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PEESPECTIVE
GEOMETRICAL
DRAWING
ADAPTED TO THE USE OF CANDIDATES FOR
SECOND AND THIRD-CLASS TEACHERS' CERTIFICATES.
BY
THOS. H. IVLcGUIRL, B.A.
Ck)MMBRciAii Master, Collingwood Collegiate Institute.
TORONTO:
WILLIAM BRIGGS, 78 & 80 KING STREET EAST.
C. W. COATES, Monteeal, Que. S. F. HUESTIS, Halifax, N.S.
1887.
Entered, according
one
Methodist
at Ottawa.
to the Act of the Parliament of Canada, in the year of our Lord
thousand eight hundred and eighty-seven, by William Briggs, Book Steward.
Book and Publishing House, in the Office of the Minister of Agriculture,
L
PEE FACE.
Drawing having been at length recognized by the Educa-
tion Department as an essential feature in High School edu-
cation, it is necessary that a work, at once simple and concise,
should be prepared on this subject. The incompleteness and
want of deliniteness in the existing works on perspective, have
induced me to place this book before High School pupils. It
consists mainly of problems, etc., that have been given from
time to time in my own classes. To obviate the necessity of
copying problems from the blackboard, I have added a num-
ber in Geometrical Drawing, which will be found useful.
Believing that this work supplies a w^ant long felt in our
schools, I have consented to place it before the public.
T. H. M.
The Ixstitfte, March, 1887.
Digitized by the Internet Archive
in 2009 with funding from
Ontario Council of University Libraries
http://www.archive.org/details/1887bperspectivegeo00mcgu
CONTENTS
, PAGE.
Introduction 9
Drawing to a Scale 15
The Point 16
Exercise on the Point 22
The Line 23
Exercise on the Line 27
Surfaces — -The Square 28
Exercise on the Square 36
The Oblong 37
Exercise on the Oblong 38
The Triangle 39
Exercise on the Triangle 45
The Hexagon 47
Exercise on the Hexagon 52
The Octagon 54
The Circle 57
Exercise on the Circle 63
Solids 64
The Cube 66
The Plinth 68
Exercise on the Plinth . . . . * 70
The Prism , 72
The Cylinder 75
The Pyramid 78
The Cone 80
Exercise on the Prism and Cylinder 81
Exercise on the Pyramid and Cone 82
The Frusta 83
Exercise on the Frusta 85, 87
vi C0NTKNT8.
Solids — ContiniiM. page.
The Sphere , 88
Exercise on the Sphere 90
Foreshortening , 92
Synthetic Perspective 93
Perspective Effect 94
Angular Perspective — Figures on Picture Plane 96
Figures within the Picture Plane 99
Exercise in Angular Perspective 101
Miscellaneous Exercises 101
Geometrical Drawing 105
To draw a perpendicular to a given line from a point on the
line or away from it 105
To describe a square on a given straight line 106
To describe a square on a given diagonal 107
To construct an oblong of given dimensions 107
To bisect a given line 107
To divide a given line into any number of equal parts 108
To draw a line parallel to a given line from an external point. 108
To divide a line proportionally to another divided line 109
To construct a triangle of given dimensions 109
To bisect a given angle 110
To trisect a right angle 110
To inscribe a circle in a given triangle .... Ill
To draw a circle through three given points Ill
To find the centre of a given circle on arc 112
To draw a tangent to a circle from a point without or on the
circumference 113
To draw an isosceles triangle of given dimensions 114
To draw an equilateral triangle of given dimensions 114
To draw^ from a point an angle equal to a given angle 115
To construct a triangle similar to a given triangle within a
given circle 115
To construct an equilateral triangle about a given circle 116
To construct a triangle similar to a given triangle about a
given circle H'^
Within a circle to draw any number of equal circles, each
touching two others and the outer circle 118
To construct a polygon on a given line 119
CONTENTS. Vll
Geometrical DnAwi-sG—Confinned. pagb.
To construct a regular polygon in a given circle 120
To construct a regular pentagon on a given line 120
To construct a regular hexagon on a given line 121
To construct a regular octagon on a given line 122
To construct a regular octagon in a given square 122
To construct an ellipse with axes and intersecting arcs 123
To construct an ellipse with concentric circles 124
To construct an ellipse when the longer axis only is given. . . . 125
To find the axes and foci of a given ellipse 125
To draw a tangent to an ellipse from a point in the curve .... 126
To draw a tangent to an ellipse from an external point 127
To draw an oval of a given width 127
To draw an involute to a circle 128
To draw a mean proportional (greater) to two given straight
lines 129
To draw a mean proportional (less) to two given straight
lines 130
To draw a circle touching two lines not parallel . 130
To draw a circle touching another circle and a given line . . . , 131
Graded Exercise on Geometrical Drawing 132
DEAWING
11 >-A RAWING may be defined as the representation of an
■^l -^y object or collection of objects on a plain or level
surface.
There are two kinds of Drawing : Perspective, or the
representation of an object as it cqypears to the eye ; and
Geometrical Drawing, as it actually is.
Our knowledge of the size of an object can be known only
by experiment. We must either see the object near the eye,
and observe its size, or know its distance from the eye.
From long practice we are enabled to tell the height of a
hill, breadth of a river, or capacity of a ship, though at a con-
siderable distance. If we hold a rule of definite length close
to the eye, and then withdraw it six or eight feet away, we
notice that it is apparently smaller. Experience teaches us
that it is not really smaller, the apparent diminution being
only the eflfect of distance.
Perspective aims, then, at measuring and representing
objects as they appear at a distance.
The horizon ahcays hounds our vision.
If we look out upon a large lake we find that the sky and
water appear to meet ; this line of apparent union is called
2
10
DRAWING.
the horizon. No matter how large an object is, if it recedes
far enough from us on a lake, it would at length appear
on the horizon as a point.
If a person stands on a level plain he can see over a range
of 60 degrees in every direction without moving his head.
The point on the horizon directly in front of him is called
the Centre of Vision.
Now, if we take this point as a centre, and join it with the
eye of the observer, and draw lines at an angle of 30° with it
from the observer's eye, we shall form a hollow cone, which
represents his range of vision : thus, if the observer's eye be
at the point X, and AB represent the horizon, and C the
point on it directly in front of X, if CX be joined, and
AX, BX be drawn at an angle of 30 degrees with CX, AB
INTKODUCTIOX.
11
will be the horizontal range of the spectator's vision. If we
describe a circle from centre C, and distance CA or CB, such
circle will be called the base of the cone of visual rays ; for it
will be observed that from point X the spectator can see just
as far as the edge of this circle. The ground as a plane is
generally supposed to extend to the horizon, and is marked
for sake of abbreviation, G. P. The horizontal line is un-
limited in length, and is written H. L. (Fig. 1.)
<
)
\
X \
w
1
'^
P
0
Fig. 2.
All representations of objects are supposed to be made on a
plane (unlimited in extent), perpendicular to, and resting on,
the ground plane, and directly in front of the spectator. This
plane is called the Picture Plane, and is written P. P.
It must be distinctly borne in mind that the bottom of the
picture plane touches the ground plane, and this line is always
at right angles to that joining the spectator's eye with the
centre of vision,
12 . DRAWING.
The line joining the spectator's eye with the centre of
vision is usually denoted L. D, (line of direction, and some-
times length of distance).
The position of the spectator's eye is called the point of
station, or P. S. (Fig. 2.)
If a spectator stand at O and observe a stick at AA, then
carry the stick back and place it at XX, parallel to A A, and
draw lines OX, OX ; it is plain the stick cannot appear as
large as AA, but will be represented in lengi;h by BB. If
again withdrawn to YY, it will appear as e^, and so on.
As we remove it the apparent length diminishes, or, in
other words, the angle made by the line with 0 constantly
decreases. (Fig. 2.)
If the stick were removed to an infinite distance, its appar-
ent length would vanish to a point S, and the lines drawn to
O would coincide, forming an angle of O degrees.
Objects appear to diminish in size as they recede from the
eye, and vice versa.
At the centre of vision all objects have no apparent magni-
tude, and must be represented by a point, and steadily in-
creased in size till they approach the eye. This is why rails
on a straight piece of railway appear to meet in the distance,
though everywhere the distance between them is the same.
Xow, if we stood on a straight piece of track, and if other
tracks were laid on each side and parallel with it, every track
would appear to vanish at the same point directly in front of
us, hence the rule :
Lines parallel to the line of direction vnll vanish at the
centre of vision. (Fig. 3.)
Thus, let C be centre of vision, and AA, BB the ends of
parallel lines : they will all meet at 0, and all the lines drawn
from C to AB will really be right angles, however different
they may appear to be. Xow draw J)jj parallel to A A, and
INTRODUCTION.
13
since the distance from A to A is always the same as from D
to D, we have the rule :
Parallel lines drawn between vanishing lines are of equal
length, and, conversely, the lines joining the extremities of
equal parallel lines will vanish to a point.
Hl
Fiff. 4.
HL
— — ""7
T
V
/
t
T
n
A
BL
0
Again, if the spectator stood on the ground at O, where the
picture plane rests, and looked towards C on the horizon, it
14 DRAWING.
is clear that he could see objects lying on the ground any-
where between AB and HL. Hence, if we take AB as the
ground line, or base of picture plane, and OC as the height of
spectator's eye above the ground, the horizontal line will pass
through C and be parallel to AB. (Fig. 4.)
The ground line is denoted G. L.
If a stake AS be placed perpendicularly in the ground
plane at A, it will necessarily touch P.P. throughout its length.
If placed as TT, still erect, it is said to be within the picture
plane, and parallel to it. Now, if the stake be same height
throughout, TT will be equal to AS ; so also will W be equal
to AS ; and being placed on the ground, their extremities will
lie in the straight lines SV, AY (being parallel). But lines
joining equal parallel lines will vanish, hence SY and AY,
being produced, w^ill meet in C.
As nothing is ever supposed to be drawn nearer to the eye
than the picture plane, we use the picture plane as a basis of
measurement, — for in the figure we can estimate the length
of TT or YY only by referring them to AS, which is drawn
on the picture plane.
Objects to the right of OC are said to be to the right;
objects left of OC, to the left. Objects which are at an equal
distance on each side are said to be directly in front.
Perspective is of four kinds : —
1. Parallel, in which some side or face of the object is per-
pendicular or parallel to the ^.P., and also to the G-.P.
2. Angular, in which a side or face makes an angle less
than a right angle with the P.P., but is parallel to the G.P.
3. Oblique, in which the sides or faces make angles less
than right angles with both P.P. and G.P.
4. Aerial, or the perspective of distinctness in a view. It
is related to shading and painting.
I
THE POINT.
15
DEAWING TO A SCALE.
Unless objects were very small, and our drawing surface
large, we could not represent the size of the object as it is.
It is usual to draw the figure in miniature, or a certain num-
ber of times smaller. Every line in the drawing must bear
this fixed proportion to the corresponding one in the figure or
object. This is called drawing to a scale.
Usually, one quarter-inch for each foot is the proportion in
perspective, but any other ratio may be used.
Thus, if a line eight feet in length were to be drawn, it
would be represented by a line {on the picture plane) two
inches in length.
HL
0
HL
ci
. \a\b
d
\\
\b
\
BL
\
\ Bl
D
A
B
Fig. 5.
Draw any horizontal line H.L., take any point O, draw OC
perpendicular to H.L. Let CD = height of spectator; through
D draw B.L. parallel to H.L. If O be spectator, C will be
centre of vision, B.L. base line, and H.L. horizontal line. 00
will be line of direction and OD length of distance. (Fig. 5.)
16 DRAWING.
In B.L. take cany points, A, B, and join them with centre
of vision (C.V.). If dab l)e drawn parallel to DAB, and
touching CD and CB, it will be equal to DB, and da = DA
and ah = AB ; also, if the distance of B to the right of D be
known, the distance of h from d is known, for db = DB ; or, in
other w^ords, every part of BC is the same distance from CD
that B is.
If we wish to find the position of a point within the plane,
we first find its distance on B.L. from D, and then join the
point marking this distance with C ; the latter line would pass
through it.
Exercise I.
1. Find a point on the ground plane, at base of j^icture
plane, 4 feet to right.
2. Find a point 3 feet above G.P., 4 feet to right, and
touching RR, — — --^^
above it. ~ ' -.— -- --
4. Find position of a point 3 feet under G.P., 3 feet to
right in P.P. produced. '
(In foregoing examples assume spectator's height to be six
feet.)
