges inclined, at 50° to picture plane and vanish to left. Station point
3 J inches above ground and 6| inches from picture plane.

Fig. 604 shows this example worked out by the " School of Art "
method. V^ and Y^ are the two vanishing points for the horizontal
edges, determined by the con-
struction explained in Art.
269, and M^ and Mo are the
corresponding measuring
points (see Art. 270). D is
the point of distance (Art.
267).

The point O', the perspec-
tive of the nearest corner on
the ground, is first determined
(Art. 267). Joining O' to V^
and V2 determines the direc-
tions of the perspectives of
the horizontal edges meeting
at O. The lengths OT' and
O'Q' of the perspectives of
these edges are then de-
termined (Art. 270). A
vertical line through O' will
be the direction of the per-
spective of the vertical edge
which has its lower extremity

at O. The length O'R'
marked off by the construction
explained in the latter part of Art.

270.

Fig. 605.

The remainder of tho

PERSPECTIVE PROJECTION

307

construction is obvious. All the construction lines are shown in the
figure.

Fig. 605 shows the same example worked out by a combination of
the " School of Art " method and the direct method explained in Art.
265. HC, the horizontal line, is taken as a plan of the picture plane,
and a plan pq of the solid is drawn, the solid being placed in its proper
position in relation to the picture plane and vertical plane. 8 is the
plan of the station point. Lines from the angular points of the plan
to s intersect HC at points which are the plans of the angular points
of the picture on the picture plane. Vertical lines through these
points on HC will contain the angular points of the picture which is
to be drawn below HC.

Fig. 606.

The vanishing points V, and Vg (Vg is outside the figure) are
determined as before. Produce po to meet HC at t and through t
draw the vertical {I to meet the ground line of the picture plane at
T. T is obviously the perspective of the point where PO produced

308 PRACTICAL GEOMETRY

meets the picture plane, and is also the point itself. The line TV^
will therefore contain the perspective of OP. The points 0' and P
are determined by vertical lines from d and p' to meet T V^ as shown.
Joining O' to Vg determines the line containing the perspective of OQ,
and the point Q' is found by dropping a vertical line from g' to meet
O'K,.

The perspectives of the vertical edges are of course vertical lines.
The length O'R' is determined as in Fig. 604 and the remainder of
the construction is obvious.

(2) A second example is shown worked out in Fig. 606 by a com-
bination of the " School of Art " and direct methods. The per-
spectives of the circular arcs are of course determined by first finding
the perspectives of a number of points in them and then joining
by fair curves. The construction lines for a point P' on one of the
curves are fully shown, others are omitted for the sake of giving
clearness to the figure.

272. Shadows in Perspective. — When a solid which is drawn
in perspective casts a shadow, the perspective of the shadow may be
drawn in the ordinary way from the shadow as determined from the
orthographic projections of the solid as shown in Chap. XXVII., and in
cases where the shadow involves the intersection of curved surfaces
this is generally the simplest method to adopt. In many cases how-
ever the perspective of the shadow can easily be determined from the
perspective of the solid, by applying the constructions explained in
the remainder of this article.

It will only be necessary to consider the perspective of the shadow
of a point. At first all the light will be assumed to come from a
point, the modifications of the constructions for parallel rays of light
being afterwards described.

Let P' be the perspective of the point from which all the rays of
light diverge, and let A' be the perspective of a point whose shadow
is to be determined. The positions of the points P and A in space are
supposed to be known.

(a) To determine the perspective of the shadow cast hy the point A on
the ground. Let M' and N' (Fig. 607) be the perspectives of the feet
of the perpendiculars from A and P on
the ground. Join M'N' and produce it
to meet PA' produced at Q'. Q' is the
perspective of the trace of PA on the
ground, and is therefore the perspective
of the shadow of A on the ground.

(6) To determine the perspective of the
shadow cast by A on a horizontal plane
ivhich is at a given height above the ground.
Produce M'N' (Fig.' 607) to meet the
ground line at T and the horizontal line
at V. Draw the vertical line TTj and

make TTj equal to the given height of the horizontal plane above the
ground. Draw TjV to cut P'A' produced at Q/. Q/ is the perspective

PERSPECTIVE PROJECTION

309

Fig. 608.

of the shadow required. TV and T^V are the perspectives of parallel
horizontal lines which are in a vertical plane containing PA, therefore
Q/ is the perspective of the trace of PA on a horizontal plane at a
height above the ground equal to TT^.

(c) To determine the perspective of the shadow cast hy A on a given
vertical plane. Let EF (Fig. 608) be the perspective of the trace of
the given vertical plane on the ground,
and let EF cut WW produced at R'.
Draw the vertical line R'Q' to meet P'A'
produced at Q'. Q' is the perspective of
the shadow required. R'Q' is evidently
the perspective of the intersection of the
given vertical plane with the vertical
plane containing PA, therefore Q' is the
perspective of the trace of PA on the
given vertical plane.

(d) To determine the perspective of the
shadow cast hy A on a given inclined plane. Let EF (Fig. 609) be the
perspective of the trace of the given plane on the grouTid and let
FFi be its trace on the picture plane.
Produce M'N' to meet the ground line at
T and the horizontal line at V. Draw
the vertical line TT^ At any convenient
height draw the horizontal line T^F, to
cut TTi at Ti and FF, at F^. Join T^V
and F^E and let these lines meet at U'.
Let TV meet EF at R'. Join R'U'.
Produce P'A' to meet R'U' at Q'. Q' is
the perspective of the shadow required.

T^V and F^E are the perspectives of
the lines in which a horizontal plane,
whose trace on the picture plane is TjFi, intersects the vertical plane
PM, and the given inclined plane respectively. U' is therefore the
perspective of a point in each of these planes; but R' is the per-
spective of another point in each of these planes, therefore R'U' is the
perspective of the line of intersection of these planes. Hence Q' is the
perspective of the trace of PA on the given inclined plane.

When the rays of light are parallel their perspectives will converge
to their vanishing point which is determined by the construction
explained in Art. 269. This vanishing point may then be taken as the
perspective of a luminous point from which the perspectives of all the
rays of light diverge, and the perspectives of the shadows are determined
as already explained. It should be observed that in this case the
perspective of the foot of the perpendicular on the ground from the
luminous point (now at infinity) is on the horizontal line.

Fig. 609.

310

PRACTICAL GEOMETRY

Exercises XXII.

1. Represent in perspective a point A on the ground plane, 1 foot to the
right of the spectator and 2 feet from the ground line ; also a point B 4 feet
to the left of the spectator, 2 feet from the picture plane, and 3 feet above the
ground plane. Join AB. AB is the nearest side of a rectangle of which two
other sides are horizontal and 4 feet long. Complete the perspective represen-
tation of the rectangle. Position of eye, 11 feet from picture plane and 5 feet
from ground. Scale | inch to 1 foot. [b.e.]

2. A rectangular stone slab, 10 feet long, 8 feet wide, and 6 inches thick,
lies with one face on the ground. The longest edges recede from the picture at
an angle of 60° towards the right, and the nearest corner is 1 foot to the left of
the spectator, and 2 feet from the ground line. Through the centre of the slab
is cut a circular hole, 6 feet in diameter. K.epresent the slab in perspective.
Position of eye and scale as in exercise 1. [b.e.]

3. Two rectangular blocks are shown at Ex. 3, Fig, 612. The lower block
lies on the ground. Draw these blocks in perspective projection. The corner
(m! is to be \ inch to the left of the spectator, and
\ inch from the ground line. The edge a6, a'h' is to
recede towards the right at an angle of 60° to the
picture plane. Position of eye, b\ inches from the
picture plane and 2i inches above the ground.

4. A solid letter R is shown at Ex. 4, Pig. 612.
Draw this in perspective. The corner aa! is to be
on the ground 1 inch to the right of the spectator
and 3 inches from the ground line. The face of the
letter is to be in a vertical plane which is to recede
from the picture plane at an angle of 50° towards
the left. Position of eye, 5^ inches from the picture
plane and 2,} inches above the ground.

5. Put into perspective a point on the ground
plane 5 feet to the right of the spectator, and 3 feet
within the picture, and, receding from the point,
draw a line 10 feet long at right angles with the
picture plane. This line is to be the lowest edge
of a square slab 1 foot thick, having its square faces
inclined to the ground at 60° towards the right. On
the upper face of the slab inscribe a circle, and,
centrally with it, draw another circle 4 feet in dia-
meter, which is to be the base of a right cone having
an altitude of 2 feet. The eye is to be 14 feet from
the picture plane, and 8 feet from the ground. Scale,
5 inch to a foot. [b.e.]

6. A right prism 3| inches high has its base,
which is a regular octagon of 1 inch side, on the
ground. A cylindrical slab 4 inches in diameter and

1 inch thick rests on the prism, the axes of the two solids being in line. Repre-
sent the two solids in perspective. One corner of the base of the prism is to be
\ inch to the right of the spectator and 2 inches from the ground line, and one
of the edges of the base adjacent to that corner is to vanish towards the right at
an angle of 50° with the ground line. Height of eye above the ground, 2^ inches.
Distance of eye from picture plane, 6 inches.

7. Fig. 610 gives the elevation and half plan of an object standing on the
ground plane. Put the whole object into perspective, with the sides of its base
inclined at angles of 45° to the picture plane, and its nearest vertical edge 2 feet
to the right of the spectator, and 2 feet within the picture. The eye is to
be 5 feet from the ground, and 12 feet from the picture. Scale \ inch to a
foot. " [B.E.]

"7(

1
1

i.P.l"'

/ATION

Fig. 610.

PERSPECTIVE PROJECTION

311

Fig. 611.

8. A vertical section of a square ceiling is shown in Fig. 611, the upper part
being of the form of a shallow square pyramid. Put the ceiling into perspective
with its horizontal edges inclined
to the picture plane at angles of
30° to the right and 60° to the
left, its nearest corner being 12
feet vertically over a point on
the ground plane 3 feet to the
left of the spectator, and 1 foot
within the picture. The eye is
to be 12 feet from the picture,
and 5 feet from the ground. Scale ^ inch to a foot. [b.e.]

9. Draw, by the method of Art. 265, the perspective projection of the solid
shown at Ex. 9, Fig. 612. The edge AB is to be on the ground and parallel to
the picture plane. The axis of the solid is to be inclined at 50° to the ground,
and the nearest point C is to be in the picture plane, and IJ inches to the right
of the spectator. Station point, 4 inches from the picture plane, and 2 inches
above the ground.

10. Draw the perspective projection of the solid shown at Ex. 10, Fig. 612.
The solid is to stand on the ground with the edge AB in the picture plane, and
1^ inches to the right of the spectator. The face ABC is to be inclined at 4.0°

e y\\ II

I \ .. Ex.13.

-r -r ^ / \ \ y ^ H

-- ,nr -::^ = } \ -rVl^T.'IIliS'

EX.3. I -af..... J'V f— t b'-^ ti

zL %..— m.^- —r- 5^-

:t: i: zri-^-zzz"-.- ". r -

H NmJmU U iffe'lm

:f E alr^ h-_-l%7jzz 'r J: - =F -

y /\ \ ^ ^ J Ex.14.

-i^i A--Z- S.J..^-=:.ii

Jl f \ ^

r'\ C! fl^ 1 ^—^

\^EX.4 - - /(-|\ ^ V

vV /NEx.iiy //|\\,«'/k \\

a W ^ '^ ' \ aW. ]i'4llx^44

"^ 7 \^ ilirk ^^^z ^ W-i^

^-K / sUp V ^

a \ / [

1 \^ 1 1 :x.i5.

Fig. 612.

In reproducing the above diagrams the sides of the small squares are to be
taken equal to half an inch.

to the picture plane. Height of station point 2 inches. The position of the
centre of vision, and the distance of the station point from the picture plane is
to be determined from the further condition that the distance between the
vanishing points for the horizontal edges is to be 7j inches.

312

PRACTICAL GEOMETRY

11. Draw the perspective of the splid shown at Ex. 11, Fig. 612. The base
of the solid is to be on the ground, and the nearest edge AB | inch behind the
picture plane and IJ inches to the right. The face ABC is to be inclined at 30^
to the picture plane. Station point 3 inches from picture plane and 2i inches
from ground.

12. A skeleton cube is shown at Ex. 12, Fig. 612. Draw this object in per-
spective when the nearest vertical edge is in the picture plane, and 1 inch to the
left, and a face containing that edge is inclined at 35'^ to the picture plane.
The station point to be 3^ inches from the picture plane and 1 inch above the top
face of the object.

13. Draw the perspective of the object shown at Ex. 13, Fig. 612. The edge
AB to be vertical, and in the picture plane, and h inch to the left. The face ABC
to be inclined at 30^ to the

picture plane. Station
point to be 4 inches from
the picture plane and 1|
inches above the point B.

14. Draw the perspec-
tive of the object shown
at Ex. 14, Fig. 612. The
edge AB to be vertical and
ix). the picture plane, and
directly opposite to the
station point. The point
C also to be in the picture
plane. Station point, 14-
inches above the level of
point C, and 3^ inches
from the picture plane.

15. A gable cross is
shown at Ex. 15, Fig. 612 ;
draw this in perspective.
The edge AB to be vertical
and in the picture plane
and ^ inch to the left.
The face ABC to be in-
clined at 45° to the picture

plane. Station point, IJ inches below the point A, and 4 inches from the picture
plane.

16. The plan and elevation of a wheelbarrow standing on the ground are
shown in Fig. 613 to the scale, J inch to a foot. Represent this object in
perspective using the scale, 1 inch to a foot. The line AB is the horizontal trace
of the central vertical plane of the wheelbarrow. The point A is to be 1 foot
to the left of the spectator and 1 foot from the ground line, and the line AB is
to vanish towards the right at 40° to the picture plane. The eye is to be
6 feet, by scale, from the picture plane and 2i feet above the ground plane, [b.e.]

17. Referring to exercise 14, let the station point be moved 1^ inches
nearer to the object, and let the picture plane be moved parallel to itself a
distance of 5 inches, so that the station point comes between the solid and the
picture plane.

18. Work the example shown in Fig. 606, p. 307, to the dimensions given.
Then take a point 3 inches to the left, 5 inches above the ground, and 3 inches
behind the picture plane. Consider this as a luminous point and determine the
perspective of the shadow cast by the solid on itself and on the ground ; all tho
rays of light to come from the luminous point.

Fig. 613.

CHAPTER XXIII

CURVED SURFACES AND TANGENT PLANES

273. Generation of Surfaces. — Surfaces may be considered as
generated by a line, straight or curved, moving in a definite manner.
Thus, a plane may be generated by a straight line moving parallel to
one fixed straight line and in contact with another fixed straight line.
Again, a sphere may be generated by the revolution of a semicircle
about its diameter which remains stationary.

The moving line which generates a surface is called the generating
line or generatrix of the surface, and a line which serves to con-
strain or direct the motion of the generatrix is called a directrix.

The same surface may be generated in numerous ways, but generally
there are only a few simple ways in which a surface may be generated.
Take the case of the surface of a right circular cylinder. There are
two simple ways in which this surface may be generated: (1) by a
straight line moving in contact with a fixed circle to the plane of
which it remains perpendicular, as shown by the oblique projection in
Fig. 614, where the moving line is shown in twelve diflferent positions :
(2) by a circle moving so that its centre remains on a fixed straight
line to which the plane of the circle is always perpendicular as shown
by the oblique projection in Fig. 615, where XX is the fixed straight
line and 1, 2, 3, 4, and 5 are positions of the moving circle. This
surface may however be generated by an ellipse which moves so that all
points on it travel along parallel lines, but the ellipse must be such

Fig. 615.

Fig. 616.

Fig. 61^

that its projection on a plana at right angles to the direction of its
motion is a circle.

314

PRACTICAL GEOMETRY

The two simple Wcays in which the surface of a right circular cone
may be generated are : (1) by a straight line which passes through a
fixed point (the vertex of the cone) and moves in contact with the
circle which is the base of the cone as shown in Fig. 616 : (2) by a
circle of changing radius which moves with its centre on the axis and
its plane perpendicular to the axis, the radius of the circle being
proportional to its distance from the vertex of the cone as shown in
Fig. 617.

A surface which may be generated by a line, straight or curved,
revolving about a fixed straight line is called a surface of revolution.
The fixed straight line about which the generating line revolves is
called the axis of the surface. Sections of a surface of revolution by
planes at right angles to its axis are circles. Sections by planes
containing the axis are called meridian sections. All meridian sections
are exactly alike.

A surface which may be generated by the motion of a straight
line is called a ruled surface. Ruled surfaces may be divided into
two classes — developable surfaces and twisted surfaces. A developable
surface may be folded back on one plane without tearing or creasing
at any point. The generating line of a developable surface moves
in such a manner that any two of its consecutive positions are in the
same plane. All ruled surfaces which are not developable are twisted
surfaces.

274. Plane Sections of Curved Surfaces. — The way in
which a curved surface is generated being known the projections of
the generating line in any number of positions can be drawn. The
intersections of a given plane with the generating line in each
of these positions can then be determined. This will give a number
of points on the inter-
section of the plane with
the surface, and a fair
curve through them will
be the complete inter-
section required.

When the curved sur-
face can be generated in
a number of simple ways,
that mode of generation
should be made use of
which has the projections
of its generating line the
simplest possible.

Example 1. A sur-
face is generated by a
horizontal line which
moves in contact with

the hne AB (Fig. 618) . ^^^^ g^g.

and the surface of the
cone VCD. To find the, section of this surface by the plane LMN.

CURVED SURFACES AND TANGENT PLANES 315

Take a section of the given cone and plane by a horizontal
plane cutting ab, a'h' at tt'. The plan of the section of the cone is
a circle and tangents to this circle from t are the plans of two
positions of the generating line. The points r and 8 in which
these tangents cut os the plan of the intersection of the assumed
plane of section with the plane LMN are the plans of points on
the section required. Projectors from r and s to meet the
horizontal through i determine / and s' . In a similar manner
any number of points on the required section may be found.

Example 2. Referring to Fig. 619, ah, ah' is a horizontal circle
3 inches in diameter, cd, c'd! is another horizontal circle IJ inches

Fig. 619.

in diameter. The line joining the centres of these circles is vertical
and 21 inches long. Two points move, one on each circle, wdth equal
velocities in opposite directions. A surface is generated by

angular

a straight line w^hich contains the above mentioned

pomts.

316

PRACTICAL GEOMETRY

Twelve positions of the generating line are shown in plan and in two
elevations.

It wiJl be found that horizontal sections of the surface at
levels IJ inches and 4^ inches above the plane of the larger circle
are straight lines e/, e'/' and gh, g'li respectively. These straight
lines are of definite lengths and are at right angles to one another.

It will also be found that horizontal sections at levels other than
that of the circles or the straight lines are ellipses. Two of these
elliptic sections are shown, one at the level m'n' and the other at
the level p'(^. The student should work out this example, full
size.

The surface described in this example is one that occurs in
the science of optics. It is obvious that the surface may also
be generated by a straight line moving in contact with the lines
ab^ a'b', cd, c'd' and one of the circles or one of the ellipses.

Example 3. A surface is generated by a circle whose plane is
horizontal and whose centre moves along the straight line AD
(Eig. 620). The radius of the circle varies so that the circle

V.T. OF PLANE

Fig. 620.

intersects the semicircle ABC. It is required to find the section
of this surface by the given plane which is parallel to XY.

Take a horizontal plane intersecting the arc ABC at M, the straight
line AD at N and the given plane in the straight line PQ. With
n as centre and radius ww describe a circle to cut pq at r and 8.
Perpendiculars to XY from r and 8 to meet pq' determine the points

CURVED SURFACES AND TANGENT PLANES 317

/ and s. rr' and ss' are points on the section required, and in like
manner any number of points may be found.

It is evident that the circle with n as centre and nm as
radius is the plan of the generating circle when it is at the
level p'q.

275. Intersection of Straight Line and Curved Surface. —
Assume a plane to contain the straight line and determine the
intersection of this plane with the curved surface by the method
described in the preceding article. The intersection of the straight
line with the intersection of the assumed plane and the curved surface
will be the intersection required. In selecting a plane to contain the
straight line choose one whose intersection with the curved surface will
have the simplest possible projections or projections which are easiest
to determine.

276. Tangent Planes. — If through a given point on a curved
surface any two lines be drawn on that surface, the plane containing
the tangents to these lines through the given point is the tangent plane
to the surface at that point. If a straight line can be drawn on the
curved surface through the given point, as can be done on all ruled
surfaces, the tangent plane at the given point will contain this line.

The normal to a surface at a point on it is the perpendicular to the
tangent plane at that point.

277. Tangent Plane to a Cone at a Given Point on its
Surface. — V (Fig. 621) is the vertex of the cone and P the given

Fig. 621. Fig. 622.

point on its surface. Join YP and produce it, if necessary, to meet
the horizontal trace of the cone at Q. Through Q draw a tangent to
the horizontal trace of the cone. This tangent will be the horizontal
trace of the required plane. The vertical trace of the plane can be
determined from the condition that the plane contains the vertex of
the cone. The straight line VQ is obviously the line of contact
between the cone and plane.

318 PRACTICAL GEOMETRY

278. Tangent Plane to a Cone through a Given External
Point. — T (Fig. 622) is the vertex of the cone and R the given point.
Join YR. The required plane will contain the straight line YR, and
the horizontal trace of the plane will pass through T the horizontal
trace of YR. The tangent TQ to the horizontal trace of the cone is
the horizontal trace of the required tangent plane and its vertical trace
may be found from the condition that the plane contains the straight
line YR.

If Q is the point of contact of the tangent from T to the horizontal
trace of the cone, then the straight line YQ is the line of contact
between the cone and plane.

When T the horizontal trace of YR falls outside the horizontal
trace of the cone there will be two tangent planes to the cone passing
through the given point R. When T falls on the horizontal trace of
the cone there will be one tangent plane only containing the given
point. If T falls inside the horizontal trace of the cone the problem
is impossible but in that case the point R would be inside the cone.

[Note. In the preceding articles on tangent planes to the cone, use
has been made of the horizontal trace of the cone ; but in practice
it may often be more convenient to take a vertical trace of the cone,
and first determine the vertical trace of the tangent plane, which will
be a tangent to the vertical trace of the cone. For example, if the
base of the cone is circular and in a vertical plane, take this vertical
plane, or a plane parallel to it, as one of the planes of projection and
use the trace of the cone on this plane in determining the tangent
plane. These remarks also apply to problems which follow on tangent
planes to cones and cylinders.]

279. Tangent Plane to a Cone Parallel to a Given Straight
Line. — Through the vertex of the cone draw a line parallel to the
given line. By the construction described in the preceding article,
determine the planes to contain the former line and touch the cone.
These planes are the planes required. In the particular case where
the given line is parallel to a generating line of the cone, there will be
only one tangent plane to the cone parallel to the given line. If the
line through the vertex of the cone parallel to the given line falls
within the cone, the problem is impossible.

280. Tangent Planes to Right Circular Cones. — The methods
described in the preceding articles on tangent planes to the cone are
applicable to any form of cone, it being assumed that the trace of the
surface of the cone on one of the planes of projection is given. If a
trace of the surface of the cone is not given there would necessarily be
sufficient data given to enable a trace on one of the planes of projection
to be found.

When the cone is a right circular cone and its axis is inclined to
one of the planes of projection its trace on that plane will be one of
the conic sections which may be drawn and the methods of the
preceding articles may then be applied. But the drawing of the conic
section which is the trace of the cone may in general be avoided by the
adoption of special constructions.

CURVED SURFACES AND TANGENT PLANES 319

Fig. 623.

Eor example, let 'pp' (Fig. 623) be a given point on the surface of a
right circular cone whose axis vo, v'd is inclined to the planes of
projection. (In general only
one projection of P will be given
and the other will have to be
found as explained in Art. 228,
p. 264). Draw the projections
of a sphere inscribed in the
cone. Join the vertex vv to
fp' and find the point rr
where the line vp, v'p touches
the sphere. A tangent plane
to the sphere at rr' will be a
tangent plane to the given cone
and the line vp, v'p' will be the
line of contact between the
cone and plane. The tangent
plane to the sphere at rr' will
be perpendicular to the radius
or, o'r' and its traces may be
determined as explained in (Art.
183, p. 217).

Further examples of special constructions to avoid drawing the non-
circular trace of a right circular cone, to which a tangent plane is
required, will be found in subsequent articles of this chapter. It
should however be noted that in many cases the drawing of the non-
circular trace of the cone is the most straightforward construction to
adopt, and, particularly when the trace is an ellipse, it will often be
found quicker and more accurate to draw the trace than to adopt
more or less elaborate constructions involving straight lines and
circles only.

281. Planes Tangential to a Given Cone and having a
Given Inclination. — Let 6 be the given inclination of the planes
which are to be tangential to the cone.

Referring to Fig. 624, the cone is given by its vertex vv' and its
horizontal trace. Determine a right circular cone having its vertex
at vv', its base on the horizontal plane, and it base angle equal to 0.
Planes tangential to these two cones are the planes required. In
Fig. 624 there are four such planes, their horizontal traces being
common tangents to the bases or horizontal traces of the cones. The
vertical traces of the planes are found from the condition that the
planes contain the common vertex of the two cones. The vertical trace
of the fourth plane is outside the limits of the £gure.

In Fig. 625 the given cone is a right circular cone whose axis
vo, v'o' is inclined to both planes of projection. The horizontal trace of
this cone may be found as in Art. 227, p. 263, and the construction just
given for Fig. 624 applied. A much simpler construction however is
as follows. Draw a sphere inscribed in the given cone, its centre
being oo'. Determine a cone enveloping this sphere, having its base

320

PRACTICAL GEOMETRY

on the horizontal plane, and its base angle equal to 6. uv! is the
vertex of this cone. Determine also a right circular cone having its

Fig. 624.

Fig. 625.

vertex at vv', its base on the horizontal plane, and its base angle equal
to 6. The two planes (1) and (2) tangential to these two auxiliary
cones are two of the planes required, their horizontal traces being
common tangents to the circles which are the bases or horizontal traces
of the auxiliary cones.

If the auxiliary cone enveloping the sphere be taken with its vertex
u{a^ below the centre of the sphere instead of above it, the other
auxiliary cone being as before, two other tangent planes (3) and (4) to
the given cone and having the inclination 6 are found. In this case
however it is the pair of tangents to the traces of the auxiliary cones
which cross one another between the circles which must be taken.
The vertical trace of plane (4) is not shown.

282. Tangent Plane to a Cylinder at a Given Point on its
Surface. — Let the cylinder be given by its horizontal trace and the
direction of its generating line (Fig. 626), and let the given point on its
surface be given by its plan p. Through^ draw pqr parallel to the plan
of the direction of the generating line, cutting the horizontal trace of the
cylinder at q and r. If P is on the under surface of the cylinder pq
will be the plan of the generating line through P and q will be its
horizontal trace. If P is on the upper surface of the cylinder pr will
be the plan of the generating line through P, and r will be its horizontal
trace. The elevations of these generating lines are found as shown.

CURVED SURFACES AND TANGENT PLANES 321

A tangent to the horizontal trace of the cylinder at q will be the
horizontal trace of the plane which is tangential to the surface of the
cylinder at P when P is on the under surface, and a tangent to
the horizontal trace of the cylinder at r will be the horizontal trace of
the plane which is tangential to the surface of the cylinder at P when
P is on the upper surface. The generating lines PQ and PR will be
the lines of contact of these planes with the surface of the cylinder,
and by making use of this the vertical traces of the planes may be
found.

Fig. 626.

Fig. 627.

If the given cylinder is a right circular cylinder its trace on one
of the planes of projection is readily found and the method just described
may then be applied. The drawing of the trace of the cylinder, when
that trace is not a circle, may be avoided as follows. Draw an elevation
of the cylinder (Fig. 627) on a vertical plane parallel to its axis. Take
a plane U V W at right angles to the axis of the cylinder cutting that
axis at 0. The section of the cylinder by this plane is a circle whose
centre is O. Draw the rabatment of this circle on the horizontal
plane as shown. The plan of the generating lines through the two
possible positions of P, P being given by its plan p as before, cuts the
rabatment of the circle at m^ and n^. Draw tangents to the rabatment
of the circle at m^ and Wi to meet V W at s and t respectively. • q and r
being the horizontal traces of the generating lines through M and N,
lines qs and rt will be the horizontal traces of the planes required.
The traces of the planes on any vertical plane of projection may be
found from the condition that each plane contains a generating line
through one of the two possible positions of P.

The theory of the foregoing construction is that MS and NT, being

322 PRACTICAL GEOMETRY

taogents to a section of the cylinder at points on the generating lines
through the two possible positions of P, must lie on the planes, one on
each, which are tangential to the cylinder along these generating lines.

283. Tangent Planes to a Cylinder through a given external
Point. — Through the given point draw a line parallel to the generating
line of the cylinder. Let Q be the trace of this line on one of the
planes of projection. Through Q draw the tangents to the trace of
the cylinder on the plane of projection containing Q. These tangents
will be the traces on one of the planes of projection of the planes
required and the other traces may be found from the condition that
the planes contain the given point. i

If the cylinder is a right circular cylinder the drawing of its trace,
when that trace is not a circle, may be avoided by a slight modification
of the construction shown in Fig. 627, p. 321. Let the line through
the given point parallel to the generating line of the cylinder intersect
the plane UVW at L and the horizontal plane of projection at K.
Find Zi the rabatment of L on the horizontal plane. From l^ draw
tangents to the rabatment of the circle which is the section of the
cylinder by the plane UVW. Let these tangents intersect VW at
s and t. K.S and K< are the horizontal traces of the planes required
and the vertical traces may be found from the condition that the planes
contain the given point.

284. Planes Tangential to a Cylinder and Parallel to a
given Straight Line. — Let AB be the given straight line.

Let the cylinder be given by its trace on one of the planes of
projection and the direction of its generating line. Through any
point C in AB draw CD parallel to the generating line of the cylinder.
Determine the traces of the plane containing AB and CD. The
planes required will be parallel to this plane. Tangents to the trace
of the cylinder parallel to the corresponding trace of the plane con-
taining AB and CD will be the traces on one of the planes of
projection of the planes required. The other traces may be found
from the condition that the required planes are parallel to the plane
containing AB and CD.

If the cylinder is a right circular cylinder, the drawing of its
trace, when that trace is not a circle, may be avoided by a modification
of the construction shown in Fig. 627, p. 321. As in the case just
considered, draw CD to intersect AB and be parallel to the generating
line of the cylinder and find the traces of the plane containing AB
and CD. Find the intersection of this plane with the plane UVW.
Let EF be this intersection. Draw the rabatment Cj/i of EF on the
horizontal plane. Draw parallel to Cj/i tangents to the rabatment of
the circle which is the section of the cylinder by the plane UVW. Let
these tangents meet VW at 8 and t. Lines through s and t parallel
to the horizontal trace of the plane containing AB and CD are the
horizontal traces of the planes required. The vertical traces may be
found from the condition that the required planes are parallel to the
plane containing AB and CD.

285. Planes Tangential to a Cylinder and having a given

CURVED SURFACES AND TANGENT PLANES 323

Inclination. — Let 6 be the given inclination, say to the horizontal
plane.

In Fig. 628 the cylinder is given by its horizontal trace and the
direction of its generating lines. Take ah, a'b' a generating line, a
being its horizontal trace. Take a point vv' in ah, a'h' as the vertex
of a right circular cone whose base is on the horizontal plane and
whose base angle is equal to 6. A tangent aL to the base of this
cone will be the horizontal trace of a plane tangential to the cone and
containing the generating line ah, a'h'. The vertical trace LM of this
plane is found from the condition that the plane contains the point vv'.
This plane will also have the given inclination 6. Tangents to the
horizontal trace of the cylinder parallel to ah will be the horizontal
traces of two of the planes required and their vertical traces will be
parallel to LM.

Two other planes (not shown) fulfilling the given conditions may
be found by using the tangent aN to the base of the cone in the same
way that aL was used.

The planes found as above will evidently have the required
inclination and they will also contain each a generating line of the
cylinder whose horizontal trace is at the point where the horizontal
trace of the plane touches the horizontal trace of the cylinder. In
Fig. 628, cd, c'd! and ef, e'f are the lines of contact with the cylinder
of planes (1) and (2) respectively.

Fig. 628.

When the given cylinder is a right circular cylinder the construc-
tion shown in Fig. 629 may be used. In this case a trace of the
cylinder is not required. Take rr' and ss' two points on the axis of
the cylinder as the centres of two spheres inscribed in the cylinder.
Envelop these spheres by cones „whose bases are on the horizontal

324

PRACTICAL GEOMETRY

plane and whose base angles are equal to 9. Tangent planes to these
cones are two of the planes required. The vertical trace of the second
plane falls outside the limits of the figure, cd, c'd' and e/, e'f the
lines of contact of the tangent planes with the cylinder pass through
the points of contact of these planes and the spheres.

Two other planes satisfying the given conditions may be found by
drawing two inverted cones enveloping the spheres and constructing
the tangent planes to them. The base angles of these inverted cones
must of course be equal to 0.

The given angle 6 must not be less than the inclination of the
generating lines of the cylinder.

286. Planes Tangential to a Sphere and containing a given
Line. — First solution. O is the centre of the sphere and AB the given
line. Draw an elevation of the
sphere and given line on a vertical
plane parallel to the line (Fig.
630). Take a plane perpendicular
to AB and containing the centre
of the sphere. LM the vertical
trace of this plane will be at right
angles to a'b' and will be an edge
view of the plane. This plane
intersects the given line at C and
the sphere in a great circle. The
plan of this great circle, which
is an ellipse, is shown but it need
not be drawn. In the plane LM
take the horizontal line OS
through the centre of the sphere.
Obtain the rabatment of the
section of the sphere by the plane
LM and also the point C by
rotating the plane LM about OS until it becomes horizontal. The
rabatment of the section of the sphere will be the circle which is the
plan of the sphere and the rabatment of C will be Cj. Through q
draw the tangent c^ei to the circle which is the plan of the sphere, e^
being the point of contact. Restore the plane LM to its original
position taking with it the point Ci determining e' and e as shown.
CE is a tangent to the section of the sphere by the plane LM, and the
plane containing CE and AB will be a tangent plane to the sphere
at E. The horizontal and vertical traces of the tangent plane will be
perpendicular to oe and o'e' respectively.

Since a second tangent can be drawn from Cj to the circle, the
above construction applied to this second tangent will lead to a
second plane tangential to the sphere and containing AB. This second
plane is not shown.

Second solution. Envelop the sphere by a cone having its vertex
in the given line AB. The planes tangential to this cone and con-
taining AB will also be tangential to the sphere. The horizontal

Fig. 630.

CURVED SURFACES AND TANGENT PLANES 325

traces of the tangent planes required will pass through the horizontal
trace of AB and will be tangential to the horizontal trace of the cone.