All lines at right angles with the base line (B.L.)., or,
parallel w^ith the direction w^e are looking (L.D) will vanish
at the centre of vision (C.V.) ; but lines parallel Avith the
base line (B.L.) or horizon (H.L.) w^ill n^ver vanish, but
always appear parallel. This is an important rule.
To find the distance of a point within the plane.
In fig. 6, let AB DC represent the face of a cube resting
on the ground, and let the given face touch the picture plane.
The near lower edge will then coincide with base line AB.
Now, if we suppose the cube placed to the left of the spec-
THE POINT.
17
tator, he will be able to see the side BD FE, or the top, if he
be as high as the cube.
We have said that all lines at right angles to AB will
vanish at point C.V. ; and since all angles of a cube are right
18 DRAWING.
angles, the line BE, if produced, will reach C.Y., and
so will line DF, for the same reason. But we want to find
the position of E, the extremity of the line BE.
Draw a line from C.V, perpendicular to AB, and produce
it indefinitely. A spectator stationed at N would see the
edge BE represented in size by the dotted line Be perpen-
dicular from B on EX. If he were placed at O, BE would
appear as Bb, perpendicular to EO from B ; also, if placed at
P, BE would appear as Ba ; and it will be noticed that, as
the spectator recedes, the apparent size of BE decreases, i.e.,
Ba is less than Bb, and Bb less than Be, and so on. (Fig. 6.)
Now from B measure lengths of Be, Bb, Ba on BE ; thus,
B/=Bc, Be = Bb, Bd = Ba, etc.
Take a point M on base line at a distance from B equal to
required length of BE ; this point may be on either side of B.
In given case, since BE --= AB or BD, make BM = AB, and
suppose M to be drawn to right of B. Join M/, Me, ^Id, etc.,
and produce them backward to horizontal line touching it in
h, k, m, etc.
Now, if M be a fixed point for a distance from B, it will be
seen that as the spectator recedes from the object the points
h, k, III, etc., recede from C.V. ; hence, if the points h, k, in,
etc., be given, and knowing position of M, we can find appar-
ent distance of BM within the plane, as shown by B/, Be, etc.,
and this is done by the following method :
Take C.V. as centre, and distances N, O, P, etc., of specta-
tors from object, as distance; and describe a semicircle cutting
horizontal line in vi, k, or h on one side, and corresponding
points on the other. The points where the semicircle cuts
the horizontal line, are called the measuring points, and are
denoted by RMP, LMP, according as they are to the right or
left of C V.
THE POINT.
19
To find the measuring points, liaving given the height nfi
spectator and his distance from the picture plane or base line.
In fig. 7, let H.L. be horizontal line, which is unlimited in
length ; take any point C. V. in it, and draw line L.D. from it
at right angles. Make C. V. 0 = 6 feet, or whatever may be
the spectator's height, and tlirough O draw B.L. parallel to
H.L. ; produce C.V. O to S.P., so that O S.P. equals distance
/
/■ '
20 DRAWING.
of spectator from base line. Then with centre C.Y. and
distance C.V. S.P. describe a semicircle, cutting H.L. in LMP
and RMP, the right and left measuring points respectively.
In fig 7, let any point A be taken, say 4 feet to the left, on
the base line ; join A C.Y.
Suppose we wish to find a point the same distance to the
left that A is, but 4 feet w^ithin the plane : we know that the
point lies somewhere on A C.V., because every part of this
line is th6 same distance from C.V. O that A is. jS^ow, we
proceed by measui-ing the required distance to right or left of
A, and joining the point thus found with the measuring point
opposite — that is, if point be taken to the right of A, as O,
we join O LMP ; if to the left, as E, we join E RMP. O
LMP and E RMP will always cut A C.V. at the same point
Bif EA = AO.
So also, if we take a lesser distance, as AD, and make
AF- AD; join F RMP and D LMP; they will intersect in c.
Now AO or AE = AB, hence B is four feet within the plane
and four feet to the left.
In practice it is not necessary to draw to both measuring
points, one (the nearest) will answer every purpose.
Example 1. — Find position of a point on the ground 6 feet
directly in front, as seen by a spectator 6 feet in height, and
4 feet from picture plane.
In fig. 8, we draw H.L. and B.L. 6 feet apart, and C.V.
S.P. perpendicular to B.L. from C.V., and make O S.P.
the spectator's distance from base line. Draw a semicircle
to cut H.L. in L^NIP and RMP. Now, since point required
is on line between C.V. and O, we measure 6 feet either way
on B.L., as A or B, and join to measuring point as B LMP
or A RMP ; they will intersect in X, the point required.
Example 2. — Find a point 3 feet (3') to right, 4 feet (4')
within the P.P., and 5 feet (5') above it. Height of spectator
5 feet 6 inches (5' 6"), and his distance from P.P. 4 feet (4'),
/?/'WP
cutting
Draw H.L. and B.L. 5' 6" apart; draw C.V. S.P
B.L. in O; make O S.P. = 4'. Find RMP (C.V. RMP must
always be equal to C.Y. S.P.); make 0A = 3', from A measure
/
22 DRAWING.
off AB equal to the required distance of point within the
plane (4'). Join A C.Y. and B RMP, intersecting in C ; draw
AD perpendicular to B.L., and make AD equal to required
height (5'). Join D C.V. Draw CX parallel to AD and
meeting D C.Y. in X. X is position required. For, since
AC = AB, and any point C in A C.V. is same distance to the
right that A is, then C is 3' to the right and 4' within P.P.
E'ow, D C.V. and A C.V. are vanishing lines, and CX and
AD are parallel lines drawn between them, then CX = AD ;
but point D is 5' high, then X is the same height, and is
vertically above C also ; therefore X is position of required
point. (Fig. 9.)
ISToTE. — All measurements must be made on the P.P.
Exercise II.
(In the following examples take height of spectator 6 feet,
and his distance from P.P. 4 feet, and make scale J inch to
one foot.)
1. Find position of a point directly in front, and 8^ feet
within P.P. on G.P.
2. Find position of a point V to left (L), A" within P.P. on
G.P.
3. Find position of a point 6' to right (R), 6' wdthin P.P.,
and 4' above G.P.
4. .Find position of a bird flying 10' to R., 12' within P.P.,
and ^ above G.P.
^ / 5. Find potion of a fish resting in water 4' beneath G.P.,
'^' to R., and f within P.P.
6. In Ex. 1, show that however far the point be away, it
must always be nearer than the point C.V.
THE LINE.
23
THE PERSPECTIVE L:
A straight line is the shortest distance between any two
points. If we know the position of any two points we can
locate the line between them.
Example 1. — Draw a staff 8' high placed erect on ground
plane, 6' to R., and 4' within P.P.; distance -i', height of
spectator 6' (H = 6'), scale ^'' = 1'.
Fi^'. 10.
Draw H.L. and B.L. 6' apart; draw C.Y. S.P., making
O S.P. (LD) 4', and find RMP as before. Take A, 6' to right
of 0, a»d erect perpendicular AB 8' in height ; join B C, Y,
J
24
DRAWING.
and A C.V. ; take C 4' to left of A, and join C RMP, cutting
A C.V. in E. Through E draw ED parallel to AB ; ED is
line required. Xow, since AE = AC, E is 6' to right and 4'
within P.P.; but ED = AB, then ED is 4' within P.P., 8'
high, and 6' to right. (Fig. 10.)
Example 2. — Draw a line 3' long lying on G.P., parallel to
P.P. (or base line), and 4' within it, near extremity of line
to left. H = 6', L.D. = 4', scale
being 2
r=r.
Draw H.L., B.L. and L.D. as before, and findi also LMP.
From O mark off OB ^ 2' (distance of near extrenmy from O),
and make BA = 3' (length of line) ; join B C.Y. and A C.V.
From B measure off BE = 4' (distance within P.P.); join
E LMP, cutting B C.V. in D ; through D draw DC to line
A C.V. and parallel to AB. CD is line required, for it is
equal to AB, and its near extremity D is same distance to
the left that B is, and BE = BD. (Fig. 11.)
Example S, — Draw a line 3' in length lying on G.P., per-
THE LINE.
pendicular to P.P. (base line), -i' to P.,
2' within P.P. H = 6', L.D. = 4', scale
251
and nearest extremity
^MP
SP
Fig. 12.
B.L., L.D.,
as before,
join C C.Y.
Fr
om C
Draw H.L., B.L., L.D., as before, and mark point RMP.
Make OC = 4', join C C.Y. From C measure off CB = 2', and
from B mark off BA = 3' (length of line) ; join B RMP and
A RMP, to cut C C.Y. in E, D. ED is the line required, for
DE is parallel to LD, and therefore perpendicular to BL.
EC^BC and DC = AC, then DE = AB, and point E is the
same distance from L.D. that C is. (Fig. 12.)
Example J/.. — Draw a line 3' in length parallel to P.P. and
4' within it, and parallel to G.P., and 4' above it, line to
be drawn with near extremity 2' to left. H = 6', LD = 4',
scale Y = 1 ',
In fig. 13 draw H.L., B.L. and L.D., and find LMP,
take B 2' to left and A 4' to left of B, also S 4' to right of
B. Erect AC and BD perpendiculars to AB, and each 4' in
height; join CD, C C.Y., D C.Y., B C.Y., and S LMP. Let
S LMP cut B C.Y. in G
3
draw GF parallel to BD, and FE .
\
w(i"H^
\
26
DRAWING.
parallel to CD : then EF is line required. For since BG =
BS, and BD = FG, then DF = BG, so also CE = DF. But
EF = CD, then EF is parallel to P.P., is same height above
G.P. as BD, and the same distance to the left as BD.
THE LINE. 27
Exercise III.
(H = 6', L.D. = 4', scale J" = 1'.)
1. Draw a line 4' in length on G.P., parallel to P.P. and 4'
within it, directly in front.
2. Draw a line 4' in leno^th parallel to G.P. and 4' above it,
touching P.P. 4' to left, and perpendicular to it.
3. Draw a line 3' in length perpendicular to G.P., and 3'
above it ; line to be 4' to right.
4. Draw a line 4' in length parallel to G.P. and 5' above it,
and in contact with P.P. ; line to have one extremity 3' to
right.
5. A line 5' in length is drawn parallel to P.P. and 6'
within it ; it is parallel to the G.P., one extremity being 3' to
riorht, the other 2' to left. It is 4' above the G.P.
28 DRAWING.
SUKFACES IN PEKSPECTIVE,
RECTANGULAR SURFACES.
The Sqiaare.
A square is a parallelogram having two adjacent sides
equal, and the included angle a right angle.
We said previously that an object appears to decrease in
size as it recedes from the eye. If, therefore, a square is
placed on the ground plane with one side touching the picture
plane, it is clear that the side most removed will appear
smaller than that touching the picture plane, and so the
square may not appear to have even one right angle.
Let AB, BC, CD, DE be all taken of equal length, and let
C be on L.D., it is plain if lines parallel to AB and of equal
length be drawn on G.P. within the P.P., they will appear
shorter than AB. Find points M, N and join BM, CM, etc.
Then AF = AB, GB = BC = AB, etc. Hence FG = AB, GH =
BC, etc., also AF = BG, BG = CH, etc., and the squares AG,
BH, CK, and DL will all be equal, and the only real right
angles will be BCH and DCH : all the other angles, FAB,
ABG, etc., though really right angles will not appear so,
(Fig. U.)
THE SQUARE.
29
30
DRAWING.
To represent a square perpeiidicailar to G.P. and P.P.
Let H.L. and B.L. be drawn, also L.D., and find X. Take
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/
C3
any points A, B, C on B.L. at equal distances, and at each
erect a perpendicular equal to AB or CD ; join EH, which
THE SQUARE. 31
will be parallel to AD; join A C.Y., B C.Y., C C.V. and
D C.V. Take a point a at a distance from A equal to AB
(on left), and join AX to cut A C.V. in K ; draw KIST parallel
to AD, intersecting the vanishing lines in L, M, N ; at K, L,
M, N, erect perpendiculars to meet the vanishing lines in Q,
P, R, S ; join QvS. I^ow, since AB = AE, AF is a square,
{a) AF is said to be drawn to right, perpendicular to G.P.,
and touching or coincident with P.P.
(6) Since Xa = AK = QE, then KE is a square drawn to
right, perpendicular to G.P., perpendicular to and touching
P.P. and parallel to L D.
(c) Since AK = AB = KL = LB, AL is a square drawn to
right, resting on G.P., perpendicular to and touching P.P.
{d) Since AB = AK = QE = PF, QF is a square drawn to
right, parallel to G.P. and raised above it, and perpendicular
to and touching P.P. and parallel to L.D.
The cube AP or BR will show the square in every position
in parallel perspective, provided it touch the picture plane.
(e) Similarly, square QL is drawn to right, perpendicular to
G.P., parallel to P.P., and within it. (Fig. 15 )
Note. — If a figure be drawn parallel to the P.P. it will
always be drawn in its true shape, but smaller, hence QL
will be a true square. In above figure, take points a, h, c, d,
etc., at equal distances to the left ; erect perpendicular de at
d, and equal to dc. Join e C.V. and d C.V., also cX, 6X,
ctXj^'to meet d C.V. in /, (/, h. Through f, g, h draw
parallels to de, meeting e C.V. in /, m, n. Then, since de, fl,
etc., are parallels between vanishing lines, they are all equal
to each other and to dc, ch, etc., for de = dc. But df= dc, and
fy = ch, etc., hence ef, ly, mn are equal squares, and are said
to be drawn perpendicular to G.P. and also to picture plane.
One of them, ef, touches the P.P., the others are within it.
It will be seen that as the squares recede they appear smaller.
s^
DRAWING.
Rrample 1. — Draw a square, side 4', 4' to R., 4' within,
lying on G.P., _L to P.P. H - 6', L.D. = 4', scale \" = V.
Construct figure as before. Take AC one inch (4') and CD
one inch, take AB to left one inch, join C C.V., D C.V.,
B MP, A MP, cutting C C.V. in F, E. Draw EK and FG,
parallel to AC, and meeting D C. V. in K, G : then FK is
square required, for FG = EK = CD, EC = CA, and EF = AB.
(Fig. 16.)
CV MP
Fig. 16.
Example 2. — Draw a square, side 4' and 6' to left, J_ to
P.P. and 3' within it, and J_ to G.P. and 1' abov5 it.
Other specifications as in last example.
Construct figure as before. Take C U" (6') to left of A,
take B 1" (4') to right of A, erect CE _\_ to AC at C and IJ"
(5') in height, and mark ofi" D J" (1') above C; join E C.V.
D C.V., C C.V., A MP and B MP. Through F, G draw FL
and GS parallel to CE, and meeting vanishing lines in L, S
and H, K respectively : LK is the figure required. For
THE SQUARE.
33
LS = HK = rG = AB = 4', SK-LH = ED = 4' and HF-DC
= 1', also FC-AC = 6'. (Fig. 17.)
Fig. 17.
Example 3. — Draw a square 4' side directly in front, paral-
lel to G.P. and 3' above it, and _j_ to P.P. and 3' within it.
Same specifications as before.
Construct figure and find M as before. From A measure
off Y both ways to B and C (BC = 4') ; erect perpendiculars
BF and CG, each |" (3'); join FG, B C.V., F C.V., and G C.V.
From B measure off BD |" (3'), and from D measure off DE
1" (4') ; join DM and EM, cutting B C.V. in H, K. Draw
HL, KN parallel to BF, and meeting FC.Y. in L, N; through
L, N draw LJ and NP parallel to FG : then LP is the
square required. For FL = BH = BD = 3', and NP = LJ = FG
= BC-4', LN = HK = DE = 4', and LH = FB-3'; also SL =
SJ = XF = XG = AB = AC = 2'. (Fig. 18.)
34
DRAWING.
^
^ /
UJ
O \
Q \
<=: a7 o
^
a.
/
CO
/ /
o
\
^
\
/
-J
/\^
^^-\
CO /
-J
/
/
/ /
/ /
/ /
/ /
/ /
/
'/
/ '
, /
'/
/
/
-J
CQ
THE SQUARE.
35
Example Jf. — Draw a square 4' side, 4' to right, lying on
G.P., J_ to P.P. and 4' within it, and within this square
place centrally a square whose sides are 2'.
Fi<f. 19.
J^OTE. — Figures are said to be placed centraUy when their
centres coincide and their like sides are parallel to each other.
Concentric circles are always placed centrally with respect to
each other.
Draw H.L., B.L., L.D., as before, and find M. Measure off
aA, ae, and AE, each one inch ; bisect AE and" ae in C and
c, and Insect CE, CA, ce and ca in D, B, d, b respectively.
Join A C.V., B C.V., C C.V., D C.V., aM, bM, cM, o?M, and
eM, cutting A C.V. in /, (/, n, k\ I ; and through these latter
points draw parallels to AE, cutting E C.V. in t, s, i\ q and j.
It will form the outer square, and XX the inner square.
For ac = BD = 2', and kg = db^2', and /A - Aa and ft = AE,
etc. (Fig. 19.)
36 DRAWING.
Note. — If aM be joined, it will always pass through / if
the figure be a square. It will also pass through X and X.
Exercise IV.
(H = 6', L.D. = 4;, scale !"=!'.)
1. Draw q^square V side, ^' to left, resting on G.P., _j_ to
and touching P.P.
2. Draw a square 5' side, directly in front, lying on G.P.,
I to and touching P.P.
3. Draw a square 6' side, J_ to G.P., _]_ to P.P., touching
both, and 6' to left.
4. Draw^ a square 3' .side, parallel to G.P. and 2' above it,
•i' to P., perpendicular to P.P. and 2' within it.
5. Draw a square 4' side, parallel to P.P. and 4' within it,
resting on ground plane, left corner touching L.D.
6. Draw a square 4' side, parallel to G.P. and 2' above it,
right corner 1' to right; square to have side _[_ to P.P. and
1' within it.
7. Draw a square 4' side, _[_ to G.P. and 1' above it, 3' to
right, _J_ to P.P. and 2' within it.
8. Draw a square 6' side, directly in front, J_ to P.P. and
2' within it, parallel to G.P. and 4' above it; and place a
square of one-fourth its area centrally within it.
9. Draw a square 3' side, 4' to right, _\_ to P.P. and 3'
within it, J_ to G.P. and just its own height beloiv it.
10. Draw a cube (edge 4') touching P.P. 4' to left, resting
on G.P.
11. Draw a cube (4' edge) directly in front, and to rest on
G.P., one edge parallel to P.P. and 2' within it.
THE OBLOXG.
37
Tine Oblong.
An oblong is a figure, whose opposite sides are equal and
parallel, and whose angles are right angles.
The drawing of the oblong differs little from that of the
square, care being required only to distinguish the sides.
Fig. 20.
Exaiwph 1. — Draw an oblong 3' x 2', lying on G.P., _L to
P.P., side 3', parallel to, and 2' within it; oblong to be 4' to
right. Specifications as before.
38 DRAWING.
Construct figure as before. Find M, and measure off AC 4'
= (distance to right), CD = 3' (length of side), and measure off
BE 2' (breadth); measure off BC-2' (distance within P.P.).
Join EM, BM, C CV., D C.V.; through H, F draw HK, FC
parallel to CD, meeting D C.V. in G, K : then FK will be
oblong required. For FG = HK = CD = 3', GK = FH = EB =
2', and HC = CB = 2'. (Fig. 20.)
Exercise V.
(H = 6', L.D. = 4', scale J" = 1'.)
1. Draw an oblong 4' x 2' J_ to G.P., and 4' to left, J_ to
P.P. and 3' within it ; oblong to rest on end.
2. Draw an oblong T x 5', standing on end on G.P., parallel
with P.P. and 2' within it, directly in front.
3. Draw an oblong 3' 6" x 4' 6", parallel to G.P. and 2' 6"
above it, right corner 3' 6" to left ; end of oblong to be paral-
lel to P.P. and 6" within it.
4. An oblong 6' x 4' is buried in the ground to a depth of
2'; it is parallel with LD and 4' to right, and 2' within P.P.
5. An oblong 6' x 4', with side resting on G.P. 4' to left,
and _j_ to it, intersects a square of 4' side, resting on G.P.
and parallel to P.P. ; the oblong divides the square into two
equal portions, and tlie square divides the oblong into parts of
4' and 2' respectively, the greater portion being nearest. The
square is 4' within the P.P.
6. Draw an oblong 4' x 2', end touching P.P. 4' to P. ;
oblong lying on G.P.
7. Draw an oblong 5' x 3', lying on G.P. directly in front;
end parallel with P.P. and 2' within it.
8. A wall, whose height is 8', begins at a point 10' to left,
and stretches inwards indefinitely ; at distances of 10' and 20'
doors 5' X 3' are made in it. Draw it.
9. Draw an oblong 6' x 4', lying on ground, sides parallel
to P.P. and 2' from it, oblong 3' to right ; and within it place
an oblong 4' x 2' centrally.
10. Draw an oblong 3' x 2', lying on G.P. 4' to left, end
parallel to P.P. and 2' from it; and about it draw an oblong
5' X 4' centrally.
yuo ^^i^Y
THE TRIANGLE.
\J
The Triangle.
A triangle is a figure enclosed
by three straight lines ; these
may be of uniform length, but
the length of any two taken
together must be greater than
the third.
The triangle cannot be con-
veniently drawn alone, Vjut is
drawn with reference to an ob-
long.
Triangles are equilateral, with
three equal sides ;
Isosceles, with two equal sides ;
Scalene, with unequal sides.
Right angled, or containing a
right angle ;
Obtuse angled, or containing
an obtuse angle ;
Acute angled, or containing
three acute angles.
To draw a triangle we first
draw a j)lan^ showing the tri-
angle and its bounding oblong.
Thus, 1 would be a plan for
an equilateral triangle ABC, in
which D A = AE = BF = FC.
2 would be a plan for a right-
angled triangle, in which the
angle hac is th eright angle, with
he as hypothenuse ; the segments
6/, fc of the hypothenuse would
determine the position of a.
40
DRAWING.
In 3, the triangle DAC being given, CE would be drawn
perpendicular to DA, and CB parallel to DA ; also DB and
AF would be drawn parallel to CE : then position of the
angles at A and C v/ould be easily determined.
Example 1. — Draw an equilateral triangle, 3' to a side,
lying on ground, one side touching P.P. near left angle, 4' to
right.
Height 6', distance 4'
scale J'
cv
HL
tr
h"
y
^
\ \^
\ ^:
1
1
N)^
\.-
^A-
\
/
v-^
■^^^ ',\
/
Ly
\/'-
^^, W
■Q
'\
/
/
/]
B
\
0
V
w
K
'\
/
/
o
Fig. 22.
^T
Draw H.L., B.L. and L.D. as before. Take A 4' to right of
S, and B 3' to right of A ; join A CV. and B C.Y.; on AB
describe equilateral triangle ABE (below AB). Draw AD,
BF at right angles to AB, and through E, draw DF parallel
to AB, and meeting AD in D and BF in F ; draw EG paral-
lel to AD. Join G C.V., and with centre A and distance
AD, describe arc DL, cutting B.L. in L, Join LM, cutting
THE TRIANGLE. 41
A C.V. in N ; through X draw NK parallel to AB, cutting
G C.Y. in C and B C.Y. in K. Join CA and CB : then
ABC is the triangle required. For AB = AE and AD = AL
= AN = CG = height of triangle ; also XC = CK = AG = GB =
DE = EF. Then AC = AE = EB = BC, for they are diagonals
of equal oblongs. (Fig. 22.)
In this figure the vertex C is directed away from the eye.
If we joined XG and KG we would have a similar triangle,
but with vertex G directed toivards the observer.
Example 2. — Draw an isosceles triangle whose sides are
2', 3' and 3' respectively, lying on ground plane, vertex
directed towards the spectator, base parallel with P.P. and 5'
away from it ; near angle 2 ' to left.
Draw B.L. and H.L. as before, find M. Take K 2' to left
of X, and G 2' to left of K ; bisect GK in H ; draw GR, HS
and KT at right angles to B.L., and 5' in length ; join RST.
Take TP 3', and with centre T and distance TP describe arc
PX, cutting HS in X^ ; join XT and XR : XRT will be the
plan of the triangle.