By selecting the point in AB for the vertex of the cone so that
the axis of the cone is parallel to one of the planes of projection, the
plane (M) of the circle of contact between the cone and the sphere is
readily found. The base angle of the cone is the inclination of the
planes required to the plane M, and the planes required may be found
by the construction of Art. 193, p. 225.

287. Cones Enveloping Two Spheres. — If one sphere lies
entirely outside the other and the spheres do not touch one another
there will be two cones which will envelop both.

^"-^■^^"^^^^^-Clrcles of contact ^ \J^ cfcordojct

Fig. 631. Fig. 632.

The projections of the cones are obtained by drawing the common
tangents to the circles which are the projections of the spheres.
These tangents intersect the projection of the line joining the centres
of the spheres at the projections of the vertices of the cones. One
cone has its vertex on the line joining the cefttres of the spheres
produced beyond the smaller sphere (Fig. 631), while the other has its
vertex between the spheres (Fig. 632). If the spheres touch one
another externally, one of the cones becomes a plane tangential to the
spheres at their point of contact. If the spheres cut one another there
is only one enveloping cone.

288. Planes Tangential to Two Spheres and having a
given Inclination. — Let the given inclination be to the horizontal
plane. Envelop the two spheres by a cone. Find by Art. 281, p. 319,
the planes tangential to this cone and having the given inclination.

The following construction follows readily from that shown in Fig.
629, p. 323. Envelop each sphere by a cone, axis vertical and base
a«igle equal to 0. Planes tangential to these two cones are the planes
required. There may be as many as eight planes satisfying the given
conditions, depending on and the relative positions of the spheres.
Denoting the cones by A and B, there are two planes for the case
where A and B are upright, two when A and B are inverted, two
when A is upright and B is inverted and two when A is inverted and
B is upright.

289. Planes Tangential to Two Spheres and containing
a given Point. — Envelop the two spheres by a cone. Find by Art.
278, p. 318, the planes tangential to this cone and containing the given
point. Or, proceed as follows. Let V be the vertex of the cone
enveloping the two spheres, and let P be the given point. Joi^ PV,
Find by Art. 286, p. 324, the planes tangential to one of the spheres

326

PRACTICAL GEOMETRY

and containing the line PV. These planes will also be tangential to
the other sphere.

290. Planes Tangential to Three Spheres. — Determine the
vertex of a cone enveloping any two of the spheres, also the vertex of
a cone enveloping another two. A plane containing these two vertices
and touching any one of the spheres will also touch the other two.

If the spheres are entirely external to one another and no two
touch one another there will be eight planes which will touch all three
spheres. Two of these will have all the spheres on the same side,
while the others will have one sphere on one side and two on
the other.

291. Tangent Plane to a Surface of Revolution at a given
Point on the Surface. — Let the axis of the surface (Figs. 633 and
634) be vertical, and let pp be the given point on the surface.

^21

Fig. 633.

Fig. 634.

All tangent planes to the surface at points on it at the same level
as pp' will have the same inclination, and the normals to the surface at
these points will meet on the axis at the same point. Hence if p'r' be
drawn parallel to XY to meet the outline of the elevation (or plane
generatrix) of the surface at /, and a line qr' be drawn perpendicular
to the tangent at r' meeting the elevation of the axis at q, q[p' will be
the elevation of the normal to the surface Q.t pp' . The plan of this
normal will be the line joining p with the centre of the circle which is
the plan of the surface.

A plane L M N through the point p^' and perpendicular to the line
fq, p'(^ will be the tangent plane to the surface a,t pp'. This plane is
also tangential to the right circular cone which envelops the surface

CURVED SURFACES AND TANGENT PLANES 327

of revolution and has for its line of contact- the horizontal circle
through pp'. MN, the horizontal trace of the tangent plane LMN, is
a tangent to the circle which is the horizontal trace of the above-
mentioned cone.

In Fig. 633 the tangent plane meets the surface of revolution at
one point only, while in Fig. 634 the tangent plane also cuts the
surface.

292. Cylinder Enveloping a Surface of Revolution. — The
vertical line a6, a'h' (Fig. 635) is the axis of a surface of revolution

Fig. 635.

whose outline is given, cd^ c'd', is a line which is parallel to the
vertical plane of projection but is inclined to the horizontal plane. It
is required to determine the cylinder which envelops the surface of
revolution and has its generatrices parallel to cd, c'd'. The cylinder is
determined when its line of contact with the surfa-ce of revolution is
found. The construction is shown for finding two points on this line
of contact, one on the upper or concave part of the surface of
revolution, and the other on the lower or convex part.

Take a point rr' on the meridian section of the surface of revolution
which is parallel to the vertical plane of projection. Draw r'o' the
normal to the elevation of the outline of this meridian section at r',

328

PRACTICAL GEOMETRY

and leb r'd meet a!V at d . With d as centre and oV as radius describe
a circle. This circle is the elevation of a sphere inscribed in the
surface of revolution. The line of contact of this sphere and the
surface of revolution is a horizontal circle whose elevation is the straight
line 5V'. A cylinder which envelops this sphere and has its axis
parallel to cd^ c'd' will touch the sphere in a circle of which s'o't',
perpendicular to c'd', is the elevation. Let s'o't' intersect 5'/ at p'y
then p' is the elevation of a point on the line of contact of the surface
of revolution and the required enveloping cylinder. The plan p of
this point is found by an obvious construction which is shown.

A line mn, m'n' through j^p' parallel to cd, c'd' is one of the generatrices
of the cylinder which envelops the sphere and is tangential to the
sphere at the point pp' ; but the surface of the sphere is tangential to
the surface of revolution at this point, therefore the line mn, m'n' is
a generatrix of the enveloping cylinder required.

The outline of a projection of the surface of revolution by projectors
parallel to cd, c'd' will be the projection of the line of contact of the

Fig. 636.

enveloping cylinder determined as above. Such a projection is shown
at (I), on a plane perpendicular to the projectors.

The projection of a surface of revolution on a plane inclined to its
axis may also be obtained directly by first drawing the projections of
a sufficient number of spheres inscribed in the surface and then
drawing the boundary line of all these projections. The " anchor
ring " shown in Fig. 636 is a solid which is easily projected by this
method. The anchor ring may be considered as generated by a sphere
rotating about the axis of the ring, and the projections of this sphere
in a sufficient number of positions being drawn, the curves bounding
them form the projection of the ring.

293. Cone Enveloping a Surface of Revolution. — The vertical
line ah, ah' (Fig. 637) is the axis of a surface of revolution whose
outline is given, cc' is a given point at the same distance from the
vertical plane of projection as ah, a'b'. It is required to determine a
cone which envelops the surface of revolution and has its vertex at cc'.
The cone is determined when its line of contact with the surface of
revolution is found. The construction is shown for finding two points

CURVED SURFACES AND TAKGENT PLANES 329

on this line of contact, one on the upper or concave part of the surface
of revolution, and the other on the lower or convex part.

As in the preceding Art. draw the elevation of a sphere inscribed
in the surface of revolution, having for its line of contact the circle
whose elevation is ^V. A cone which envelops this sphere and has its
A^ertex at the point cc' will touch the sphere in a circle of which the
straight line s't' is the elevation. The point j?' where s't' intersects gV
is the elevation of a point on the line of contact of the surface of
revolution and the required enveloping cone. The plan p of this
point is obtained by an obvious construction which is shown. A line
mw, m'n' through cc' and pp is a generatrix of the cone which envelops

Fig. 637.

the sphere and is tangential to the sphere at pp' . But the surface of the
sphere is tangential to the surface of revolution at this point, therefore
the line mw, w'w' is a generatrix of the enveloping cone required.

294. The Spheroid. — A spheroid may be generated by the
revolution of an ellipse about one of its axes, and is called a prolate or
oblate spheroid according as the axis of revolution is the major or minor
axis of the ellipse.

A spheroid may also be generated by a circle of varying diameter
which moves so that its plane is perpendicular to, and its centre in, a
fixed straight line, the diameter of the circle being regulated by an
ellipse which has one axis coinciding with the fixed straight line. The
fixed straight line is the axis of the spheroid.

330

PRACTICAL GEOMETRY

Fig. 638 shows the plan and elevation of a prolate spheroid whose
axis is vertical. Eight meridian sections and seven circular sections at
equal intervals are shown. A section by a plane LM perpendicular to
the vertical plane of projection and inclined to the horizontal plane is
also shown. All plane sections of a spheroid other than sections at
right angles to its axis are ellipses.

Fig. 639 shows the same spheroid tilted over so that its axis is
inclined at 45° to the horizontal plane but remains parallel to the
vertical plane of projection. The same meridian and circular sections

Fig. 638.

Fig. 639.

which are shown in Fig. 638 are shown in Fig. 639. The student
should draw these projections full size taking the major and minor axes
of the generating ellipse 4 inches and 3 inches long respectively. The
plan in Fig. 639 may be projected directly from the elevation in
Fig. 638 by turning the ground line through 45°. This will save the
labour of redrawing the elevation in the inclined position.

It should be noted that all the curves in the plan, Fig. 639, includ-
ing the boundary line are ellipses. The boundary ellipse in the plan,
Fig. 639, is the horizontal trace of a vertical cylinder enveloping the

CURVED SURFACES AND TANGENT PLANES 331

Hypcrboloid of revolution
of one sheet.

Fig. 640.

Hyperboloid of revolu-
tion of two sheets.

Fig. 641.

spheroid, the straight line i'i' being the elevation of the line of contact
between the spheroid and the enveloping cylinder.

The axes of the various ellipses having been found the curves may
be drawn by the paper trammel method.

When the two axes of the revolving ellipse are equal the spheroid
becomes a sphere.

295. The Hyperboloid of Revolution. — The Tiyperholoid of
revolution may be generated by the revolution of an hyperbola about one
of its axes. If the axis
of revolution is the con-
jugate axis BBi of the
hyperbola (Fig. 640) each
branch of the hyperbola
describes the same sur-
face which is the hyper-
boloid of revolution of one
sheet. If the axis of re-
volution is the transverse
axis AAi of the hyper-
bola (Fig. 641) the two
branches of the hyper-
bola describe separate surfaces and the two surfaces form the hyper-
boloid of revolution of two sheets. Of these two hyperboloids of revo-
lution the one of one sheet is the more important, and when an
hyperboloid of revolution is referred to the one of one sheet will be
understood.

In Fig 642, a' a/ is the transverse axis of an hyperbola whose plane is
parallel to the vertical plane of projection and whose conjugate axis
b'bi is vertical, m'o'm' and n'o'n' are the asymptotes of the hyperbola, and
/' and // are the foci. As the hyperbola and its asymptotes revolve
about the vertical conjugate axis, the hyperbola describes an hyperboloid
of revolution and the asymptotes describe a cone which is asymptotic to
the hyperboloid. Sixteen positions of the moving hyperbola are shown
in the plan (u) and elevation (v).

Sections of the hyperboloid by planes perpendicular to the axis of
revolution are circles, the smallest of which has a diameter equal to the
transverse axis of the hyperbola and is called the collar or throat or
gorge of the hyperboloid.

At (w) in Fig. 642 is shown a plan of the hyperboloid when the
axis of revolution is inclined at 30° to the horizontal plane. In this
new plan only the visible meridian and circular sections shown in the
plan (m) and elevation (v) are shown.

A section of the hyperboloid by a plane is of the same kind, as the
section of the asymptotic cone by that plane or by a plane parallel to
it. For example if a plane cuts the asymptotic cone in an ellipse all
planes parallel to that plane will cut the hyperboloid in ellipses. The
same is true for parabolic and hyperbolic sections, and it is obviously
true for circular sections.

A plane containing the axis of revolution cuts the hyperboloid in an

332

PRACTICAL GEOMETRY

hyperbola and the asymptotic cone in two straight lines which are the
asymptotes of the hyperbola.

A plane parallel to the axis of revolution and tangential to the
collar of the hyperboloid cuts the asymptotic cone in an hyperbola and
the hyperboloid in two straight lines P and Q which intersect on the
collar and which are the asymptotes of the hyperbola. These two
straight lines P and Q are also parallel to the straight lines Pj and Q^
respectively in which the asymptotic cone is cut by a plane containing
the axis of revolution and parallel to the plane of P and Q. Also all
these straight lines P, Pj, Q and Qi are inclined at the same angle to

HYPERBOLOID
OF REVOLUTION
OF ONE SHEET

Fig. 642.

the axis of revolution but if P and Pi slope forwards Q and Qi slope
backwards.

It follows that from any point S on the surface of the hyperboloid
two straight lines can be drawn on the surface, because two planes can
be taken parallel to the axis of revolution and containing S and touch-
ing the collar and each of these planes will intersect the surface in a
pair of straight lines and one of each pair will pass through S.

There are therefore two systems of straight lines which can be
drawn on the surface of the hyperboloid. No two lines of the same
system are in the same plane, but any line of the one system and any
line of the other are in the same plane. The plane which contains a
line of one system and a line of the other is the tangent plane to the
surface at the point of intersection of the lines.

CURVED SURFACES AND TANGENT PLANES 333

Since any line of one system and any line of the other are in the
same plane it follows that any line of one system will intersect every
line in the other system if the lines are produced far enough. Hence
the hyperboloid of one sheet may be generated by the motion of a
straight line which moves in contact with any three lines of either
system.

The hyperboloid may also evidently be generated by the revolution
of a line belonging to one or other of the systems above referred to,
the axis of revolution being the axis of revolution already used. From
this mode of generation the hyperboloid of revolution is also called the
twisted surface of revolution. This mode of generation is illustrated by

HYPERBOLOID
OF REVOLUTION
OF ONE SHEET

Fig. 643.

Fig. 643 which shows the same hyperboloid represented in Fig. 642,
but instead of different positions of the revolving hyperbola different
positions of the revolving straight line are shown. It may be left
as an exercise for the student after drawing the views as shown
in Fig. 643 to add the second system of straight lines on the
hyperboloid.

The hyperboloid of two sheets cannot be generated by the motion of
a straight line, it is therefore not a ruled surface.

296. Hyperboloids of Revolution in Rolling Contact. —
Referring to the plan (u) and elevation (v) in Fig. 644, o'a' is the eleva-
tion and the point o^a is the plan of a vertical axis, o'b' is the elevation
and o^h is the plan of another axis which is parallel to the vertical

334

PRACTICAL GEOMETRY

plane of projection but is inclined to the horizontal plane. The point o'
is the elevation and o^o.^ is the plan of the common perpendicular to these
two axes. o'c' is the
elevation and oc is the D_
plan of a straight line
which is parallel to the
vertical plane of projec-
tion, o'c' makes an angle
a with o'a' and an angle
P with oh'.

If an hyperboloid of
revolution be described
by the revolution of OC
about the vertical axis
and another hyperboloid
be described by the re-
volution of OC about
the inclined axis, these
two hyperboloids will be
in contact along the line
OC, and if one hyper-
boloid be made to rotate
about its axis the other
will also rotate if the

Fig. 644.

frictional resistance between the hyperboloids is sufficient.

The construction of the hyperboloids when the axes and the line of
contact are given is clearly shown in Fig. 644. The view (w) is a pro-
jection of the inclined hyperboloid on a plane perpendicular to its axis.

The common normal to the surfaces of the hyperboloids at C will
be at right angles to the tangent plane at C and this tangent plane
will contain the line OC. Since the line OC is parallel to the vertical
plane of projection the vertical trace of the tangent plane containing
OC will be parallel to o'c', therefore the elevation of the common
normal will be perpendicular to o'c'. But this common normal must
intersect the axis of each hyperboloid. Hence a line m'c'n' at right
angles to o'c' and intersecting o'a' at m' and o'b' at n' will be the eleva-
tion of the common normal to the surfaces at C. The plan men of this
normal is easily found as shown.

The relative angular velocities of the two hyperboloids, when one
rotates the other, will now be determined. To facilitate reference
denote the vertical hyperboloid by A and the inclined hyperboloid by
B. Let a>i and wg be the angular velocities of A and B respectively.
Let Vi be the linear velocity of the point O considered as a point on
the throat of A, and let Vg be the linear velocity of the point O con-
sidered as a point on the throat of B. Let r^ and r^ be the radii of the
throats of A and B respectively.

Draw o'D, o'E, and o'F at right angles to o'a'^ o'V, and o'd respec-
tively. Make o'D equal to Vj and draw DFE at right angles to o'F
meeting o'F at F and o'E at E. Then o'E will be equal to Va , because

CURVED SURFACES AND TANGENT PLANES 335

o'F and ED are evidently the components of Vi along and perpen-
dicular to OC respectively, and since there is rolling motion at O the
component of Vg at right angles to OC must also be equal to o'F. In
addition to the rolling of the one hyperboloid on the other there is
also a sliding motion along the line of contact and the velocity of this
sliding is represented by DE.

<oi o'D . o'E o'D To

o'E

But the triangle Do'E is similar to the

o'D

triangle m'o'n', therefore -p^ =

-7—7. Also irom the plan (u) — = — •
on ^ ^ ^ Vi cm

_, , en c'n' ^^ w, o'm' c'n' c'n' c'm! sin 8

But — = -r-,. Hence — = —n • -r~, = -7-1-. — r~f =

sin a

On o'a' make o'H = w^, and draw HK parallel to o'h' to meet o'c' at

K. Then angle o'KH = 8. and ^=--- = -?^5_r but this has been shown

UK sm a'

to be equal to ^, therefore since o'H = Wi, HK must be equal to 0J2.

This gives the solution of the most common problem in connection
with rolling hyperboloids which is, given the two axes and the relative
angular velocities to find the line of contact and then construct the
hyperboloids. For if on o'a' o'H is made equal to Wi and HK is drawn
parallel to o'h' and made equal to (02 then o' being joined to K the eleva-
tion of the line of contact is found. To find the plan oc draw m'c'n'
perpendicular to o'c' to meet o'a' at m' and o'b' at n'. A projector from
n' to 02?> fixes n and mn being drawn a projector from c' to meet mn
determines c. Then co parallel to o.J) is the plan required and the
hyperboloids may be drawn as shown.

These hyperboloids are the pitch surfaces of what are called skew
bevel wheels, which are used to transmit motion directly from one shaft
to another when the axes of the shafts do
not intersect and are not parallel to one
another. In practice generally only com-
paratively short frusta of the hyperboloids
are used as the pitch surfaces of the
wheels as shown in Fig. 645. If the
frusta are taken at the bases of the hyper-
boloids, as A and B, they are approxi-
mately conical, and if at the throats, as
Ai and B^, they are approximately cylin-
drical. In these wheels the teeth have
line contact.

297. The Geometry of Cross-Rolls. — Referring again to Fig.
644, if the axes of the hyperboloids are given and it is also given that
the line of contact OC is parallel to the axis of the inclined hyper-
boloid, then o'c' would coincide with o'h' and the inclined hyperboloid
would become a cylinder. But the point nn' would coincide with the
point cc' and the cylinder would have no diameter, that is, it would
become a straight line only.

Fig. 645.

336

PRACTICAL GEOMETRY

If it is desired that the hyperboloid which has become a straight
line should be a cylinder of definite diameter, then another surface of
revolution will have to take the place of the other hyperboloid. It
will now be shown how this other surface of revolution may be deter-
mined.

Referring to Fig. 646, hob is the plan and h'o.Jb' is the elevation of
the axis of the cylinder which is assumed to be horizontal, aoa is the
plan and o/ is the elevation of
the axis of the other surface or
solid which is assumed to be
horizontal and perpendicular to
the vertical plane of projection.
This other solid will be called
the roll, o is the plan and o^o'o^
is the elevation of the common
perpendicular to the two given
axes. Making o^o' equal to the
radius of the cylinder determines
o^o' the radius of the roll at the
throat. The plane of the throat
circle of the roll intersects the
cylinder in an ellipse which
touches the throat circle at oo'.
Any other plane parallel to the
plane of the throat circle will
intersect the cylinder in an
ellipse and the roll in a circle,
and the ellipse and circle will
touch one another. Moreover
all such sections of the cylinder
will be exactly alike.

HT is the horizontal trace
of a plane parallel to the plane
of the throat circle of the roll.
This plane intersects the cylin-
der in an ellipse of which m'v!
is the elevation. A circle with

o/ as centre drawn to touch the ellipse w'w' is the elevation of the
circular section of the roll by the plane HT, and this determines the
points e and / on the horizontal meridian section of the roll. A pro-
jector from s', the point of contact of the ellipse and circle, to HT
determines s a point on the plan uov of the curve of contact between
the cylinder and roll.

The radius of the circular section of the roll at HT and the point
of contact ss' may be found without drawing the ellipse m!n'. It is
evident that if the ellipse m'n! be moved to the right a distance equal
to w on the plan it will then coincide with the ellipse c'd' which is the
elevation of the section of the cylinder by the plane of the throat
circle of the roll. And if the circle e'f be moved an equal distance to

Fig. 646.

CURVED SURFACES AND TANGENT PLANES 337

the right it will then be in the position e/// and will touch the ellipse
c!d' at s/. Hence the circle e//i' may be drawn first and then the circle
e'f which is equal to it. Projecting from s/ to Si and making s-^s equal
to w determines th^ point s. In like manner, by taking other positions
for the plane HT any number of points on the horizontal meridian
section of the roll and any number of points on the plan uov of the
curve of contact may be determined. The ellipse cd' is therefore the
only ellipse which need be drawn but it should be constructed as
accurately as possible, to ensure a good result.

298. The Paraboloid of Revolution. — The paraboloid of revo-
lution may be generated by the revolution of a parabola about its axis.
Sections of the surface by planes parallel to or containing the axis are
equal parabolas. Sections by planes perpendicular to the axis are
circles. All other plane sections are ellipses.

299. The Ellipsoid. — The ellipsoid may be generated by a vari-
able ellipse which moves so that its plane is parallel to a fixed plane
and the extremities of its axes are on two fixed ellipses which have one
common axis and which have their planes at right angles to one another
and to the fixed plane.

Referring to Fig. 647, let the horizontal plane be the fixed plane.
Let oa, o'a' and oh, o'h' be the semi-axes of one fixed ellipse whose plane
is horizontal, and let oh, o'h
and oc, o'c' be the semi-
axes of the other fixed
ellipse whose plane is pa-
rallel to the vertical plane
of projection. A moving
variable ellipse whose axes
are horizontal chords of
these two fixed ellipses
generate an ellipsoid.

The plan of the hori-
zontal fixed ellipse is the
plan of the ellipsoid, and
the elevation of the other
fixed ellipse which is pa-
rallel to the vertical plane
of projection is an elevation
of the ellipsoid. def is
the half plan and d'e'f is
the elevation of the moving
variable ellipse in one
position. OG on the half
plan is equal to o/e' on the

part elevation (y^ which is a projection on a vertical plane at right
angles to the plane of the elevation (m).

All plane sections of the ellipsoid are either ellipses or circles.

If g'o'd', a diameter of the ellipse which is the elevation of the
ellipsoid, be taken as the vertical trace of a plane which is perpendicular

ELLIPSOID

Fig. 647.

338 PRACTICAL GEOMETRY

to the vertical plane of projection, the elliptic section of the ellipsoid
by this plane will have one semi-axis equal to o'^' and the other
semi-axis equal to oa. If o'^ is equal to oa then the section is
evidently a circle. This suggests the construction for finding the
plane containing the centre of the ellipsoid and cutting the ellipsoid in
a circle. There are evidently two such planes.

All plane sections parallel to one circular section are circles.
In Fig. 647, g'o'd! is the elevation of one circular section of the
ellipsoid and the chords of the ellipse parallel to g'o'd' are the
elevations of other circular sections. The elevations of the centres
of these circles lie on the diameter wVw' which is conjugate to the
diameter g'o'd!.

As the plane of a circular section moves away from the centre of
the ellipsoid the circle gets smaller and smaller until the plane becomes
tangential to the ellipsoid at mm! or nn' when the circle becomes a
point which is called an umhilic. There are evidently four umbilics on
an ellipsoid.

At (w) in Fig. 647 is shown a projection of the ellipsoid on a plane
parallel to planes of circular sections. On this projection nine circular
sections are shown.

The curve of contact between an ellipsoid and an enveloping
cylinder or an enveloping cone is a plane section of the ellipsoid. The
outline of the projection (w) in Fig. 647 is the trace of a cylinder
which envelops the ellipsoid and is perpendicular to the plane of the
projection ; the diameter i'o'i' of the ellipse (u) is the elevation of the
curve of contact.

The tangent plane to the ellipsoid at any point on it contains the
tangents at that point to any two plane sections of the ellipsoid
through the point. When two of the three axes of an ellipsoid are
equal it becomes a spheroid.

300. The Hyperboloid of One Sheet. — The Jiyperloloid of
one sheet may be generated by a variable ellipse which moves so
that its plane is always parallel to a fixed plane and has the
extremities of its axes on two fixed hyperbolas whose planes are
perpendicular to one another and to the fixed plane and which have a
common conjugate axis'.

Referring to Fig. 648, OA and OB are the semi-transverse axes
of two fixed hyperbolas and OC is a common semi-conjugate axis.
OA and OB are horizontal and OC is consequently vertical. OA
is parallel to the plane of the elevation (v). The moving variable
ellipse is in this case always horizontal, and it will be smallest
when its plane coincides with AOB, and it is then the throat
ellipse or principal elliptic section of the hyperboloid. In this
article when " hyperboloid " is mentioned " hyperboloid of one sheet "
will be understood.

The moving ellipse remains similar to the throat ellipse, hence if
DD and EE are the axes of the moving ellipse in one position de is
parallel to ah.

At (w) is shown an elevation of the hyperboloid on a plane

CURVED SURFACES AND TANGENT PLANES 339

parallel to OB. The true form of one fixed hyperbola is shown
in the elevation (v) and the true form of the other is shown in the
elevation (zc).

The asymptotic cone of the hyperboloid is generated by a variable
ellipse which moves so that its plane is parallel to the throat ellipse
and has the extremities of its axes on the asymptotes of the fixed
hyperbolas. The vertex of this cone is at O. GG and HH are the
axes of this moving variable ellipse in one position. This generating
ellipse is similar to the throat ellipse, hence gh is parallel to ab. The
ellipse which is the section of the asymptotic cone by a plane parallel

/ /
/ /
/ /
/ /

HYPERBOLOID /'/

OF ONE SHEET/'/

Fig. 648.

to the plane of the throat ellipse and at a distance from it equal to
OC is evidently equal to the throat ellipse.

Sections of the hyperboloid and of the asymptotic cone by the same
or by parallel planes are curves of the same kind.

If the transverse axes of the fixed hyperbolas are equal the hyper-
boloid becomes an hyperboloid of revolution.

The- plane of a circular section of the hyperboloid may be found as
follows. Let OA be greater than OB. Then on the elevation (w)
take o' as centre and with a radius equal to oa describe an arc to cut
the hyperbola b'k'e' at k' ; then k'o'k' is the vertical trace of a plane,
perpendicular to the plane of the elevation (w), which will cut the

340

PRACTICAL GEOMETRY

hyperboloid in a circle whose radius is equal to o'h'. All sections
by planes parallel to this plane will be circles. There are evidently
two systems of parallel planes which will cut the hyperboloid in
circles.

The hyperboloid of one sheet may also be generated by a moving
variable hyperbola having a fixed conjugate axis, and having the
extremities of its transverse axis on a fixed ellipse whose plane bisects
at right angles the fixed conjugate axis of the moving hyperbola. The
fixed ellipse is the throat ellipse of the hyperboloid.

Referring to the plan (w) and elevation {v) Fig. 648, the straight
lines RSR and TST lie on the hyperboloid. These lines are in a
vertical plane and their plans coincide and are tangential to the plan
of the throat ellipse. The hyperboloid may therefore be generated by
the motion of one or other of these straight lines. These two lines
belong to two different systems. A line of one system will never meet
any other line of that system but it will intersect every line of the
other system if the lines are produced far enough. Hence if a straight
line moves in contact with any three lines of one of the systems it will
generate the hyperbol'oid.

301. The Hyperboloid of Two Sheets. — The hjperbohid of
two sheets may be generated by a variable ellipse which moves so
that its plane is always parallel to a fixed plane and has the extremi-
ties of its axes on two fixed hyperbolas whose planes are perpendicular
to one another and to the fixed plane and which have a common
transverse axis.

Fig. 649.

Referring to Fig. 649, which is a pictorial projection of one quarter
of an hyperboloid of two sheets, OB and OC are the semi- conjugate
axes of two fixed hyperbolas and OA is a common transverse axis.
OA and OB are horizontal and OC is vertical. The plane containing
OB and OC is the fixed plane referred to in the above definition.
ADE is part of one branch of one fixed hyperbola having OA for its
semi-transverse axis and OC for its semi- conjugate axis ; the plane of
this hyperbola is vertical. AFH is part of one branch of the other fixed
hyperbola having OA for its semi-transverse axis and OB for its semi-

CURVED SURFACES AND TANGENT PLANES 341

conjugate axis ; the plane of this hyperbola is horizontal. The corre-
sponding parts of the other branches of these fixed hyperbolas are
shown to the right in Fig. 649. 1, 2, and. 3 are three positions of the
moving variable ellipse which is always parallel to the fixed plane
BOC. The generating ellipse in any position is a section of the
hyperboloid by a plane parallel to the plane BOC.

The asymptotic cone of the hyperboloid of two sheets is generated
by a variable ellipse which moves so that its plane is parallel to the
fixed plane BOC and has the extremities of its axes on the asymptotes
of the fixed hyperbolas. The vertex of the cone is at O. This
generating ellipse in any position is a section of the asymptotic cone
by a plane parallel to the plane BOC.

All sections of the hyperboloid of two sheets, and of the asymptotic
cone, by planes parallel to the plane BOC are ellipses similar to the
ellipse having OB and OC as semi-axes. Straight lines such as KL
and MN are therefore parallel to the straight line BC.

Any plane containing the common transverse axis of the fixed
hyperbolas cuts the hyperboloid in an hyperbola, and the asymptotic
cone in straight lines which are the asymptotes of the hyperbola.
APQ is such an hyperbolic section, OR being an asymptote. The
semi-conjugate axis of this hyperbolic section is OT the semi-diameter
of the ellipse BTC which is in the plane of section.

The hyperboloid of two sheets may therefore be generated by a
moving variable hyperbola having a fixed transverse axis, and having
the extremities of its conjugate axis on a fixed ellipse whose plane
bisects at right angles the fixed transverse axis of the moving
hyperbola. The fixed ellipse is the ellipse whose axes are the conjugate

axes of the fixed hyperbolas.

The hyperboloid of two sheets cannot be generated by a straight
line and it is therefore not a ruled surface.

302. The Elliptic Tai.YSiholoid.— The elliptic paraboloid may he
generated by a parabola which moves with its vertex in a fixed
parabola, the two parabolas

having their axes parallel,

their planes at right angles

to one another, and their

concavities turned in the

same direction.

Fig. 650 is a pictorial

projection of one quarter of

an elliptic paraboloid. ABE

is the fixed parabola of

which AN is the axis and

whose plane H.P. is hori-
zontal. 1, 2, 3, and 4 are

different positions of the

moving parabola whose plane

is vertical and whose axis is

parallel to AN, and whose vertex is on the parabola ABE.

Fig. 650.

342

PRACTICAL GEOMETRY

Sections of the elliptic paraboloid by planes perpendicular to AN
are ellipses. Four such sections are shown at 1', 2', 3', and 4'. These
ellipses are similar to one another so that lines such as BC and ED
are parallel. Hence the elliptic paraboloid may also be generated by
a variable ellipse which moves with its centre on, and its plane
perpendicular to, the common axis AN of two fixed parabolas ABE
and ACD whose planes are perpendicular to one another, the
extremities of the axes of the ellipse being on the parabolas.

303. The Hyberbolic Paraboloid. — The hyperhoUc paraboloid
may be generated by a parabola which moves with its vertex on a
fixed parabola, the two parabolas having their axes parallel, their
j)lanes at right angles to one another, and their concavities turned in
opposite directions.

Fig. 651 is a pictorial projection of one quarter of a hyperbolic
paraboloid. APC is the fixed parabola of which the vertical line AN
is the axis and whose

plane is the vertical plane ' .r V.P.

V.P. 1, 2, 3, etc. are
different positions of the
moving parabola whose
plane is at right angles
to the plane V.P. and
whose axis is parallel to
AN, and whose vertex
is on the parabola APC.

A section of the para-
boloid by a horizontal
plane through A, the
vertex of the fixed para-
bola, is two straight lines
of which AL is one.
Vertical planes contain-
ing these two straight
lines are asymptotic
planes of the paraboloid.
All other horizontal sec-
tions of the paraboloid
are hyperbolas the
asymptotes of which are the lines of intersection of the planes of
section with the asymptotic planes. Horizontal sections are shown
at the levels C, P, A, and B. It will be found that the hyperbolic
sections above A are on opposite sides of the asymptotes to those
below A.

The projections of the horizontal hyperbolic sections on a horizontal
plane are hyperbolas of which the horizontal traces of the asymptotic
planes are the asymptotes.

It follows that the hyperbolic paraboloid may be generated by a
variable hyperbola which moves with its centre on, and its plane
perpendicular to, the line aAN containing the axes of the two fixed

Fig. 651.

CURVED SURFACES AND TANGENT PLANES 343

parabolas APC and ATB whose planes are at right angles to one
another, whose asymptotes are in fixed planes, and having the
extremities of their transverse axes on the corresponding fixed
parabolas.

Sections of the hyperbolic paraboloid by planes parallel to the
asymptotic planes are straight lines. RS and RT are two such
sections, rs the projection of RS on the plane H.P. is parallel to al
the horizontal trace of one of the asymptotic planes and it will be
found that Rs' the projection of RS on the plane V.P. is tangential to
the parabola APC at R. Also, rt the projection of RT on the plane
H.P. is parallel to the horizontal trace of the other asymptotic plane
and E< the projection of PvT on the plane V.P. is also tangential to
the parabola APC at R. It will also be found that the projections of
these straight lines RS and RT on the plane of the parabola ATB are
tangential to that parabola.

Hence the hyperbolic paraboloid may also be generated by either
of two systems of straight lines, one system being parallel to one fixed
plane and the other parallel to another fixed plane. No two lines of
the same system intersect, but each line of one system will intersect
every line of the other system if they are produced far enough.
These results lead to the following definition. The hyperbolic para-
boloid may be generated by a straight line which moves parallel to a
fixed plane and intersects two straight lines which are not parallel
and do not intersect. From this mode of generation the hyperbolic
paraboloid has been called the twisted plane.

304. Tortuous Curves. — A tortuous curve is one in which no
definite part is in a plane.

If Q, P, and R be points on a tortuous curve, P being between Q
and R, a plane may be found containing these three points. If Q and
R be moved nearer to P then as Q and R move up to P the ultimate
position of the plane containing Q, P, and R is the osculating plane of
the tortuous curve at P, and the ultimate position of the straight line
joining Q and R is the tangent to the tortuous curve at P. The
tangent at P is evidently in the osculating plane at P.