Through X draw ITXQ parallel to RT, cutting GR in U
and KT in Q ; with centre K and distance KQ describe arc
QL, cutting B.L. in L ; and with same centre and distance
KT describe arc TY, cutting B.L. in Y. Join G C.Y..
H C.Y. and KC.Y.; and join also LM and YM to cut
K C.Y. in E and C. Through E draw EAF, and througli
C draw CDB, each parallel to B.L., and cutting H C.Y. in A
and D respectively ; join BA and AC : then ABC will be the
triangle required. For CK = KY = KT = 5', and EC = LY =
QT = XS = height of triangle, and EK = KL = KQ. Hence
AD = XS and AH = HX. But since tri;ingle is isosceles, GH
is made equal to HK ; hence BD = DC = GH = HK, then CA
= X^T = XR = BA. (Fig. 23.)
Example 3. — Draw an equilateral triangle, each side 4', one
side on ground plane parallel to L.D. ; triangle placed J_ to
ground plane and touching P.P. 3' to right.
Draw H.L., B.L. and L.D., and find M. Take point D 3'
to P., and B 3' to right of D ; on DB describe the equilateral
triangle DGB. Bisect DB in E, join EG ; through G draw
FH parallel to B.L., and through D, B draw DF, BH parallel
to EG, cutting FH in F, H. Join B C.V., draw EM, DM,
cutting B C.Y. in X and A. With centre B and distance BH
describe arc HK, cutting BK, a perpendicular on B.L., at K ;
THE trtant;le.
48
join K C.V. Through A and X draw parallels to BK, meet-
ing K C.y. in L and C ; join AC, CB. Then ABC is the
triangle required. For LA = CX = KB = BH = EG = altitude
of the triangle; and AX = NB = DE = EB, and AB = DB =
length of side ; then C corresponds to G, and AC, CB to DG
and GB respectively. (Eig, 24.)
Example Jf. — Draw a right angled-triangle whose hypothe-
nuse is 4' and one of the other sides 3' : the right angle is
44
DRAWING.
directed away from the observer, and is 5' to the left, and
the hypothenuse is 3' to the left. The triangle lies on the
ground with the hypothenuse in contact with the B.L.
Draw H.L., B.L. and L.D. as before. Find M, and take a
point D, 5' to left, C, 3' to left, and B, 4' to left of C. On
BC describe the semicircle BGC. (The angle BGC in a semi-
circle is always a right angle.) Join DG, B C.V., D C.V. and
C C.V.; through G draw EF parallel to BC, and through B, C
draw BE, CF perpendiculars on EF, at E and F. With
centre C and distance CF, describe arc FL, cutting B.L. in L ;
join LM, cutting C C.V. in H ; through H draw KAH
parallel to BC, cutting D C.V. in A ; join AB and AC : then
THE TRIANGLE. 45
ABC is the triangle required. For AB = BG and AC = CG,
and AD = DG = CF = CL, and K A = BD and AH = DC ; then
KH = BC, and A corresponds to G ; then angle BAC cor-
responds to angle BCG. (Fig. 25.)
Exercise YI.
(H = 6', LD = 4', scale I" = 1'.)
1. Draw an equilateral triangle, 3' side, lying on G.P., one
side parallel to L.D., vertex directed to the left and distant
from the L.D. 4' ; the triangle touches P.P.
2. Draw an equilateral triangle parallel to P.P. and 3' from
it, _\_ to G.P., near angle 2' to left, triangle 3' to side, one
side on ground.
3. Draw a triangle whose sides are 4', 5', 6', respectively,
6' side on ground, __[_ to P.P. and 2' within it ; triangle 3' to
right.
4. Draw an isosceles triangle whose base is 3' and each
equal side 4', lying on ground plane directly in front, vertex
directed towards P.P. and 1' from it.
5. Draw a right-angled triangle whose hypothenuse is 5',
and the perpendicular on it from the right angle divides it
into segments of 3' and 2'. The triangle is parallel with G.P.
and 4' above it, and the vertex is directed away from P.P. at
a distance of 4'. The right angle is directly in front, and the
larger segment is to the right.
6. Draw a triangle, each side being 3', _[_ to G.P. and 1'
above it, with a side parallel to it. The triangle is 4' to left,
I to P.P. and 2' within it.
7. Draw an isosceles right-angled triangle, the equal sides
being 3'; one equal side is parallel to G.P. and 2' above it.
The triangle is 2' to right, parallel to P.P., and 2' within it.
8. Draw an isosceles triangle, base 4', equal sides 3' each,
directly in front, parallel to P.P. and 4' from it ; vertex touches
ground, and base is parallel to it.
9. An equilateral triangle, each of whose sides is 4', lies on
the ground, vertex directed away ; one side parallel to P.P.
46 DRAWING.
;ind 2' from it. Triangle 2' to right. Within this place cen-
trally a similar triangle whose sides are 2'.
10. An equilateral triangle, each of whose sides is 4', is j^
to ground and also to P.P., which it touches at a point 5' to
left ; the vertex of the triangle is directed downwards, and
one side is horizontal. The triangle is buried one-fourtli in
the ground. Draw it.
THE HEXAGON,
47
Thie He:Ka.gon.
A hexagon is a figure bounded by six sides, \yiien the
sides and angles are all equal the hexagon is regular. Only
regular hexagons will be considered here.
In drawing a hexagon, a plan is made somewhat similar to
that for the triangle. The following is the general form of
the plan.
k' DL D h C C E
1
\ V
-K-
\ /
■
^
/
/
\
\
N
N
^\
/
V
y
Let B.L. =base line, and AB given side of hexagon, and in
given position. Bisect AB in C, make AD and BE each equal
BC ; on DE describe equilateral triangle DEE, and construct
oblong DL. Bisect DK in G ; draw GH parallel to KL ; join
AG, GM, BH and HN. With centre D and distances DG
and DK describe arcs to cut B.L. in X and P respectively.
Like triangles, hexagons may lie flat on the ground, with a
side parallel to the P.P. or to L.D., or they may be drawn
perpendicular to the ground plane, with a side or an angle
touching it ; and a side or an angle may touch the P.P. or be
within it. (Fig. 26.)
If a side touch the P.P., the extremity of it determines the
48
DRAWING.
distance of the hexagon to the rifrht or left When an angle
touches the P.P., that point determines the distance ; so also
when the figure is within the P.P.
Example 1. — Draw a hexagon, each side 2', lying on G.P.,
one side coincident with P.P. and 4' to R. H=: 6', L.D. = 4',
scale \" = 1 '.
Let H.L., B.L. and L.D. be drawTi. Take C.V., tind 0 and
MP; take AB 4' to R. of Q, make AB = 2'. On AB con-
struct equilateral triangle ABH, and produce to Y, Z, making
HZ and HY each equal BH ; join AY and YZ, and produce
YZ both ways. Through H draw GHW parallel to YZ ;
through C draw CE parallel to AY ; draw also DWF parallel
to AY. Join C C.V., A C.V., B C.V. and D C.Y. ; also
L MP and K MP, cutting C C.Y. in X and P. Through
N, P draw NT and PV : parallel to AB ; join SP, PA. TY
and VB. then STYBAP is the hexagon required. For PC =
THE HEXAGON.
49
CK = CG, and NP = LK = GE. Then SP = GY, also PA =
AG = GY; similarly, TV = VB = BW = WZ = AB, and QA
= 4' and AB = 2'. (Fig. 27.)
Example 2. — Draw a hexagon same as in Example 1, but
! to G.P., and with an angle touching P.P. 4' to L.
Fig. 28.
Here take E 4' to left, and for plan of the hexagon we
may proceed as follows : — Bisect EQ (4') in Z, also EZ, QZ
in A, B respectively; with centre A and distance AB
describe arc BC ; similarly, draw arc AD. Draw EC and
QD, perpendiculars to AB at E and Q respectively, meeting
arcs in C and D; produce EC to F, making CF = EC.
Erect perpendicular EN and make it equal EF, and make
50
DllAWrXG.
ES = EC; join N C.V., S C.Y. and E C.Y., also AMP,
B MP and Q MP, cutting E C.Y. in K, L, M. Through
K, L, M draw parallels to EN, cutting X C.Y. in O, P, R ;
join OS, SK, LT and TP : then OPTLKS will be hexagon
required. For ES = EC and EN = EF, also KL = AB and EM
- EQ. Then OS = SK = AB ; and PT = TL = BD. (Fig. 28.)
Example 3. — Draw a hexagon, 2' side, resting on G.P. _|^
to P.P., ha\dng one side coincident with it and 4' to right.
H = 6', L.D. = 4', scale l" = r.
CV
HL
H'P
S' h
Fi^. 29.
We proceed as follows : — After drawing H.L. and B.L. and
finding positions of 0, MP and C. Y., take A i' to right of S ;
at A erect perpendicular AC = 4', and measure off AE = 1' and
DE = 2'. Produce CA to B, making AB = AC and FX = DE ;
make FG = FX and GH= GA ; join HKMP, GLMP, A C.Y.,
thp: hexagon.
51
I) C.V., E C.V. and C C.V. At K, L ereol perpendiculars to
meet vanishing lines in N and M ; join ^IR, MD, LE and
LP : then RMDELPR will be hexagon required. For AK =
AH and AL = AG = FX = 2', and DE = RP and ML = NK =
CA. Then RM = MD = DE = 2', etc. (Fig. 29.)
Example 4-— Draw a hexagon, 2' side, lying on ground,
near side parallel with P.P. and 2' within it; hexagon to be
directly in front. H = 6', L.D. = 4', scale V = 1'.
52 DRAWING.
Draw H.L., B,L., and find O C.V. and MP as before.
Make SY and ST each = 1 ', and make TX and VA each = 1' ;
make AF = 2', and draw it _i_ to AX; draw XP, TY, YZ
parallel to AF. Make RG = RL or TY, and GZ also = RL,
and complete the oblong MH. Make Aa = AF, A6 = AG and
AE = AH; join EMP, 6MP, aMP, and A C.Y., Y C.Y.,
T C.Y. and X C. Y. Through intersecting points D, C, B, draw
parallels to AX, meeting X C.Y. in e, /, ff, respectively ; join
cf, fk, dC and Cm, completing the hexagon : then AB = Aa =
2', and BC = CD = a6 = 6E = GH, and hence Cjji^GZ and dC
= YG=2', etc. (Fig. 30.)
Exercise V^II.
In these examples take H = 6', L.D. = 4', and scale J' = 1' ;
but a scale of V = 1' may be used if thought more convenient.
1. Draw a hexagon, side 2', lying on ground plane, one side
perpendicular to P.P., and an angle touching it at a point 4'
to right.
2. Draw a hexagon, side 3', standing on edge, _l_ to ground
plane and P.P., and ha^TJig an angle touch the P.P. 3' to left.
3. Draw a hexagon, side 2', standing on edge, parallel to
P.P. J_ to ground plane, directly in front, and 3' away.
4. Draw a hexagon, 2' side, lying on G.P., one side parallel
to P.P. and 3' away ; hexagon 4' to left.
5. Draw a hexagon, 3' side, resting on an angle _L to
G.P. and P.P., one side parallel to P.P. 4' to right and 4'
within it.
6. Draw a hexagon whose edge is coincident with that of a
square, and lying in same plane. The square is 2 ' to the side,
and is placed J^ to P.P. and G.P. 2' to right and 2' within,
7. Draw a hexagon, 3' side, parallel to G.P. and 4' above
it, 4' to right, 3' within P.P., and one side parallel to P.P.
8. Draw a hexagon 3' to side, 4' to right, one angle touch-
ing P.P. ; hexagon to be J to P.P. and parallel to G.P.
THE HEXAGON. 58
9. Draw a hexagon 3' to a side, directly in front, lying on
ground plane, one angle touching P.P., and sides _\_ to it.
10. Draw a hexagon about an equilateral triangle lying on
G.P., vertex directed away from observer; the triangle is 3'
to a side, and one side is parallel to P.P. and 3' within it.
The vertex of the triangle is 4' to left.
11. Draw a hexagon, 2' side, placed _|_ to P.P. and G.P.,
4' to right, lower side parallel to ground plane and 4' above
it.
12. Draw a hexagon, 4' side, lying on G.P., near side paral-
lel to P.P. and 2' within it; hexagon to be 4' to right.
Within this draw (centrally) another hexagon, whose sides
shall be 2' in length.
13. Draw a hexagon, side 2', parallel to P.P. and 2' within
it, lying on G.P. directly in front.