A plane through a point P on a tortuous curve at right angles to
the tangent at P is called the normal plane to the curve at P. Any
straight line passing through P and lying in the normal plane to the
curve at P is a normal to the curve at P. The line of intersection of
the normal and osculating planes of the curve at P is called the
vrincipal normal of the curve at P. The normal through P which is
perpendicular to the osculating plane is called the hinormal to the
curve at P.

The projection of the tangent at a point P of a tortuous or a plane
curve is the tangent to the projection of the curve at p the projection
of P.

344

PRACTICAL GEOMETRY

Exercises XXIII

1. A circle 4 inches i-n diameter is in the V.P. A vertical straight line is 4 inches
in front of the V.P. and its elevation touches the elevation of the circle. A
surface is generated by a horizontal straight line vsrhich moves in contact with the
given line and circle. Draw the elevation of the section of the surface by a plane
parallel to the V.P. and 2 inches in front of it.

2. A right circular cone, base 4 inches in diameter, and altitude 4 inches,
stands with its base on the HLP. A right circular cylinder 2 inches in diameter
and 4 inches long stands with its base on the H.P. and its axis 3 inches away
from the axis of the cone. A surface is generated by a horizontal straight line
which moves in contact with the surface of the cone and the axis of the cylinder.
Draw the elevation of the line of intersection of this surface with the surface of
the cylinder, the plane of the elevation to be inclined at 45° to the plane of the
axes of the cylinder and cone.

3. Three straight lines AB, CD, and EF are given by their projections in Fig.
652. A surface is generated by a straight line which moves in contact with each

g»

■^S

•^v.

««

7 1

7l'

Y X''

T
h

7,'

■y

■^

/I'"

■*

•^

■^

—

...

Fia. 652.

Fig. 653.

of the given lines. Draw the elevation of the intersection of this surface with the
vertical plane whose horizontal trace is H.T. Take the squares of the squared
background in Fig. 652 as of \ inch side.

4. The semicircles ahc and a'h'c' (Fig. 653) are the projections of the half of an
ellipse, de and d'e' are the projections of a vertical line. A horizontal straight
line moves in contact with the curve ABC and the line DE. Draw the plan and
elevation of the curve of intersection of this surface and the plane whose traces
are given. Find also the point of intersection of the line tun, vi'n' with the sur-
face generated as above. Take the squares of the squared background in Fig. 653
as of \ inch side.

5. The circles ahcd and a'h'c' d' (Fig. 654) are the plan and elevation respectively
of an ellipse. A conical surface is generated by a straight line which moves in
contact with the ellipse and passes through the fixed point vv'. Determine the
horizontal trace of the conical surface. Show also the traces of the planes which
are tangential to the conical surface and contain the point rr'»

CURVED SURFACES AND TANGENT PLANES 345

6. xi'a'h' (Fig. 655) is the elevation of a right circular cone and vs is the plan
of its axis, v'f is the elevation of a straight line on the front surface of the cone.

Fig. 654

Fig. 656.

Without drawing an ellipse determine the traces of a plane which touches the
cone along the line VR.

7. The plan and elevation of a cone enveloping a sphere are given in Fig. 656.
Determine the traces of a plane inclined at 60° to the horizontal plane and
tangential to the cone.

8. Determine the traces of a plane tangential to the cone given in Fig. 656 and
inclined at 45° to the vertical plane of projection.

9. Referring to Fig. 656, ^ is the plan of a point lying on the upper surface of
the given cone. Draw the traces of the plane which is tangential to the cone at
the point P.

10. A right circular cone whose vertical angle is 50° lies with its slant side on
the horizontal plane and the plan of its axis perpendicular to XY. Draw the
traces of a plane which touches the cone in a line whose inclination to the
horizontal plane is 40°.

11. A right circular cone, base 2*5 inches diameter and axis 2*5 inches long,
has its base on the vertical plane of projection and its axis 1"5 inches above the
horizontal plane. Draw the traces of the four planes which are tangential to this
cone and inclined to the horizontal plane at 60°.

12. The plan of the vertex of a cone is on the circumference of its horizontal
trace which is a circle 2 inches in diameter. The height of the vertex above the
horizontal plane is 2-5 inches. Draw the scale of slope of a plane tangential to
this cone and inclined at 60° to the horizontal plane.

13. The circle ah (Fig. 657) is the horizontal trace of a cylinder and c^d^^ is
the indexed plan of its axis, p is the plan of a point on the upper surface of the
cylinder. Draw the traces of the plane which is tangential to the cylinder at P.

14. Draw the traces of a plane which is tangential to the cylinder of the
preceding exercise and inclined at 70° to the horizontal plane.

15. Draw the traces of a plane which is tangential to the cylinder of exercise
13 and which contains the line tn^n^ (Fig. 657).

16. Os&2o (Fig. 658) is the indexed plan of the axis of a right circular cylinder
1-5 inches in diameter, r is the plan of a point on the lower surface of the
cylinder. Draw the scale of slope of a plane which is tangential to the cylinder
atR.

17. Draw the scale of slope of a plane which is tangential to the cylinder of
the preceding exercise and which contains the point S33 (Fig. 668).

346

PRACTICAL GEOMETRY

r.J^a E% ^^-^ '" ^^? ^^'^^ P^^"",^^ ^ '*^^^S^* 1^^« ^^ and a sphere whose

fP^alf !?• f^'^'l*^^ *'^'^' °^ *^^ P^^^^^ ^^i^^ ^^ntain the line and are
tangential to the sphere.

llllllll'^^^

^ » AZ Y

-^ '' z

^^ ^^/ V

y -.2v 3

^ ^7 4 ^

-? J 4

^ ^ Jn

r 2 ^

\ .4

/•

yrv

.,-, Y

\ t

7fl

" i

^ it

;^ A

J~2^

^^ J

: t

Fig. 657.

Fig. 658.

a,!

Fig. 659.

In reproducing the above diagrams the sides of the small squares are to be taken
equal to a quarter of an inch. The unit for the indices is 0*1 inch.

19. Draw two circles, one 2 inches and the other 1 inch in diameter. The
centres of the circles to be 2 inches apart. These circles are the plans of two
spheres. The centre of the smaller sphere is 1 inch and the centre of the larger
sphere is 2-5 inches above the horizontal plane. Draw the scales of slope of all
the planes which are tangential to the two spheres and are inclined at 70"^ to the
horizontal plane.

20. AB is a straight line whose plan is 3 inches long. A is one inch and B is
2 inches above the horizontal plane. Represent a plane which passes between A
and B, is inclined at 60° to the horizontal plane, and is at perpendicular distances
of 1 inch and 1*5 inches from A and B respectively.

21. vh and he are two straight lines at right angles to one another and each 3
inches long. V is the vertex and VH the axis of a right circular cone whose
vertical angle is 40°. H is on, and V is three inches above the horizontal plane.
C is the centre of a sphere 2 inches in diameter which rests on the horizontal
plane. Represent a plane which is tangential to the cone and sphere.

22. abc is a triangle, ab = 1*5 inches, be = 1"3 inches, and ac = 2*5 inches,
A is 2 '5 inches, B is 1 inch, and G is 2*4 inches above the horizontal plane. AB
is the axis of a right circular cylinder 1"6 inches in diameter. C is the centre of
a sphere 1-4 inches in diameter. Draw the scale of slope of a plane which passes
over the cylinder and under the sphere and is tangential to both.

23. A, B, and C are the centres of three spheres whose radii are, 0"8 inch, 0*4
inch, and 0*6 inch respectively, ab = 1*7 inches, be = 1'3 inches, and ac = 1'2
inches. The heights of A, B, and C above the horizontal plane are, O'S inch,
0*6 inch, and 1-6 inches respectively. Represent a plane which passes over the
spheres A and C and under the sphere B and is tangential to all three.

24. The elevation of a solid of revolution, whose axis MN is parallel to the
vertical plane of projection, is given in Fig. 660. Draw the horizontal trace of a
vertical cylinder which envelops this solid, and show the elevation of the curve of
contact, s' is the elevation of a point on the front surface of the solid. Represent
a plane tangential to the surface at S.

25. The circle ab (Fig. 661) is the plan of a prolate spheroid whose axis is
vertical and 3 inches long, its lower end being on the horizontal plane, cd is the
plan of a straight line, G being 4 inches above and D on the horizontal plane. A
cylinder whose axis is parallel to GD envelops the spheroid. Show in plan and
elevation the curve of contact between the cylinder and the spheroid, r is the
plan of a point on the lower surface of the spheroid. Draw the traces of the plane
which is tangential to the spheroid at R.

CURVED SURFACES AND TANGENT PLANES 347

26. Taking the spheroid and the point C as in the preceding exercise, draw in
' plan and elevation the curve of contact between the spheroid and a cone

-/-i^. "

<!.^

' H

Uj^

'^'

•

Fig. 661.

Fig. 662.

In reproducing the above diagrams the sides of the small squares are to he taken
equal to half an inch.

enveloping it, the vertex of the cone being at C. Draw also the traces of a plane
containing the point C, touching the spheroid, and inclined at 70° to the
horizontal plane.

27. In Fig. 663, aa, &&, and cc are the plans and a'a', h'b', and c'c' are the
elevations of three straight lines, pp' is a point on AA. Determine a point qq'
on BB such that the line PQ shall be 1-5 inches
long and sloping downwards from P to Q. Find
a point rr' in CC such that PR + QR = 3-5
inches.

28. ach (Fig. 662) is the plan of a straight
line inclined at 45° to the horizontal plane and
sloping upwards from A to B. o is the plan of
a vertical axis. A twisted surface of revolution
is generated by the revolution of ACB about the
vertical axis to which it is connected by a hori-
zontal line of which oc is the plan for the position
of ACB shown. Draw in plan and elevation 16
positions of the generating line at equal intervals.
r is the plan of a point on the lower part of the
surface generated. Draw the traces of the plane
which is tangential to the twisted surface at R.
de is the plan of a horizontal line whose height
above the horizontal plane is 1-5 inches. Show
in plan and elevation the points of intersection
of DE with the twisted surface.

29. ah, and cd (Fig. 664) are the plans of two
straight lines. The points B and D are on the
horizontal plane. A is 3 inches and C is 2 inches above the horizontal plane.
An hyperboloid is generated by the revolution of CD about AB. Show the true
shape of the section of the hyperboloid by a vertical
plane of which HT is the horizontal trace, r is the
plan of a point on the lower surface of the hyper-
boloid. Taking Hd as a ground line draw the traces
of the plane which is tangential to the hyperboloid
atR.

30. The axes of two hyperboloids A and B which
are in rolling line contact are horizontal. The axis
of B is 1-2 inches above that of A. The angular
velocity of B is 1*5 times that of A. The ends of A
are each 2 inches distant from its throat circle.
Draw the plan of the hyperboloids and an elevation on a plane perpendicular to
their line of contact.

31. The throat of a cross-roll (Fig. 646, p. 336) is 3*5 inches in diameter. The
effective length of the roll is 7 inches. The cylinder is 2-5 inches in diameter and

Fig. 663.

Fig. 664.

348

PRACTICAL GEOMETRY

its axis is inclined at 45° to the axis of the roll. Draw, half size, a plan of the
roll and cylinder, their axes being horizontal, showing the curve of contact.

32. OA, OB, and OC are the semi-axes of an ellipsoid. OA = 1-75 inches,
OB = 1-25 inches, and 00 = 1 inch. OA is horizontal and OB is perpendicular
to the vertical plane of projection. The centre O is
1 inch above the horizontal plane. The plan is shown
in Fig. 665. Draw a projection of the ellipsoid on a
plane parallel to the plane of a circular section.

mn (Fig. 665) is the plan of a straight line which
passes through O and is inclined at 45° to the hori-
zontal plane. A cylinder enveloping the ellipsoid has
its generating line parallel to MN. Show in plan and
elevation the curve of contact between the cylinder
and ellipsoid.

V (Fig. 665) is the plan of a point which is 3-5
inches above the horizontal plane. A cone having its
vertex at V envelops the ellipsoid. Draw in plan and
contact between the cone and ellipsoid.

33. OA, OB, and 00 are three straight lines mutually perpendicular, 00 being
vertical. OA = 1-25 inches, OB = 1 inch, and 00 = 0-75 inch. OB and 00 are
the semi-conjugate axes of two hyperbolas having OA as a common transverse
axis. These hyperbolas are axial sections of an hyperboloid of two sheets lying
between two planes perpendicular to OA and each 4-25 inches from 0. Draw a
plan of both sheets of the hyperboloid and an elevation on a plane parallel to OA.
Draw also an elevation on a plane perpendicular to OA. On each projection
show the elliptic sections by planes perpendicular to OA at intervals of, say, 0*75
inch. Show also on each projection a number of axial sections.

34. An elliptic paraboloid is given in Fig. 666 by two elevations on planes at
right angles to one another. Draw these elevations to the dimensions given, and

Fig. 665.
elevation the curve of

Fig. 666.

from the elevation {v) project a plan. Show on each projection the parabolic and
elliptic sections indicated. V.T. is the vertical trace of a plane which is perpen-
dicular to the plane of the elevation (v) ; show on che plan
and on the elevation (w) the section of the paraboloid by this
plane.

35. An hyperbolic paraboloid is given in Fig. 667 by two
elevations on planes at right angles to one another. Draw
these elevations to the dimensions given and from the
elevation {v) project a plan. Show on each projection the
parabolic and hyperbolic sections indicated. Show also on
each projection a number of straight lines, say ten, which
lie on the paraboloid, half the number to belong to one
system and thei other half to belong to the other system of
straight generators of the paraboloid.

36. A tortuous curve ARC is shown in plan and elevation
in Fig. 668. Draw the projections of the tangent to the
curve at the point R. Show the traces of the normal and
osculating planes of the curve at R. Draw also the pro-
jections of the principal normal and the binormal to the curve at R.

CHAPTER XXIV

DEVELOPMENTS

305. Development of a Surface. — A surface is said to be
developed when it is laid out on a plane, and the figure so obtained is
called the development of the surface. As has already been pointed
out there are certain surfaces which cannot be developed, as for
example the surface of a sphere, but approximate developments of
such surfaces may be obtained by dividing them up into a number of
parts. The determination of the developments of developable surfaces
will first be studied and then the method of determining approximate
developments of undevelopable surfaces will be considered.

In general when the surface of a prism, pyramid, cylinder, or cone
is referred to, the bases or ends will be excluded, but these bases or
ends, being plane figures, may easily be added to the development of
the remainder of the surface, in order to complete the development, if
required.

306. Development of the Surface of a Prism. — First consider
the case of a right prism. Draw the plan and an elevation of the
prism when the ends are horizontal as shown at (a) and {a!) in Fig.
6b9. The vertical faces of the prism are rectangles and these placed
side by side in a plane form the required development which is a
rectangle. This development is most conveniently drawn as shown.
On the horizontal line LM parts 1 2, 2 3, 3 4, etc. are marked off equal
to the sides 1 2, 2 3, 3 4, etc. respectively of the base. Lines through
the points 1, 2, 3, etc., on LM at right angles to LM and equal in
length to the height of the prism are the positions of the vertical
edges of the prism on the development LN.

Now suppose this prism to be converted into an oblique prism by
having for the elevations of its ends the lines c'd! and e'f. The
development of the surface lying between the ends of the new prism
will be a portion of the rectangle LN obtained as follows. From the
ends of the vertical edges of the new prism draw horizontals or
parallels to LM to meet the lines on the development which are the
positions of these edges on the development. For example, from /
the elevation of the upper end of the edge 3 draw the horizontal r'R
to meet the vertical 3 on L^T. Other points such as R are determined

350

PRACTICAL GEOMETRY

in the same way and consecutive points being joined as shown, the
required development is determined.

The student should now have no difficulty in following the con-
struction shown for determining the part of the development which
must be removed due to a hole in the prism, the elevation of the hole
being the triangle s'p'q. Observe that six additional vertical lines are
required on the faces of the prism ; these lines pass through the points
where the horizontal internal edges of the hole meet the faces of the
prism, and their elevations coincide in pairs. The positions of these
additional lines on the development are easily found, and they contain
important points in the complete development.

If any line or figure be drawn on the development of the surface
of the prism, it is obvious how the elevation of the line or figure when

Fig. 669.

the development is wrapped round the solid, may be obtained by
working backwards from the development to the elevation of the
prism.

Again, given, by their projections, two points on the surface of the
prism, the projections of the shortest line lying on the surface of the
prism and joining the given points, are easily obtained when it is
noticed that the required line will on the development be a straight
line. It is necessary to observe however that there are two ways of
joining the given points by lines whose developments are straight lines,
one line goes round the surface in one direction and the other in the
other, but one of these will generally be shorter than the other.

307. Development of the Surface of a Pyramid. — Excluding
the base, the surface of a pyramid is made up of a number of triangles
having a common vertex, and if the true forms of these triangles be

DEVELOPMENTS

351

found and placed together so that those angular points which are
made to coincide coincide when on the surface of the solid, the resulting
figure will be the development of the surface of the pyramid. All the
triangles which make up the development will have a common vertex.

The plan and elevation of a pyramid are given in Fig. 670, the
base of the solid being horizontal. The true lengths of the sloping
edges are conveni<3ntly found as follows. On the elevation of the base
produced make t^^ai, vj)i, v^c^, etc. equal to va^ vh, vc, etc. respectively.
Draw the vertical «;iVi equal to the altitude of the pyramid. Join
«!, bi, Ci, etc. to Vi, then V,cii, V^fej, Y-^c^, etc. are obviously equal to
the true lengths of VA, VB, VC, etc. respectively. The true forms

Fig. 670.

of the triangular faces of the pyramid may now be found and put
together as shown to form the development aiBiCiDiEiAiVjaj.

Considering the line p'q' on the elevation of the pyramid as the
elevation of a plane section of the solid, the construction is shown for
finding the boundary line between the developments of the surfaces
of the two parts of the solid into which it is divided by the plane of
section. Consider the angular point of the section which is on the
edge YE and which has the point r' for its elevation. Draw the hori-
zontal rVj to meet Vjgj at r^. With centre V^ and radius Y^r^ describe
the arc rjlj^ to meet V^Ei at Rp R^ is the position of the point R on
the development. The positions of the other angular points of the
section are determined in like manner.

A straight line M^SiN, is drawn on the development and the
construction is shown for finding the plan and elevation of this line

352

PRACTICAL GEOMETRY

when the development is wrapped round the solid. Consider the
point Sj where the straight line cuts ViBj. With centre V^ and radius
VjSi describe an arc cutting Y^b^ at s^. Draw the horizontal s^s' to
meet v'b' at s'. A vertical through s' to meet vh at s determines the
plan of S. Or 8 may be found as follows. Let s^s' meet Y-^v^ at o.
Make vs equal to osj. The latter construction for finding the plan of
a point on one of the sloping edges of the pyramid is to be preferred
when the plan of that edge is nearly perpendicular to the ground line.
In determining the point m' observe that Mi7% is a line parallel to
a^Bi and is not an arc of a circle with centre Vj.

308. Development of the Surface of a Cylinder. — Since a
cylindrical surface is one described by a straight line which moves so
that it is always parallel to a fixed line, it is
evident that the development of a strip of the
surface lying between two positions of the
generating line which are not far apart will be
a four-sided figure of which two opposite sides
are straight and parallel and the other two oppo-
site sides will be either straight or approximately
straight. The two opposite sides which are either

I 12 li 10 9

Fig. 671.

straight or approximately straight are equal to the arcs of the ends
of the cylinder lying between the two positions of the generating line
considered. This suggests the method of finding the development
of the surface of a cylinder, which is, to divide the surface into a
number of strips by a number of positions of the generating line
and then to determine the true forms of these strips which added
together give the required development.

In the case of a right cylinder the ends are perpendicular to the
generating line and each of the strips mentioned above is a rectangle
and all the strips when put together on a plane form a rectangle
whose height is equal to the height of the cylinder and whose base
has a length equal to the circumference of one end.

Referring to Fig. 671 (a) is the plan and (a!) the elevation of a
right circular cylinder whose axis is vertical. The circumference of
the circle which is the plan of the cylinder is shown divided into

DEVELOPMENTS

353

twelve equal parts at the points 1, 2, 3, etc. These points are the
plans of twelve positions of the generating line of the surface of the
cylinder and from these the elevations are projected. The straight
line LM. in line with the elevation of one end of the cylinder is made
equal to the circumference of one end and is then divided into twelve
equal parts and numbered as shown. The length LM may be obtained
by calculation but it will generally be sufficiently accurate to obtain it
by stepping out with the dividers from L twelve divisions each equal
to the chord of one of the twelve equal arcs on the plan (a). The
rectangle LN in which MN is equal to the height of the cylinder is
the development of the surface of the cylinder. If c'd! is the elevation
of a plane section of the cylinder, the form and position of the outline
of this section on the development is found as in the case of the prism
and is fully shown. Considering the circle e'f as the elevation of a

_^-^

DE\

—a

^ELOrmc

NT-
6—

*^ ^^

•^

/ \

I.I2

y \

■\

^ \

i^/

—

Fig. 672.

cylindrical hole in the given cylinder the forms and positions of the
ends of this hole on the development are found as shown.

Fig. 672 shows how the development of the surface of an oblique
cylinder having parallel circular ends may be conveniently obtained.
The plan and elevation are first drawn as shown. The base circle is
divided into twelve equal parts at the points, 1, 2, 3, etc. Considering
these points as the lower ends of the lines which are positions of the
generating line of the surface of the cylinder the elevations of theSe
lines are then drawn. On the development these lines are not, how-
ever, at equal distances apart, but the distances 12, 2 3, 3 4, etc. on
the development are each equal, very nearly, to one of the twelve
equal chords on the base circle. The points 1, 2, 3, etc. on the
development lie on the lines which are at right angles to the elevation
of the axis of the cylinder and which are projected from the elevation
as shown. The points, 1, 2, 3, etc. on the development are quickly
obtained by stepping ofi" with the dividers, starting at the point 1.

309. Development of the Surface of a Cone. — Since the

2 a

354

PRACTICAL GEOMETRY

surface of a cone may be described by a straight line which, while
moving, passes through a fixed point which is the vertex of the cone,
it is evident that if two straight lines be taken which are tw^o positions
of the generating line including a small angle, the portion of the
surface of the cone lying between these straight lines will when
developed be a three sided figure of which these straight lines will be
two sides and the third side or base will be nearly straight and of a
length equal to the arc of the base of the cone lying between the two
positions of the generating line
considered. The surface of the
cone may therefore be divided into
a number of parts whose develop-
ments are very approximately tri-
angles. , The true lengths of the
sides of these triangles are easily
found and the triangles being drawn
and placed together with their ver-
tices at the same point and those
sides coinciding which correspond
to those which coincide on the sur-
face of the cone, the development
of the surface is determined.

Two examples will serve to
illustrate the procedure in deter-
mining the development of the sur-
face of a cone.

In the first example (Fig. 673)
the cone is a right circular cone.
The base is horizontal and its plan
is therefore a circle whose centre v
is the plan of the vertex of the
cone. The elevation is the isosceles
triangle a'v'h'. Since the line which
describes the surface of this cone
has a constant length equal to v'a'
it is evident that the development
of the surface will be a sector of a
circle v'a'FH whose radius is v'a'
and having the arc a'FH. equal in length to the circumference of the
base of the cone. The length of the arc a'FH is conveniently laid oflT
with sufficient accuracy for practical purposes by dividing the circum-
ference of the base of the cone into, say, twelve equal arcs and stepping
the chord of one of these arcs out with the dividers on the arc a'FH
twelve times.

If a line on the surface of the cone is given, the form and position
of this line on the development is found by a method similar to that
used in the case of the pyramid. For example let h'c' be the elevation
of a plane section of the cone, it is required to show the form and
position of the outline of the section on the development. Draw the

Fig. 673.

DEVELOPMENTS 355

elevations of the lines which join the vertex of the cone to the points

1, 2, 3, etc. which divide the circumference of the base into, say,
twelve equal parts. (The plans of these lines need not be drawn.)
Join V to the points 1, 2, 3, etc. which divide the arc a'FH into the
same number of equal parts as the circumference of the base is divided
into. Consider the point / which is the elevation of the points on the
section which lie on the lines from the vertex to the points 5 and 9 on
the base. Draw rW^ parallel to Vob to meet v'd at r^. vV^ is evidently
the true distance of the points of which / is the elevation from the
vertex of the cone. Hence if with «?' as centre and i^'r^ as radius an
arc be described to cut the lines «;'5 and «;'9 on the development, points
RR on the required curve are found. In like manner the points where
the required curve cuts the other radial lines on the development are
found and a fair curve through them is the curve required. The figure
lying between this curve and the arc rt'FH is the development of the
surface of the frustum of the cone lying between the base and the
plane of section. If the plane of section is parallel to the base and
has for its elevation the line ci'e' then the development of the surface
of the frustum is the figure lying between the arc a'FH and the arc
e'KL struck from the centre v' . This figure is the development of a
lamp shade whose elevation is a'h'd'e'.

Fig. 673 also shows how to determine p'^' the elevation of a line on
the surface of the cone which on the development is a straight line PQ.
The construction is simply the converse of that already given for find-
ing on the development a line whose elevation is given and which lies
on the surface of the cone.

In the second example now to be considered the cone is an oblique
cone. The plan ^ and elevation of the cone are given to the left in
Fig. 674. The horizontal trace of the surface of this cone is a circle
but the procedure would be the same if this horizontal trace were an
ellipse or any other curve. Divide the horizontal trace of the surface
of the cone into a number of equal parts, say twelve, at the points 1 ,

2, 3, etc. Draw the plans and elevations of the straight lines joining
vv' the vertex of the cone to the points of division on the horizontal
trace. The next step is to find the true lengths of these lines. This
is conveniently done as follows. Let i\ be the foot of the perpendicular
from v' on XY". Set ofi* on XY to the right of v^ the distances v^i,
Vi2, Vi3, etc. equal to the distances v\, v2, v3, etc. respectively on the
plan. Join v' to the points 1, 2, 3, etc. on XY to the right of v^.
These lines give the true lengths of the lines joining the vertex of the
cone to the points of division on the horizontal trace of its surface.
With centre ?/ and these true lengths as radii describe arcs as shown.
Now set the dividers to one of the equal divisions on the horizontal
trace of the surface of the cone (shown on the plan) and starting at,
say, the point A on the arc whose radius is the true length v'l step
out from arc to arc as shown and obtain the points 2, 3, 4, etc. which
will lie on the curve ABC which together with the straight lines v'A
and v'C form the outline of the development of the surface of the cone.

^ To economise space the plan has been placed above the elevation in Fig. 674.

356

PRACTICAL GEOMETRY

It is not absolutely necessary that the parts 1 2, 2 3, 3 4, etc. on the
plan be equal and it may happen that in the solution of some problem
on the cone, say, the determination of its intersection with another
cone, plane sections through the vertex of the cone may have been
used giving a series of lines whose horizontal traces do not divide the
horizontal trace of the surface of the cone into equal parts. In such
a case it is not necessary to draw a fresh set of lines arranged as in
Fig. 674, but the parts 12, 2 3, 3 4, etc. on the curve ABC must be
made equal respectively to the parts 1 2, 2 3, 3 4, etc. on the plan.

a? 6

9 10

Fig. 674 also shows the curve EFH, which is the form and position
of the outline on the development of a cylindrical section of the surface
of the cone. The construction lines for two points Rj and Rg only on
this curve are shown. The constructiop is obvious and needs no
description.

310. Approximate Developments of Undevelopable Sur-
faces. — It has already been stated (Art. 273, p. 313) that a developable
surface must be capable of generation by a straight line and that any
two consecutive positions of the generating line must be in the same
plane. A surface which is undevelopable may however be divided
into parts of which approximate developments may be drawn, the
degree of approximation being greater the more numerous the parts
into which the surface is divided. Three examples will be taken to
illustrate the method of determining approximate developments of
undevelopable surfaces.

The sphere will be taken as the first example. The surface of a

DEVELOPMENTS

357

sphere may be conveniently divided into zones. A zone of the surface
of a sphere is the portion lying between two planes which are per-
pendicular to an axis of the sphere. The surface of a sphere may also
be conveniently divided into lunes. A lune of the surface of a sphere
is the portion lying between two planes which contain an axis of the
sphere.

Fig. 675 shows in plan and elevation one-eighth of a sphere. The
surface is divided into three zones. Each zone is assumed to be the
surface of a frustum of a cone and its development A is found in
the usual way as shown. The plan is shown divided into three equal
sectors and these are the plans of three equal lunes. The figure B is
the approximate development of the half of one of these equal lunes of

Fig. 675.

Fig. 676.

the surface of the sphere. In addition to the information given in
Fig. 675, it is only necessary to state that the distances 1 2, 2 3, and
3 4 on the figure B are equal to the arcs 1 2, 2 3, and 3 4 on the
elevation.

The second example (Fig. 676) is the quarter of an annulus. The
surface is divided into zones and lunes which are developed as in the
case of the sphere. Any surface of revolution may be treated, in the
same way.

In the last example (Fig. 677) the surface is one traced by a straight
line which moves in contact with a straight line ae, a'e and a curved
line //, /'/'. The straight line ae, a'e is divided into a number of equal
parts at the points hb', cc, and dd'. The curved line is divided into
the same number of equal parts at the points gg\ hh', and hh'. The

358

PRACTICAL GEOMETRY

straight lines joining the points on the straight line to the points on
the curved line as shown divide the surface into a number of three-
sided figures whose true
forms are found in the
usual way and are put
together to form the
approximate develop-
ment AELF.

Exercises XXIV

Note. The student is
advised to cut out the de-
velopments in the follow-
ing exercises, after he has
drawn them. He should
then fold them up so as to
exhibit the forms of the
solids. Where a develop-
ment has to be folded to
give a sharp edge, the line
corresponding to that edge
on the development should
be scratched with a needle
or perforated at short intervals
edges of the development to form overlaps for
development is folded up to the form of the solid.
In Fig. 678, which is the development of the
surface of a cube, the strips referred to are
shown at a, a . . . ^

1. A plan and an elevation of a right prism
are shown in Fig. 679. p'g' is the elevation of a
plane section of the solid. Draw the develop-
ment of the surface of the prism and show on it
the boundary line of the section. Draw also the
elevation of the shortest line lying on the surface
of the prism and joining the points M and N.

2. Two intersecting prisms are shown in Fig.
680, one being vertical and the other inclined. Draw the development of
surface of that part of the inclined prism which lies outside the other.

Fig. 677.

Strips may also be left adjoining certain of the
gumming together when the

a a

Fig. 678.

the

Fig. 679.

Fig. 680.

Fig. 681.

vlZ'l'

1Z\ ^vIS

'I 7

t Z5 aS

\ j^

^
V

z Z ^[^

t^ L

nz : J.

^ z zk .

-W

:g: li ^.

133 ^z

\ -

-^ Ts

■ 7^ 7"=?:

-~"=^ 7- Z

4 Z -

^ t% 17

^4 ^ ^

/ /

T\ ^ ^

^ \1- t Li

I t ^

^-^

LE fkz

: \l ^^ti

■■Jl S

Fig. 682.

In reproducing the above diagrams the sides of the small squares are to be
taken equal to half an inch.

DEVELOPMENTS

359

3. An oblique prism standing on a square base is shown in Fig. 681. There
is a rectangular hole in this prism which is shown in the elevation only. Draw
the development of the surface of the prism showing on it the outline of the
hole.

4. Draw the development of the frustum of a square pyramid shown in
Fig. 682.

5. A pyramid is shown in Fig. 683. y'q^ is the elevation of a plane section of
the solid. Draw the development of the surface of the pyramid with the outline
of the section on it.

6. Fig. 684 shows a pyramid in plan and elevation. A square hole in the
solid is shown in the elevation only. Draw the development of the surface of the
pyramid with the outline of the hole on it.

/V\^\ I

Jz.\

^^v

/ \

' \

3 §fc

Fig. 683.

Fig. 684.

Fig. 685.

Fig. 686.

In reproducing tJie above diagrams the sides of the small squares are to be
taken equal to half an inch.

7. Draw the development of the surface of the pyramid given in Fig. 685.
r's' is the elevation of a plane section of the solid. Show the outline of the
section on the development. Draw, in plan and elevation, the shortest line
lying on the surface of the pyramid and joining the points M and N.

8. A solid is shown in plan and elevation in Fig. 686. The base of the solid
is a square and all the other faces are triangles. Draw the development of the
surface of this solid.

9. The shaded part of Fig. 687 is the elevation of a portion of a right circular
cylinder whose axis is vertical. Draw the development of the surface of this
portion of the cylinder.

Fig. 687

10. Fig. 688 shows two views of a sheet metal pipe B with two sheet metal
branches A and C. Draw, one quarter of full size, the development of the surface
of B.

11. Two pipes A and C (Fig. 689), of circular cross section and having their
axes parallel, are connected by a third pipe B as shown. Draw to a scale of 1
inch to 2 feet the development of the surface of the pipe B.

360

PRACTICAL GEOMETRY

Fig. 689.

Fig. 690.

Fig. 691.

12. Draw to a scale of 1 inch to 2 feet the
development of the surface of the part of a
sheet metal uptake for a boiler shown in Fig.
690.

13. The plan and part elevation of a
dome, horizontal sections of which are regular
octagons, are shown in Fig. 691. Complete
the elevation. Draw a development of one
of the eight curved faces of the dome.
Scale 1 inch to 5 feet. - [b.e.]

14. Fig. 692 represents a pipe made from
sheet metal. Draw, to a scale of |, a de-
velopment of the pipe showing the shape of
the sheet from which it is bent. [b.e.]

15. A right circular cone is cut by a plane
which bisects its axis and is inclined at 45°
to it. Draw the development of the surface
of the frustum. Diameter of base of cone
3 inches, altitude 3 inches.

16. The elevation of a can is given in
Fig, 693. Show the shapes to which the
sheet metal must be cut, when fiat, to form
the two conical parts of the can. Omit the
allowances for overlap at the seams.

17. Fig. 694 shows the elevation of a
right circular cone, whose axis is vertical,
penetrating two circular cylinders whose axes
are perpendicular to the axis of the cone.
Draw the development of the surface of that part of the cone
which lies between the two cylinders.

18. Fig. 695 shows the elevation of two frusta of two
cones of revolution enveloping the same sphere. Draw the
developments of the surfaces of the frusta.

19. Two vertical circular pipes of different diameters are
connected by another which is conical, as shown in plan and
elevation in Fig. 696. Draw the development of the surface
of the conical pipe.

20. The solid shown in plan and elevation in Fig. 697
has a base which is a quadrant of a circle. Of the remain-
ing six faces, four are plane triangles and two are conical sur-
faces. Draw the complete development of the surface of this solid

Fig. 692.

rbs

'■J.

Fig. 693.

DEVELOPMENTS

361

1 ^_\ —

Fig. 694.

Fig. 695.

Fig. 696.

Fig. 697.

In reprodticing the above diagrams the sides of the small squares are to be
taken equal to half an inch.