14. Represent a hexagon, side 3', half buried vertically in.
the ground, one side parallel to G.P. ; hexagon _|_ to P.P.
and 3' within it, 4' to left.
n[
DRAWING.
Tl^ie Octagon.
The drawing of an octagon differs but little from that of
the hexagon, we shall, therefore, merely show the plmi. The
following are methods of drawing the plan: —
Fisr. 31.
Let AB = given side (on B.L.) ; bisect AB in O. and draw
OG _L to OA and equal it, and describe semicircle AGE ;
THE OCTAGON
55
join AG and GB : with centres A, B, and distances equal to
AG, describe arcs to cut AB produced in C and D ; on CD
describe square CDFE. With centre A and distance AB
describe arc to cut CE in H ; Hnd P similarly : with centres
H and P and distances equal to HA, describe arcs to cut CE
and DF in K and N respectively, a-nd wath same distances
describe arcs to cut EF in L and M ; join AH, KL, MX and
BP, which will complete the hexagon. (Fig. 31.)
Another way :
is
-y
~~^\
/
\
/
\
\
F
i/
0
B
!
\
i
} /
\
1
y
Fig. 3i
56 DKAWING.
Let AB = given side ; take centre O, and with distance AO
describe circle ACD ; draw diameter CD _L *o ^^- Join
CA and produce it. With centre A and distance AB describe
arc to cut CA produced in E ; then draw EF _l_ to BA pro-
duced, and produce FE, making EG = EA, etc. (Fig. 32.)
Note. — In parallel perspective a hexagon or an octagon
must be supposed to have a side parallel or perpendicular to,
the picture plane.
THE CIRCLE. 57
Tl^e Circle.
Hitherto we have been dealing e:s;clusively with straight
lines, in so far as the appearance of figures is concerned ; we
now proceed to represent curved lines in perspective. It is
evident that a curve cannot be correctly represented, without
the aid of straight lines.
There is only one position in which a circle will appear true
to the eye, and that is, when the eye is in a line exactly per-
pendicular to its plane, at its centre. In all other positions it
will appear an ellipse, varying from a circle to a line. If, for
instance, we place a hoop on the ground, and look at it
directly, it will appear true, but if turned on an imaginary
axis it will assume the form of an ellipse. The height or
diameter of the hoop corresponding to the imaginary axis will
remain the same, while the diameter at right angles to it, or
the revolving axis, will diminish, till at length it is a mere
point. Hence, to know the appearance of a circle not viewed
directly, we must know the angle the eye makes with its
plane, or its appearance in relation to some figure easy of
representation, contained by straight lines.
Now, a square answers admirably for this purpose, for if we
draw the diameters of a square, and then draw a circle so as
to touch its sides at the extremities of the diameters, we can
without much difficulty represent the circle, for we will have
four points as guides. If, however, the diagonals also, of the
square be drawn, the four points where they cut the circum-
ference of the circle will furnish additional points, so that we
will have altogether, eight points for guidance in drawing the
circle. Thus, —
58
DRAWINTJ.
Let KLMO = given square, draw diameters and diagonals,
arid inscribe circle cutting the diagonals in A, C, G, E;
join AC and GE, and produce them to meet KL in S, S ; join
THE CIRCLE.
59
K C.Y., S C.Y., etc.; also make KR = KB and RX-BO or
KB. Join X MP and R MP, and where they meet K C.V.
draw parallels to KL : then bf will represent the diameter
BF, and it will be easily seen that a, b, c, d, e,/, g, h will cor-
respond with A, B, C, D, E, F, G, H respectively, and the
curve traced between them will represent the circle (in this
case) lying on the ground plane. (Pig- --^S.)
Bxample 1. — Draw a circle, diameter 4!, lying on G.P., cen-
tre 4' to right and 2' within P.P. H = 6', L.D. = 4', scale
i" = r.
Fiff. 34.
Here ED = 4, EA = 2, DB = 2'. Describe semicircle ASD,
draw AG and BM _J_ to AB, and draw GSM through S,
60
DRAWING.
parallel to AB ; make AE = AG and AF- AB ; join F MP
and E MP, also A C.V., D C.Y., B C.V. Join DG and DM,
cutting curve in H and K. Draw HC and KL parallel to
AG; join C C.V. and L C.V. ; complete square NABP, draw
diagonals; then on the eight points thus shown draw^ the curve
required. (Fig. 34.)
Example 2. — Draw a circle touching P.P. 4' to left, stand-
ing on G.P. and J_ to it and P.P. : circle to be 4' in diameter.
H = 6', L.D. = 4', scale i" = l'.
Fig. 35.
Here take A 4' to left, bisect it in G ; describe semi-
circle, and complete oblong APQB ; join GP, GQ, cutting
curve in R, S ; draw^ RF, SH J_ to AB ; join A C.V. Erect
at A the perpendicular AC = AB, and bisect it in X ; join
C C.V., X C.V., also F MP, G MP, H MP, B MP. At points
of section K, L, M, E draw parallels to CA : draw diagonals
THE CIRCLE.
61
CE, AD, and trace curve between the eight marked points.
(Fig. 35.)
Example 3. — Draw a circle, diameter 4', directly in front,
lying on G.P., centre 4' within P.P. H =6', L.D. = 4', scale
62
DRAWING.
Here take OA, OB. each = 2'; upon AB describe square
ACDB: bisect BD in E, and draw EM parallel to CD;
describe semicircle and find points F, G, as already shown.
Make BK = BD, BH = BE and KL = DE: join L MP,
K MP, etc.; also A C.Y., G C.V., etc.; and on eight points
thus formed describe circle required. For BX = BH = BE
= 2', andXZ = HL = BD = 4. etc.
Example 4- — Draw a circle, diameter = 4', _[_ to G.P.,
parallel with P.P. and 2' within it ; centre of circle 5' to
right. H = 6\ L.D. - 4'. scale \" = 1'.
CV
MP
Here take A 5' to right of 0 ; erect the perpendicular AB.
2' = radius of given circle; join B C.V. and A C.Y. ; take C
2' to left of A; join C MP, cutting A C.Y. in D. Draw DE
parallel to AB ; then with centre E and distance ED describe
circle required. For AD = AC = BE: then E is 2' within
P.P. and 5' to right, also DE = AB = 2', etc. (Fig. 37.)
THE CIRCLE. 63
Exercise VIII,
(H = 6', L.D. = 4';scalei" = l'.)
1. Draw a circle, diameter 4', resting on ground plane and
touching P.P. at a point 4' to right,
2. Draw a circle, diameter 4', resting on G.P., centre 4' to
left and 4' within P.P.
3. Draw a circle, diameter 4' its plane perpendicular to
G.P. and touching it at a point 4' to left ; the circle is per-
pendicular to P.P. and touches it.
4. Draw a circle, diameter 4', lying on ground plane directly
in front, centre 5' wdthin P.P.
5. Draw a circle, diameter 4', parallel to G.P. and 2' above
it, placed with centre 4' to right and 4' within P.P.
6. Draw a circle, diameter 4', coincident with P.P. and
touching G.P., centre 4' to right.
7. Draw a circle, diameter 4', plane perpendicular to G.P.
and P.P. ; the centre of the circle is 5' to left and 3' within
P.P.
8. Draw a circle, diameter 4', parallel to G.P. and 9' above
it, directly in front, centre 4' within P.P.
9. Draw a circle, diameter 4', parallel to P.P. and 6' within
it; centre of circle 1' to right and 3' below G.P.
10. Draw a circle, diameter 6', lying on G.P., centre 4' to
right and 4' within P.P., and within it draw a concentric
circle of 3' diameter.
11. Draw a quadrant, radius 2', lying on G.P., vertex
directed away, and placed 4' to left and 4' within P.P.; the
radii make an angle of 45° with P.P.
12. Draw a circle, diameter 4', buried vertically in the
ground to a depth of 1'; the circle is perpendicular to P.P.,
and its centre is placed at a point 5' to left and 3' within
P.P.
64 DRAWING.
SOLIDS.
It is expected that the pupil will have drawn all the figures
mentioned in the exercises. Unless the problems have been
thoroughly understood, comparatively little progress can be
made in the perspective of solids.
Solids may be classified thus :
I. Those contained by plane surfaces.
II. Those partially or wholly contained by convex surfaces.
They are sometimes classified as solids with Developable, or
with Undevelopable surfaces.
Those belonging to Class I. are Cubes, Plinths, Parallelo-
pipeds. Prisms, Pyramids, Wedges, and Frusta.
Of those contained by convex surfaces in part, are Cones,
Cylinders, Hemispheres, and frusta of Cones.
Those contained wholly by convex surfaces are Spheres,
Spheroids, Ellipsoids, Cylindroids, Spindles, etc.
All the latter have undevelopable surfaces, i.e., they cannot
be straightened out to a plane surface.
Solids contained by plane surfaces may be subdivided into :
1. Those rectangular throughout.
2. p partly rectangular.
3. II wholly oblique.
The latter class of solids cannot be readily drawn in per-
spective, and will not be treated of, here.
Of (1) are Cubes and Plinths or Parallelopipeds.
A cube is a solid contained by six equal squares, and all
its angles are right angles.
A plinth is a solid contained by three pairs of equal and
similar oblongs. Each pair of surfaces may be equal or
SSOLIDS. 65
unequal to one or both of the other pairs, but the angles are
right angles.
(2) A prism is a solid contained by two regular polygons
whose planes are parallel to each other, and whose like sides
are joined by rectangular planes.
A pyramid is a solid formed by joining the angles of a
triangle, square, etc., with some external point. If the exter-
nal point be vertically above the centre of the pyramid, the
pyramid is said to be right ; if in any other position, oblique.
A frustum of a pyramid is the part remaining after a
smaller pyramid is cut off by a plane parallel to the base.
Of solids with convex surfaces : —
A sphere is formed by the revolution of a semicircle around
the diameter, which remains fixed.
A cone is formed by the revolution of a right-angled
triangle around one of the containiiig sides, which remains
fixed.
A cylinder is formed by the revolution of an oblong around
one side, which remains fixed.
A spheroid is formed by the revolution of a semi-ellipse
around one of the axes, which remains fixed.
If the fixed axis be major, the spheroid is prolate ; if
minor, oblate.
66
DRAWING.
The Cu-be.
The drawing of the cube is so simple that a single example
will suffice for its explanation.
THE CUBE. 67
Example 1. — Draw a cube, edge 4', placed on G.P., one side
parallel to P.P. and 2' within it, near angle 3' to right.
H = 6', L.D. = 4', scale i" = l'.
Draw B.L,, H.L., and O, C.Y. and find ]SIP as before.
Take A 3' to right, and B 4' to right of A ; take C 2' to left
of A, and D 4' to left of C ; join A C.Y., D MP and
C MP, cutting A C.V. in F and E. On AB describe square
ASMB ; join S C.V., M C.V. and B C.Y.; through PJ and F
draw parallels to AS, meeting S C.Y, in H and G ; through
G, H draw parallels to SM, meeting M C.Y. in K and L.
Draw LN from L _L to MB, and ifiiST from E parallel to AB.
This will complete the required cube. For XA = 3', and
AE = AC = 2', AF = AD and EF = CD = AB = 4'; also FG =
EH = AS = 4', and GK = HL = EN = AB = 4': thenEF = GH
= KL=4', etc. (Fig. 38.)
68
nUAWlNci.
Tbie Pliratin.
The plinth differs from
a cube only in the rela-
tion of its dimensions;
the principle employed
in drawing them is the
same, but a particular
side of the plinth is men-
tioned in reference to
the P.P. or G.P. In the
cube this is quite un-
necessary, as all the sides
are equal.
d Example 1. — Draw a
.rf plinth whose dimensions
are 4' x 3' x 2' (4' long,
3' wide and 2' thick),
the side 4' x 3' rests on
the ground plane, and
side 3' X 2' is parallel to
P.P. and 2' from it; the
plinth is 2' to the left.
H = 6',L.D. = 4',scale
■ r=i'.
Draw H.L., B.L. and
tind C.V.,0 and MP as
before. Take B, 2' to
left, and A, 3' to left
of B; also M, 4' to
On AB
right of O.
THE PLINTH.
69
construct the oblong ABDC, 3x2'; join C C.V., D C.V.,
A C.V., B C. V. ; also 0 MP and M MP. From points K, H,
where B C.V. intersects 0 MP and M MP, erect KX and HG
parallel to BD ; and through N, G draw NE and GF paral-
lel to CD. Draw EL _L to EN and KL _L to EL, which
will complete the plinth. For BK = BO and KH = OM ; then
EF = NG = KH = OM = 4', and FG = EN = LK = AB = 3', GH
= NK = DB = 2', etc.
Example 2. — Draw a flight of four steps, each step 4' x 1'
X 1'. The ends of the steps are coincident with the P.P., and
4' to right. H = 6', L.D. = 4', scale J" = 1 '.
Fig. 40.
Here take A, 4' to right of E, and B 4' to right of A : on
AB describe square ABDC, and divide it into sixteen equal
squares ; join each angle, as shown in figure, with the C.V. ;
join also E MP, cutting A C.V. in F ; draw FG parallel to
AC, GS \__ to GF, etc. Then FG = AK-r, and GS = KN
= 1', etc.; and AF = AE = AB = 4'. (Fig. 40.)
70
DRAWING.
Example 3. — Draw same, with ends perpendicular to P.P.
one step being coincident with it and 4' to left.
Fie:. 41.
Draw H.L., B.L., and find C.V. and MP as before. Take
A, i' to left, and B, 4' to left of A : on AB construct square
ABCD : join O MP ; and on AE, complete square ADFE :
then draw steps similar to preceding example.
Exercise IX.
(H = 6', L.D. = 4', scale Y=l'.)
ithin P.P. parallel to it, and
1. Draw a cube, edge 4
4' to right.
2. Draw a cube, edge 3', directly in front, at a distance of
3' from P.P., resting on G.P.
3. Draw a cube, edge 5', parallel to G.P. and 2' above it:
cube 3' to right, parallel to P.P. and touching it.
4. Place two cubes, each 4' edge, on a line parallel to L D.,
cubes to be 4' apart, and nearest "4' from P.P. and 4' to left.
5. Draw a cube, edge 4', touching P.P. and 4' to right; and
place a cube, 2' edge, centrally upon it.
THE PLINTH. 71
6. Draw a plinth 6' x 4' x 2', side 4' x 2' on ground, side
6' X 2' parallel Avith P.P. and 4' to left : figure to be 2' within
P.P.
7. Draw a slab 4' x 2' x 2', lying on ground directly in
front, side 4' x 2' parallel to P.P. and 4' from it.
8. Draw a slab 5'x5'xl' lying flat on G.P., side S'xl'
parallel to P.P. and 3' from it; slab to be 4' to left. Place
centrally on this slab a cube whose edge is 3'.
9. Draw a cube, 2' edge, on each side of L.D., 2' from it,
and touching P.P.; on these cubes place a slab 6' x 2' x 1'
coincident with the cubes.
10. A wall 8' high and 2' thick starts from- a point on the
P.P. 4' to the left, and runs straight forward to the horizon ;
at distances of 6' and 12' doors 5' x 3' are placed.
11. Draw a cross whose beams are 7' x 1' x 1' and 5' x 1' x 1'
respectively; the cross-beam is placed at a height of 3'. The
cross stands erect, its cross-beam parallel to P.P. and 4' within
it ; the foot of the cross is 4' to right.
12. Draw same, with end of cross-beam coincident with
P.P., 4' to left.
13. Draw same, lying on G.P., cross-beam J^ to P.P. and
its end coincident with it, 3' to right.
14. Draw same, lying on ground directly in front, cross-
beam directed away, end of main beam coincident with P.P.
15. A circular table 4' in circumference is supported by
four legs 2 ' high, which proceed from the edge of the table ;
the legs form a square whose side is parallel to P.P. and 3'
within it. The centre of the table is 4' to right. Thickness
of neither table nor legs, taken into account.
16. Draw a set of four steps, each 4' x 1' x 1', ends parallel
to P.P. and 2' within it ; to be 5' to left, facing toward right.
17. Draw same, ends perpendicular to P.P. and 2' within
it, and 3' to right.
1 8. Draw same, with steps descending as they recede ; back
coincident with P.P. and 2' to left.
19. Draw same, directly in front, steps ascending as they
recede, and 2' within P.P.
DRAWINO.
Xhie F*risrrL.
Prisms are square, triangular, hexagonal, etc., according to
their ends or bases.
Tlie square prism may be considered as a mere modification
of the plinth.
To draw a prism, we have only to draw the two surfaces
forming its ends, and join similar angles.
Example 1. — Draw a triangular (equilateral) prism, length
6', side of base 2', lying on G.P., one end perpendicular to
P.P. and 3' to left; prism to touch P.P. H = 6', L.D. = 4',
scale Y =^Y .
r/ip CV
Take B, 1' to left, A, 3' to left, and D, 6' to left of A. On
AB describe equilateral triangle ABC ; draw CL J_ to AB
THE PRISM.
73
and bisecting it. Draw AE _j_ to AB and equal to CL ; simi-
larly draw DF; join F C.V., E C.V. and A C.Y., also B MP
and L MP ; through M draw MG parallel to AE, meeting
E C.V. in G; draw GH parallel to EF; join HD, AG and
GK, completing the prism. Then AK = AB = AC = BC = 2'.
Hence AG - KG = AK, and GM = EA = LC = required height,
and HG = FE = DA = 6', etc. (Fig. 42.)
Example 2. — Draw a hexagonal prism (edge of base 2')
whose length shall be 4', one side touching P.P. 4' to right;
prism to stand on end. H = 6'; L.D.'=4'; scale, \" = 1-
Fig. 43.
74 DRAWING.
Draw the plan in proper situation, as already explained;
then draw the hexagons, one on G.P., the other -4' above it;
then join similar angles in each, forming the required hexagon.
(Fig.-lS.)
THE CYLINDER.
75
Tine Cylinder.
The drawing of the cylinder differs from that of the prism,
only in the plan. Draw the circles, forming the ends, in the
proper positions, and then draw tangents to them, forming
the cylinder.
Example 1. — Draw a cylinder, length 4' and diameter 4',
lying on ground plane parallel to P.P. and 2' within it; the
end of cylinder 4' to left.
Fig. 44.
76
DRAWING.
Example 2. — Draw a cylinder lying on G.P., 4' to right,
having end perpendicular to P.P.. and toucliing it; cylinder
8' long and 3' in diameter.
Take A, 4' to right and E, 4' to left; make AB= H', and
describe circle; join B C.V. and A C.V., also E MP, and from
C, draw CD parallel to AB, and describe smaller circle; then
draw common tangents, completing the cylinder (Fig. 45.)
THE CYLINDER.
77
Example 3. — The figure shows how to draw a common pail,
showing staves and hoops.
Fijf.
78
DRAWING.
Tine Pyramid.
We now come to consider solids, which are not wholly rec-
tangular; they are cones and pyramids and their frusta.
In speaking of the height of a pyramid or cone, we mean
the distance from the vertex perpendicularly to the base.
This is important, especially in frusta, where the slant height
might be mistaken for the real height of the solid.
Example 1. — Draw a pyramid, 8' high with square base,
each side of which is 4', and touches P.P. 4' to right. H = 6',
L.D. = 4', scale \' = '.
Take A, 4' to right and B, 4' to right of A; complete the
square ABDC; draw diagonals intersecting in O; join O with
THE PYRAMID. 79
C. V. and produce it backward to meet base line in E ; at E
erect perpendicular, 8' in heiglit to M; join M C.V.; through
O, draw OX, parallel to EM; join XA, XB, XC, completing
the pyramid. Now, OX = EM = 8', and this represents the
vertical height. (Fig. 47.)
It is not absolutely necessary to join O with C.V. We
may draw it to any point on the H.L., as N or 8., and pro-
duce it backward to F or A, and erect a perpendicular from
either of these points; but it must be carefully remembered,
that the so- found point K or L, must be joined to S or N
respectively. 8uch lines, KS, LN, M C.V., etc., will all pass
through same point X, which may be considered as a locus
for all such lines. For convenience, however, the line O C.V.
should be used, unless the solid be directly in front.
80
DRAWING.
Xhie Cone.
The drawing of the cone does not differ materially from
that of the pyramid. The circle forming the base being
drawn, and the position of the vertex found, it is only neces-
THE CONE. 81
sary to draw the tangents from it to the circle. We give a
particular example : —
Draw a cone whose base = 6' in diameter and slant height 6.
The cone lies on its side; plane of base, i_ to P.P., and the
line joining the centre of base w4th the vertex is parallel to
the P.P. and 4' from it. The cone is 4' to the right. H = 6',
L.D. = 4', scale I" = V.
Take A, 4 to right and C, 6' to right of A; on AC describe
equilateral triangle ADC; draw DB perpendicular to AC;
join A C.Y., B C.V., C C.Y.; take T and E, 3' from O; join
E MP, O MP and Y MP; through F, G, H draw parallels to
AB; through S, R, K draw parallels to DB; join PE, NG,
MH; then in square PH descHbe circle; produce GR to L,
and from L draw tangents LX, LX to circle, completing the
cone. (Eig. 48.)
Exercise X. — On the Prism and the Cylinder
(H = 6', L.D. = 4', scale 1"=1'.) .
1. Draw a prism 6' in length, triangular base, each side of
which is 2'. The prism stands on end 4' to riglit, w4th one
side coincident with P.P.
2. Draw same, 3' to left, one side perpendicular to P.P.
3. Draw same, directly in front, one side parallel to P.P.
and 2' from it, vertex away.
4. Draw same, lying on ground plane, perpendicular to P.P.,
3' to right and 3' within P.P.
5. Draw a hexagonal prism 6' high, each side of base 2',
standing on end directly in front, one side touching P.P.
6. Draw same, lying on ground parallel to P.P. and 2' within
it; one end projecting 2' to right, and opposite end 4' to left.
7. Draw same, lying on ground perpendicular to P.P., 3' to
right and 3' within P.P.
8. Draw a cylinder, diameter of base 4', height 6', lying on
ground, parallel to P.P. and 4' within it; left end just in line
with L.D.
82 DRAWING.
9. Draw a cylindrical vessel 4' feet in height and 4' in
diameter, standing on end, touching P.P. 4' to left; show 4
hoops at distances of 1 ' from each other.
10. Draw a hollow cylinder, 4' in length, outer diameter 4',
inner diameter 3', lying on ground perpendicular to P.P., 2' to
right and touchincr P.P.
Exercise XI. —On the Puramid and Cone.
(H = 6', L.D. = 4', scale J" = 1'. )
1. Draw a square pyramid 5' high, each side of base 3',
standing on ground, directly in front, touching P.P.
2. Draw same, 4' to right, parallel to P.P. and 4' within it.
3. Draw same, standing on a 3' cube, parallel to P.P. and
2' within it, 4' to left.
4. Draw same, 3' above ground plane and parallel with it,
4' to right and touching P.P.
5. Draw same, with vertex downwards, base parallel with
ground and P.P., vertex, 4' to left and 3' within P.P.
6. Draw a cone, height 5', diameter of base 4', standing on
ground 4' to right and 4' within P.P.
7. Draw same, touching P.P., 3' to left.
8. Draw same, directly in front, 3' from P.P.
9. Draw same, standing on a cylinder 4' in diameter and ^
high, 3' to right and touching P.P.
10. Draw same, placed centrally on a cylinder of 5' in diam-
eter, directly in front and touching P.P.
11. Draw a cone, base 4', slant height 4', lying on side;
base perpendicular to P.P. and touching it, 4' to left; vertex
directed toward left.
FRUWA.
cS3
Kru-Sta.,
The dimensions of a frustum may be given by stating dimen-
sions of each end, and vertical height.
84
DRAWING.
Example 1. — Draw the frustum of a square pyramid, whose
bases are 6' and 4' square, respectively ; the frustum touches
P.P. 4' to right, height 4'. (Fig. 49.)
From the above the method of drawing may be easily
understood.
Example 2. — A pyramid with square base, each side of
which is 4', stands on the ground plane 4' to the left, touch-
ing P.P. The pyramid is 8' in height ; 3' from the vertex
the pyramid passes through a square plinth 4' x 4' x 1' placed
parallel to the ground plane! The pyramid cuts the plinth
centrally. H = 6', L.D. = 4', scale \' = 1'. (Fig. 50.)
Fig. 50.
By a careful observation of the lines drawn above, the
method may be easily seen.