21. A cone, base 27 inches diameter, height 2-35 inches, has its axis inclined at
■10^. ' A curve is traced on the cone which, in development, would be a circle of 1
inch radius touching the base of the cone. Draw the plan of the cone, and of the
curve traced on it, touching the base of the cone at its highest point. [b.e.]

22. The radius of a sphere is 1*5 inches. Draw the approximate development
of a luno of the surface of this sphere. Angle of lune 30°.

23. A surface is described by the revolution of an ellipse about its minor axis.
The major and minor axes of the ellipse are 3 inches and 2 inches long respectively.
Draw the approximate de-
velopment of a lune of
this surface. The lune
to lie between two planes
containing the axis of re-
volution and including an
angle of 30°.

24. An elbow pipe is
6 inches in diameter and
the radius of its centre
line is 8 inches. Draw
the approximate develop-
ment of a lune of the sur-
face of this pipe. Angle
of lune 22i°. Scale \.

25. A sheet - metal
hood is square at the top
and circular at the bot-
tom as shown in Fig. 698.
Show the shape to which
the flat sheet of metal,
must be cut to form the
hood.

26. ABC (Fig. 699) is a sheet-metal pipe, the portions A and C of circular and
rectangular section respectively. By developing B set out the shape to which the
fiat sheet of metal forming it must be cut. Omit all allowances for overlap at
th« seams. [b.e.]

Fig. 698.

Fig. 699.

CHAPTER XXV

HELICES AND SCREWS

311. The Helix. — The lielix is the curve described by a point
which moves with uniform velocity along a generating line of a right
circular cylinder while the generating line revolves with uniform angular
velocity about the axis of the cylinder.

The axial pitch of a helix is the distance between one turn of the
helix and the next, measured parallel to the axis of the cylinder upon
which it is traced. Or, the axial pitch is the distance travelled by the
describing point along the cylinder while it moves once round the
cylinder. The normal pitch of a helix is the distance between one turn
and the next, measured along the shortest line on the surface of the
cylinder. If several helices of the same pitch be traced on the surface
of the same cylinder, at equal distances apart, the distance between two
adjacent helices is called the divided pitch. When the term " pitch "
is used without any qualification, " axial pitch " is understood. The
diameter of a helix is the diameter of the cylinder upon which it is
traced.

The construction for drawing the projection of a helix on a plane
parallel to the axis of the cylinder follows at once from the definition of
the curve and is shown in the right hand portion of Fig. 700. The
axis of the cylinder is assumed to be perpendicular to the vertical
plane of projection. Divide the circle which is the elevation of the
cylinder into a number of equal parts, say twelve, at the points
1, 2, 3, etc.

It is evident from the definition of a helix that if the generating
point moves round any fraction of the circumference of the cylinder, it
will at the same time move in the direction of the axis of the cylinder
a distance equal to the same fraction of the pitch. Thus, if the point
move round the cylinder a distance shown in the elevation by the arc
12, that is, through 1-1 2th of the circumference, it will at the same time
move parallel to the axis a distance equal to 1-1 2th of the pitch. In like
manner, in moving round another 1-1 2th of the circumference, it will
move parallel to the axis another distance equal to 1-1 2th of the pitch.
Hence the following simple construction.

Divide the pitch ah into as many equal parts as the circle in the
elevation is divided into, in this case twelve, at the points 1, 2, 3, etc.

HELICES AND SCREWS

363

Through these points draw perpendiculars to ah to meet projectors from
the points on the circle as shown. The points of intersection of these
two sets of lines carrying the same numbers are points on the plan of
the helix and a fair curve through them is the projection required.
aeh is the plan of one turn of the helix and hfd is the plan of the next
turn. The plan of the second turn of the helix may be obtained in the
same way as the tirst, or it may be determined from the first by
measuring from it, along the plans of the generating lines, a constant
length equal to the pitch of the helix.

On the development of the surface of the cylinder the helix becomes
a straight line. In Fig. 700 tlje straight line
AEB is the development of one turn of the helix
and the straight line CFD parallel to AEB is
the development of the next turn. A straight
line MN at right angles to AB and CD is the
development of a helix at right angles to the one
already considered. The perpendicular distance

RS between the lines AB and CD is the normal pitch of the first
helix.

The inclination of the helix or the pzVc^ angle of the helix is the
angle on the development in Fig. 700, and is the complement of the
angle which the tangent to the helix at any point makes with the
generating line of the cylinder through that point. If d is the diameter

p. If a

7rd

of the cylinder and p the pitch of the helix then tan

second helix on the same cylinder is perpendicular to the first and p' is
its pitch and 0' its pitch angle, then ^' = 90° — ^ and p' =

The helix shown in Fig. 700 is right-handed. If the full and dotted
parts of the plan of the helix shown in Fig. 700 be made dotted and
full respectively the helix would become left-handed.

312. Helix of Increasing Pitch. —A point which moves round

364

PRACTICAL GEOMETRY

a cylinder with uniform angular velocity and at the same time moves
along the cylinder with an increasing velocity describes a curve w^hich
is generally called a helix of increasing pitch. The curve is however
not a helix. The development of a helix of increasing pitch is a
curved line while the development of a true helix is a straight line.

A helix of increasing pitch is shown in Fig. 701. In this example
the describing point is supposed to move along the cylinder with

Fig. 701.

uniform acceleration while its angular velocity about the axis of the
cylinder is constant.

The development of the curve is the parabola KPM, having KN
for its axis and K for its vertex. If a tangent PQ be drawn to the
parabola at P, then the inclination of PQ to KL is the pitch angle
of the helix of increasing pitch at the point whose elevation is p' and
the corresponding pitch is NR obtained by drawing MR parallel to
PQ to meet NK at R.

313. Screw Surfaces. — A screw surface is generated by a
straight line which slides with uniform velocity along a fixed straight
line or axis with which it makes a constant angle, and at the same
time revolves about that a^xis with uniform angular velocity. It is
obvious that any point in the generating line describes a helix.

Fig. 702 shows the portion of a screw surface generated by a
straight line which starts from the position ao in plan and a'd' in
elevation and makes half a revolution about a vertical axis. The
generating line is shown in thirteen positions in plan and elevation.
The outer end of the generating line describes the helix whose
elevation is a'b'c'.

The curve t's'c' is the elevation of the section of the screw surface
by the vertical plane whose horizontal trace is LM, and ore is the plan
of the section of the screw surface by the horizontal plane whoso
vertical trace is PQ. The constructions for one point ss' in the former
section and one point rr' in the latter are shown.

It should be observed that the boundary line of the elevation of

HELICES AND SCREWS

365

the left-hand portion of the screw surface, apart from the curve a'h' is
not straight but is a curved line (not shown) which touches the
elevations of different positions of the generating line. Also if the
surface be extended upwards beyond c'f the boundary line of the right-
hand portion of the screw surface will not be the straight line c'f but
a curved line touching the elevations of different positions of the
generating line.

Fig. 703 shows the surface generated by a quadrant of a circle
which slides along a vertical axis with uniform velocity and at the same
time revolves about that axis with uniform angular velocity, ao is the
plan and a'd'o' the elevation of the generating figure in its initial
position. The generating figure makes half a revolution. The helices

Fig. 703.

described by four points on the moving arc are shown. The generating
figure is also shown in nine positions in plan and elevation. These
two sets of contour lines give to the elevation a pictorial effect and
represent the surface more clearly.

The remarks on the boundary line in Fig. 702 apply also to Fig. 703
except that in the latter Fig. the boundary line on the left-hand
portion of the elevation has been added.

314. Screw Threads. — If the edges of a screw thread are sharp
they form true helices. In practice the edges are often slightly
rounded, as in the Whitworth standard Y-thread, and this rounding
can only be shown on the complete projection of the thread by shading
or contouring. When the rounding of the edges is small it is generally
neglected on drawings except in the case of a cross section.

Ordinary screws, such as are found on bolts, are generally

366

PRACTICAL GEOMETRY

represented in a conventional way on technical drawings. The helices of
such screws are of such small pitch compared with their diameter that
their projections on a plane parallel to the axis of the screw are approxi-
mately straight lines. It is only when the pitch angle of the helix of
a screw is comparatively large that the projection of the helix is drawn
correctly. For conventional methods of representing screw threads the
student is referred to any text-book on machine drawing.

Fig. 704 shows at {a) a projection of a right-handed Y-threaded
screw on a plane parallel to its axis. The section of the thread by a
plane containing the axis is an equilateral triangle. The top and
bottom edges of the thread are helices of the same pitch on two co-

=m.

1 1

— crr'^^^I

k^r"^l>v, I

--s^^^S

\ 1
\ 1

Fig. 704.

Fig. 705.

axial cylinders. Although the outer edge of this thread is quite sharp
it will be observed that in the projection (a) it has at the right and
left of the figure the appearance of being rounded. The boundary lines
connecting the projections of the outer and inner helices at (a) are
tangential to these projections and they are practically straight lines.
A section of the nut for this screw by a plane containing its axis is
shown at (h).

Fig. 705 shows at (c) a projection of a right-handed square-threaded
screw on a plane parallel to its axis. The section of the thread by a plane
containing the axis is a square. The top and bottom edges of the thread
form four helices of the same pitch, two on one cylinder and two QO.

HELICES AND SCREWS

367

riiiother, the cylinders being co-axial. A section of the nut for this
screw by a plane containing its axis is shown at (d).

A screw thread of reetangular section is shown in Fig. 706. In
this case the thread is thin compared with its depth and pitch.

A quadruple- threaded screw or worm is shown in Fig. 707. At the
top left-hand corner of the figure is shown the form of the section of
the threads by a plane containing the axis of the screw.

Fig. 706.

Fig. 707.

gr 315. Pitch and Lead of a Multiple-threaded Screw.— It is
usual to take the pitch of one of the threads of a multiple-threaded
screw as the " pitch " of the screw, and the distance between the centres
of two adjacent threads (measured parallel to the axis of the screw) is
then called the " divided pitch " of the screw, as already stated for
helices (Art. 311, p. 362). In America however it is a common
practice to call the pitch of one of the threads of a multiple-threaded
screw the "lead" of the screw and reserve the term "pitch" to what
has been called the " divided pitch." The lead of a screw is of course
the axial distance through which the nut moves for one revolution of
the screw.

316. Helical Springs. — "While a screw has a screw thread in one
piece with a cylinder from which it projects, a helical spring is a screw
thread without the attached cylinder. It therefore follows that in the
case of the spring the absence of the cylinder will expose to view a
greater part of the thread than is seen on a screw.

Fig. 708 shows a helical spring of which the section by a plane
containing the axis of the spring is a square.

A helical spring made of round wire is represented in Fig. 709.
The centre line of the wire forms a helix and the boundary lines of the
projections of the spring are obtained by considering the spring as the

368

PRACTICAL GEOMETRY

envelope of a sphere whose diameter is equal to that of the wire and
which moves with its centre on the helix. On the half plan of the
spring is shown a section by a horizontal plane whose vertical trace is
LN. The outline of the section is tangential to the horizontal sections
of the moving sphere by the given plane of section. A section of the
spring by a plane containing the axis is nearly but not quite circular.

Fig. 708.

Fig. 709.

317. Axial and Normal Sections of Screw Threads.— By

an axial section of a screw thread is meant a section by a plane contain-
ing the axis and by a normal section is meant a section by a plane at
right angles to the central helix of the thread. When the pitch of a
screw thread is small compared with its diameter there is little differ-
ence between an axial section and a normal section, but when the pitch
is large the difference is considerable.

Fig. 710 shows a projection of a part of a screw thread or helical
spring whose central helix has a large pitch angle, the projection
being on a plane parallel to the axis. It is evident that where the
projections of the various helices cut the projection of the axis these
projections are straight and are inclined to the projection of the axis
at angles which are the complements of the pitch angles of the helices.
Also at the points considered the helices are parallel to the plane of
projection.

Still referring to Fig. 710, the axial section of the thread is the
square S, oah is the projection of the central helix of the thread, and ot
is its tangent at o. LoN, at right angles to ot is the edge view of a
plane which is perpendicular to the helix at O. The figure EFHK is
the true form of the normal section at LN. The edges EF and KH
are nearly straight but are arcs of ellipses, being parts of an oblique
section of two cylindrical surfaces. The edges EK and FH are sections
of screw surfaces and are very approximately straight lines. It will be

HELICES AND SCREWS

369

seen that the thickness of the thread tapers distinctly at a normal
section.

Fio. 710.

Fig. 711.

Fig. 711 shows the construction for finding S the true form of an
axial section when the noi:nial section EFHK is of uniform width, that
is, when the thickness of the thread at a normal section is uniform,

318. Handrails. — The making of handrails for stairs is considered
to be one of the most difficult parts of the craft of the joiner and a
knowledge of the geometry of handrails is essential to their correct and
economical construction. The complete treatment of the subject of
handrailing in all its technical details is beyond the scope of this work
but the fundamental principles and their applications to a few examples
may be studied here with advantage.

The construction of a straight handrail presents no difficulty.
Bamps and level casings are also easily drawn and made. A ramp is the
part of a handrail whose centre line is curved in a vertical plane only.
A level easing is the part of a handrail whose centre line is curved in a
horizontal plane only. A level easing whose centre line is a quarter of
a circle is called a level quarter. The part of a handrail whose centre
line is curved vertically and horizontally is called a wreath. A straight
piece of handrail formed on the end of a wreath is called a shank. It is
the construction of wreaths which is the difficult part of handrailing
and their geometry will now be considered.

The geometry of the centre line of the wreath will first be studied.
Referring to Fig. 712, ah, a'h' and cd, c'd' are the straight centre lines
of two straight pieces of handrail which are to be connected by a
wreath whose centre line is, in plan, a quarter of a circle he. The lines
ah, a'h' and cd, c'd' are called the central tangents of the wreath and in
order that the wreath may be cut economically from a plank it is usual
to arrange that these tangents shall intersect and therefore lie in a
plane. In Fig. 712 the tangents ah, a'h' and cd, c'd' intersect at the

370

PRACTICAL GEOMETRY

point it! and in the example considered the tangents are equally inclined
to the horizontal plane.

The centre line of the wreath lies on the surface of a vertical
circular cylinder, and to the left of the elevation (U) is shown the
development of the vertical surface containing the centre line of the
wreath and the tangents, A^Bj and C^Dj being the developments of the
tangents. The development of the centre line of the wreath is shown
as a straight line BjC^, but it will be seen that AjB^ and C^Di are not
in the same straight line with BjC^. To produce a graceful centre line
" easing curves " should be introduced between the centre line of the
wreath and the tangents. These easing curves may be entirely on the
surface of the cylinder or entirely outside of it or they may be partly
on the cylinder and partly ontside of it.

Fig. 712.

Taking the straight line B^Cj as the development of the centre line
of the wreath the elevation of this centre line will be a helix obtained
by the usual construction or by projectors from the plan and the
development as shown. The plane containing the tangents ah, aV and
cdj c'd' is an important one. If a'b' be produced to meet XY at h' and
a perpendicular h'h be drawn to XY to meet ab produced at h, the
horizontal trace of the tangent ah, a!h' on a plane at the level of cc' is
determined, and he is the horizontal trace of the plane of the tangents
ah, a'h' and cd, c'd'. Any line in the plane of the tangents which is
parallel to he will be a horizontal or level line of the plane. When the
tangents are equally inclined to the horizontal it is easy to prove that
he is parallel to to and that to bisects the angle hoc. This is true
whatever be the magnitude of the angle hoc provided the tangents
intersect and have equal inclinations to the horizontal.

HELICES AND SCREWS

371

An elevation of the tangents on a pl^ne perpendicular to Tic or to
will show the two tangents in one straight line since this elevation is
an edge view of the plane containing them. Such an elevation is
shown to the right in Fig. 712, the plan having been redrawn and
turned round for convenience so that ot is perpendicular to XY. The
plane of the tangents will intersect the cylinder, on the surface of
which the centre line of the wreath is situated, in an ellipse whose
major axis is m'w' and whose semi-minor axis is the radius of the
cylinder. The true form of this ellipse is shown with the tangents of
the wreath in their proper positions in relation to the ellipse. The
student who has reached this stage will not require to be fold how to
construct the ellipse and the lines connected with it shown to the right
in Fig. 712. The line ABECD is the centre line of what is called the
face mould of the wreath, the use of which will be explained later.
Lines o'^ and o'C drawn from the centre of the ellipse to the points B
and C where the tangents meet the ellipse are called the springing lines
of the wreath. The plane containing the tangents will not contain the
true centre line of the wreath but in most cases the true centre line of
the wreath will lie very near to this plane. In the case shown in Fig.
712 the true centre line of the wreath lies partly on one side of the
plane of the tangents and partly on the other crossing it at the point
ee'. In the right hand part of Fig. 712 the elevation of the true centre
line of the wreath is not shown.

The case where the tangents are in vertical planes at right angles
to one another and one of the
tangents

cd.

c'd' is horizontal is
illustrated in Fig. 713. The
student should have no diffi-
culty in dealing with this case
after having mastered the more
general case which has just
been considered. It will be
noticed that the plane of the
tangents has dct for its hori-
zontal trace and the elevation
(U) is an elevation on a plane
perpendicular to the plane of
the tangents ABCD the
centre line of the face mould is
made up of the two tangents
AB and CD and BC the quarter
of an ellipse. B^C^ the develop-
ment of the centre line of the
wreath has been drawn with
an easing curve joining it to
CjDi the development of the
horizontal tangent, this easing curve being entirely on the develop-
ment of the surface of the cylinder on which the central line of the
wreath is situated.

372

PEACTICAL GEOMETRY

If the centre line of the face mould, determined as explained for
Figs. 712 and 713, be taken as the true centre line of the wreath, the
student should have no difficulty in drawing the development and any
required projections of it. The construction of this new centre line is
shown in Fig. 714 for the case illustrated in Fig. 712. The plan and
the positions of the tangents ah, a'b' and cd, c'd' are supposed to be given
as before. The plane of the tangents and the edge view of this plane
shown to the right in Fig. 714 are determined exactly as in Fig. 712.
From the edge view of the plane of the tangents and the plan beneath
it the development AiBiCjDi is projected as shown. The elevation (U)
is projected from the plan and the development, or from the plan and
the edge view as shown.

In making a wreath it is first formed of rectangular or approxi-
mately rectangular cross-section, and in this form it is called a squared
loreaih. The squared wreath is moulded to the required cross-section
by hand with suitable tools. The geometry of the handrail is mainly
the geometry of the squared wreath, and the correctness of the form of
the finished wreath very largely depends on the skill and care with
which the squared wreath is constructed.

The squared wreath resembles to some extent the thread of a
square-threaded screw of large pitch. In one system of constructing
squared wreaths the surfaces of the wreath are described by the sides
of a rectangle which moves with its centre on the centre line of the
wreath, the plane of the rectangle intersecting the centre line at right
angles, and two sides of the rectangle are horizontal. The horizontal
sides of the moving rectangle sweep out the upper and lower surfaces

HELICES AND SCREWS

373

of the wreath but even when the centre line of the wreath is a helix
these surfaces are not true screw surfaces because the generating lines
do not intersect the axis of the cylinder. Also, the other sides of the
moving rectangle do not lie exactly on the cylindrical surfaces. It may
however be assumed with sufficient accuracy for all practical purposes
that the sloping sides of the moving rectangle sweep out portions of
cylindrical surfaces which become the inner and outer vertical surfaces
of the wreath. In that case the developments of these surfaces will be
figures of uniform width. These developments of the inner and outer
vertical surfaces of the wreath are called the inner and outer falling
moulds of the wreath. The development of the vertical section of

the wreath which contains its centre line is called the central falling
mould.

The surface which contains the centre line and the centre lines of
the inner and outer vertical surfaces of the wreath will be described by
the horizontal centre line of the moving rectangle which, when produced',
intersects the axis of the cylinder. This gives the construction for find-
ing the centre lines of the inner and outer falling moulds from the centre
line of the central falling mould.

Fig. 715 shows the constructions for determining the falling moulds
and projections of the squared wreath whose centre line is the same as
that shown in Fig. 712. The tangents ah, a'V and cd, c'd' are first
drawn intersecting at it'. The shanks are then drawn in plan and
elevation. The plan may then be completed. The centre line of the
central falling mould is next drawn as in Fig. 712 and from this the
centre lines of the inner and outer falling -mould are determined as

374

PRACTICAL GEOMETRY

shown. Upon these centre lines the falling moulds are constructed of
uniform width. The elevation (U) may now be projected from the
plan and the falling moulds as shown.

Eig. 716 shows an elevation of the wreath on a plane pei^endicular
to the plane of its central tangents and from this the face mould is
shown projected. The /ace mould is the section of the wreath by the
plane containing the central tangents. The determination of the centre
line of the face mould has already been considered in connection with
Figs. 712 and 713. The inner and outer curves of the face mould are
the elliptic sections of the inner and outer cylindrical surfaces of the
wreath by the plane containing the central tangents.

Thickness of Plank

Fig. 716.

It is not necessary to draw the elevation of the wreath shown in
Fig. 716 in order to determine the face mould. Having found the
centre line ABECD, as previously explained, OE being the semi-minor
and OF the semi-major axis of the ellipse of which the curve EEC is a
part, make EH equal to half the horizontal width of the wreath.
Join EF and draw HK parallel to EF. Then OH is the semi-minor
and OK is the semi-major axis of the ellipse of which the inner curved
portion of the face mould is a part. In lij^e manner the semi-minor
and semi-major axes of the ellipse of which the outer curved portion of
the face mould is a part may be determined, and the elliptic arcs may
then be drawn by the trammel method. The ends of the face mould
are perpendicular to the tangents AB and CD, and the other straight

HELICES AND SCREWS

375

edges are tangents to the ellipses and are also parallel to the central
tangents respectively, as shown.

On the elevation of the wreath in Fig. 716 is shown the thickness
of the plank from which the wreath may be cut.

The wreath is cut from the plan either by the " bevel cut " shown at
(m) Fig. 716, or by the " square cut " shown at (n) Fig. 716. The latter
is considered to be the best as it involves less labour and requires less
material.

The angle marked ^ at (w) and (n) in Fig. 716 is called the
" bevel " for that end of the wreath and this angle must be determined
for each end of the wreath before the shanks can be formed and the
'' twist " given to the wreath. The angle ^ is simply the angle
between the vertical faces of a shank and the plane of the central
tangents, and it may be con-
veniently found as follows. Re-
ferring to Fig. 717, at, at' and td,
t'd', intersecting at tt\ are the
central tangents of the wreath, ah
being the ground line for these
projections, hd is the horizontal
trace' and ha' is the vertical trace
of the plane of these tangents.
By the usual construction (Art.
164, p. 200) ^A is the angle be-
tween the plane of the tangents
and the vertical plane containing
the tangent at, a't'. Taking td
as a ground line t'^d is a second
elevation of the tangent td, t'd'
and it is also the new vertical
trace of the plane of the tangents.
The angle <^i„ which is the angle
between the plane of the tangents
and the vertical plane containing the tangent td, t'd', can now be found
as shown. If the given tangents are equally inclined to the horizontal
it is evident that ^^ is equal to ^d-

In Figs. 715 and 716 the central, inner, and outer falling moulds
are all of the same width, measured at right angles to their centre
lines, and the resulting wreath has everywhere an approximately
rectangular cross-section, the cross-section being taken at right angles
to the centre line of the wreath. In another system of construction,
sections of the wreath by planes containing the axis of the cylinder
are rectangles, and the sides of these rectangles which lie on the upper
and lower surfaces of the wreath are horizontal ; the falling moulds
are then of different widths as will be seen from Figs. 718 and 719. Fig.
718 shows the falling moulds, on this second system, for the wreath
illustrated by Fig. 715, and Fig. 719 shows the falling moulds, on this
second system, for the case where the lower central tangent is hori-
zontal. The central falling mould is constructed in each case as

376

PRACTICAL GEOMETRY

before and from this the outer and inner falling moulds are projected
as shown. It will be seen that at corresponding vertical lines the
widths of the moulds, measured veriicalhj, are the same. Having con-

FiG. 718.

\ /\

"'\

.2. -^

__^

.^rfC ^

y
/ /
1^^

,,>

-^-^

>'
-?

L<1_.

—

■

—

— 1

—

IdT-

- —

---

—

t^i

\ — ;

)--€

—

K4 5 6

^<57rfna^ laUxng Mould

Outside FalUng Mould

Fig. 719.

Inside Falling Mould

structed the falling moulds, any required projections of the wreath
may be drawn as before. But it will easily be seen that only the
central falling mould need be drawn for the pui'pose of obtaining the
projections of the wreath.

319. Screw Propellers. — The geometry of a screw propeller
blade and the construction of the working drawings of it will be more
easily understood after a brief reference to the usual method of pre-
paring the mould for a propeller in the foundry.

Referring to Fig. 720, ABC is a templet, which when laid out fiat
is a rightrangled triangle having the right angle at B. The templet
ABC, having AB vertical, forms part of a vertical cylindrical surface
of which DE is the axis. If a straight line, intersecting the axis DE
and making a constant angle with it, moves in contact with the sloping
edge AC of the templet ABC it will generate a true screw surface.
AE and CD are the extreme positions of the moving line and three
intermediate positions are also shown.

In the foundry a rough bed of brickwork and loam is built up, the
upper face of this bed roughly conforming to the screw surface required.
A vertical spindle is fixed so that its axis coincides with DE. A
sleeve attached to one end of a horizontal arm fits on the spindle and
can slide or rotate freely on it. To the arm is attached a board, called

HELICES AND SCREWS

377

a loam hoard, the lower edge of which, guided by the sloping edge of
the templet ABC, sweeps out the screw surface on the loam as the arm
rotates about the axis of the spindle. The screw surface is then
smoothed by hand. Sharp iron pins attached to the lower edge of the
loam board at intervals cut helices FHJ, KLM, etc. on the screw
surface of the loam. The work described so far is for one blade and
this has to be repeated for each of the other blades if all the blades are
cast together.

The boss of the propeller may be moulded from a pattern inserted
in the usual way or it may be swept out of the loam with a suitable
loam board. The mould thus far prepared is dried and blackwashed.
The next step is to form with loam a piece of the same shape and
size as the required blade. Strips of wood EHJN, PLRQ, etc. are

cut out to the shapes of the required cylindrical sections of the blade
at different radii. One of these strips laid out flat is shown at (a).
These strips or thickness pieces are bent and fixed on edge to the
mould, their lower edges coinciding with the corresponding helices
FHJ, KLM, etc. which have been previously traced on the mould.

The surface of the blade having been outlined on the mould this
surface is covered with loam to the depth of the thickness pieces
FHJN, PLRQ, etc. After drying and blackwashing, the upper part
of the mould can now be built up on the lower part. The upper part
of the mould is taken off and then smoothed and dried. The loam
thickness piece is now removed and the finished halves of the mould
are put together.

A screw surface cannot be developed, but the face of the blade of a
screw propeller is generally a small portion of a complete convolution
of a screw surface and it may be developed approximately.

378

PRACTICAL GEOMETRY

It may here be pointed out that the helices on the face of the
blade are approximately elliptic arcs each being a portion of a plane
section of the surface of a cylinder whose axis is the axis of the screw,
the inclination of the plane being the same as that of the helix. This
is made use of in obtaining the projections of the blade from its
approximate development in the manner to be shown presently.

The shape and area of the developed face of the blade, also the
pitch of the screw surface and the general dimensions of the propeller
are settled by the naval architect in consultation with the marine
engineer. It will now be shown how the drawings of a propeller may
be made, the necessary particulars being given.

The case where the face of the blade is a true screw surface gene-

zLB

TRUE SCREW, LEFT-HAND
PITCH = 1-3 X DIAMETER

Fig. 721.

rated by a straight line moving at right angles to the axis will first be
considered. Referring to Fig. 721, the dotted curve ABC is the given
outline of the developed face of the blade drawn on the end elevation
about the vertical line dz which bisects the developed surface in the
neighbourhood of the boss.

Take a point e on o'^' and with centre d and radius dd describe
the arc dtid cutting the horizontal through d at d. Make do equal to
•p
^ where P is the pitch of the screw surface of the blade. Join od

and produce it. The portion of this straight line od which falls within
the plan of the blade will be, for all practical purposes, the plan of
that part of the helix lying on the face of the blade at the radius dd.

HELICES AND SCREWS 379

Further, od is the horizontal trace of a vertical plane which cuts
the surface of the cylinder of which e'n'd is a part elevation, and the
section is an ellipse an arc of which practically coincides with the por-
tion of the helix just referred to. Obviously od is the length of the
semi-major axis and o'e' is the length of the semi-minor axis of this
ellipse. Make o'D equal to od and taking o'D and o'e' as the semi-
major and semi-minor axes respectively construct DNe' a part of the
ellipse. This ellipse cuts the outline of the development of the face of
the blade at N. A horizontal through N to eut the circle through e'
at n' determines a point on the end elevation of the edge of the blade,
and a vertical from n' to meet od determines the plan of this point.
These points n' and n may be more accurately found as follows. From
N drop a perpendicular to meet the horizontal through o at Wq. Make
on equal to on^. Draw the vertical nn' to meet the horizontal through
N at n'.

Instead of drawing the ellipse it will be sufficient to draw e'N as a
circular arc struck from the centre of cu^ature of the ellipse at e.
The construction for finding this centre of curvature has been given
in Art 51, p. 48, and is shown to the right in Fig. 721. DF is
parallel to o'e' and e'F is parallel to o'D. Fs perpendicular to e'D
meets e'o' produced at s. A circular arc struck from the centre 8 with
a radius se' will practically coincide with as much of the ellipse as lies
within the development of the face of the blade.

Other points on the plan and end elevation of the edge of the
blade are found in the same way. For the side elevation shown to
the left in Fig. 721, make »h/«i' equal to mn and similarly for other
points.

On the elevation to the left in Fig. 721 is shown the maximum
thickness section of the blade. This is not a true plane section of the
blade, but its width at any distance from the axis of the screw is the
maximum thickness of the blade at a radius equal to that distance.

The developed sections of the blade at different radii are shown
between the two elevations. Referring to the second section from
the axis, the straight base NjHi is made equal in length to the elliptic
arc Ne'H. These developed sections give the sizes and shapes of the
thickness pieces such as PLRQ (Fig. 720) referred to in describing
the moulding of a propeller. These developed sections are generally
superimposed on one of the elevations of the blade, but in Fig. 721
they have been moved to one side for the sake of clearness.

A second example is illustrated by Fig. 722 which shows a detach-
able blade with a flange for bolting to a boss keyed to the shaft. In
this example the blade is set back by inclining the generating line of
the screw surface to the axis at an angle less than a right angle.
This is shown on the elevation to the left in Fig. 722 by the tilting
over of the maximum thickness section. Points on the generating
line will describe helices just as before but the various helices will be
displaced in an axial direction each by a definite amount depending on
its distance from the axis. For example, the helix at the radius o'e'
will be displaced by the amount d^^Oi. Hence the plan of this helix

380

PRACTICAL GEOMETRY

will cub the plan of the axis at e such that oe is equal to dj'e/. The
slope 6 of the helix may be obtained as before but in Fig. 722 the
triangle for finding 6 is shown on the elevation instead of on the plan.
neSy the plan of the helix on the face of the blade at radius o'e' is drawn
at right angles to e'T. The plan and the two elevations of the edge
of the blade are otherwise drawn as before, the construction lines being
clearly shown.

It will be seen that for the same diameter of screw the true length

RIGHT- HANDED SCREW
PITCH s|-38x DIAMETER

Fig. 722.

of the blade is slightly increased by giving it a set back, but this is
generally neglected in applying the developed face of the blade to the
elevation as shown in Fig. 722.

In the examples so far considered all the helices on the face of the
blade have had the same pitch. Frequently however the helix at the
root has a less pitch than the helix at the tip and the pitch increases
uniformly from the root to the tip. The screw is then said to have a
radially increasing pitch. Fig 723 shows clearly how the slopes of the

HELICES AND SCREWS

381

helices at different radii are obtained when the blade has no set back.
After the slopes of the various helices have been found, the plan, shown
in Fig. 723, and the elevations (not shown) are determined as before
from the developed face of the blade.

For a screw which has a radially increasing pitch the moulder

Ri= Radius at Tip
Ri= Radius at Root
Pi = Pitch at Tip
Pz' Pitch at Root

Fig, 723.

requires a templet for the root as well as one for the tip for the purpose
of guiding the loam board in sweeping out the face of the blade. The
loam board must also be connected to the arm in such a way that it
can alter its inclination as the arm revolves, or the arm may be simply
forked to embrace the central spindla

Ri= Radius at Tip
Rz- Radius at Root
Q-' Set Back at Tip
Pi = Pitch at Tip
P2= Pitch at Root

Fig. 724.

The slopes of the various helices will not be altered by giving the
blade a set back and the construction given in Fig. 723 will therefore
still apply in finding these slopes but the plans of the helices on the
face of the blade will not now pass through the point o but through
points determined as explained in connection with Fig. 722.

382

PRACTICAL GEOMETRY

A convenient construction for finding at the same time the slopes
and the positions of the plans of the various helices on the face of the
blade is clearly shown in Fig. 724 and needs no further description.

Still another modification of the design of the face of the blade of

Leading Edge

A 2. 3

Ri = Radius at Tip
R2= Radius at Root
Pi = PiUh of Leading Edge
P3= Pitch of ToUoimg Edge

ij'iG. 725.

a screw propeller has to be mentioned. The part of the face of the
blade towards the leading edge may have a less pitch than the remain-
ing part towards the following edge. The face of the blade is then
said to have an axially increasing pitch. The necessary modification in

Fig.

^^^Radias at Tip
^-^Radius at Root
Q^ Set Back at Tip
, - Pitch of Leading Edge at Tip
?i= Pitch of Leading Edge at Root
?i=-Pitch of Following Edge at Tip
Pa'^ Pitch (^ Following Edge at Root

726.

the construction of the plan of the blade is shown in Figs. 725 and
726. In Fig. 725 the face of the blade has an axially increasing pitch
but a radially constant pitch and no set back. In Fig. 726 there is
an axially increasing pitch, a radially increasing pitch, and also a set
back.

HELICES AND SCREWS 383

Exercises XXV

1. A right-handed helix of 2 inches pitch, and a left-handed helix of 1 inch
pitch are traced on a vertical cylinder 2 inches in diameter. Draw the elevation
of two turns of the first helix and four turns of the second. The lower ends of
the two helices to be at opposite extremities of a diameter of the cylinder which
is perpendicular to the ground line.

2. Two helices are traced on the surface of a vertical cylinder 2 inches in
diameter. The helices are at right angles to one another at their intersection
and one of them, which is right-handed, has a pitch of 4 inches. Show as much
of the two helices as is contained in 4 inches length of cylinder. Draw the
elevations of the tangents to these helices at points which are 1 inch, 2 inches,
and 3 inches above the lower end of the cylinder.