FRUSTA. 85
Exercise XII.
(H. = 6', L.D. = 4', scale i" = r.)
1. A pyramidal frustum with square base, and vertical
height 4', touches P.P. 4' to right, resting on ground plane ;
the sides of the square are 3' and 5' respectively.
2. Draw a pyramidal frustum same as No. 1, 3' to left, and
3' within P.P.
3. Draw a triangular frustum (equilateral), sides of base
5' and 3' respectively, height 4' ; on G.P. 4' to right and 4'
within P.P., vertex away, one edge parallel to P.P.
4. Draw a square pyramidal frustum, height 4', sides of
square 2' and 4 respectively, standing on G.P. reversed,
directly in front, 1 ' within P. P.
5. Draw a conical frustum, height 5', diameters 5' and 3'
respectively, on G.P., and touching P.P. 4' to right.
6. Draw No. 5, 6' to left and 3' within P.P.
7. Draw same, directly in front, 3' within P.P. and 3'
above G.P.
8. Draw a hexagonal frustum, height 5', sides of bases 3'
and 2' respectively, touching P.P., resting on G.P. 4' to right.
9. Draw same, 4' to left and 4' within P.P.
10. Draw an octagonal frustum, height 5', edges of bases
2' and 1' respectively, resting on G.P., and touching P.P.
4' to left.
S()
DRAWING.
Fig. 51 shows a method of laying out a plan for a frustum
of a cone.
FRUSTA. 87
Exercise XIII.
(H = 6', L.D. = 4', scale i" = r^)
1. 4' to right and 2' within P.P. draw a frustum of a
square pyramid, edges of squares 4' and 2' respectively, lieicrht
4'.
2. Draw same, touching P.P. 2' to left.
3. Drav,- same, directly in front, parallel to G.P. and 3'
above it.
4. Draw a conical frustum, height 5', diameters 3' and 2'
respectively ; frustum rests on ground, with centre of base
3' to right and 3' within P.P.
5. Draw a frustum of a triangular. pyramid, edges of ends
3' and 2' respectively, height 4'; it rests on G.P. with one
edge coincident with P.P., and 3' to left.
6. Draw a frustum of a hexagonal pHsm, edges of bases
3' and 2' respectively, height 4' ; one edge of frustum is per-
pendicular to P.P., and an angle touches it at a point, 3' to
the right.
7. A cone, whose height is 8' and diameter of base 4',
touches the P.P. 4' to left; it is encircled by a rectangular
collar whose dimensions are 4' x 4' x 1', placed centrally over
it, 4' above the ground. The cone rests on the ground.
A frustum of a square pyramid, whose bases are 5' and
3' respectively, and whose height is 4', supports a cone placed
centrally upon it ; the diameter of the cone is 3' and its height
3'. The edge of the base touches the P.P. 2' to right.
88
DRAWING.
TThie Sphere.
The perspective of the sphere must necessarily be repre-
sented by a true circle, and little difficulty will be experienced
in drawing a complete sphere. However, w^hen a hemisphere
K
THE SPHEilE.
89
is to be represented, an apparent fallacy appears, owing to
the representation of the circle that shows the section of the
sphere. A sphere must always be supposed to be drawn with
the radius of the circle as distance. However, as the per-
spective of the circle, \dewed in any oblique position, shows
diameters of varying length, care must be used in drawing the
curve of the hemisphere at the greatest apparent diameter,
and this diameter cannot be definitely determined in perspec-
tive, if drawn in any but a direct view. The sphere rests on the
ground at a point directly beneath the centre, and it touches
P.P. at a' point perpendicular to the vertical, from the centre.
Example 1. — Draw a sphere, radius 2', resting on ground
at a point 3' to right and 2' within P.P. H = 6', L.D. = 4',
scale J" = 1 '.
Here FA = 3', BA = AD = 4' and AC = ED = 2'. (Fig. 52.)
Example 2. — On centre of the top of a cube of 4' edge, placed
4' to left and touching P.P., place a sphere of radius 1 J'.
90
DRAWING.
Here CA = AB = AD = AE = 4', EF = FM = 2', CtF=11'
hence K is centre at intersection of diagonals and HK = GF.
(Fig. 53.)
Note. — The sphere ^vill not touch the P.P. unless HK =
KF.
Example S. — Draw a cube, edge 4', touching P.P. 4' to
right, and in this place a sphere whose radius = 2'. Here the
sphere ^^'ill touch the centre of each side.
Fig. 54.
Draw cube, and diagonals of those sides, perpendicular to
P.P., join intersections, and bisect this horizontal line as
sho^\^l. The circle drawn on this line as diameter will repre-
sent the sphere, and touch the centre of each side. (Fig. 54.)
Exercise XIY.
(H = 6', L.D. = 4', scale \" = 1.)
1. Draw a sphere, radius 2', placed centrally on a cylinder
(on end), touching P.P. 4' to left, cylinder 3' high and diame-
ter 4',
THE SPHERK. 91
2. Draw a sphere, cliameter 3', 8' high, 6' to right and 6'
within P.P.
3. Draw a sphere, diameter 4', directly in front, touching
P.P.
4:. Draw a .sphere, radius 2', buried completely beneath the
ground, centre of sphere 6' to right and 6' within P.P.
5. Place a sphere in a cubical box of 4' edge, diameter of
sphere 4'; cube to be 4' to left, 4' within P.P., parallel to, and
2' above G.P.
6. A cylinder whose height is 4' and diameter 3' stands on
end, touching P.P. 3' to right ; this cylinder passes centrally
through. a sphere whose diameter is '4' and whose centre coin-
cides with that of the cylinder.
92
DRAWING.
KoreshLorten.ing.
Foreshortening consists in re-
presenting the apparent length of
a visible object. It depends on
the distance of the object, and its
position with regard to the eye.
Thus, a lead-pencil may be so
turned as to show only the end,
or it may be placed so as to show
its whole, or greatest length.
Again, if AB represent a line of
definite length, and O, the ob-
server, the apparent length of
AB as seen from O will be AD.
(Fig. 55.)
This representation of a line
AB by AD, which is always less
than AB, is called " foreshorten-
ing."
Note. — AD is always perpen-
dicular to the longest side OB.
SYNTHETIC PERSPECTIVE.
93
Taking an object " out of '"' perspective means, that when an
object is drawn, and the position of the observer's eye given,
the size and position of the object may be determined.
Fig. 56.
Here, the figure BX only, would be given, and it would be
assumed to touch the picture plane. We first produce the
vanishing lines DX and AC to meet in C.V., then draw
C.V, S perpendicular to B.L., and make it equal to height of
spectator; then draw H.L. parallel to B L. through C.V.;
next take a point MP at a distance to the left equal to the
height of spectator and his distance away, combined ; then
join MP with C, and produce it to base at E. Then, scale
being given, find BA, AD and AC, the dimensions of the solid,
and AS will show its distance to the left. (Fig. 56.)
94
DRAWING.
F*erspecti\"e Effect.
This consists in showing merely the appearance of an object
when placed in a certain position. The dimensions and dis-
tance of the object are not taken into account.
PEESPECTIVE EFFECT. 95
It will be remembered that if an object of less height than
the observer, be placed on the ground, the observer will be
able to see the upper side of it, and if placed above him he
will see the under side ; if placed on his right, he will see
the front and left sides ; if placed directly in front, he will
see front side and upper or lower sides, according to the
height of the object.
Take a cube, for instance, placed on
the ground parallel with the P.P.
Now, if the cube be lower than the
observer's eye, the upper face will be
\'isible ; if raised above, the lower face,
and so on.
Figure 57 will illustrate perspective
efiect.
(1) Shows object above and left of eye.
(2) Above and directly in front of eye.
(3) Above and to right of eye.
(4) Level with and to left of eye.
(5) Level with and directly in front.
(6) Level with and to right.
(7) Below and to left.
(8) Below and directly in front.
(9) Below and to right.
If an object, as for instance a cube, is
to be drawn, say to right and above the
eye, draw first a square, then take a
point to left and below, draw the van-
ishing lines to this point, and mark off
CT ^ lines for thickness, etc.
crzDi
96 DRAWING
ANGULAE PEESPECTIYE.
We now come to consider the rules pertaining to angular
perspective, or the perspective of two vanishing points. If a
rectangular object, as a cube, rests on the ground parallel to
P.P., it is evident that its sides, if produced, will appear to
vanish directly in front, at the point called the centre of
vision. If we move the cube by even a small amount
from the parallel position, its sides will no longer vanish at
the centre of vision, but at a point to the right or left of it,
and at a distance from it, depending on the angle which the
sides make with the P.P. Now, in parallel perspective we
deal with only one vanishing point — the centre of vision; but
there are really two: for all lines parallel to the P.P., if pro-
duced to an infinitely great distance, will appear to meet at a
point to right or left. Hence in parallel perspective only one
vanishing point is of practical utility. However, when the
cube is moved out of its parallel position, this apparently-
hidden vanishing point appears, and strikes the horizon at a
distance from the centre of vision, depending on its angle, as
already explained. Thus : —
In Fig. 59, AB on left side shows the base of a cube in
parallel perspective, while in Fig. 60 AB has been moved
around to position of AD, and AC will not now vanish to
C.V., but to a point Y to right of it; so also AD will not
vanish at a point parallel to AB, but at X, a point in the
Fig. 60.
horizontal line to left of C.V. We will now proceed to
ascertain the positions of these points.
Example 1. — Draw a square (side 4') lying on ground; sides
make an angle of 45° with picture plane, and the angle
touches the P.P. at a point 4' to the right; scale ^ = ^ •
DRAWING.
ANGULAK PERSPECTIVE. 99
Here draw H.L. and B.L. as before, and take P.S. at given
distance; then draw a straight line through P.S. parallel to
B.L. and on each side of L.D. ; lay off the sides at required
angle (in this case 45^); produce these lines till they meet
the horizon in R V.P. and LV.P. (the vanishing points) ; with
L y.P. as a centre, and P.S. as distance, describe an arc to
cut H.L. in R MP.; similarly find L MP. These are called
measuring points. Now take A, 4' to right and draw A L V.P.
and A R V.P.; take 4' on each side of A, namely, B and C,
and draw B R MP and C L MP to cut vanishing lines in E
and D. Draw D L V.P. and E R V.P. to cut in F. Then
ADFE will be the square required. For AD and EF are
parallel to P.S. R V.P. placed at given angle, and AE and
DF are parallel to P.S. L V.P. also placed at given angle;
and DF = AE = AB = 4', and EF = AD = AC = 4'. (Fig. 61.)
Example 2. — Draw a cube, edges 4', right face at an angle
of 60° and left face at an angle of 30° with P.P. H = 6',
L.D. = 4', scale \" =1'. Cube to have an angle 4' to left and
2' within P.P.
Draw H.L., B.L. and 00 as before; find also C.V., R V.P.,
L V.P., R MP, L MP and MP (parallel perspective), as already
shown. Take A, 4' to left; join A C.V.; erect AK = 4'; join
K C.V.; join also B M.P.; through E draw EL parallel to
AK; join E RMP and E LMP, and produce them backwards
to meet B.L. in Z and X. Mark off XC = 4', also ZD = 4';
join D RMP and C LMP, also L R V.P. and LLV.P.;
through F draw FM, and through G draw GH, parallel to
EL; join M L V.P. and H R V.P. to meet at N, completing
the required cube. For FM = EL = HG = AK = 4', and EG =
EF = BC = AD = 4' ; so also MN = HL = GE, etc. (Fig. 62.)
Note. —The point E must always be determined by parallel
perspective; lience necessity for finding MP.
100
DRAWING.
MISCELLANEOUS EXERCISES. .101
Exercise XV. — Figures in Angular FerspecfAve.
(In the following consider H = 6', L.D. = 4', scale I" = 1'.)
1. A square whose sides are 4', lies on ground; an angle
touches P.P. 2' to right; angle 45''.
2. Draw same, 3' to left and 2' within P.P.; right side makes
angle of 60°.
3. Draw same, touching P.P. directly in front; angle 45"^.
4. Draw a cube, of 4' edge, touching P.P. at a point 2' to
right; angle 45°.
5. Draw same, 4' to left and 1' within P.P.; angle 45°,
6. Draw a square pyramid, edge of base 4', height 8', angle
45°; touches P.P. 4' to left.