3. Show one turn of a helix of 3 inches pitch on a vertical cylinder 1 inch in
diameter. Draw the locus of the horizontal trace of a moving tangent to the
helix, and determine the true form of a vertical section of the surface described
by the moving tangent, the plane of section to be at a distance of 1 inch from the
axis of the cylinder.

4. A cylinder 1-5 inches in diameter and 3 inches long has three helices of
3 inches pitch traced on its surface at equal distances apart. Draw the plan of
the cylinder and helices when the axis of the cylinder is inclined at 45^^ to the
horizontal plane.

5. Work the example on the helix of increasing pitch illustrated by Fig. 701,
p. 364, having given, — diameter of cylinder 2 inches, height of cylinder 3 inches.
On the development draw the curve whose ordinates are equal to the pitch at
each point of the helix of increasing pitch.

6. A straight line 1-5 inches long is at right angles to a fixed vertical axis and
has one end on that axis. This straight line describes a screw surface of 3 inches
pitch. Draw the elevation of one complete turn of this screw surface, showing
on it the generating line in positions whose plans are at intervals of 15^. Show
also the helices described by points on the generating line which divide it into
three equal parts in addition to the helix described by the outer end.

7. Same as the preceding exercise except that the generating line is 2*4 inches
long and is inclined at 45° to the axis.

8. Eeferring to the example illustrated by Fig. 703, p. 365, draw the elevation
of one complete convolution of the surface generated by the quadrant of a circle,
having given, — radius of quadrant 2 inches, axial advance per revolution 3 inches.
Show the generating arc in positions whose plans are at intervals of 15'^. Also,
in addition to the helix described by the lower end of the generating arc, show
the helices described by three intermediate points. Lastly draw the elevation of
a section of the surface by a plane parallel to the vertical plane of projection and
0-5 inch in front of the axis.

9. An equilateral triangle of 2*5 inches side moves with one side on a vertical
axis so that the other two sides describe screw surfaces of 1*5 inches pitch. Draw
the elevation of the surfaces described by four revolutions of the triangle. On
the lower half of the elevation show the generating lines in positions whose plans
are at intervals of 15*^. Show the helix which is the intersection of the two
screw surfaces.

10. Same as the preceding exercise except that the pitch is 2*5 inches instead
of 1-5 inches. [In this case the two screw surfaces do not intersect.]

11. Full size axial sections of various screw threads are shown in Fig. 727.
In each the diameter over the threads is 3 inches, the screws are all right-
handed and single threaded. Draw for each a projection on a plane parallel to
the axis of the screw, showing as much of the screw as falls within a length of
4 inches measured parallel to the axis. Show also for each a section of the nut
by a plane containing the axis. Height of nut in each case 3 inches.

12. A right-handed treble threaded worm 3*5 inches external diameter and

384

PRACTICAL GEOMETRY

4 inches long has threads of the form and dimensions shown by the axial section
{d) Fig. 727. Draw a projection of this worm on a plane parallel to its axis.

(a)

'^^iii^^

Section on A A

13. The following particulars relate to three helical springs. — No. 1. Axial
section a rectangle 0*5 inch x 0-125 inch with the longer sides parallel to the
axis. External diameter 2 inches. Pitch 1 inch. No. 2. Axial section a rect-
angle 0-5 inch x 0-125 inch with the longer side perpendicular to the axis.
External diameter 3 inches. Pitch 1 inch. No, 3, Normal section a circle 0*5
inch diameter. External diameter 3-5 inches. Pitch 1-75
inches.

14. A cylinder 3 inches in diameter and 4 inches long
has a left-handed helical groove cut in it. A section of the
groove by a plane at right angles to the axis of the cylinder
is 1 inch wide and 1 inch deep, the sides being parallel to
and the bottom at right angles to the mid-radial line of the
section. Draw a projection of the cylinder and groove on
a plane parallel to the axis of the cylinder and determine
the true forms of axial and normal sections of the groove,

15. A portion of a 1| inch twist drill is shown in Fig,
728, consisting of a cylinder with a conical end, cut with
two spiral grooves each of 12 inches pitch and of the form
defined by the given sectional elevation. Complete the
plan, showing the curve BB, and the helical' grooves cor-
rectly projected for a distance of 6 inches from the line AA.

[B,E.]

16. A circle 1^ inches in diameter, whose plane
is vertical, revolves with uniform angular velocity
about a vertical axis in its plane, describing an annulus
or anchor ring whose mean radius is 1^ inches. While
the circle is revolving a point P on its circumference
moves round the circle with uniform velocity. The
point P travels sixteen times round the circle while
the circle revolves three times about the vertical
axis. Draw the plan of the curvd traced by the point
P. The result is shown to a reduced scale in Fig. 729.
Note that the curve is endless.

17. ABO is a right angled triangle. The sides AC
and BC which contain the right angle are 3*5 inches
and 1-5 inches respectively. The triangle revolves
with uniform angular velocity about AC which is ver-
tical. A point P starting at A moves along AB with uniform velocity and

HELICES AND SCREWS

385

reaches B while AB makes two revolutions. Draw the plan and elevation of
the path of P. Draw also the development of the surface described by AB and
show on it the curve traced by P.

18. The radius of the base of a right cone is 1-25 inches, the axis, which is
4-5 inches long, being inclined at 40^ to the H.P. The apex is above the base
and pointing to the right. From the highest point in the base, a cord 0-25 inch
in diameter is coiled round the cone in a right-handed spiral. The clear distance
between the coils, measured in a straight line parallel to the surface of the cone
is 1-25 inches. Draw the plan of the cone showing two turns of the cord, [b.e.]

19. The centre line CDDO of a portion of a handrail is shown in plan and
elevation in Fig. 730, the parts CD, CD being straight, and the part DD being an
elliptic arc. Draw the elevation of the centre line on X'Y'. Find the true shape
of the figure CDDC. [b.e.]

20. Referring to Fig. 730, draw the development of the surface of which cddc
is the plan and show on it the centre line CDDC.

Fig. 730. Fig. 731. Fig. 732.

The above Fig's, are to be reproduced double size.

21. The plan and part bt an elevation of the centre line ABCD of a portion of
a handrail are shown in Fig. 731. Assuming that the straight and curved parts
are in the same plane complete the elevation and draw another elevation on X'Y'.
Show also the true form of the centre line ABCD.

22. Fig. 732 shows the plan and part of an elevation of the centre line ABCDE
of a portion of a handrail. The parts AB and DE are straight. ST is the tangent
to the centre line at and this tangent meets AB produced at S and ED produced
at T. Assuming that BC is in the plane containing AS and SC and that CD is in
the plane containing CT and TE, complete the elevation of the centre line and
draw another elevation on X'Y'. Draw also the development of the vertical
surface containing the centre line ABCDE showing on it that centre line.

23. Fig. 733 gives the plan and part of the elevation of a portion of a handrail
of rectangular cross-section (as prepared for moulding) ; complete the elevation of
the rail. Draw a second elevation of the rail on X'Y'. Set out the " face mould "
for this rail. [b.e.J

386

PRACTICAL GEOMETRY

24. Fig. 734 gives the plan and part of the elevation of a portion of a handrail
of rectangular cross-section (as prepared for moulding) ; complete the elevation of
the rail. Set out the "face mould" for this rail. Draw also the " falling mould "
for the central vertical section CDDC of the rail. [b.e.]

Fig. 733. Fig. 734. Fig. 735.

The above Figs, are to be reproduced double size.

25. Stair rail (Fig. 735). To be made in two parts, cut from planks and con- *
nected at C by a joint perpendicular to FG. A plan and part elevation are
given. Complete the elevation of the centre line ABODE and draw a second
elevation of this line on X'Y'. Show the plan and two elevations of the joint at
C. Draw the "face mould" for the wreath ABC, and set out the "bevels" for
its ends.

Plot the " falling moulds " for the central section and the inner and outer
surfaces of the wreath ABC. Complete the elevation of the wreath ABC. [b.e.]

26. In Fig. 736 are given the flange and the maximum thickness section of
one blade of a three-bladed propeller ; the dotted curve is the developed face of
the blade. The face of the blade is a true right-handed screw surface whose pitch
is one and a quarter times the diameter of the propeller. Draw the figure double
size and then construct a plan and two elevations together with sections of the
blade as shown in Fig. 721, p. 378.

27. Same as the preceding exercise except that the blade is to have a set back
which, measured at the tip, is 0*06 of the diameter of the propeller ; also the
screw is to be left-handed instead of right-handed.

28. The flange and the maximum thickness section of one of the four de-
tachable blades of a screw propeller are shown in Fig. 737. The dotted curve is
the developed face of the blade. The diameter of the propeller is 19 ft 6 in. at
the tip and 5 ft. 2 in. at the root. The pitch of the face towards the leading edge
increases from 19 ft. at the root to 21 ft. at the tip, and the pitch of the face
towards the following edge increases from 19 ft. at the root to 23 feet at the tip.

HELICES AND SCREWS

387

Draw the plan and two elevations of this blade to a scale of one inch to a foot,
and determine the developed sections of the blade by cylindrical surfaces of 3A,
44, 5i, 6i, 7^ and 8.^ ft. radii.

t*- 18 — ^

1 "^^

1 \

1 tu

! ^

■^ i

1 u.

1 .'

\
\

1 /
1 /

AXIS OF

SHAFT i

DIMENSIONS
IN INCHES

T""1 1

Fig. 736.

AXIS OF SHAFT
Fig. 737.

CHAPTER XXVI

INTERSECTION OF SURFACES

320. General Method. — The general method of finding the
intersection of two surfaces is as follows. Let A and B denote two
given surfaces whose intersection with one another is required. Cut
the surfaces A and B by a third surface C, the latter surface being so
chosen and employed that the projections of its intersections with A
and B are lines, such as straight lines or circles, which can easily be
drawn. Let A' denote the line of intersection of C and A, and let B'
denote the line of intersection of C and B. Let the lines A' and
B' meet at a point P. Then the point P lying on the intersection
of C and A, lies on A. Also since the point P lies on the intersection
of C and B, it lies on B. The point P therefore lies on A and B, and
therefore it must be a point on the intersection of A and B, By
moving the surface C into diflPerent positions, or by cutting A and B
by other surfaces similar to C, any number of points on the intersection
of A and B may be determined.

On the intersection of two surfaces there are generally certain
important points which should be first determined. For example if
the projection of the intersection meets a boundary line of the pro-
jection of one of the surfaces, the meeting point will be an important
one to determine and in the case of a curved surface this will be a
tangent point.

In moving the cutting surface in one direction from a position
which gives points on the required intersection to a position which
gives no such points, there will be an intermediate position which
gives the last of the points obtained by moving the surface in that
direction and these points are usually important.

As a general rule after the important points have been determined
only a few others are necessary to fix the required line of intersection.

It is a good plan to number or letter the cutting surfaces in order,
and the points determined by them, each point having the same number
or letter as the cutting surface on which it is situated.

321. Intersection of Two Cylinders. — The general method
of finding the intersection of the surfaces of two cylinders is to cut the
surfaces by planes parallel to their axes or to their generating lines.
These planes will intersect the surfaces of the cylinders in straight

INTERSECTION OF SURFACES

389

f? -s^^^

lines, the intersection of which with one another will determine points
on the intersection required.

Example 1. — The horizontal trace of a vertical cylinder is an
ellipse, 32 inches X 2-4 inches, the major axis being parallel to XY.
A horizontal cylinder, 1 '6 inches in diameter has its axis parallel to
the vertical plane and 0'3 inch in front of the axis of the vertical
cylinder. It is required to show the elevation of the intersection of
the cylinders.

There will be two identical curves in this case and in Fig. 738
only enough of each of the cylinders is shown for determining one of
these curves.

Cutting planes parallel to the axes of both cylinders will in this
case be vertical planes parallel to the vertical plane of projection.
HT is the horizontal trace of one of
these planes. This plane will cut the
vertical cylinder in two vertical straight
lines, one of which will have the point a
for its plan, and a perpendicular to XY
from a for its elevation. This same
plane will cut the horizontal cylinder in
two horizontal lines of which ah will be
the plan. To find the elevations of
these lines, take an elevation of the
horizontal cylinder on a plane perpen-
dicular to its axis, this elevation will be
a circle, but only half of it need be
drawn, as shown. The cutting plane
now being considered will have a trace
on this new vertical plane which will
coincide with its horizontal trace. The
point 6/ where this trace cuts the circle,
or semicircle, just mentioned will be the
end elevation of AB one of the lines in Fiq, 733.

which the horizontal cylinder is cut by

the cutting plane HT, and the length hhi will be the vertical distance
of these lines above and below the horizontal plane containing the axis
of that cylinder. Hence the required elevations a'6', a'b' will be at
distances equal to 66/ above and below the elevation of the axis of the
horizontal cylinder.

The intersections of a'h' and a'b' with the vertical line through a
determine points on the intersection required.

In like manner, by taking other planes, other points can be found.

The most important points on the curves of intersection in this
case are those determined by the following cutting planes. (1) The
two planes which touch the horizontal cylinder, these determine the
points 1' and 5' on the elevation. (2) The plane which contains the
axis of the horizontal cylinder, this determines the points marked 3'.
(3) The plane which contains the axis of the vertical cylinder, this
determines the points marked 4'.

390

PRACTICAL GEOMETRY

The foregoing example may also be worked by using horizontal
cutting planes, and the student would do well to work the problem in
this way.

Example 2. — A vertical tube, external diameter 2-75 inches,
internal diameter 1-5 inches, has a horizontal cylindrical hole bored
through it 1-25 inches in diameter. The axis of the hole is 0-25 inch
from the axis of the tube. It is required to draw an elevation on a
vertical plane inclined at 35° to the axis of the hole.

First determine the intersection of the boring cylinder with the
outside of the tube exactly, as in the preceding example. The fact of
the horizontal cylinder being inclined to
the vertical plane will make no difference
in the working, as the heights of all the
lines in which the cutting planes inter-
sect the horizontal cylinder will remain
the same whatever be its inclination to
the vertical plane.

Next in like manner the intersection
of the boring cylinder with the interior
surface of the tube is determined. It will
be found that in this example this
interior intersection consists of one line
only.

The cutting planes which give the
most important points on the curves of
intersection in this example are the same
as those in example 1, with the addition
of the plane which touches the inner
surface of the tube.

The result of this example is shown
in Fig. 739, but the construction lines
have been omitted.

This example, like the preceding one, may also be worked by using
horizontal cutting planes.

Example 3 (Fig. 740).— a6 a'h' is the axis of a cylinder whose
horizontal trace is a circle 2*25 inches in diameter, cd c'd' is the axis
of a cylinder whose horizontal trace is an ellipse (major axis 3 inches,
minor axis 2 inches), whose minor axis is parallel to XY. It is required
to show, in plan and elevation, the intersection of the surfaces.

In this example the planes which will intersect the surfaces of both
cylinders in straight lines will be inclined to both planes of projection.

To find the directions of the traces of the cutting planes take a
point j>p' in ab a'h'. Through this point draw pq p'^ parallel to cd c'd'.
The line aq which joins the horizontal traces of ah a'h' and pq p'^ will
be the horizontal trace of a plane which contains the axis of one
cylinder and is parallel to the axis of the other, and this plane, and
planes parallel to it will, if they cut the cylinders at all, cut them in
straight lines parallel to their respective axes. In this particular
example the horizontal traces of these planes are parallel to XY.

Fig. 739.

INTERSECTION OF SURFACES

391

Seven planes whose horizontal traces are shown, and numbered
from 1 to 7, will determine all the important points, and no other
planes need be taken.

Fig. 740.

Consider plane number 3, This plane cuts each cylinder in two
straight lines whose plans are parallel to the plans of the axes of the
respective cylinders, and whose horizontal traces are at the points
where the horizontal trace of the plane cuts the horizontal traces of
the cylinders. The elevations of these lines will be parallel to the

392 PRACTICAL GEOMETRY

elevations of the axes of the respective cylinders upon which they lie.
These four lines intersect at four points in plan and in elevation and
determine the four points on the intersection of the cylinders which
have the number 3 attached to them.

It will be observed that plane number 1 touches both cylinders.

As already stated the horizontal traces of all the necessary cutting
planes are shown, but the remaining construction lines are only shown
for planes 1, 3, and 6, for the sake of making the figure clearer.

The portion of the curve which is seen in plan, and which must
therefore be put in as a full line, is found from the intersection of those
lines which lie on the upper surface of one cylinder with those which
lie on the upper surface of the other, that is, from the intersection of
the lines whose horizontal traces lie on the part fgh of the ellipse with
those whose horizontal traces lie on the part Imn of the circle. The
remainder of the curve in plan will be hidden by one or other or both
of the cylinders, and will therefore be put in as a dotted line.

In like manner the part of the curve which is seen in the elevation
is found from the intersection of the lines which lie on the front of one
cylinder with those which lie on the front of the other, that is from
the intersection of the lines whose horizontal traces lie on the halves
of the horizontal traces of the cylinders which are furthest from XY.

Instead of using the horizontal traces of the cylinders and the
horizontal traces of the cutting planes, the vertical traces may be
employed if the vertical traces of the cylinders come within a con-
venient distance on the paper. When one cylinder has a vertical trace
but no horizontal trace within a convenient distance, and the other
has a horizontal trace but no vertical trace within a convenient
distance, it is necessary to use both the horizontal and vertical traces
of the cutting planes.

322. Intersection of Cylinder and Cone. — The general
method of finding the intersection of a cylinder and cone is to cut the
surfaces by planes parallel to the axis of the cylinder and passing
through the vertex of the cone. These planes will cut the surface of
the cylinder in straight lines parallel to its axis, and the surface of the
cone in straight lines passing through its vertex. The intersection of
the lines on the cylinder with the lines on the cone determine points
on the intersection required.

Example 1 (Eig. 741). A right cone, base 2*9 inches diameter, axis
3-8 inches long, has its base horizontal. A cylinder 2 inches in
diameter has its axis horizontal and 1*2 inches above the base, and 02
inch from the axis of the cone. It is required to draw the plan of the
intersection of the surfaces and an elevation on a vertical plane in-
clined at 30° to the axis of the cylinder.

The general method which has just been described can easily be
applied to this example, but in this case the cutting planes may be
horizontal, because horizontal sections of the cone are circles and hori-
zontal sections of the cylinder are straight lines. In Fig. 741, hori-
zontal cutting planes have been taken.

Draw an elevation of the cylinder and cone on a plane perpendicular

INTERSECTION OF SURFACES

393

to the axis of the cylinder. In Fig. 741 this elevation is shown brought
round into the plane of the other elevation.

Consider the cutting plane of which L'M' is the vertical trace.
This plane intersects the cone in a circle of which the diameter is equal
to a/6/. The plan of this circle is a circle concentric with the plan of
the cone. This same cutting plane intersects the curved surface of the
cylinder in two straight lines of which the points 6/ and c/ are the end
elevations. The plans of these lines are parallel to the plan of the
axis of the cylinder, and at distances from it equal to half of t/c/. The
points h and e where these lines on the plan meet the circle already
mentioned, are the plans of points on the curve of intersection, and

Fig. 741.

perpendiculars from b and c to XY to meet L'M' at h' and c' determine
the elevations of these points. In like manner any number of points
on the curve of intersection may be found.

In Fig. 741 eight cutting planes are shown numbered from 1 to 8,
and these should be sufficient.

The most important points are on the planes 1, 3, 4, 5, 7, and 8.
The position of the plane number 5 is obtained as follows, jp/^/ is the
vertical trace of a plane which touches the cylinder and passes through
the vertex of the cone, p/ being the end elevation of the line of contact
of this plane and the cylinder, o^p^ is of course perpendicular to
2?/^/. The vertical trace of plane number 5 passes through p^'. The

394

PRACTICAL GEOMETRY

lines in which the plane p^ql cuts the cone are also shown and as
these lines are tangents to the curve of intersection they assist in the
correct drawing of the curve at important points.

Example 2 (Fig. 742). vv' is the vertex and va v'a' the axis of a
cone whose vertical angle is 30°. he h'c is the axis of a cylinder 2 inches
in diameter. The dimensions which fix the positions of the axes of
the cone and cylindei' are given at the top right hand corner of

Fig. 742.

Fig. 742. It is required to show, in plan and elevation, the inter-
section of the surfaces of the cone and cylinder.

This example is worked by using cutting planes passing through
the vertex of the cone and parallel to the axis of the cylinder which,
as already stated, is the general method of finding the intersection of
the surfaces of a cone and cylinder.

The first step is to determine the horizontal traces of the cylinder
and cone by Arts. 219 and 226 respectively. The construction lines for
this are omitted in Fig. 742.

All planes which are parallel to the axis of the cylinder and pass

INTERSECTION OF SURlFACES

395

through the vertex of the cone will contain the line vt v't' drawn
through v>v' parallel to he h'c . Hence the horizontal traces of all the
cutting planes will pass through t, the horizontal trace of vt v't'.

There are eight planes required to determine all the important
points but only three of these planes, numbered 1, 4, and 8, are shown
in Fig. 742. Plane number 1 touches the cylinder and* cuts the cone,
and plane number 8 touches the cone and cuts the cylinder ; each of
these planes will therefore determine two points only on the inter-
section required. Plane number 4 cuts both surfaces and will
determine four points on the intersection required. The omitted
planes 2, 3, 5, 6, and 7 will each determine four points.

323. Intersection of Two Cones. — The general method in this

Fig. 743.
•ase is to use cutting planes passing through the vertex of each cone.
These planes will cut the surface of each cone iu straight lines, and

396

PRACTICAL GEOMETRY

they will all contain the line joining the vertices of the cones. Hence
the horizontal traces of all the cutting planes will pass through the
horizontal trace of the line joining the vertices of the cones. Also, the
vertical traces of the cutting planes will pass through the vertical
trace of the line joining the vertices of the cones.

Example (Fig. 743). v{ol is the vertex of a cone whose horizontal
trace is a circle 3 inches in diameter, v^v.^ is the vertex of a cone
whose vertical trace is a circle 2*5 inches in diameter. The other
dimensions are given on the figure. It is required to show, in plan
and elevation, the intersection of the surfaces of the two cones.

Seven cutting planes passing through the vertex of each cone
will determine all the important points in this example, but in Fig. 743
only four of these planes, numbered 1, 2, 3, and 4, are shown. Both
traces of each cutting plane are employed. All the horizontal traces
of the cutting planes pass through v^, and all the vertical traces pass
through v/.

Consider plane number 1. This plane touches one cone and cuts
the other. The horizontal trace of the plane is drawn first, and if the
point where this trace meets the ground line comes within the paper
the vertical trace is obtained by joining «?/ to this point. If however
the point where the horizontal trace meets the ground line is off
the paper, then the construction given in Art. 11, p. 9, may be used
for drawing the vertical trace. The intersections of the line in which
plane number 1 touches the one cone with the lines in which it cuts
the surface of the other cone determine the points numbered 1 on
the intersection required. In like manner the other points are
determined.

324. Intersections of Cylinders and Cones enveloping
the same Sphere. — If two cylinders envelop the same sphere their

Fig. 744.

Fig. 745.

Fig. 746.

intersection will be plane sections of both and will therefore be
ellipses. Fig. 744 shows a projection of two such cylinders on a plane
parallel to their axes. The straight lines ah and cd form the projection
of the intersection of the cylinders.

The same remarks apply to the cone and cylinder (Fig. 745), and
to two cones (Fig. 746). Referring to Fig. 745, if the vertex of the

INTERSECTION OF SURFACES

397

cone is placed on the curved surface of the cylinder, one of the
ellipses will become a straight line. Referring to Fig. 746, it is
obvious that if the cone v.2ad be turned round so as to make V2d more
nearly parallel to v^a, the straight line cd will become more nearly
parallel to v^a and v^, and when v./l is parallel to v^a, cd will also be
parallel to v^a and v.^, and the intersection cd will then be a parabolic
section of each cone. Continuing the motion of the cone v.^ad in the
same direction, the intersection cd will become a hyperbolic section of
each cone.

325. Intersection of Cylinder and Sphere. — Since the sur-
face of a sphere is a particular form of a surface of revolution this
problem is a particular case of that discussed in Art. 326, p. 398. If

3^*diam.

Fig. 747.

the cylinder is also a surface of revolution, that is, a right circular
cylinder, then this problem is also a particular case of that considered
in Art. 328, p. 401. But instead of using the method described in
Art. 326, or the method of Art. 328 cutting planes parallel to the axis
of the cylinder and perpendicular to one of the planes of projection
may be employed, because such planes will cut the sphere in circles
and the cylinder in straight lines.

If the axis of the cylinder is inclined to both planes of projection
then an auxiliary elevation on a vertical plane parallel to the axis of

398 PRACTICAL GEOMETRY

the cylinder should be drawn, and the cutting planes being vertical
and parallel to the axis of the cylinder they will cut the sphere in
circles which will appear as circles in the auxiliary elevation.

Example (Fig. 747). The plan and elevation of a cylinder and
sphere are given, the dimensions being marked on the figure. It is
required to show, in plan and elevation, the intersection of the surfaces
of the cylinder and sphere.

The method used here is that of cutting the surfaces by horizontal
planes.

The upper right hand portion of Fig. 747 is a projection of the
cylinder and sphere on a plane perpendicular to the axis of the cylinder
and it is from this projection that the positions of the cutting planes
which give the important points on the required intersection arc
determined.

326. Intersection of Cylinder and Surface of Revolution. —
In all the problems hitherto considered on the intersection of surfaces,
the auxiliary cutting surfaces which have been used, in order to find
points on the required intersection, have been planes. The only simple
plane sections of a surface of revolution are, in general, sections at right
angles to its axis, which are always circles. Now it is evident that
only in very particular cases would a plane which cuts the surface of
revolution in a circle cut the surface of the cylinder in a circle or in
straight lines. For instance, a plane which cuts the surface of revolu-
tion in a circle may cut the cylinder in an ellipse, and it would clearly
be a laborious process to construct an ellipse for each cutting plane.
This objection may however be got over in the present problem by using
a tracing of the section of the cylinder by a plane which cuts the surface
of revolution in a circle, in a manner to be explained at the end of this
article.

Instead of cutting the given surfaces by planes they may be cut by
the surfaces of cylinders and points on the required intersection obtained
by means of circles and straight lines. Let a circle which is a section
of the surface of revolution be taken and let a straight line move in
contact with this circle and also remain parallel to the axis of the given
cylinder. This moving line will describe the surface of a cylinder which
intersects the surface of revolution in a circle and the surface of the
given cylinder in straight lines, and the points in which these straight
lines cut the circle will be points on the intersection required.

In order that the projections of the circular sections of the surface
of revolution may be circles and straight lines the axis of revolution
must be arranged perpendicular to one of the planes of projection.

Referring to Fig. 748, mn, m'n' is the axis of a cylinder which
intersects a surface of revolution whose axis is vertical. The con-
struction for the outline of the elevation of the surface of revolution is
shown at (c).

The ellipse which is the horizontal trace of the cylinder is deter-
mined as explained in Art. 219, p. 253.

ab, a'b' is a circular «ection of the surf acfe of revolution, oo' being the
centre of this circle. o<, o't' is a line through oo' parallel to mn, m'n'.

INTERSECTION OF SURFACES

399

ot, o't' is the axis of an auxiliary cylinder of which the circle a&, a'h' is
one horizontal section, and all horizontal sections of this cylinder will
be circles of the same diameter, it' being the horizontal trace of the
axis of this auxiliary cylinder the horizontal trace of its surface will be
a circle whose centre is t and radius equal to oa. The horizontal trace

f t

■7

Fig. 748.

of this auxiliary cylinder intersects the horizontal trace of the given
cylinder at s and r. The auxiliary cylinder intersects the given
cylinder in straight lines sjp, s'p' and r^, r'q' which are parallel to wm,
m'n'. pp' and qq the intersection of the lines 8p, s'p' and rq, r'q' with

400 PRACTICAL GEOMETRY

the circle ah, a'b' are points on the line of intersection of the given
cylinder and surface of revolution. By taking other auxiliary cylinders,
any number of points on the required intersection may be found.

The student should notice that each line of intersection of an
auxiliary cylinder and the given cylinder intersects the corresponding
circle on the surface of revolution in one point only although its plan
may cut the plan of the circle in two points.

A convenient and practical method of solving the problem which
has just been considered is to take horizontal sections of both the given
surfaces. The sections of the cylinder are ellipses but these ellipses are
all of the same size and if one of them be drawn and a tracing of it
made, it only remains to draw a sufficient number of circular sections
of the surface of revolution and apply the tracing of the ellipse to each
to find points in the required intersection. The position of the centre
of the ellipse corresponding to a particular circular section of the surface
of revolution is where the plane of that section cuts the axis of the
cylinder. For the circular section afe, a'b' (Fig. 748) e is the position
of the centre of the ellipse in the plan and the tracing of the ellipse is
placed so that the centre is at e and the major axis on mn. The points
where the ellipse cuts the circle ah are points on the plan of the inter-
section required and these points may be pricked through. The eleva-
tions of the points are of course perpendicularly over their plans and on
the elevation of the corresponding circular section.

327. Intersection of Cone and Surface of Revolution. —
Placing the surface of revolution so that its axis is vertical, horizontal
sections of it will be circles, but except in the special case where the
cone is a right circular cone and its axis is vertical, horizontal sections
of the cone will not be circles or straight lines and all the horizontal
sections will be different. The tracing paper method which is applicable
to the intersection of a cylinder and a surface of revolution and which
was described in the latter part of the preceding Art., is therefore not
suitable in the case of the intersection of a cone and a surface of
revolution.

Auxiliary cones are taken which have their vertices at the vertex
of the given cone and for their directrices they have circular sections
of the surface of revolution. These auxiliary cones will intersect the
given cone in straight lines and the intersection of these straight lines
with the corresponding circles on the surface of revolution will deter-
mine points on the intersection required.

Referring to Fig. 749, vn, v'n' is the axis of a cone whose vertical
angle is 30° and which intersects a surface of revolution whose axis is
vertical. The dimensions are given on the figure.

The horizontal trace of the cone is determined as explained in Art.
226, p. 259. ah, a'h' is a circular section of the surface of revolution,
oo' being the centre of this circle, vt, v't' is a line passing through oo',
the centre of the circle, and the vertex of the cone, vt, v't' is the axis
of an auxiliary cone of which the circle ah, a!h' is one horizontal section.
All horizontal sections of this auxiliary cone will be circles but they
will be of different diameters. The point U' being the horizontal trace

INTERSECTION OF SURFACES

401

of the avxis of the auxiliary cone it will be the centre of the circle which
is the horizontal trace of that cone. The radius of this circle is equal
to /'W where m' is found by joining v' to a' and producing it to meet XY
as shown. The horizontal trace of the auxiliary cone and the horizontal
trace of the given cone intersect at s and r, and the auxiliary cone
intersects the given cone in straight lines vr, v'r' and t's, v's'. pp and

Fig. 749.

qcl the points of intersection of the lines vr, v'r' and vs, v's' with the
circle ah, a'h' are points on the intersection of the given cone and
surface of revolution. By taking other auxiliary cones any number of
points on the required intersection may be found.

328. Intersection of Two Surfaces of Revolution whose
Axes are Parallel. — Since all sections of a surface of revolution by
planes perpendicular to its axis are circles, it follows, that if the axes
of two surfaces of revolution are parallel, a plane which is perpendicular

402

PRACTICAL GEOMETRY

to the axis of one will be perpendicular to the axis of the other, and
this plane, if it cuts both surfaces, will cut them in circles the inter-
section of which with one
another determines points
on the intersection required.

The surfaces of revolu-
tion should be arranged so
that their axes are perpen-
dicular to one of the planes
of projection. The projec-
tions of the circles men-
tioned above will then be
circles and straight lines
which are easily drawn.

An example is shown in
Fig. 750. An " anchor ring "
whose surface is described
by the revolution of a vertical
circle about a vertical axis is
shown penetrated by a cone
of revolution whose axis is
also vertical. One cutting
plane is shown intersecting
the anchor ring in two cir-
cles ah, a'b' and cd, c'd'.
This same plane intersects -p^^^ rj^

the cone in the circle mn,

m'n'. From the plan it is seen that the circle mn on the cone only inter-
sects the circle ah on the anchor ring determining the two points p
on the plan of the intersection of the cone and anchor ring. It will
be seen that the cone touches the surface of the anchor ring at the
point qq'.

329. Intersection of Two Surfaces of Revolution whose
Axes Intersect. — Place the surfaces so that the axis of one of them
is vertical and the axis of the other is parallel to the vertical plane of
projection. Referring to Fig. 751, vn v'n' is the axis of a cone of
revolution which intersects a surface described by the revolution of an
arc of a circle about a vertical axis as shown. The axes of the two
surfaces intersect at oo'. h'a'd', the elevation of a sphere is shown, the
centre of this sphere being at the intersection of the axes of the given
surfaces of revolution. This sphere intersects the surface whose axis is
vertical in a circle whose elevation is the horizontal line a'h' and whose
plan is the circle ah having its centre at o. This sphere also intersects
the other given surface of revolution in a circle whose elevation is the
straight line c'd'. These two circles lie on the sphere and one being on
one of the given surfaces of revolution and the other on the other, their
points of intersection are points on the intersection required. ^>', the
point of intersection of a'h' and c'd', is the elevation of the points of
intersection of the two circles and their plans are determined by a

INTERSECTION OF SURFACES

403

projector to intersect the
circle ah. Observe that the
plan of the circle of which
c'd' is the elevation, and
which would be an ellipse,
need not be drawn. Taking
other spheres with their
centres at oo any number
of points on the inter-
section of the given sur-
faces may be found.

330. Intersection of
Two Spheres.— Since the
surface of a sphere is a
surface of revolution the
methods described in the
two preceding articles for
determining the intersec-
tion of two surfaces of
revolution are applicable
to the case of two spheres.
But, since the intersection
of two spheres is a circle,
the projections of this cir-
cle may be found by a
simpler method.

Referring to Fig. 752,
o^ and 0.2 are the plans of
the centres of two spheres
whose radii are r^ and n
respectively. An elevation
is drawn on a ground line
XY parallel to o^o^. The
common chord a'h' of the
circles which are the ele-
vations of the two spheres
on the ground line XY is
an elevation of the circle
which is the intersection
of the spheres. The plan
of this circle is an ellipse
whose minor axis is ab and
whose major axis is equal
to a'h'. From the plan and
elevation shown an eleva-
tion on any other ground
line is easily determined.

The intersection of the
surfaces of the two spheres

Fig. 752.

404 PRACTICAL GEOMETRY

referred to above is the locus of a point whose distances from the points
Oy and O^ are 7\ and r.> respectively.

In like manner the intersection of a third sphere whose centre is
0;{ and radius r.^ with the second sphere whose centre is O2 and radius
r.j is the locus of a point whose distances from Oo and O.. are n and r..,
respectively. The plan of this third sphere and the ellipse which is
the plan of its intersection with the second sphere are shown in Fig.
752. The second ellipse is found in the same manner as the first from
an elevation on a ground line parallel to 0.^0.^. The circle which is the
intersection of the first and second spheres intersects the circle which
is the intersection of the second and third spheres at points whose plans
are p and q. The points P and Q are evidently points whose
distances from 0^, Og and O3 are r^, ^2 and r^ respectively.

If a'h' is inclined to XY at an angle which is nearly a right angle
the positions of p' and </', which determine the distances of P and Q
from the horizontal plane of projection, are best obtained as follows.
With centre and radius equal to o'a' describe an arc of a circle PjcQi.
Through p and q draw ^l^^ and ^Q^ perpendicular to oc to meet this
arc at Pi and Qi ^nd oc at d and e respectively. Then o'p' is equal to
dJP, and o'q' is equal to eQi. The theory of this construction is obvious
when it is observed that the arc PicQ^ is part of the circle of inter-
section of the first and second spheres turned into a horizontal position
about its horizontal diameter.

In Fig. 752, the centre of the third sphere is at the same level as
the centre of the first sphere, hence the plane of the circle which is the
intersection of the first and third spheres is vertical and the plan of
the circle is a straight line.

331. Intersection of a Curved Surface with a Prism or
a Pyramid. — Since the faces of a prism or a pyramid are portions of
planes their intersections with any curved surface are plane sections
or portions of plane sections of that surface and these may be deter-
mined by the methods described in Chapters XVIII and XXIII. If
these plane sections meet one another, they will intersect at the points
where the edges of the prism or the pyramid meet the curved surface.
The latter points may be found separately by the method explained in
Art. 275, p. 317.

332. Intersection of Prisms and Pyramids with one
another. — Since the faces of prisms and pyramids are portions of
planes, their intersections with one another are straight lines which
are determined by the rules for the intersection of two planes (Art.
169, p. 204). The theory of the constructions for finding the inter-
sections of prisms and pyramids with one another is therefore very
simple, but in some cases the solution of the problem may be very com-
plicated owing to the large number of lines which may be required,
and considerable skill is necessary to obtain an accurate result. More-
over the student will find in some cases ample opportunity for select-
ing the simplest and most accurate method of finding the intersection
of the different pairs of plane faces of the solids which may be given.

In a complicated example it may be desirable to proceed in some

INTERSECTION OF SURFACES 405

systematic way such as follows. Denote the faces of one of the solids
by A^, Bi, Cj, etc., and the faces of the other by A^, B^,, C^, etc.
Consider the faces A^ and A.,. Find the intersection of these two
faces if they do intersect, noting that only that part of the line of
intersection of their planes which lies within both faces is required.
Decide at once whether this line is visible or not in plan and in eleva-
tion, and line it in distinctly. Next consider the faces A^ and B^ in
the same way, then Aj and C.^ and so on until the intersections of the
face A, of the one solid with each of the faces of the other solid have
been found. In like manner deal with the face Bj of the one solid and
each of the faces of the other in turn, and so on until the intersections
of all the faces have been found. Of course a face of the one solid
may not intersect any of the faces of the other or it may only intersect
one or two faces of the other. Very often an inspection of the figure
will show which faces intersect especially after the intersections of one '
or two pairs of faces have been found.

The student is recommended to work out the three cases of the
following exercise. — The square ah
(Fig. 753), of 2 inches side, is the
horizontal trace of a prism whose long
edges are parallel to e/, e'f. The
square cd, of 2*5 inches side, is the
horizontal trace of a prism whose long
edges are parallel to gk, g'k'. e is the
centre of the square ah, and g is the
centre of the square cd. Show in plan
and elevation the intersection of the
surfaces of the prisms, (1) when h = 0,
(2) when h = 0'5 inch, and (3) when
h is such that the line joining h and
c is parallel to XY.

The solution of case (2), without the construction lines, is shown
in Fig. 754.

Exercises XXVI

Note. Where the surfaces in the following exercises are developable, the student
should, in a selected number of cases, draw the developments and show on them the
lines of intersection. He should also, in some cases, cut out the developments and
construct m,odels of the solids with them.

1. The axes of two cylinders are horizontal and at right angles to one another.
One cylinder is 2^ inches in diameter, and the other is 2\ inches in diameter.
Draw the plan, and an elevation on a vertical plane inclined at 30° to the axis
of the larger cylinder, showing the intersection of the surfaces of the cylinders,
(a) when the axes intersect, (6) when the axis of the smaller cylinder is | inch
above the axis of the other, (c) when the axis of the smaller cylinder is \ inch
above the axis of the other.

2. The same as the preceding exercise except that the angle between the axes
is 60° instead of 90°.

3. A cylinder, 2-5 inches in diameter, has its axis perpendicular to the V.P.
Another cylinder, 2 inches in diameter has the elevation of its axis inclined at

406

PRACTICAL GEOMETRY

45° to XY. The angle between the plans of the axes is 90^, Draw the plan,
showing the intersection of the surfaces of the cylinders, (a) when the axes
intersect, (6) when the axes are 0-25 inch apart, (c) when the axes are 05 inch
apart.

4. The same as the preceding exercise except that the angle between the plans
of the axes is 60^^ instead of 90^.

Fig. 754.

5. A vertical tube, having an external diameter of 3 inches and an internal
diameter of 2 inches, has a cylindrical hole through it, 1-5 inches in diameter. The
axis of the hole is inclined at 45° to the horizontal plane and its perpendicular
distance from the axis of the tube is 0-25 inch. Draw an elevation of the tube
on a vertical plane which is parallel to the axis of the hole.

INTERSECTION OF SURFACES

407

6. The same as the preceding exercise except that the elevation is to be on a
vertical plane which makes 45^ with the plan of the axis of the hole.

7. AB (Fig. 755) is the axis of a cylinder whose horizontal trace is a circle
2-25 inches in diameter. CD is the axis of a

cylinder whose horizontal trace is an ellipse (major
axis 3 inches, minor axis 2 inches) whose minor
axis is parallel to XY. The axes of the cylinders
are parallel to the vertical plane. Show the inter-
section of the cylinders in plan and elevation in
each of the following cases, (i) when h = 0, (ii)
when h = 1 inch, (iii) when h = \ inch.

8. Same as example 3, p. 390, except that h
(Fig. 740) is to be instead of I inch.

9. Same as example 3, p. 390, except that h
(Fig. 740) is to be \ inch instead of f inch.

10. The axis of a cylinder 2 inches in diameter
is inclined at 60° to the ground. The axis of a
second cylinder 2^ inches in diameter is inclined
at 45° to the ground. The angle between the plans
of the axes is 80°, and the common perpendicular

to the axes has a true length equal to k. Draw the Fig. 755.-

plan, and an elevation on a vertical plane parallel

to the axis of the first cylinder, showing the intersection of the surfaces, for each
of the following cases, (i) k = 0, (ii) k = 0*25 inch, (iii) k = 0-5 inch.

11. abd is an equilateral triangle of 2*5 inches side, c is a point within the
triangle, 1-75 inches from b and 1-25 inches from d. a, &, c, and d are the plans
of points whose heights above the ground are 0*5 inch, 1 inch, 2*5 inches, and 1*5
inches respectively. A circular cylinder has its axis parallel to the line AD, and
its surface contains the four points A, B, C, and D. A second circular cylinder
has its axis parallel to the line BC, and its surface also contains the four points
A, B, C, and D. Draw the plan showing the intersection of the surfaces of the
cylinders.

12. A right cone having a base 3*5 inches in diameter, and an altitude of 35
inches, stands with its base on the ground. A cylinder, 2 inches in diameter lies
on the ground and penetrates the cone. The axis of the cone is at a distance
h from the axis of the cylinder. Draw the plan, and an elevation on a vertical
plane inclined at 80° to the axis of the cylinder, showing the intersection of the
surfaces, (i) when h = 0, (ii) when h = 0'25 inch, and (iii) when h is such that the
curved surface of the cone touches the curved surface of the cylinder.

13. A right circular cone passes through a cylindrical tube. The axis of the
cone intersects the axis of the tube at right angles. Internal diameter of tube,
3 inches. Vertical angle of cone 15°. Diameter of cone at centre of tube 1*5
inches. Determine the development of the surface of that part of the cone which
is within the tube.

14. The elevation
of two horizontal
cylinders is shown
in Fig. 756, also the
elevation of a right
circular cone, whose
axis is perpendicular
to the axes of the
cylinders. Determine
the development of
the surface of that
part of the cone which
lies between the cy-
linders.

15. Determine the intersection of the given cone
(Fig. 757), and draw the development of the surface of
which is within the cylinder.

Fig. 756.

Fig. 757.

and hollow cylinder
that part of the cone

408 PUACTICAL GEOMETRY

16. Same as example 2, p. 394, except that av is to make 25° with XY instead
of 23^

17. abc is an equilateral triangle of 4 inches side. A circle is described on ab
as diameter, and another is described on ac as diameter. These circles are the
horizontal traces of two cones, c is the plan of the apex of the cone of which
the circle ab is the horizontal trace, and b is the plan of the apex of the other
cone. Each apex is at a height of 4 inches above the horizontal plane. Show
the plan of the intersection of the surfaces of the two cones and add an elevation
on a ground line parallel to ab.

18. Same as the example on p. 396, except that the distance h in Fig. 743 is
to be 2-25 inches instead of 1-75 inches.

19. Same as the example on p. 396, except that the distance h in Fig. 743 is
to be such that the plane which contains the vertices of the two cones and is
tangential to one of the cones shall also be tangential to the other cone.

20. Vioa and v^ob are two straight lines at right angles to one another.
VyO = 2*5 inches, v^o = 3 inches, v^oa is the plan of the axis of a cone, semi-
vertical angle 25°, which lies with its slant side on the ground, v^ being the plan
of its vertex, v^ob is the plan of the axis of a second cone also lying with its slant
side on the ground, v.^ being the plan of its vertex. The axes of the cones inter-
sect. Both cones are right circular cones. Draw the plan showing the intersec-
tion of the surfaces, and add an elevation on a vertical plane parallel to the axis
of the first cone. Note. These cones will envelop the same sphere and their
intersection will be two ellipses.

21. A circle 3 inches in diameter is the elevation of a sphere. Another circle
2 inches in diameter is the elevation of a cylinder. The centres of the circles lie
on a line inclined at 60° to XY, and are at a distance h from one another. The
horizontal plane which contains the axis of the cylinder is above the centre of
the sphere. Show the plan of the intersection of the surfaces of the sphere and
cylinder, (a) when h = '^ inch, (&) when h = I inch, (c) when /i = ^ inch.

22. A circle 3 inches in diameter is the elevation of a sphere. A line inclined
at 40° to XY, and at a perpendicular distance of 0-5 inch from the centre of the
circle is the elevation of the axis of a cylinder 1*5 inches in diameter which is
parallel to the vertical plane. The cylinder penetrates the sphere and touches
its surface internally. Draw in plan and elevation the complete curve of inter-
section of the surfaces.

23. A solid of revolution is generated by an ellipse, 4 inches by 2*5 inches,
revolving about its major axis which is vertical. A cylinder 2 inches in diameter
has its axis situated so that its plan and elevation are inclined at 45° to XY. The
elevation of the axis of the cylinder passes through the centre of the elevation of
the solid of revolution and the plan is at a perpendicular distance of 0'15 inch
from the centre of the plan of the solid of revolution. Draw the plan and
elevation of the two solids showing the intersection of their surfaces.

24. Eeferring to Fig. 748, p. 399, draw the plan and elevation of the given
cylinder and surface of revolution, showing their intersection, keeping to the
dimensions given except that : case I, h ~ 0; case II, h = j inch ; case III,
h = ^ inch.

25. Taking the particulars of exercise 23 except that the axis of the cylinder
is to be made the axis of a cone having the elevation of its vertex at a distance of
4 inches from the centre of the elevation of the solid of revolution and having a
vertical angle of 24°. Draw the plan and elevation of the solids showing the
intersection of their surfaces.

26. Referring to Fig. 749, p. 401, draw the plan and elevation of the given
cone and surface of revolution, showing their intersection, keeping to the
dimensions given, except that : case I, h = 0; case II, h = ^ inch ; case III,
h= 1 inch.

27. A cone of revolution, base 3 inches in diameter, axis 2'5 inches long, has
its axis vertical. A cylinder of revolution 2 inches in diameter, has its axis
parallel to the axis of the cone and 0-5 inch distant from it. Draw an elevation
on a plane inclined at 45° to the plane containing the axis of the solids showing
the line of intersection of their surfaces.

28. A circle, 3 inches in diameter, is the plan of a right circular cone, altitude

INTERSECTION OF SURFACES

409

4 inches, standing on the ground. A circle, 2^ inches in diameter, is the plan of
a sphere also standing on the ground. The centres of these circles are at a
distance h from one another. Show the plan of the solids and their intersection
and an elevation on a vertical plane which makes 30° with the plane containing
the axis of the cone and the centre of the sphere, (1) when h = \ inch, (2) when
h = l inch, and (3) when h is such that the curved surfaces of the solids touch one
another.

29. Work the example illustrated by Fig. 750, p. 402, when the axis of the
cone is moved until it is 0*5 inch distant from the axis of the anchor ring.

30. Work the example illustrated by Fig. 751, p. 403, when the vertical angle
of the cone is increased until the straight boundary lines of the plan of the cone
are tangential to the plan of the surface of revolution.

31. abc is a triangle, ah = 2 inches, he = 1-25 inches, ca = 1-5 inches, a is
the plan of a point which is 0*5 inch below the
H.P. h and c are the plans of points which are
1'25 inches and 2 inches respectively above the
H.P. Determine the plans of the two points which
are 275 inches distant from each of the points
A, B, and C, and state their distances from the
H.P.

32. The plan of a sphere with a vertical tri-
angular hole in it is shown in Fig. 758. Draw an
elevation on a ground line parallel to ah.

33. The equilateral triangle ahc (Fig. 759) of
2.| inches side is the horizontal trace of a pyramid
of which V is the plan of the vertex. The height
of the vertex above the H.P. is 4 inches. The
square defg of 2:] inches side is the horizontal
trace of a prism. Draw the plan and
elevation of the pyramid and prism
showing the intersection of their sur-
faces, — (1) when the long edges of the
prism are vertical, (2) when the long
edges of the prism are parallel to the
V.P. and inclined at 60° to the H.P.,
sloping upwards from right to left in
the elevation.

34. a6 (Fig, 760) is a regular hexa-
gon of 1^ inches side, cde is an equi-
lateral triangle of 2 inches side. The
hexagon is the horizontal trace of a prism whose long edges are parallel to a/,
«'/'. The triangle is the horizontal trace of a prism whose long edges are

Fig. 758.

Fig. 759.

Fig. 760.

Fig. 761.

parallel to dg, d'g'. Draw the plan and elevation of these solids showing the
intersection of their surfaces.

410

PRACTICAL GEOMETRY

35. ab (Pig. 761) is a regular hexagon of 1\ inches side and whose centre is li.
cd is a square of 2 inches side and whose centre is k. The hexagon is the horizontal

Fig. 763.

Fig. 764.

trace of a prism whose long edges are parallel to ae, a'e'. The square is the hori-
zontal trace of a pyramid whose vertex V is 5 inches above the H.P. Draw the
plan and elevation of these solids showing the inter-
section of their surfaces.

36. Certain solids are shown in Figs. 762, 763, and
764 in plan and elevation. For each solid draw the
plan as shown and add an elevation on a ground line
inclined at 45"^ to the centre line ss without drawing
the elevation given. Use the dimensions marked but
make no measurements from the illustrations given.

37. Draw the curve of intersection of the given
cone (Fig. 765) with the helical surface (of uniform
pitch) generated by the revolution of the horizontal
line VH about the axis of the cone, the line de-
scending to the base during one anti-clockwise
turn.

The given point P will lie on the required curve.
Determine the tangent to the curve at P. Also draw
the normal and osculating planes at P. [b.e.]

Fig. 765.

CHAPTER XXVII

PEOJECTION OF SHADOWS

333. Theory of Shadows. — In a homogeneous medium light
travels in straight lines, and in this chapter it will always be assumed
that the medium through which the light passes is homogeneous. The
rays of light may be parallel, or they may diverge from a point or con-
verge to a point. If the light comes from a great distance, as from the
sun, the rays are practically parallel. If the light comes from a point,
which is practically the case when the luminous body is very small, the
rays diverge from that point in all directions. Rays of light may be
made to converge to a point by means of a reflector or a lens of suitable
form.

If an opaque body be placed before a source of light, part of the sur-
face of the body will be illuminated and the remainder left in darkness.
Also a portion of the light from the luminous body will be intercepted,
and a portion of the space behind the opaque body will be in darkness.
This dark space behind the opaque body is called the shadow of that
body. The surface which bounds the shadow is the shadoio surface, and
the line on the surface of the opaque body which separates the illumined
from the unillumined part is the sJiade line. It is evident that the
shadow surface is a ruled surface, and that its directing line is the
shade line.

The intersection of the shadow surface with any other surface which
it meets is the cast shadow of the opaque body on that surface, but
generally the cast shadow is called simply the shadow.

The outliue of the cast shadow of an opaque body is evidently the
cast shadow of its shade line. In general the cast shadow is best deter-
mined by first finding the shade line, and then the cast shadow of that
line ; especially is this the case when the surface upon which the shadow
is cast is other than a single plane.

It will always be assumed that the beam of light is large enough
to embrace the whole of the object whose shadow is to be determined.
It will also be assumed that any surface upon which a shadow is cast
is an opaque surface. Hence a point can only have one shadow with
one system of illumination.

334. Cast Shadow of a Point. — To determine the shadow cast
by a given point on a given surface, draw the projections of the line
which represents the ray of light which is intercepted by the point,

412

PRACTICAL GEOMETRY

and determine the point of intersection of this Hne with the given
surface upon which the shadow is cast. This point of intersection is
the shadow required. The Hne which represents the ray of light may
intersect the given surface at more than one point, but only that point
of intersection which is nearest to the given point which casts the
shadow is to be taken as the cast shadow of the point.

Three examples are shown at (a), (b), and (c), Yig. 766. In each
case pjp' is a given point
and rr' is the ray of light ,

intercepted by pp. ^^^^p'

At (a), L'MN is a given J^yv'

oblique plane and ^qPo' is ^^' /• »

the shadow cast by pp' on
this plane, p^pl is found
by using the vertical plane
containing the ray rr' . The
intersection of this plane
with the given plane inter-
sects rr' at p^p^.

At (6) is shown the
shadow cast by the point
pi>' on the surface of a ver-
tical cylinder. The con-
struction in this case is Fig. 766.
obvious.

At (c), oc> is the centre of a sphere upon the surface of which the
point pp casts the shadow p^p^. The vertical plane containing rr' is
taken. This plane intersects the sphere in a circle. Turning this
plane with the ray and circle in it about the vertical axis of the sphere
until the plane is parallel to the vertical plane of projection, the inter-
section of the ray and circle in the new position is found. Turning
this plane with the ray and circle and their point of intersection back
to its original position, p^p^ is determined as shown.

335. Cast Shadow of a Line. — If the line is a straight line,
then, whether the rays of light are parallel or proceed from a point, it
is evident that all the rays which meet the line are in the same plane.
Hence the cast shadow of the line on any surface will be a portion of the
intersection of this plane with the surface. The extremities of the cast
shadow of the line will be the cast shadows of its extremities. The
cast shadow of a straight line on a single plane is the straight line
joining the cast shadows of the extremities of the line.

Fig. 767 shows the shadow cast by the straight line «&, dV on the
plane L'MN and on the horizontal plane of projection, the rays of light
being parallel to rr' . The traces of the plane containing the line ah a'h'
and parallel to rr' are first determined. The required cast shadow is
made up of a part of the horizontal trace of this plane and a part of
the intersection of this plane with the plane L'MN as shown.

If the line whose cast shadow is required is a curved line, and the
rays of light are parallel, all the rays which meet the line will lie on a

PROJECTION OF SHADOWS

413

cylindrical surface ; but if the rays all proceed from a point those which
meet the curved line will lie on a conical surface. The intersection of
the fore-mentioned cylindrical or conical surface with the surface upon
which the shadow is cast is the cast shadow required.

Generally when the line is curved its cast shadow is determined by
first finding, by Art. 334, the cast shadows of a sufficient number of
points in the line and then drawing a fair curve through these. Fig.
768 shows the cast shadow of a curved line ah a'b' the rays of light
being parallel to rr'. The shadow is cast partly on the surface of a
horizontal cylinder and partly on the horizontal plane of projection.

V. ^

Fig. 767

Fig. 769.

Fig. 770.

A simple case of importance is that of the cast shadow of a circle
on a plane parallel to the plane of the circle. The cast shadow in this
case is a circle whose centre is at the cast shadow of the centre of the
original circle. If the rays of light are parallel (Fig. 769) the diameter
of the cast shadow is equal to that of the original circle. If the rays
of light proceed from a point (Fig. 770) the diameter of the cast
shadow is greater than that of the original circle and is found by an
obvious construction.

Two other simple cases of importance relate to the cast shadows of
parallel lines on a plane.
If the rays of light are
parallel then the cast
shadows of parallel lines
on a plane are themselves
parallel. If the rays of
light proceed from a
point then the cast
shadows of parallel lines
on a plane will, produced
if necessary, meet at a
point. The latter case is
illustrated by Fig. 771
where ah a'h' and cd c'd' Fig. 771.

are parallel lines which
cast their shadows on the inclined plane L'MN, the light coming from

414

PRACTICAL GEOMETRY

the point ss. The point W to which the cast shadows converge is the
trace on the plane L'MN of the line sf s't' which is parallel to the lines
ah a'b' and cd c'd'.

One other case may be mentioned. If a straight line casts a shadow,
partly on one plane and partly on another plane parallel to the first,
the shadow on the second plane is parallel to the shadow on the first.

336. Shadow of a Solid having Plane Faces. — When a solid
having plane faces is placed in a beam of light which is large enough
to embrace the whole of the solid, it is evident that one part of a face
of the solid cannot be in light and another part in shade, unless the
solid has re-entrant angles, in which case one part of the solid may
cast a shadow on another part. Hence the shade line must be made
up of edges of the solid. Those edges which make up the shade line
can generally be determined by inspection. A particular edge is
part of the shade line if a line representing a ray of light meeting a
point on the edge in question does not enter the solid at that point.

Fig. 772.

Fig. 773.

Having determined the shade line, its cast shadow, which is the
outline of the cast shadow of the solid, may be determined by the con-
structions of the two preceding Arts.

Instead of first determining the shade line, the cast shadow of each
edge of the solid may be found. The resulting figure is either a
parallel or a conical projection of the solid according as the rays of
light are parallel or proceed from a point. The boundary line of this
projection is the cast shadow of the solid.

Eig. 772 shows the shadow cast by a short chimney on a roof and
also the shadow cast by the coping on the chimney itself. The rays of
light are parallel to rr. All the necessary construction lines are
shown.

Eig. 773 shows the shadows cast by a hexagonal pyramid and a tri-
angular prism on the horizontal plane and also the shadow cast by the
pyramid on the prism. The rays of light are parallel to rr'. All the
necessary construction lines are shown.

PROJECTION OF SHADOWS

415

337. Shadow of a Cylinder. — A cylinder which casts a shadow
i^^enerally has its shade line made up of two straight lines and two
curved lines. The straight lines are the lines of contact of the two
tangent planes to the cylinder, these tangent planes are parallel to the
rays of light or pass through the luminous point. Conceive a plane to
contain these straight lines. This plane will divide each end of the
cylinder into two segments, and the curved boundary line of one of
these segments on each end will constitute the curved parts of the
shade line. Those segments of the ends which must be taken in get-
ting the curved parts of the shade line will be evident from inspection.

In the particular case where the rays of light are parallel to the
axis of the cylinder the surface of the cylinder is itself the shadow
surface, and in the particular case where the rays of light diverge from
a point within the cylinder produced, the outline of that end of the
cylinder which is nearest to the luminous point is the shade line.

Fig. 774.

Having determined the shade line, its cast shadow, which is the
outline of the cast shadow of the cylinder, may be determined by the
constructions of Arts. 334 and 335.

Fig. 774 shows the shadow cast by a right circular cylinder on the
horizontal plane. The axis of the cylinder is horizontal, and the rays
of light are parallel to rr'. All the necessary construction lines are
shown.

Fig. 775 shows the cast shadow of a right circular cylinder whose
axis is vertical. The light proceeds from the point sr\ and the shadow
is cast partly on the horizontal plane and partly on the inclined plane
L'MN. All the necessary construction lines are shown.

338. Shadow of a Cone. — A cone which casts a shadow gene-
rally has its shade line made up of two straight lines and one curved
line. The straight lines are the lines of contact of two tangent planes
to the cone, these tangent planes are parallel to the rays of light or

416

PRACTICAL GEOMETRY

pass through the luminous point. A plane containing these sti-aight
lines will divide the base of the cone into two segments and the curved
boundary line of one of these will be the curved part of the shade line.
The segment of the base which must be taken in getting the curved
part of the shade line will be evident from inspection.

In the case where the line which represents the ray of light
through the vertex of the cone falls inside the cone, the shade line will
consist simply of the whole outline of the base of the cone.

Figs. 776 and 777 show two cases of the shadow cast on the hori-
zontal plane by a right circular cone, the axis of the cone being per-
pendicular to the vertical plane of projection. In Fig. 776 the rays
of light are parallel to rr, while in Fig. 777 they diverge from the
point ss. it' is the trace on the plane of the base of a line through the

Fig. 776.

Fig. 777.

vertex of the cone parallel to rr' in Fig. 776 and through ss' in Fig. 777.
Lines drawn from the vertex of the cone to the points of contact of
the tangents from W to the base of the cone are the straight portions
of the shade line. The remainder of the construction is obvious.

339. Shadow of a Sphere. — A sphere which casts a shadow,
when the rays of light are parallel, will have for its shade line the
circle of contact of the enveloping cylinder whose axis is parallel to the
rays of light. This cylinder and its line of contact with the sphere
are determined by the construction of Art. 221, p. 256.

If the rays of light proceed from a point, the sphere which casts a
shadow will have for its shade line the circle of contact of the envelop-
ing cone whose vertex is at the luminous point. This cone and its
circle of contact w4th the sphere are determined by the construction
of Art. 227, p. 263.

The intersection of the surface of the enveloping cylinder or

PROJECTIOlSr OF SHADOWS

enveloping cone with any given surface will be the cast shadow of the
sphere on that surface.

Two examples on the shadow cast by a sphere are illustrated by
Figs. 778 and 779. The results only are shown, all the construction
lines being omitted. The student should work out these examples to
the dimensions given, which are in inches.

Fig. 778 shows the shadow cast by a sphere partly on the horizontal
plane and partly on a concave cylindrical surface. The part of the

Fig. 778.

Fig. 779.

sphere in shade is also indicated. The rays of light are parallel
to rr'.

Fig. 779 shows the shadow cast by a sphere partly on the horizontal
plane, partly on the vertical plane of projection and partly on the
vertical plane L'MN. The part of the sphere in shade is also indicated.
The rays of light proceed from the point ss'.

340. Shadow of a Solid of Revolution. — If the rays of light
are parallel determine, by Art. 292, p. 327, a cylinder to envelop the
surface of the solid of revolution, the generatrices of the cylinder being
parallel to the rays of light. The trace of this cylinder on the surface
upon which the shadow is to be cast will determine the required cast
shadow.

If the surface of the enveloping cylinder does not intersect the
surface of the solid (it may touch at one part but cut at another) then
the line of contact of the surface of the solid and the enveloping
cylinder will be the line of separation between light and shade on the
surface of the solid. If the surface of the enveloping cylinder also
intersects the surface of the solid, then this intersection must be found
in order to complete the shadow on the solid itself.

418

PRACTICAL GEOMETRY

The solution of an example, which the student should work out
carefully, is shown in Fig. 780. The solid, which is in the form of a
vase, has its axis vertical. The rays of light are parallel to rr\ and the
shadow of the vase is cast on the horizontal plane of projection.

The chief difficulty is with the line of separation on the upper or
concave part of the vase. a!h'dd' and e'/'gltJIi! are the elevations of the

lines of contact of the cylinder or cylinders enveloping the upper part
of the solid. A part of the line CD casts a shadow MG on the neck of
the vase, A part of CD and a part of the solid in the neighbourhood
of H cast a shadow on a part of the solid below H and the outline of
this shadow is the curve HN. The curves MG and HN are obtained
by the construction of Art. 334, p. 411.

If the rays of light proceed from a point the problem may be solved
by the application of the construction of Art. 293, p. 328.

PROJECTION OF SHADOWS

419

Exercises XXVII

1. The projections of a rectangle ABCD and a square MNOP are given in Fig.
781. Determine the shadow cast by the square on the rectangle and the shadows
cast by both figures on the planes of projection. The rays of light are parallel
to rr'.

2. Determine the shadows cast by the triangle ahc a'h'd (Fig. 782) on the
planes of projection. The rays of light proceed from the point ss'.

»' Ex.l. i

...

\ 1

. Ex.2.

T—

EX.3.1

]

EX.4.

f^>\

■^

S-

r ^

Fig. 781.

Fig. 782.

Fig. 783.

Fig. 784.

In reproducing the above diagrams take the small sqtiares as of half inch side.

3. The semicircles abc and a'b'c' (Fig. 783) are the plan and elevation re-
spectively of a certain curve. Show the shadows cast by this curve on the planes
of projection when the rays of light are parallel to rr'.

4. The plan ab and elevation a'b' of a horizontal circle are given in Fig. 784.
An oblique plane L'MN is also given. Determine the shadows cast by the circle
on the oblique plane and on the horizontal plane when the rays of light are
parallel to rr'.

5. ab, the plan of a straight line makes 45° with XY. a and b are ^ inch 1^
inches respectively below XY. A and B are 2 inches and 1 inch respectively
above the H.P. The shadow of AB on the V.P. is a horizontal line 3 inches long.
(1) Determine the directions, in plan and elevation, of parallel rays of light which
would cast this shadow. (2) The shadow being cast by light from a luminous
point which is in a plane perpendicular to XY and which contains the point A,
determine the plan and elevation of the luminous point. Show the shadow in
each case.

6. A cube of 1^ inches edge has its base horizontal and 1 inch above the H.P.
Determine the shadow cast by the cube on the H.P. when the rays of light are
parallel to a diagonal of the solid.

7. A solid of the form of the letter T stands on the floor in the angle of two ,
vertical walls as shown in Fig. 785. Determine the shadows cast by the solid on
the floor and on the walls, the rays of light being parallel to rr'.

8. Fig. 786 represents a square pyramid penetrating a square prism. Determine
the shadows cast by these solids on one another and on the planes of projection,
the rays of light being parallel to rr'.

9. Determine the shadow cast by the object shown in Fig. 787 on itself and on
the horizontal plane. The rays of light are parallel to rr'.

10. Determine the shadow cast by the object shown in Fig. 788 on itself and
on the horizontal plane. The rays of light are parallel to rr'.

11. o is the centre of a circle 2 inches in diameter, oab is a straight line
cutting the circle at c. oa = 0*5 inch, o6 = 2 inches. The circle is the plan of a
right circular cylinder standing on the H.P. Height of cylinder, 2 inches, ab is
the plan of a straight line which touches the upper end of the cylinder at G and
has the end B on the H.P. Show in plan and elevation the shadow cast by tha

420

PRACTICAL GEOMETRY

line on the cylinder and on the H.P. The ground line for the elevation makes
75° with ab. The plans of the rays of light make 45° with the ground line and
30° with ab, and the elevations of the rays are perpendicular to their plans.

EX.7 1

y//. ■////.

'?W

Y/<

■/,

''(,

\ A EX.8

■ \.

7 I

XL 3y

^-5^

2w^

^ ! \A\ r

/"[">

^ /N^^-^

^N-r-

-*.

1 1 EX.10 1

•^

^-.

Fig. 785. Fig. 786. Fig. 787. Fig. 788.

In reproducing the above diagrams take the small squares as of half inch side.

12. The solid shown in Fig. 789 is illuminated by rays of light which are
parallel to rr'. Determine the parts of the surface of the solid which are in
shadow.

13. A solid made up of two cylinders is shown in Fig. 790. Determine the
shadow cast by this solid on the V.P. when the rays of light proceed from the
point ss'. Also, indicate the parts of the surface of the solid which are in shadow.

14. A cone and cylinder are shown in ?'ig. 791. Determine the shadows cast
by these solids on the H.P., also the shadow cast by the cone on the cylinder.
Indicate the parts of the surfaces of the solids which are not illuminated. The
rays of light are parallel to rr'.

15. A right cone, base 3 inches diameter, axis 3*5 inches long, lies on the
ground, one of the generating lines of the cone being the line of contact. The
rays of light being parallel, inclined at 45° to the ground, and their plans inclined

^Z\ *^

uv lyy*

L-^-\

^5^

(^t2j^ ^

I t^

|5^-=/2y ^

^ ^ :^

:,, 3S yy

----- i

Ex.l2^

■~, ^ ...

£ 1

W\ \W'^ }\k

vi ^r' W

^^ X /0(

y ^^q^-^rA:^^^ V

/'^■-5>v

Vx 1/ V

S EX.16 Z ZS

i^zm^i

VZ Z 2

^^2 ^2^

^^r:-?^

Fig. 789. Fig. 790. Fig. 791. Fig. 792.

In reproducing the above diagrams take the small squares as of half inch side.

at 35° to the plan of the axis of the cone, and the base of the cone being in shadow,
determine the shadow cast by the cone on the ground.

16. Fig. 792 shows a hemispherical cup with a cross bar at the top. Show on
the plan the shadow cast on the inside surface of the cup. The rays of light are
parallel to rr'.

17. The curved surface of a bowl is a zone of a sphere. Larger diameter of
zone, 2-75 inches. Smaller diameter, 1*75 inches. Height, 1 inch. The bowl
stands with its base on the ground and is illuminated from a point which is 2-25
inches above the ground and whose plan is 2 inches from the centre of the plan

PROJECTION OF SHADOWS

421

of the bowl. Determine the shadow of the bowl on the ground and the portions
of the bowl in shade, both in plan and elevation. The ground line for the elevation
makes 30° with the line joining the plan of the source of light and the centre of
the plan of the bowl. The thickness of the bowl is to be neglected.

18. a is the centre of a circle 3"25 inches in diameter, b is the centre of a
circle 2 inches in diameter, ab = 1*75 inches, ca is a line making the angle
cab = 10°. The larger circle is the plan of a hemispherical hole in the ground.
The smaller circle is the plan of a sphere resting on the ground. Determine the
plan of the shadows cast in the hole and on the sphere by parallel rays of light.
The plans of the rays of light are parallel to ca and the rays are inclined at 35°
to the ground.

19. Front and side elevations of a corbel projecting from a vertical wall are
given in Pig. 793. Show on the front elevation the shadow cast by the corbel on
the wall, also the part of the corbel in shade. The direction of the parallel rays
of light is given.

::^

EX.19

}

; N

.Si. -

EX.20

■"

EX.21

ijXii

{

Fig. 798.

Fig. 794.

Fig. 795.

In reproducing the above diagrams take the small squares as of half inch side.

20. Two elevations of a wall bracket and a hanging conical lamp shade are
given in Fig. 794. Show on the left hand elevation the shadow cast by the
bracket and shade on the wall. One of the parallel rays of light is shown.

21. Fig. 795 shows a suspended sphere and a niche in a vertical wall. The

Fig. 796.

Fig. 798.

TJiese diagrams are to be reproduced twice this size.

422

PRACTICAL GEOMETRY

surface of the niche is cylindrical. Determine the shadows cast in the niche
and the part of the sphere in shade. The rays of light are parallel to rr' .

22. The circular arc ABC (Fig. 796) rotates about its chord AG. Draw the
elevation and plan of the figure generated. Determine the shadow cast on
the horizontal plane by parallel rays of light, one of which K is given. Show the
margin of light and shade on the surface of the solid. [b,e.]

23. The elevation and half plan of a solid of revolution are given in Fig. 797,
the axis being vertical. Find the projections of the limits of light and shade on
the solid, and the complete outline of the shadow thrown by the solid on the
horizontal plane. The arrows indicate the direction of the parallel rays of
light. [B.E.]

24. The line LL (Fig. 798) rotates about the axis AA. Draw the plan and
elevation of the figure generated. Determine the shadow cast on the horizontal
plane by parallel rays of light, one of which R is given. [b.e.]

25. An annulus or anchor ring is cut in two by an axial plane. One of the
halves is shown in plan (Fig. 799), with its section ends resting on the horizontal
plane. Draw the elevation of this semi-annulus on XY. Determine also the
shadow cast on the horizontal plane by rays of light, parallel to the vertical
plane, and inclined at 30° to the horizontal plane, one of which is shown in plan
at r. And show the projections of the limits of light and shade on the surface of
the solid. [b.e.]

Fig. 799.

Fig. 800.

26. Two elevations of a pipe bend, with two circular flanges, are shown in
Fig. 800. Reproduce these elevations three times the size given, and determine
the projections of the limits of light and shade on the solid when it is illuminated
by parallel rays of light inclined as shown. Show also on the right hand eleva-
tion the shadow cast by the whole solid on the vertical plane containing the face
of. the vertical flange. To avoid the interference of the left hand elevation with
the shadow cast on the vertical plane the two elevations may be placed further
apart.

27. The plan of an anchor ring resting on the horizontal plane is two con-
centric circles, the larger being 3'1 inches in diameter, and the smaller 1-1 inches
in diameter. The smaller circle is also the plan of a vertical right cylinder
(height 1*6 inches) standing on the horizontal plane. Determine the shadow
cast on the horizontal plane, and the unilluminated portion of the anchor ring,
taking a point as a source of light which is 2-7 inches above the horizontal plane
and whose plan is 2-2 inches from the centre of the plan of the ring.

CHAPTEE XXVIII

MISCELLANEOUS PKOBLEMS IN SOLID GEOMETRY

341. The Regular Dodecahedron. — The regular dodecahedron
is one of the five regular solids and has twelve faces all equal and
regular pentagons. This solid is shown in its simplest position, in
relation to the planes of projection, in Fig. 801. One face ABODE is
on the horizontal plane and AB,
one edge of that face, is at right
angles to XY. Let o be the centre
of the circumscribing circle of the
regular pentagon abcde. The sides
of the pentagon ahcde are the
horizontal traces of five of the
faces of the dodecahedron, and
these five faces meet in straight
lines whose plans pass through the
angular points a, h, c, cZ, and e,
and if produced these plans pass
through o.

Consider the two faces whose
horizontal traces are ah and fee;
these faces meet in a line whose
plan gh, when produced, passes
through 0. To find the point g,
imagine the face ABG to revolve
about AB in the clockwise direction
until it is in the horizontal plane.
It will then evidently coincide with
the pentagon ahcde and the plan
of the point G will travel in the
line gc perpendicular to ah Hence
a line through c perpendicular to
ah to meet a radial line ohg at g
determines the point g.

The plan may now be completed as follows. With centre o and
radius og describe a circle. Divide the circumference of this circle into
ten equal parts at the points /, g, h, etc. Draw the circumscribing
circle of the pentagon ahcde. Draw the radial lines fm, hn, etc. and
join the various points thus found, as shown.

Fig. 801.

424

PRACTICAL GEOMETRY

To draw the elevation : with centre a' and radius a'd! describe the
arc d'f to cut the projector from / at f. Join a'f. The line a'f is the
elevation of the face whose horizontal trace is ah. A projector from g
to meet a'f at g' determines the elevation of the point G. Points
whose plans are the alternate angular points of the outer polygon,
beginning with g, have their elevations at the same level as g', and the
points whose plans are the remaining angular points of this outer poly-
gon have their elevations at the same level as /'. Points such as M
and N" on the top face are at a height above F equal to the height of G
above the horizontal plane.

From the plan and elevation thus determined other projections may
be drawn in the usual way. For example, if a ground line be taken
perpendicular to m'd\ the elevation of one of the axes of the solid, a
plan of the solid, when that axis is vertical, may be projected from the
elevation already drawn. But the student is recommended to try and
draw directly the plan and an elevation of the dodecahedron when its axis
is vertical, without first drawing it with one face on the horizontal plane.

When an axis of the solid is vertical it should be noticed that the
planes of the three faces
meeting at an extremity of
that axis will be equally
inclined to the ground, and
being equally inclined to
one another, their lines of
intersection will in plan
make 1 20° with one another.

As an exercise in draw-
ing the projections of the
dodecahedron the edges of
the solid may be conveni-
ently taken, say, 1'25 inches
long.

342. The Regular
Icosahedron. — The regu-
lar icosahedron is another
one of the five regular solids
and has twenty faces all
equal equilateral triangles.
Fig. 802 shows a regular
icosahedron, in plan and
elevation, when one face
ABC is on the horizontal
plane and AB one edge of
that face is perpendicular
to XY. The plan of the
top face, which is also hori-
zontal, is an equilateral triangle inscribed in the same circle, centre o,
as the triangle ahc, the sides of the one being parallel to the sides of
the other, as shown.

Fig. 802.

MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 425

A property of the icosahedron, which leads to simple constructions
for drawing it, is, that any angular point of the solid is the vertex of
a right pentagonal pyramid, the five faces meeting at the vertex being
faces of the icosahedron. Thus the faces meeting at the point F are
the faces of a right pyramid whose base is the regular pentagon
ABEDE, and the sides of the pentagon are edges of the icosahedron.

To complete the plan proceed as follows. Draw on ah as base the
regular pentagon ah^J)^%. Through Ej draw E^e perpendicular to
ah to meet the radial line oae at e. With centre o and radius oe
describe a circle! A regular hexagon inscribed in this circle, one
angular point being at e, is the boundary line of the plan, which is
completed by joining the angular points of the hexagon to the angular
points of the two central equilateral triangles, as shown. The pentagon
at El Dj El is the rabatment, on the horizontal plane, of the pentagon
of which ahede is the plan. Keeping this in mind the construction of
the elevation easily follows.

The student should also draw directly a plan and an elevation of
the icosahedron when an axis of the solid is vertical. The extremities
of the axis in question will be the vertices of two of the right pen-
tagonal pyramids previously mentioned. The pentagonal bases of
these pyramids will be horizontal and their plans will be inscribed in
the same circle.

As an exercise in drawing the projections of the icosahedron the
edges may be conveniently taken, say, 1*75 inches long.

343. Solids Inscribed in the Sphere. — A solid is said to be
inscribed in a sphere when all its angular points are on the surface of
the sphere.

Fig. 803 shows the constructions
for finding the length of an edge of
each of the five regular solids when
inscribed in a sphere of given diameter.

AB is the diameter of the sphere.
On this a semicircle is described.

Take BD equal to one-third of AB.
Draw DE at right angles to AB to
meet the semicircle at E. Join AE
and BE. AE is the edge of the in-
scribed tetrahedron, and BE is the
edge of the inscribed cube.

C being the middle point of AB,
draw CF at right angles to AB to
meet the semicircle at F. Join AF.
octahedron.

Draw AG at right angles to AB and make AG equal to AB.
Draw CG cutting the semicircle at H. Join AH. AH is the edge of
the inscribed icosahedron.

On EA make EK equal to half of BE. Join BK. With centre K
and radius KE describe an arc to cut BK at L. BL is the edge of the
inscribed dodecahedron.

Fig. 803.

AF is the edge of the inscribed

426 PRACTICAL GEOMETRY

The following facts should be kept in view when working problems
on the regular solids inscribed in the sphere.

The plane containing one edge of a tetrahedron and the centre of
the circumscribing sphere bisects the opposite edge at right angles.

The rectangle which has for its diagonals two of the diagonals of a
cube is inscribed in a great circle of the circumscribing sphere.

The octahedron can be divided into two square pyramids in three
different ways, and the square bases of all these pyramids will be
inscribed in great circles of the circumscribing sphere.

344. Solids Circumscribing the Sphere. — A solid is said to
circumscribe a sphere, or a sphere to be inscribed in a solid, when all
the faces of the solid are tangential to the surface of the sphere.

A plane bisecting the angle between any two faces of the solid
passes through the centre of the inscribed sphere.

In the case of any one of the five regular solids the centres of the
inscribed and circumscribing spheres coincide.

345. The Sphere, Cylinder and Cone in Contact. — If two
spheres touch one another, the point of contact and the centres of
the spheres are in the same straight line.

If a sphere touches a cylinder or cone it will do so at a point, and if
the cylinder or cone be a right circular cylinder or a right circular
cone, the centre of the sphere the point of contact, and the axis of the
cylinder or cone are in the same plane.

If two cylinders or two cones or a cylinder and a cone touch one
another, they will do so either along a straight line or at a point. If
the cylinders or cones are right circular cylinders or right circular
cones and they touch one another along a straight line, the line of
contact and the axes of the surfaces are in the same plane ; the axes
will therefore either be parallel or they will intersect.

When two right circular cylinders touch one another at a point
the common perpendicular to their axes passes through the point of
contact.

Two surfaces which touch one another have a common tangent
plane.

346. Projections of Four Spheres in Mutual Contact. —
Denote the spheres and their centres by A, B, C, and D, and let their
radii be r^, r^j rg, and r^ respectively. Assume that the spheres A, B,
and C are resting on the horizontal plane and that the line joining the
centres A and B is parallel to the vertical plane of projection. The
elevations of the spheres A and B (Fig. 804) are circles touching one
another and XY and after these are drawn the plans may be pro-
jected from them as shown. Next determine the horizontal distances
between the centres A and C, and B and C as shown on the elevation.
This determines the centre cc' and the plan and elevation of the sphere
C may then be drawn.

The centre D of the fourth sphere will be at a distance r^ + r^ from the
centre A of the first sphere, and at a distance ra + r^ from the centre B
of the second sphere. If the triangle ADB be conceived to rotate
about the side AB, the point D will describe a circle which will be

MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 427

the locus of the centre of a sphere of radius r^ which touches the spheres
A and B.

Remembering that ah, a'V is parallel to the vertical plane of pro-
jection, with centres a' and V and radii equal to r^ + r^ and r^, + r^^
respectively, describe arcs intersecting at d^. Draw d/e' to intersect a'h'
at right angles at e'. Then ^d{ is the radius of the circle described by
the point D when the triangle ADB is rotated about AB. The plan of
this circle is an ellipse whose semi-minor axis is ed^, the projection

Fig. 804.

of e'd^ on ah. The semi-major axis ef is at right angles to ah and equal
to e'd-l. A quarter of this ellipse is shown.

Again, the centre D of the fourth sphere will be at a distance
fi + r^ from the centre A of the first sphere, and at a distance r^ + ^4 from
the centre C of the third sphere. If the triangle ADC be conceived to
rotate about the side AC, the point D will describe a circle which will
be the locus of the centre of a sphere of radius r^ which touches the
spheres A and C. The circle whose centre is c^ is the elevation of the
sphere C when that sphere is brought round into the position in which
it touches the sphere A and the horizontal plane and has its centre in
a plane parallel to the vertical plane of projection and containing the
centre A of the first sphere. With centres a' and c^ and radii r^ -f r^
and r^ -f r^ respectively, describe arcs intersecting at d.2. Draw d.20.2 to
intersect aJc^ at right angles at O2'. Then Oa'^a' is the radius of the circle

428 PRACTICAL GEOMETRY

described by the point D when the triangle ADC is rotated about AC.
The plan of this circle is an ellipse whose semi-minor axis on is on ac
and is obtained as shown. The semi-major axis oh is at right angles to
ac and equal to o.Jd^. A quarter of this ellipse is shown.

The two circles which have been referred to as described by the
point D intersect at two points one of which has the point d for its
plan, d being on the two ellipses which are the plans of the circles.
The elevation d' is in e'd^' and in the projector from d. This determines
the centre of the fourth sphere and the plan and elevation of that
sphere may now be drawn.

The two circles referred to as described by the point D are on the
surface of a sphere whose centre is A and radius r^ -f- r^. The ellipses
which are the plans of these circles may intersect at four points, but
only two of these are plans of the points of intersection of the circles.
The student should have no difficulty in deciding on which of the points
of intersection of the ellipses are to be taken.

347. Spherical Roulettes. — If two cones be placed in line
contact with their vertices coinciding and if one cone be made to roll
on the other, which is fixed, any point carried by the rolling cone will
describe a spherical roulette. The describing point carried by the roll-
ing cone may be outside, or inside, or on the surface of that cone and
it will be at a constant distance from the common vertex of the two
cones ; hence the describing point will move on the surface of a fixed
sphere.

When the two cones are right circular cones, and the describing
point is on the surface of the rolling cone, the spherical roulette
becomes a spherical epicycloid or a spherical hypocycloid according as the
rolling cone rolls outside or inside the fixed cone.

Spherical roulettes are tortuous curves, and no single projection of
a tortuous curve can show its true form. A spherical roulette has
therefore to be represented by two projections.

Projections of a spherical epicycloid and a spherical hypocycloid are
shown in Fig. 805. The description which follows applies to either
curve, v'a'a' is the elevation of the fixed cone the axis of which is
vertical, and the semicircle aba is the half plan of this cone, v'a'e' is
the elevation of the rolling cone when its position is such that its axis
is parallel to the vertical plane of projection.

Let the rolling cone start from the position in which vb, v'V is the
line of contact and Iqt the point of contact hV of the bases of the cones
in this position be the initial position of the describing point. Next
suppose that the rolling cone rolls into the position in which the line of
contact has vm for its plan. Draw a circle mnr^ having a radius equal
to the radius of the base of the rolling cone and touching the plan of
the fixed cone at m. Consider this circle to be the rabatment of the
base of the rolling cone on the plane of the base of the fixed cone when
the rolling cone is in the position now being considered. The rabat-
ment of the describing point will be rj, the position of r^ being such
that the arc mnr^ is equal to the arc mh. Restoring the base of the
rolling cone to its inclined position the plan r of the describing point

MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 429

will be in the line through r^ parallel to mv. To fix the position of r
in the line r^r let the circle mnr^ be carried round about the centre v into
the position asiC^, the diameter ae.^ being in line with va and parallel to
the ground line, r^ will move to s^. Draw the projector s^s/ to meet
the ground line at s/. With centre a' and radius a's/ describe the arc
SjV to meet a'e' at s'. Draw the projector s's to meet s^s parallel to aci
at 8. With centre v and radius vs describe the arc sr to cut r^r at r.

i SPHERICAL
EPICYCLOID

SPHERICAL
HYPOCYCLOID

Fig. 805.

Draw the projector rr' to meet the horizontal line s'r' at r'. Then rr' is
a point on the spherical epicycloid or on the spherical hypocycloid. In
like manner any number of points maybe found. The length of curve
shown is that due to half a revolution of the rolling cone.

The points such as r^ lie on a curve, shown dotted, which is either
a plane epicycloid or a plane hypocycloid.

The tangent to the spherical epicycloid or spherical hypocycloid at
a point R lies on the tangent plane to the sphere whose centre is V
and radius YA or YR at the point R. It also lies on the tangent
plane to the sphere whose centre is M and radius MR at the point

430

PRACTICAL GEOMETRY

R. The tangent is therefore the line of intersection of these two
planes.

The spherical epicycloid and the spherical hypocycloid occur in
connection with the formation of the teeth of bevel wheels.

348. Reflections. — When a ray of light impinges on a polished
surface at a point Q, it is reflected so that the incident and reflected
rays and the normal to the reflecting surface at Q are in the same
plane ; also the incident and reflected rays are equally inclined to the
normal.

Referring to Figs. 806 and 807, let the plane of the paper be the
plane containing the incident ray PQ and the normal QT to the surface
whose section by the plane of the paper is AB, Then the reflected
ray QR will be in the plane of the paper and will make with QT an
angle 6 equal to the angle between PQ and QT.

Fig. 806.

Fig. 803.

If the reflecting surface is a plane (Figs. 807 and 808) it is obvious
that all incident rays from a point P will be reflected from the plane
as if they came direct from a point P^ on the other side of the plane and
in the normal PN to the plane, P^N being equal to PN. A knowledge
of this fundamental theorem makes the solution of problems on
reflections from plane mirrors quite simple. The point Pj is called the
image of the point P in the mirror AB. It should be noticed that
although the image of P is in PN produced it is not necessary that the
mirror should extend to the point N in order that the image may be
seen from the point R.

Fig. 809 shows how to trace the path of a ray of light which
passes from a fixed point P and is reflected in turn from a number of
plane mirrors AB, BC, CD, and DE and then passes through a fixed
point S, the mirrors being at right angles to the plane of the paper.
Pi is the image of P in AB, Pg is the image of Pi in BC, P3 is the
image of Pg in CD, and P4 is the image of P3 in DE. The ray which
is reflected from DE and passes through S must evidently be in the
line P4S. The incident ray on DE is the reflected ray from CD and
must therefore be in the straight line from Pg, and so on the ray is
traced back to the point P as shown.

Fig. 810 illustrates the general case of the problem : given the
incident ray and a plane from which it is reflected, to determine the
reflected ray. PQ is the given incident ray and H.T. and V.T. are the

MISCELLANEOUS PEOBLEMS IN SOLID GEOMETRY 431

traces of the given plane. Find Q the point of intersection of PQ and
the plane. From a point P in PQ draw the normal PNPj to the plane
and find N the point of intersection of this normal and the plane.
Make NP^ equal to NP. Join PiQ and produce it. QR the produced
part of this line is the reflected ray required.

p^r:

P3^-

Fig. 809.

Fig. 810.

If the given surface from which a given incident ray is Reflected is
a curved surface, the point of intersection of the incident ray and the
surface must be determined by the method for finding the intersection
of a straight line and the particular curved surface given. The normal
to the surface at the point where the incident ray strikes it must then
be found and the plane containing it and the incident ray determined.
It will then generally be necessary to obtain a projection of the
incident ray and the normal on a plane parallel to their plane, or to
obtain a rabatment of the incident ray and normal into one of the
planes of projection, in order to draw the reflected ray which must
make with the normal an angle equal to the angle between the incident
ray and the normal.

ELEVATION

-''

UECT
PLA

IMAG

Fig. 811.

Fig. 812.

An example on the projection of an object and its image in a plane
mirror is illustrated by Figs. 811 and 812. The object and its image
and the mirror are shown in plan and elevation to the left in Fig. 811,

432 PRACTICAL GEOMETRY

the mirror being at right angles to the planes of projection. The same
Fig. shows to the right an elevation on a second ground line X^Y^
inclined to the first ground line XY. The pictorial projection in Fig.
812 illustrates further the relative positions of the object, mirror, and

Exercises XXVIII

1. Draw a circle, centre o and radius 1-75 inches. Take a point a 1 inch from
0. Draw a straight line ax making an angle of 60^ with oa. The circle is the
plan of a sphere and a is the plan of a point A on its upper surface. A is one
corner of a cube inscribed in the sphere and ax is the direction of the plan of
one edge containing the point A. Complete the plan of the cube.

2. Referring to the preceding exercise, make oa = 15 inches and the angle
oax = 60°. a is the plan of one angular point of a tetrahedron inscribed in the
sphere and ax is the direction of the plan of one edge containing the point A.
Complete the plan of the tetrahedron.

3. Same as exercise 2, except that the solid is an octahedron instead of a
tetrahedron.

4. The base of a pyramid is an equilateral triangle ABC of 2 5 inches side.
The vertex V of the pyramid is in a line through A at right angles to the plane
of ABC, and VA is 2*5 inches long. Placing the pyramid with its base on the
ground and AB parallel to XY, draw the projections of the Id scribed and circum-
scribing spheres.

5. A right circular cone, base 2*5 inches in diameter and altitude 2*5 inches,
stands with its base on the ground. A cylinder 2 inches in diameter lies on the
ground in contract with the cone, the axis of the cylinder being horizontal. A
sphere 1-25 inches in diameter rests on the ground in contact with the cone and
cylinder. Draw a plan of the group and an elevation on a vertical plane parallel
to the axis of the cylinder showing the projections of the points of contact of the
solids.

6. Three spheres, two of them 1-5 inches in diameter, and one 2 inches in
diameter, rest on the horizontal plane, each in contact with the other two. A
cylinder 2 inches in diameter rests, with its axis horizontal, on top of the spheres,
touching each of them. Draw the plan of the group.

7. Two straight lines ab and cd bisect one another at" right angles, ab = 4:
inches and cd = 3-5 inches. These lines are the plans of the axes of two cylinders
whose diameters are equal and which touch one another. The heights of the
points A, B, C, and D above the horizontal plane are 0-7 inch, 2*8 inches, 2*4
inches, and 4*5 inches respectively. Draw the plan of the cylinders and an
elevation on a ground line parallel to ab, showing the point of contact.

8. abc is a triangle, ab = 2 inches. Angle bac = angle abc = 30°. a, is the
plan of the centre of a sphere of 1*3 inches radius which rests on the horizontal
plane, b is the plan of the centre of a second sphere which rests on the horizontal
plane and touches the first sphere, c is the plan of a point on the upper surface
of the first sphere. Draw the plan of a third sphere which touches the other two,
c being its point of contact with the first sphere.

9. A cone of revolution, vertical angle 64°, of indefinite length, lies with its
curved surface on the ground ; draw ite plan. Determine a sphere of 0*6 inch
radius, which rests on the ground and touches the cone at a point 2-5 inches
from its vertex. Show the indexed plan of the point of contact, and determine
the common tangent plane at this point. [b.e.]

10. A right circular cone, base 2-5 inches in diameter and axis 3 inches long,
lies with its slant side on the ground. A sphere, 2 inches in diameter, moves in
contact with the ground and the surface of the cone. Draw the plan of the locus
of the point of contact.

11. vab is a triangle, va = 3*4 inches, vb = 2'4 inches, and ab = 2 inches, a
is the plan of the centre of a sphere of 2-8 inches diameter, A being 25 inches

MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 433

above the ground, h is the plan of the centre of a sphere of 1-75 inches diameter,
B being 1*5 inches above the ground, v is the plan of the vertex of a cone whose
semi-vertical angle is 20" and w^hich touches the two spheres, V being 2 inches
above the ground. Draw the plan of the group and an elevation on a ground line
parallel to av, showing the points of contact of the cone and spheres.

12. Same as preceding exercise except that a cylinder 1-5 inches in diameter
is to take the place of the cone, V being a point on the axis of the cylinder.

13. A cone, in elevation v'h'k' (Fig. 813) lies on the horizontal plane, its axis
VC parallel to the vertical plane. A cylinder of 1 inch radius, its axis parallel to
the line de, d'e', touches the cone at a point whose height is represented by the
horizontal g'h' . Draw the plan of the surfaces, and the projections of their point
of contact, showing the trace of the cylinder on the horizontal plane, and the
projections of its generator in contact with the cone. [b.e.]

60
50
^0°

dto

Ob -0-7"

^0\ .

Fig. 813.

-30
Fig. 815.

-.J

14. The straight line AB (Fig. 814) is tangential to a sphere of 1 inch radius
whose centre is on the straight line CD. Draw the figured plan of the lines and
sphere and show the point of contact of the sphere and the line AB. Unit for
indices O'l inch.

15. Two lines AB and CD are given in Fig. 815 by their figured plans, and a
plane is given by its scale of slope. Determine a sphere of 1*25 inches radius
touching the lines and plane. Unit 0*1 inch. The sphere is intended to be below
the plane. [b.e.]

16. v'a'o' (Fig. 816) is the half elevation of a fixed right circular cone (1 ) whose
axis VO is vertical and whose vertex is
V. v'a'e' is the elevation of a second
right circular cone (2) in line contact
with cone (1) and having its vertex at
V. The cone (2) rolls on the cone (1)
and a point on the circumference of
the base of the rolling cone describes
a spherical epicycloid of which a'c'h'
(above a'o') is the elevation, or a spheri-
cal hypocycloid of which a'c'b' (below
a'o') is the elevation. Draw the plan
and elevation of these spherical rou-
lettes.

Next take a right circular cone (3)
which has the base of the cone (1) for
a circular section and u' for the eleva-
tion of its vertex, the angle v'a'u' being
a right angle. Take the spherical
roulette, of which a'c'h' is the eleva-
tion, to be the directing curve of a
cone (4) with its vertex at V. Deter-
mine the curve of intersection (5) of the cones (3) and (4). Develop the surface
of the cone (3) with the curve (5) on it. a'CB is a sketch of this development.

2 F

Fig. 816.

434

PRACTICAL GEOMETRY

Lastly, draw on the circular arc a'B as base a plane epicycloid and a plane
hypocycloid with a rolling circle 1-8 inches in diameter (the diameter of the base
of the rolling cone) for comparison with the curves a'CB on opposite sides of
the arc a'B.

17. Taking the fixed and rolling cones (1) and (2) as given in the preceding
exercise and Fig. 816, draw the plan and elevation of the spherical epitrochoid
described by a point on a diameter of the base of the rolling cone produced, the
describing point to be 1-4 inches from the centre of the base of the rolling cone.

18. The ray of light rr' (Fig. 817) impinges on the horizontal plane and is
reflected on to the vertical plane of projection from which it is again reflected.
Show the path of the ray in plan and elevation.

19. A ray of light rr' (Fig. 818) impinges on the horizontal plane and is
reflected on to the vertical plane HTV from which it is again reflected. Draw
the plan and elevation of the path of the ray. Show also in plan and elevation the
path of a ray which after reflection from the horizontal plane and the plane HVT
passes through t-he point nn', the incident ray being parallel to rr'.

•2- SIT

\-W . :=

:ni 53n

-' \ ^

^A% t-,<-

^i ^v

AI. ^Z _

/ \

ISIy is- !S

_ii y X

. Sy «JL _

-L Jy

„ Ti

""r^^-

3 ^x

-«• -s-.^t

h>.

. V /

^^: /A\x \\\

t -a-

y \h , r>j I fe»^i

^ A

^^mr .J,

Ml \A\\

I J

Fig. 817.

Fig. 818.

Fig. 819.

Fig. 820.

20. A polished right circular cylinder (Fig. 819) has its axis vertical. The
ray of light rr' strikes this cylinder and is reflected on to the vertical plane of
projection. Draw the plan and elevation of the path of the ray.

21. The surface of a shield (Fig. 820) is spherical. A projectile strikes this
shield and is deflected, rr' is the line of flight before impact. Show the line of
flight after impact, assuming that the projectile is deflected like a ray of light.

22. A right circular cone, diameter of base 2 inches, altitude 2 inches, stands
with its base on the ground. A ray of light is parallel to the ground line and
0"75 inch above the ground, and its plan is at a perpendicular distance of 0-25
inch from the centre of the plan of the cone. This ray of light impinges on the
surface of the cone and is reflected. Draw the plan and elevation of the reflected
ray.

23. An object and a mirror are given in Fig. 821. Draw an elevation of the

Fig. 821.

Fig. 822.

MISCELLANEOUS PROBLEMS IN SOLID GEOMETRY 435

object and its image in the mirror on a ground line inclined at 60° to XY.
Assume that the mirror is large enough to show the
whole of the image in the elevation asked for.

24. Fig. 822 represents a hanging lamp shade and a
tilted mirror. Draw the two elevations and the plan of
the image of the conical shade in the mirror. From the
plan project on a ground line inclined at 70° to XY the
elevation of the mirror, the shade, and as much of the
image as would be seen within the boundary of
the mirror. [b.e.]

25. The point cc' (Fig. 823) is the centre of a polished
sphere of 1 inch radius. A ray of light parallel to rr'
impinges on the sphere and is reflected, the reflected ray
passing through the point 2>p'. Draw the projections of
the path of the ray.

Fig. 823.

APPENDIX

MATHEMATICAL TABLES

The Mathematical Tables on the following five pages are those
supplied to Candidates at the Examinations of the Board of Education,
and are reprinted by permission of the Controller of H.M. Stationery
Office.

MATHEMATICAL TABLES

437

Angle. 1

Chord.

Sine.

Tangent.

Co-tangent.

Cosine.

De-
grees.

Radians.

0°

1^414

r5708

90"

89
88
87
86

1
2
3
4

•0175
•0349
•0524
•0698

■017
•035
•052
•070

•0175
•0349
•0523
•0698

•0175
•0349
•0524
•0699

57^2900
28^6363
19-0811
14-3007

•9998
-8994
-9986
-9976

r402
1^389
r377
1364

r5533
1-5359
1-5184
1-5010

•0873

•087

•0872

•0875

11^4301

•9962

1-351

1-4835

7
8
9

•1047
•1222
•1396
•1571

•105
•122
•140
•157

•1045
•1219
•1392
•1564

•1051
•1228
•1405

•1684

9^5144
8-1443
7-1154
6-3138

•9945
•9925
•9903
•9877

1338
1-325
1-312
1-299

1-4661
1-4486
1-4312
1-4137

81
83

82
81

•1745

•174

•1736

•1763

5-6713

•9848

1-286

1-3963

11
12
13
14

•1920
•2094
•2269
•2443

•192
•209
•226
•244

•1908
•2079
•2250
•2419

•1944
•2126
•2309
•2493

5-1446
4-7046
4-3315
4-0108

•9816
•9781
•9744
•9703

1-272
1-259
1-245
1-231

1-3788
1-3614
1-3439
1-3265

79
78
77
76

•2618

•261

•2588

•2679

3-7321

•9659

1-218

1-3090

16
17
18
19

•2793
•2967
•3142
■3316

•278
•296
•313
•330

•2756
•2924
•3090
•3256

•2867
•3057
•3249
•3443

34874
3-2709
3-0777
2-9042

•9613
•9563
•9511
•9455

1-204
1-190
1-176
1-161

1-2915
1-2741
1-2566
1-2392

74
73
72
71

•3491

•347

•3420

•3640

2-7475

•9397

1-147

1-2217

21
22
23
24

•3665
•3840
•4014
•4189

•364
•382
•399
•416

•3584
•3746
•3907
•4067

•3839
•4040
•4245
•4452

2-6051
2-4751
2-3559
2-2460

•9336

-9272
-9205
•9135

1-133
1-118
1-104
1-089

1-2043
1-1868
1-1694
1-1619

69
68
67
66

•4363

•433

•4226

•4663

2-1445

•9063

1-075

1-1345

26
27
28
29

•4538
•4712

•4887
•5061

•450
•467

•484
•501

•4384
•4540
•4695
•4848

•4877
•5095
•5317
•5543

2-0503
1-9626
1-8807
1-8040

•8988
•8910
•8829
■8746

1-060
1^045
1^030
1^015

1-1170
1-0996
1-0821
1-0647

64
63
62
61

•5236

•518

•5000

•5774

1-7321

•8660

1-000

1-0472

31
32
33
34

•5411
■5585
■5760
•5934

•534
•551
•668
■585

•5150
•5299
•5446
•6592

•6009
■6249
•6494
■6745

1-6643
1-6003
1-5399
1-4826

-8572
-8480
•8387
•8290

•985
•970
-954
-939

1-0297

1-0123

-9948

•9774

59
58
57
56

•6109

■601

■5736

•7002

1-4281

•8192

-923

-9599

36
37
3S
39

•6283
•6458
•6632
•6807

•618
•635
■651
•668

•5878
•6018
•6157
•6293

•7265
•7536
•7813
•8098

1-3764
1-3270
1-2799
1-2349

•8090
-7986

-7880
-7771

-908
-892

-877
-861

•9425
•9250
•9076
•8901

54
53
52
51

•6981

•684

•6428

•8391

1-1918

-7660

-845

•8727

49
48
47
46

45°

41
42
43
44

•7156
•7330
•7505
•7679

•700
•717
•733
•749

•6561
•6691
•6820
•6947

•8693
•9004
•9325
•9657

1-1504
1-1106
1-0724
1-0355

-7547
-7431
-7314
-7193

•829
•813

•797
•781

•8552
•8378
•8203
•8029

45°

•7854

•765 i '7071

1^0000

1-0000

•7071

•765

•7854

Cosine

Co-tangent

Tangent

Sine

Chord

Radians

Degrees

Angle 1

438

MATHEMATICAL TABLES
Logarithms

12 3 4

6 7 8 9

0000

0043

0086

0128

0170

0212

0253

0294

0334

0374

4 9 13 17
4 8 12 16

21
20

26 30 34 38
24 28 32 37

0414

0453

04S2

0531

0569

0607

0645

0682

0719

0755

4 8 12 15
4 7 11 15

19
19

23 27 31 35
22 26 30 33

0792

0828

0864

0899

0934

0969

1004

1038

1072

1106

3 7 11 14
3 7 10 14

18
17

21 25 28 32
20 24 27 31

13
14

1139

1173

1206

1239

1271

1303

1335

1367

1399

1430

3 7 10 13
3 7 10 12

16
16

20 23 26 30
19 22 25 29

1461

1492

1523

1553

1584

1614

1644

1673

1703

1732

3 6 9 12
3 6 9 12

15
15

18 21 24 28
17 20 23 26

1761

1790

1818

1847

1875

1903

1931

1959

1987

2014

3 6 9 11
3 5 8 11

14
14

17 20 23 26
16 19 22 25

2041

2068

2095

2122

2148

2175

2201

2227

2253

2279

3 5 8 11
3 5 8 10

14
13

16 19 22 24
15 18 21 23

2304

2330

2355

2380

2405

2430

2455

2480

2504

2529

3 5 8 10
2 5 7 10

13
12

15 18 20 23
15 17 19 22

2553

.2577

2601

2625

2648

2672

2695

2718

2742

2765

2 5 7 9
2 5 7 9

12
11

14 16 19 21
14 16 18 21

2788

2810

2833

2856

2878

2900

2923

2945

2967

2989

2 4 7 9
2 4 6 8

11
11

13 16 18 20
13 15 17 19

3010

3032

3054

3075

3096

3118

3139

3160

3181

3201

2 4 6 8

13 15 17 19

21
22
23
24

3222
3424
3617
3802

3243
3444
3636
3820

3263
3464
3655
3838

3284
3483
3674
3856

3304
3502
3692

3874

33^4
3522
3711
3892

3345
3541
3729
3909

3365
3560
3747
3927

3385
3579
3766
3945

3404

3598
3784
3962

2 4 6 8
2 4 6 8
2 4 6 7
2 4 5 7

10
10
9
9

12 14 16 18
12 14 15 17
11 13 15 17
11 12 14 16

3979

3997

4014

4031

4048

4065

4082

4099

4116

4133

2 3 5 7

10 12 14 15

26
27
28
29

4150
4314
4472
4624

4166
4330

4487
4639

4183
4346
4502
4654

4200
4362
4518
4669

4216
4378
4533
4683

4232
4393
4548
4698

4249
4409
4564
4713

4265
4425
4579

4728

4281
4440
4594
4742

4298
4456
4609
4757

2 3 5 7
2 3 5 6
2 3 5 6
13 4 6

8
8

10 11 13 15
9 11 13 14
9 11 12 14
9 10 12 13

4771

4786

4800

4814

4829

4843

4857

4871

4886

4900

13 4 6

9 10 11 13

31
32
33
34

4914
5051
5185
5315

4928
5065
5198
5328

4942
5079
5211
5340

4955
5092
5224
5353

4969
5105
5237
5366

4983
5119
5250
5378

4997
5132
5263
5391

5011
5145
5276
5403

5024
5159
5289
5416

5038
5172
5302
5428

13 4 6
13 4 5
13 4 5
13 4 5

7
7
6
6

8 10 11 12
8 9 11 12
8 9 10 12
8 9 10 11

5441

5453

5465

5478

5490

5502

5514

5527

5539

5551

12 4 5

7 9 10 11

36
37
38
39

5563
5682
5798
5911

5575
5694
5809
5922

5587
5705
5821
5933

5599
5717
5832
5944

5611
5729
5843
5955

5623
5740

5855
5966

5635
5752
5866
5977

5647
5763

5877
5988

5658
5775
5888
5999

5670
5786
5899
6010

12 4 5
12 3 5
12 3 5
12 3 4

6
6
6
5

7 8 10 11
7 8 9 10
7 8 9 10
7 8 9 10

6021

6031

6042

6053

6064

6075

6085

6096

6107

6117

12 3 4

6 8 9 10

41
42
43
44

6128
6232
6335
6435

6138
6243
6345
6444

6149
6253
6355
6454

6160
6263
6365
6464

6170
6274
6375
6474

6180
6284
6385
6484

6191
6294
6395
6493

6201
6304
6405
6503

6212
6314
6415
6513

6222
6325
6425
6522

12 3 4
12 3 4
12 3 4
12 3 4

5
5
5
5

6 7 8 9
6 7 8 9
6 7 8 9
6 7 8 9

46
47
48
49

6532

6542

6551

6561

6571

6580

6590

6599

6609

6618

12 3 4

6 7 8 9

6628
6721
6812
6902

6637
6730
6821
6911

6646
6739
6830
6920

6656
6749
6839
6928

6665
6758

6848
6937

6675
6767
6857
6946

6684
6776
6866
6955

6693
6785
6875
6964

6702
6794
6884
6972

6712
6803
6893
6981

12 3 4
12 3 4
12 3 4
12 3 4

5
5

4
4

6 7 7 8
6 6 7 8
5 6 7 8
5 6 7 8

6990

6998

7007

7016

7024

7033

7042

7050

7059

7067

12 3 3

5 6 7 8

n^^^i^

MATHEMATICAL TABLES
Logarithms

439

3 4

6 7 8 9

51
52
53
54

7076
7160
7243
7324

7084
7168
7251
7332

7093
7177
7259
7340

7101
7185
7267
7348

7110
7193
7275
7356

7118
7202

7284
7364

7126
7210
7292
7372

7135

7218
7300
7380

7143
7226
7308
7388

7152
7235
7316
7396

2
2
2
2

2
2
2

3
3
3
3

4
4

5
5
5

6 7 8
6 7 7
6 6 7
6 6 7

7404

7412

7419

7427

7435

7443

7451

7459

7466

7474

5 6 7

56
57
58
59

7482
7559
7634
7709

7490
7566
7642
7716

7497
7574
7649
7723

7505

7582
7657
7731

7513
7589
7664

7738

7520
7597
7672
7745

7528
7604
7679
7752

7536
7612
7686
7760

7543
7619
7694
7767

7551
7627
7701
7774

2
2

2
2
2
2

3
3
3
3

4
4

5
5

5 6 7
5 6 7
5 6 7
5 6 7

7782

7789

7796

7803

7810

7818

7825

7832

7839

7846

5 6 6

61
62
63
64

7853
7924
7993
8062

7860
7931
8000
8069

7868
7938
8007
8075

7875
7945
8014
8082

7882
7952
8021
8089

7889
7959
8i)28
8096

7896
7966
8035
8102

7903
7973
8041
8109

7910
7980
8048
8116

7917
7987
8055
8122

2
2
2
2

3
3
3
3

4
3
3
3

5 6 6
5 6 6
5 5 6
5 5 6

8129

8136

8142

8149

8156

8162

8169

8176

8182

8189

5 5 6

66
67
68
69

8195
8261
8325
8388

8202
8267
8331
8395

8209
8274
8338
8401

8215
8280
8344

8407

8222
8287
b351
8414

8228
8293
8357
8420

8235
8299
8363
8426

8241
8306
8370
8432

8248
8312
8376
8439

8254
8319
8382
8445

2
2

3
3
3
2

3
3
3
3

5 5 6
5 5 6
4 5 6
4 5 6

8451

8457

8463

8470

8476

8482

848,s

8494

8500

8506

4 5 6

71
72
73
74

8513
8573
8633

8692

8519
8579
8639
8698

8525
8585
8645
8704

8531
8591
8651
8710

8537
8597
8657
8716

8543
8603
8663
8722

8549
8609
8669

8727

8555
8615
8675
8733

8561
8621
8681
8739

8567
h627
8686
8745

2
2
2
2

2
2

3
3
3

4 5 5
4 5 5
4 5 5
4 5 5

8751

8756

8762

8768

8774

8779

8785

8791

8797

8802

4 5 5

76
77
78
79

8808
8865
8921
8976

8814
8871
8927
8982

8820
8876
8932
8987

8825
8882
8938
8993

8831
8887
8943
8998

8837
8893
8949
9004

8842
8899
8954
9009

8848
8904
8960
9015

8854
8910
8965
9020

8859
8915
8971
9025

2
2
2
2

3
3
3
3

4 5 5
4 4 5
4 4 5
4 4 5

9031

9036

9042

9047

9053

9058

9063

9069

9074

9079

4 4 5

81
82
83

9085
9138
9191
9243

9090
9143
9196
9248

9096
9149
9201
9253

9101
9154
9206
9258

9106
9159
9212
9263

9112
9165
9217
9269

9117
9170
9222
9274

9122
9175
9227
9279

9128
9180
9232
9284

9133
9186
9238
9289

2
2
2
2

3
3
3
3

4 4 5
4 4 5
4 4 5
4 4 5

9294

9299

9304

9309

9315

9320

9325

9330

9335

9340

2
2
2
2

2
2
2

4 4 5

86
87
88
89

9345
9395
9445
9494

9350
9400
9450
9499

9355
9405
9455
9504

9360
9410
9460
9509

9365
9415
9465
9513

9370
9420
9469
9518

9375
9425
9474
9523

9380
9430
9479

9528

9385
9435
9484
9533

9390
9440
9489
9538

3
3
3
3

4 4 5
3 4 4
3 4 4
3 4 4

9542

9547

9552

9557

9562

9566

9571

9576

9581

9586

3 4 4

91
92
93
94

9590
9638
9685
9731

9595
9643
9689
9736

9600
9647
9694
9741

9605
9652
9699
9745

9609
9657
9703
9750

9614
9661
9708
9754

9619
9666
9713
9759

9624
9671
9717
9763

9628
9675

9722
9768

9633

9680
9727
9773

2
2
2
2

2
2
.2
2

3
3
3
3

3 4 4
3 4 4
3 4 4
3 4 4

96
97
98
99

9777

9782

9786

9791

9795

9800

9805

9809

9814

9818

3 4 4

9823
9868
9912
9956

9827
9872
9917
9961

9832
9877
9921
9965

98:'6
9881
9926
9969

9841
9886
99:0
9974

9845
9890
9934

9978

9850
9894
9939
9983

9854
9899
9943
9987

9859
9903
9948
9991

9863
9908
9952
9996

2
2
2

2
2
2
2

3
3
3
3

3 4 4
3 4 4
3 4 4
3 3 4

440

MATHEMATICAL TABLES
Antilogarithms

•00

♦01
•02
•03
•04

12 3 4

6 7 8 9

1000

1002

1005

1007

1009

1012

1014

1016

1019

1021

0011

1222

1023
1047
1072
1096

1026
1050
1074
1099

1028
1052
1076
1102

1030
1054
1079
1104

1033
1057
1081
1107

1035
1059
1084
1109

1038
1062
1086
1112

1040
1064
1089
1114

1042
1067
1091
1117

1045
1069
1094
1119

0011
0011
0011
0111

1
1
1
1

1222
1222
1222
2222

•05

1122

1125

1127

1130

1132

1135

1138

1140

1143

1146

0111

2222

•06
•07
•08
•09

1148
1175
1202
1230

1151
1178
1205
1233

1153
1180
1208
1236

1156
1183
1211
1239

1159
1186
1213
1242

1161
1189
1216
1245

1164
1191
1219
1247

1167
1194
1222
1250

1169
1197
1225
1253

1172
1199
1227
1256

0111

111
111
111

1
1
1
1

2222
2 2 2 2
222 3

2 2 2 3

•10

1259

1262

1265

1268

1271

1274

1276

1279

1282

1285

111

2 2 2 3

•11
•12
•13
•14

1288
1318
1349
1380

1291
1321
1352
1384

1294
1324
1355

1387

1297
1327
1358
1390

1300
1330
1361
1393

1303
1334
1365
1396

1306
1337
1368
1400

1309
1340
1371
1403

1312
1343
1374
1406

1315
1346
1377
1409

111
111
111
111

2
2
2

2 2 2 3
2 2 2 3
2 2 3 3
2 2 3 3

•15

1413

1416

1419

1422

1426

1429

1432

1435

1439

1442

111

2 2 3 3

•16
•17
•18
•19

1445
1479
1514
1549

1449
1483
1517
1652

1452
1486
1521
1556

1455
1489
1524
1560

1459
1493
1528
1563

1462
1496
1531
1567

1466
1500
1535
1570

1469
1503
1538
1574

1472
1507
1542
1578

1476
1510
1545
1581

111
111
111
111

2
2
2

2 2 3 3
2 2 3 3
2 2 3 3
2 3 3 3

•20

1585

1589

1592

1596

1600

1603

1607

1611

1614

1618

111

2 3 3 3

•21
•22
•33
•24

1622
1660
1698
1738

1626
1663
1702

1742

1629
1667
1706
1746

1633
1671
1710
1750

1637
1675
1714
1754

1641
1679
1718
1758

1644
1683
1722
1762

1648
1687
1726
1766

1652
1690
1730
1770

1656
1694
1734
1774

112
112
112
112

2
2
2
2

2 3 3 3
2 3 3 3
2 3 3 4
2 3 3 4

•25

1778

1782

1786

1791

1795

1799

1803

1807

1811

1816

112

2 3 3 4

•26
•27
•28
•29

1820
1862
1905
1950

1824
1866
1910
1954

1828
1871
1914
1959

1832
1875
1919
1963

1837
1879
1923
1968

1841
1884
1928
1972

1845
1888
1932
1977

1849
1892
1936
1982

1854
1897
1941
1986

1858
1901
1945
1991

112
112
112
112

2
2
2
2

3 3 3 4
3 3 3 4
3 3 4 4
3 3 4 4

•30

1995

2000

2004

2009

2014

2018

2023

2028

2032

2037

112

3 3 4 4

•31
•32
•33
•34

2042
2089
2138
2188

2046
2094
2143
2193

2051
2099
2148
2198

2056
2104
2153
2203

2061
2109
2158
2208

2065
2113
2163
2213

2070
2118
2168
2218

2075
2123
2173
2223

2080
2128
2178
2228

2084
2133
2183
2234

112
112
112
112 2

2
2
2

3 3 4 4
3 3 4 4
3 3 4 4
3 4 4 5

•35

2239

2244

2249

2254

2259

2265

2270

2275

2280

2286

112 2

3 4 4 5

•36
•37
•38
•39

2291
2344
2399
2455

2296
2350
2404
2460

2301
2355
2410
2466

2307
2360
2415
2472

2312
2366
2421
2477

2317
2371
2427
2483

2323
2377
2432

2489

2328
2382
2438
2495

2333
2388
2443
2500

2339
2393
2449
2506

112 2
112 2
112 2
112 2

3
3
3
3

3 4 4 5
3 4 4 5
3 4 4 5
3 4 5 5

•40

2512

2518

2523

2529

2535

2541

2547

2553

2559

2564

112 2

4 4 5 5

•41
•42
•43
•44

2570
2630
2692
2754

2576
2636
2698
2761

2582
2642
2704
2767

2588
2649
2710
2773

2594
2655
2716
2780

2600
2661
2723
2786

2606
2667
2729
2793

2612
2673
2735
2799

2618
2679
2742
2805

2624
2685
2748
2812

112 2
112 2
112 3
112 3

3
3
3
3

4 4 5 5
4 4 5 6
4 4 5 6
4 4 5 6

•45

2818

2825

2831

2838

2844

2851

2858

2864

2871

2877

112 3

4 5 5 6

•46
•47
•48
•49

2884
2951
3020
3090

2891
2958

3027
3097

2897
2965
3034
3105

2904
2972
3041
3112

2911
2979
3048
3119

2917
2985
3055
3126

2924
2992
3062
3133

2931
2999
3069
3141

2938
3006
3076
3148

2944
3013
3083
3155

112 3
112 3
112 3
112 3

3
3
4

4 6 5 6
4 5 5 6
4 5 6 6
4 5 6 6

MATHEMATICAL TABLES
Antilogarithms

441

•50

12 3 4

6 7 8 9

3162

3170

3177

3184

3192

3199

3206

3214

3221

3228

112 3

4 5 6 7

•51
•52
•63
•54

3236
3311

3388
3467

3243
3319
3396
3475

3251
3327
3404
3483

3258
3334
3412
3491

3266
3342
3420
3499

3273
3350
3428
3508

3281
3357
3436
3516

3289
3365
3443
3524

3296
3373
3451
3532

3304
3381
3459
3540

12 2 3
12 2 3
12 2 3
12 2 3

4
4
4
4

5 5 6 7
5 5 6 7
5 6 6 7
5 6 6 7

•55

3548

3556

3565

3573

3581

3589

3597

3606

3614

3622

12 2 3

5 6 77

•56
•57
•58
•59

3631
3715

3802
3890

3639
3724
3811
3899

3648
3733
3819
3908

3656
3741

3828
3917

3664
3750
3837
3926

3673
3758
3846
3936

3681
3767
3855
3945

3690
3776
3864
3954

3698
3784
3873
3963

3707
3793
3882
3972

12 3 3
12 3 3
12 3 4
12 3 4

4
4
4
5

5 6 7 8
5 6 7 8
5 6 7 8
5 6 7 8

•60

3981

3990

3999

4009

4018

4027

4036

4046

4055

4064

12 3 4

6 6 7 8

•61
•62
•63
•64

4074
4169
4266
4365

4083
4178
4276
4375

4093

4188
4285
4385

4102
4198
4295
4395

4111
4207
4305
4406

4121
4217
4315
4416

4130
4227
4325
4426

4140
4236
4335
4436

4150
4246
4345
4446

4159
4256
4355
4457

12 3 4
12 3 4
12 3 4
12 3 4

5
5
5

6 7 8 9
6 7 8 9
6 7 8 9
6 7 8 9

•65

4467

4477

4487

4498

4508

4519

4529

4539

4550

4560

12 3 4

6 7 8 9

•66
•67
•68
•69

•70

4571

4677
4786
4898

4581
4688
4797
4909

4592
4699
4808
4920

4603
4710
4819
4932

4613
4721.
4831
4943

4624
4732
4842
4955

4634
4742
4853
4966

4645
4753
4864
4977

4656
4764
4875
4989

4667
4775
4887
5000

12 3 4
12 3 4
12 3 4
12 3 5

5
5
6
6

6 7 9 10

7 8 9 10
7 8 9 10
7 8 9 10

5012

5023

5035

5047

5058

5070

5082

5093

5105

5117

12 4 6

7 8 9 11

•71
•72
•73

•74

5129
5248
5370
5495

5623

5140
5260
5383
5508

5152
5272
5395
5521

5164
5284
5408
5534

5176
5297
5420
5546

5188
5309
5433
5559

5200
5321
5445
5572

5212
5333

5458
5585

5224
5346
5470
5598

5236
5358
5483
5610

12 4 5

13 4 5
13 4 5

6
6
6
6

7 8 10 11

7 9 10 11

8 9 10 11
8 9 10 12

•75

5636

5649

5662

5675

5689

5702

5715

5728

5741

13 4 5

8 9 10 12

•76
•77
•78
•79

5754
5888
6026
6166

5768
5902
6039
6180

5781
5916
6053
6194

5794
5929
6067
6209

5808
5943
6081
6223

5821
5957
6095
6237

5834
5970
6109
6252

5848
5984
6124
6266

5861
5998
6138
6281

5875
6012
6152
6295

13 4 5
13 4 5
13 4 6
13 4 6

7
1

7
7

8 9 11 12
8 10 11 12

8 10 11 13

9 10 11 13

•80

6310

6324

6339

6353

6368

6383

6397

6412

6427

6442

13 4 6

9 10 12 13

•81
•82
•83
•84

6457
6607
6761
6918

6471
6622
6776
6934

6486
6637
6792
6950

6501
6653
6808
6966

6516
6668
6823
6982

6531
6683
6839
6998

6546
6699
6855
7015

6561
6714
6871
7031

6577
6730
6887
7047

6592
6745
6902
7063

2 3 5 6
2 3 5 6

8
8
8

9 11 12 14
9 11 12 14
9 11' 13 14
10 11 13 15

•85

7079

7096

7112

7129

7145

7161

7178

7194

7211

7228

2 3 5 7

10 12 13 15

•86
•87
•88
•89

7244
7413
7586
7762

7261
7430
7603

7780

7278
7447
7621
7798

7295
7464
7638
7816

7311

7482
7656
7834

7328
7499
7674

7852

7345

7516
7691

7870

7362
7534
7709
7889

7379
7551

7727
7907

7396
7568
7745
7925

2 3 5 7
2 3 5 7
2 4 5 7
2 4 5 7

8
9
9
9

10 12 13 15

10 12 14 16

11 12 14 16
11 13 14 16

•90

7943

7962

7980

7998

8017

8035

8054

8072

8091

8110

2 4 6 7

11 13 15 17

•91
•92
•93
•94

8128
8318
8511
8710

8147
8337
8531
8730

8166
8356
8551
8750

8185
8375
8570
8770

8204
8395
8590
8790

8222
8414
8610
8810

8241
8433
8630
8831

8260
8453
8650
8851

8279
8472
8670
8872

8299
8492
8690
8892

2 4 6 8
2 4 6 8
2 4 6 8
2 4 6 8

10
10
10

•11 13 15 17
12 14 15 17
12 14 16 18
12 14 16 18

•95

8913

8933

8954

8974

8995

9016

9036

9057

9078

9099

2 4 6 8

12 15 17 19

•96
•97
•98
•99

9120
9333
9550
9772

9141
9354
9572
9795

9162
9376
9594

9817

9183
9397
9616
9840

9204
9419
9638
9863

9226
9441
9661

9886

9247
9462
9683
9908

9268
9484
9705
9931

9290
9506
9727
9954

9311
9528
9750
9977

2 4 6 8
2 4 7 9
2 4 7 9
2 5 7 9

11
11
11
11

13 15 17 19
13 15 17 20

13 16 18 20

14 16 18 20

INDEX

Numbers refer to pages

Abscissa, 128

Acceleration in harmonic motion, 156
Accuracy in drawing, 4
Acute angle, 2
Addition, 10

of vectors, 83, 84

Advance or lead, 151

Altitude of tetrahedron, 187

Amplitude, 149

Analysis, harmonic, 158

Anchor ring, 328

Angle between two planes, 200, 224

Angles, 1

', trigonometrical ratios of, 3

Antiparallel, 7

Approximate developments of undevelop-
able surfaces, 356

solutions, 61 et seq.

Arc of a circle, 1
Arch, linear, 123

, three-hinged, 121

Arched ribs, 122

Archimedean spiral, 140

Arithmetic, graphic, 10

Arithmetical mean, 11

Asymptotes, 13

Asymptotic cone of hyperboloid, 331, 339,

341
Auxiliary circle, 39, 149

projections, 191

of a point, 191

of solids, 193

Axial and normal sections of screw threads,
368

■ pitch of helix, 362

Axis of a conic, 30

, radical, 25

Axometric projection, 295

Bending moment diagrams,
Binormal, 343
Bow's notation, 92

Catenary as a roulette, 76
Centre of curvature, 13

of a conic, 47

of a roulette, 66

107

Centre of parallel forces, 95

of pressure, 98

of stress, 98

of vision, 301

, radical, 26

Centres of gravity* 96

■ of similarity, 14

of similitude, 14, 25

Centroid of a triangle, 7
Centroids, 96
Chord of a circle, 1

of a conic, 34

Chords, scale of, 3
Circle, 1, 16 et seq.

, equation to, 132

of curvature, 12

, properties of the, 16

Circles in contact, 21

Circular arcs, rectification of, 61

measure of an angle, 2

Circumference of a circle, 1
Circumscribed circle of spherical triangle.

237
Collar of hyperboloid, 331
Collinear points, 6
Complementary angles, 2
Composition of harmonic motions, 151,
154, 155

of vectors, 86

Concurrent straight lines, 6
Coney clic points, 6
Cone, 183

enveloping sphere, 263

surface of revolution, 328

, oblique, 266

, projections of, 187, 258

■ , sections of, 259

, tangent plane to, 317

Cones enveloping two spheres, 325

Conical projection, 167

Conic sections, 30 et seq. •

Conies, general properties of, 35

Conjugate axis, 38

Contour lines, 283

Contouring a surface from its equation,

284

444

INDEX

Co-ordinate geometry, plane, 128 c^ sea.

planes, 168

Coplanar forces, 91

Cosecant of an angle, 4

Cosine of an angle, 4

Cotangent of an angle, 4

Couples, 107

Co versed sine of an angle, 4

Cross-rolls, geometry of, 335

Cube, 182

Cubic equation, 144

Curtate cycloid, 69

Curvature, centre of, 13

, circle of, 12

, radius of, 13

Curve, harmonic, 149

y -a-\- bx», 137

y = ae'^"", 139

y = a logg bx, 140

y = bx'\ 136

y == x», 135

yx^ = c, 138

Curved surfaces and tangent planes, 313
et seq.

, sections of, 314

Curves, similar, 14

, tortuous, 343

Cusp, 13
Cycloid, 67
Cycloidal curve, 66
Cylinder, 182, 252

, elliptical, 255

enveloping sphere, 256

surface of revolution, 327

, oblique, 265

, projections of, 187, 253

, sections of, 253

Cylinders, intersection of, 388

Dams, masonry, 100
Definitions of solids, 181
Deflections of braced frames, 1 18
Descriptive geometry, 166
Developable surfaces, 314
Development of a surface, 349

of surface of cone, 353

of cylinder, 352

of prism, 349

of pyramid, 350

Developments, 349 et seq.
Dihedral angle, 200
Directrix, 313

of a conic, 30

Divided pitch of helices, 362

of screw, 367

Division, 11
Dodecahedron, 182, 423

Eccentricity of a conic, 30

Elevation, 168

, sectional, 241 ,

Ellipse, 30, 32, 38-45, 48, 49

, equation to, 133

Ellipsoid, 337
Elliptic paraboloid, 341
Elliptical cylinder, 255
Envelope, 13

glissette, 78

roulette, 77

Epicycloid, 70

, spherical, 428

Epitrochoid, 72

Equation, graph of, 129

Equations, graphic solution of, 142

Equiangular spiral, 141

Error in measuring, 4

Escribed circles of a ti

Evolute of a conic, 49

of a curve, 13

Exponential curve, 139
equation, 139

Face mould, 371, 374
Falling moulds, 373
Figured plans, 279
Focus of a conic, 30
Forces, coplanar, 91

, specification of, 91

Fourier series, 158
Frequency, 148
Funicular polygon, 92

Generating line, 313
Generation of surfaces, 313
Generatrix, 3i3
Geometrical mean, 11
Glissettes, 78
Gorge of hyperboloid, 331
Graph of equation, 129
Graphic arithmetic, 10

solution of equations, 142,

statics, 91 et seq.

Gravity, centres of, 96
Great circle of a sphere, 249
Ground line, 168, 301

Handrails, 369

Harmonic analysis, 158

curve, 119

motion, 148

motions, composition of, 161, 154,

155
Harmonical mean, li
Helical springs, 367
Helices and screws, 362 et seq.
Helix, 362

, axial pitch of, 362

, inclination of, 363

, left-handed, 363

, normal pitch of, 362

of increasing pitch, 363.

, pitch angle of, 363,

INDEX

445

Helix, ri^ht-handed, 363

Hip rafter, 245

Hipped roof, 245

Horizontal projection, 279 et seq.

Hyperbola, equation to, 134

, properties of, 45

, rectangular, 47

Hyperbolic paraboloid, 342
Hyperboloid of one sheet, 338
— — of revolution, 331

of two sheets, 340

Ilyperboloids of revolution in rolling con-
tact, 333
Hypocycloid, 73
— "■ — , spherical, 428
Hypotrochoid, 74

Icosahedron, 182, 424

Image, 9, 430

Inclination of a line to a plane, 223

of a plane, 280

of helix, 363

Inclined plane, 1 99

Indexed plan, 279

Inertia, moment of, 110

Inferior epitrochoid, 72

hypotrochoid, 74

trochoid, 69

Inflexion, point of, 13

Inscribed circle of spherical triangle, 237

of a triangle, 20

Intersection of cone and surface of revolu-
tion, 400

of cones, 395

of curved surface with prism or

pyramid, 404

of cylinder and cone, 392

of c'vlinder and sphere, 397

of cylinder and surface of revolution,

398

of cylinders, 388

of line and plane, 216, 283

of plane and contoured surface, 284

of i)risms and pyramids, 404

of spheres, 403

of straight line and cone, 265

of straight line and cylinder, 258

of straight line and curved surface,

317

• of straight line and sphere, 252

of surfaces, 388 et seq.

of surfaces of revolution, 401, 402

of two planes, 204, 282

Involute of a circle, 76

of a curve, 13

Involution, 12
Isometric projection, 291
scale, 292

Jack rafter, 245

Lag and lead, 150
Latus rectum of a conic, 34
Lead and lag, 150
Left-handed helix, 363
Level easing, 369

quarter, 369

Linear arch, 123

line of resistance, 101

Lines, I

Link polygon, 92

Locors, 82

Locus, 1

Logarithmic curve, 140

spiral, 141

Lune of a sphere, 249

Masonry dams, 100
Mean proportional, 11 •
Measuring point, 305
Medians of a triangle, 6
Meridian sections, 314
Method of sections, 1 1 7
Modulus figure, 99
Moment of a force, 104

of inertia, 110

Moments, principle of, 107
Motion, periodic, 148 et seq.
Moulding, raking, 244
Mouldings, 243

, sections of, 243

Multiple point, 13

Multiple -threaded screw, pitch and lead

of, 367
Multiplication, 11

Node, 13

Normal pitch of helix, 362

to a curve, 12

to a surface, 317

Notation, Bow's, 92
in projection, 170

Oblate spheroid, 329
Oblique cone, 266

co-ordinates, 128

cylinder, 265

parallel projection, 295

plane, 199

Obtuse angle, 2
Octahedron, 182

, projections of, 187

Orthocentre of a triangle, 7
Orthocentric triangle, 7
Orthogonal projection, 167
Orthographic projection, 167
Osculating plane, 343

Parabola, equation to, 133

, properties of, 36

Paraboloid, elliptic, 341
, hyperbolic, 342

446

INDEX

Paraboloid of revolution, 337
Parallel forces, centre of, 95

projection, 167

Parallelepiped, 182
Parameter, 37
Pedal triangle, 7
Perimeter of a triangle, 21
Periodic motion, 148 6-^ seq.

time, 148

Perpendicular plane, 199

projection, 167

Perspective projection, 167, 300 et seq.

Pictorial projections, 290 et seq.

Pitch and lead of multiple-threaded screw,

367
Pitch angle of helix, 363
Plan, 168

• , sectional, 241

Plane co-ordinate geometry, 128 et seq.

, inclined, 199

, oblique, 199

■ of projection, 166

, perpendicular, 199

section of a sphere, 249

, traces of, 198

Planes, representation of, 198
Point of distance, 302

of inflexion, 1 3

Polar co-ordiuates, 128

triangle, 234

Pole and polar, 24, 49
Polygon, funicular, 92
Polyhedron, 181
Pressure, centre of, 98
Principal elliptic section of hvperboloid,
338

normal, 343

section of oblique cone, 266

— of oblique cylinder, 265

Principle of moments, 107
Prism, 182

, projections of, 183

■ , section of, 241

Projecting surface of a line, 167
Projection, 166 <?^ seq.

, axometric, 295

, horizontal, 279 et seq.

, isometric, 291

, oblique parallel, 295

of shadows, 411 et seq.

, perspective, 167, 300 et seq.

Projections, auxiliary, 191

of a cone, 187, 258

of a cylinder, 187, 253

of a prism, 183

of a pyramid, 185

of a solid right angle, 229

of a sphere, 187

of octahedron, 187

of points and lines, 171 et seq.

of solids, 181

Projections, pictorial, 290 ct seq.
Projectors, 166, 172
Prolate cycloid, 69

spheroid, 329

Properties of a circle, 16
Proportion, 10
Protractor, 2
Pyramid, 182

, projections of, 185 •

; section of, 242

Rabatment of oblique plane, 221
Radial projection, 167
lladian, 2
Radical axis, 25

centre, 26

Radius of curvature, 1 3

vector, 128

Rafter, common, 245

■ , hip, 245

, jack, 245

Rafters, geometry of, 245
Raking moulding, 244
Ramp, 369
Reciprocal figures, 93
Rectangular co-ordinates, 128

hyperbola, 47

Rectification of circular arcs, 61
Refiections, 430
Representative crank, 149
Resolution of vectors, 86
Resultant moment of forces, 106
Right angle, 1
Right-handed helix, 363
Roof, hipped, 245
Rotors, 82
Roulettes, 56, 66 et seq.

, spherical, 428

Ruled surfaces, 314

Scalars, 82

, Scale, isometric, 292

of chords, 3

of slope, 279

Screw propellers, 376

surfaces, 364

threads, 365

, axial and normal sections of,

368
Secant of an angle, 4
Section of a prism, 241

of a pyramid, 242 '

Sectional elevation, 241

plan, 241

Sections, conic, 30 et seq.

, method of, 117

of cone, 259

of curved surfaces, 314

of cylinder, 253

of mouldings, 243

of solids, 241 et seq.

INDEX

Sector of a circle, 1
Segment of a circle, 1
Shade line, 411
Shadow of a cone, 415

of a cylinder, 415

of a line, 412

of a point, 411

of a solid having plane faces, 414

of a solid of revolntion, 417

of a sphere, 416

snrface, 411

Shadows in perspective, 308

, projection of, 411 et seq.

, theory of, 411

Shearing-force diagrams, 107
Similar curves, 14

rectilineal fignres, 9

Similarity, centres of, 14
Similitude, centres of, 14, 25
Simple harmonic motion, 148
Sine curve, 149

of an angle, 4

Skew bevel wheels, 335

Slope, scales of, 279

Small circle of a sphere, 249

Solid right angle, projections of, 229

Solids circumscribing sphere, 426

, definitions of, 181

inscribed in sphere, 425

, projections of, 181

, sections of, 241 et seq^.

Specification of forces, 91
Sphere, 183, 249

, cone enveloping, 263

, cylinder enveloping, 256

, cylinder, and cone in contact, 426

, great circle of, 249

, lune of, 249

■ — — , plane sections of, 249

, projections of, 187

, small circle of, 249

, solids circumscribing, 426

, inscribed in, 425

, zone of, 249

Spheres in mutual contact, 426
Spherical epicycloid, 428

hypocycloid, 428

roulettes, 428

triangles, 233

Spheroid, 329

Spiral, archimedean, 140

curves, 140

, equiangular, 141

, logarithmic, 141

Springs, helical, 367

Squared wreath, 372

Square root, 1 1

Squaring the circle, 63

Statics, graphic, 9\ et seq.

Stereographic projection, 268 (Ex. 4)

Straight line and plane, 211 et seq.

centre of, 98

diagrams, 113, 115

Sub-contrary section of oblique cone, 267

of oblique cylinder, 266

Subnormal, 37
Subtraction, 10

of vectors, 85

Summation of vectors, 86
Superior epitrochoid, 72

hypotrochoid, 74

trochoid, 69

Supplementary angles, 2

trihedral angles, 234

Surface of revolution, 314

of revolution, cone enveloping, 328

of revolution, cylinder enveloping,

327

Surfaces and solids, 249

, generation of, 313

Syramedian of a tiiangle, 7

point of a triangle, 7

Symmetrical curves, 58

Tangent of an angle, 4

plane to cone, 317

to cylinder, 320

, to surface of revolution, 326

planes, 317

to sphere, 324

to a circle, 1

to a curve, 12

Tetrahedron, 182

, altitude of, 187

Theory of shadows, 411
Three-hinged arch, 121

Throat ellipse of hyperboloid, 338

of hyperboloid, 331

Tortuous curves, 343
Traces of a line, 1 74

of a plane, 198

Tracing-paper problems, 54 ct seq.
Trammel method of drawing ellipse, 41
Transversal, 6
Transverse axis, 38
Trigonometrical ratios of angles, 3
Trihedral angles, 233

, supplementary, 234

Trochoid, 69
Twisted plane, 343

surface of revolution, 333

surfaces, 314

Umbilic, 338

Undevelopable surfaces, approximate de-
velopments of, 356

Vanishing point, 303
Vector geometry, 82 et seq.

polygon, 84

Vectorial angle, 128

448

INDEX

Vectors, 82

— , addition of, 83, 84

, resolntion of, 86

, subtraction of, 85

Velocity in harmonic motion, 156
Versed sine of an angle, 4
Vertex of a conic, 30

Vertical plane, 301
Vision, centre of, 301
Volute, 147 (Exs. 31 and 32)

Wreath, 369

Zone of a sphere, 249

THE END

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