7. Draw same, 4' to right and 2' within P.P.; angle 45°.
8. Draw same, 2' to left and 2' within P.P.; left and right
angles, 30° and 60° respectively.
9. Draw a triangular prism, each edge of base 3' and 5'
long; standing on end, angle touching P.P. 4' to left; angles
60° and 60°.
10. Draw a square pyramid, edge of base 3', placed cen-
trally on a cube of 4' edge; angle 45°; touches P.P. 3' to left.
Height of pyramid 4'.
11. Show perspective effect (angular) of a pyramid to left
and above the eye.
12. Show angular perspective effect of a square pyramid
placed centrally over a cube of smaller base, to right and below
eye.
MiSCELLANKOUS ExERCISES.
(Unless otherwise stated, consider H = 6', L.D. =4', and
scale |" = r.)
1. Two circles, whose diameters are 4', and intersect at
right angles, having their common diameter perpendicular to
ground and touching it at a point 4' to left and 4' within P.P.
2. Draw an equilateral triangle, lying on ground plane, side
3', vertex directed away, one side parallel to P.P. ; vertex 4'
to right and 4' within P.P.
3. Draw a hexagon, each side 2', standing on ground, _[_ to
P.P., one side touching it 4' to left.
102 DRAWING.
4. Draw a circle, diameter 4', touching G.P. and P.P.. and
perpendicular to both, 4' to right.
5. A rod is placed obliquely in the ground, and its outside
length is 5' ; it makes an angle of 30° with the ground and
60° with the P.P. The rod descends toward the left, and
lower point is 6' within P.P. and 6' to right. Draw it.
6. Draw an octagon, side 2', lying on ground, touching
P.P. 4' to left.
7. Within a circle, diameter 4', lying on ground plane, 4' to
right and 4' within, describe a square whose side shall be
parallel to P.P.
8. Draw a triangular prism, length 6', edges 2', parallel to
P.P. and 3' within it, one end 4' to right, other 2' to left.
9. Draw a cone, diameter 4', height 4', standing on ground
plane, touching P.P. directly in front.
10. Draw a pyramidal frustum (square), edges 2' and 4',
height 4', touching P.P. 4' to left.
11. Draw same, in angular perspective, angle 45°, 4' to left,
touching P.P.
12. Draw a sphere, diameter 4', half buried in ground, 6'
to right and 6' within P.P.
13. Draw a hemisphere, plane directed towards right and
perpendicular to P.P. and G.P.. touching each : hemisphere to
be 4' diameter and 4' to left.
14. Draw a cylinder on end, diameter 4', height 4', touch-
ing P.P. 4' to right, and on this place a hemisphere centrally,
4' diameter, convex surface upward.
15. Draw a sphere touching sides of a cubical box of 4' edge,
box on ground parallel to P.P., 4' to left and 4' within P.P.
16. Draw a pyramid, base 4' square, 4' high, 4' to right
and 3' within P.P.
17. Draw an equilateral triangle, sides 3', in angular per-
spective; angle 60°; 4' to left, 3' within P.P., on G.P.
18. A square, sides 4', stands on ground plane perpendicular
to it, making an angle of 45° with P.P. and 2' from it at
nearest lower point ; it is 3' to left.
19. Draw a triangular pyramid, height 8', each side of base
4', presenting an angle of 60° to the P.P. 4' to left.
20. A square prism, length 4', edge 2', stands on end, an
angle touches P.P. directly in front; sides at 45°. This prism
MISCELLANEOUS EXERCISES. 103
supports a pyramid placed evenly upon it, uf equal Ijase and
4' in height.
21. Draw a plinth 6' x 4' x 2', side 6x4' on ground, placed
at angles of 60"^ and 30^, 4' to left, touching P.P.
22. Draw an ordinary Roman cross, beams 6' and 4' in
length, and 1' square at ends, at an angle of 45"^, 4' to right
and 4' within P.P.
23. Place a cube of 4' edge on top of a cylinder (on end), of
4' diameter, centres coincident ; cylinder touches P.P. directly
in front.
24. Draw middle zone of a sphere whose radius is 4', height
of zone 2', plane parallel to ground plane, centre of zone 4' to
right, 4' within P.P. and 4' above it.' H = 6', L.D. = 6', scale
i" = r.
25. Draw four pyramids, each in contact at bases, 4' square
and 4' in height, standing on ground plane at an angle of 45°
with P.P., 4' to left and 2' within P.P. H = 6', L.D. = 6',
scale \" = r.
26. Draw a frustum of a cone, height 5', diameters 3' and
2' respectively, touching P.P. 4' to right.
27. A cone whose slant height is 6' and diameter of base.
6', rests on ground plane, slant touching gcound, parallel to
P.P. and 6' within it, vertex directed towards left ; cone to
be 4' to left.
28. A cube of 4' edge contains a cylinder of equal diameter
and height ; the cube makes an angle of 45° with P.P., and
touches P.P. 4' to right. The cylinder is vertical.
29. A pyramid, whose base is 4' square and whose height is
6', presents an angle to the P.P. 4' to left, the inclination of
the sides being 45°. This pyramid passes centrally through a
plinth 4' X 4' X 1', placed horizontally upon it at a height of 3'.
30. A cube, whose edges are 4', is su.spended from an angle
so as to just touch the ground directly in front, while another
angle touches the P.P. directly in front.
31. Draw a stove-pipe elbow, diameter of ends 6", length of
each half (outside) 1'. The elbow rests on the ground, one
end touching P.P. 2' to left, the other bending towards the
right and parallel to P.P. H = 6', L.D. = 4', scale 1 " = 1 '.
32. Show perspective effect of a pipe lying on ground paral-
lel to P.P. and to left.
104 DRAWING.
33. Show same, standing on end to right and bolo^v eye.
34. Show angular perspective effect of a square pyramid to
riglit, and below eye.
35. Show perspective effect of a cylinder on end, to right
and below.
36. Show perspective effect of a pail with three lioops, below
and dh-ectly in front.
37. Show perspective effect of a water pitcher, to left and
below; lip to right.
38. Show perspective effect of a chair, straight back, directed
away, angle to right and below the eye.
39. Show hollow pipe lying on ground, perpendicular to
P.P., to right.
40. Show perspective effect of an ink bottle (conical).
41. Show perspective effect of an ink bottle in form of
pyramidal frustum, angle to left, below the eye.
42. Show perspective effect of a plinth to left and below,
angular.
43. Show perspective effect of a teacup below the eye.
44. Show i^erspective effect of a sphere placed centrally on
a cube, directly in front, below the eye.
45. Show angular perspective effect of a table below and to
right.
46. Show perspective effect of a triangular prism on end,
one side perpendicular to P.P., below and to right.
47. Show perspective effect of a reversed cone, below the
eye.
48. Show perspective effect of a hexagon on one side, per-
pendicular to ground, to right.
49. Show perspective effect of a hollow conic frustum lying
on ground parallel with P.P., larger end to right, smaller end
to left.
50. Show perspective effect of a door in three different
positions, revolving round an axis through the hinges :
(a) When shut, parallel to P.P.
(b) When opened at an angle of 45° with P.P.
(c) When opened perpendicularly to P.P.
GEOMETKICAL DRAWING.
105
GEOMETKICAL DBA WING.
To construct the following figures, pupils should provide
themselves with a pair of good compasses with pen attach-
ment, and a ruler, with marks for inches and fractions of an
inch. No proof is necessary, but it will be well to investi-
gate the methods as far as possible, many of which are but
modifications .of the Euclidian.
No. 1. — To draw a perpendicular to a given line (a) from a
point on the line.
Fig. 64.
Fig. 63.
Let AB be the given line, and let B be a point at which
the perpendicular to AB is to be drawn. In first, take any
point C above, and with distance CB describe circle, cutting
AB in E; join EC and produce to meet circumference in D;
join DB, which will be the perpendicular required. In second,
take any point C above, and with B as centre and BC as dis-
tance describe arc, cutting AB in A; then with C as centre
and CB as distance describe arc, cutting former arc in E;
with E as centre and same distance describe arc, cutting in D \
8
106
DRAWING.
join DB, which will be perpendicular to AB, at B. (Figs. 63
and 64.)
(b) From a point above or below AB.
Fig. 65.
Let AB be the given line and C given point above; with
centre C describe an arc to cut AB in D and E; with centre
D and distance greater than half of DE describe an arc; with
centre E and same distance describe an arc to cut former arc in
F; join CF, which will cut AB at right angles at G. (Fig, 65.)
iVo. 2. — To describe a square (a) on a given line.
Fig. 66.
In first, let AB be the given line; erect at A a perpendicular
and make it equal to AB; with centre C and distance CA
describe arc AD; with centre B and distance BA describe
GEOMETRICAL DRAWING.
107
arc to meet former arc in D; join CD and BD, completing
the square. (Fig. 66.)
(b) On a given diagonal AB.
With centre A and distance greater than half of AB, de-
scribe arc CD, and with centre B and same distance describe
an arc to cut former arc in C and D; join CD and produce
both ways; with centre E and distance EA or EB describe a
circle cutting diameter in F and G; join FA, FB, GA and
GB, completing the square. (Fig. 67.)
iVo. 3. — To construct an oblong of given dimensions.
Fig. 68.
Let AB represent the greater side; erect AC perpendicular
to AB at A, and with centre C and distance equal to AB
describe an arc ; with centre B and distance equal to AC
describe an arc cutting former arc in D; join DC and DB,
completing the required oblong. (Fig. 68.)
iVo. 4- — To divide a given line into (a) two equal parts.
108
DRAWING.
Let AB be the given line ; with centre A and distance equal
to more than half of AB describe an arc, and with centre B
and same distance describe an arc cutting former in C and D ;
join CD, cutting AB in E into two equal parts. (Fig. 69.)
(b) Into any number of equal parts.
Note. — Before this can be done it is necessary to show how
to draw a line parallel to another from an external point.
Cx ~~^ D
Fig. 70.
Let AB be the given line and C the external point; at any
point B and distance BC describe an arc to cut AB; with C
as centre and CB as distance describe an arc, and with B as
centre and distance equal to AC describe an arc cutting
former arc in D ; join CD, which will be parallel to AB.
(Fig. 70.)
(b) To divide a line into any number of equal parts.
C
Fig. 71.
Let AB be the given line ; draw DE parallel to AB on
either side, and on this line set off the required number of
GEOMETRICAL DRAWING.
109
equal distances (in this case five); then join each point of
section with A, which will divide AB into the same number
of equal parts. (Fig. 71.)
(c) To divide a given line proportionally to another line.
Let it be required to so cut AB, that the smaller part
shall be to the greater, as the greater is to the whole line.
Let any line CD not equal to AB, be cut in G, so that
CG:GD::GD:CD; place CD parallel to AB, and join CA
and DB; produce them to meet in E; join EG to cut AB in
F; then will AF : FB : : FB : AB. (Fig. 72.)
Ro. 5. — To construct a triangle of given dimensions.
Fig. 7:
Let AB, CD and EF be the given sides, no two of which,
taken together, are equal to, or less than, the third. Take
one of them AB, and with centre B, and distance equal
110
DRAWING.
to CD describe an arc; with centre A and distance equal to
EF describe an arc cutting former arc in G; join GA and
GB, completing the triangle. (Fig. 73.)
No. 6. — To bisect a cjiven ano^le.
Fig. 74.
Let BAG be a given angle ; with centre A and any distance
AB describe an arc BC ; with centre B and any distance less
than half of BC describe an arc; with centre C, and same dis-
tance, describe an arc to cut former arc in D; join AD, which
will bisect the angle. (Fig. 74.)
No, 7. — To trisect a right angle.
Fig. 75.
Let ABC be the given right angle; with B as centre and
at any distance BA describe the quadrant AC; with centre A
and distance equal to AB describe an arc to cut AC in E; and
with centre C, and distance equal to CB or BA describe an
GEOMETRICAL DRAWING.
Ill
arc to cut arc in D. Then D, E will be points of trisection,
and lines from D and E to B will trisect the angle. (Fig. 75.)
iVo. 8. — To inscribe a circle in a oriven triangle.
Let ABC be the given triangle; bisect the angle at A by
AD, and the angle at B by BD, cutting AD in D; draw DE
perpendicular to AB at point E ; then with centre D and
distance DE describe the circle. (Eig. 76.)
iVb. 9. — To draw a circle through three given points, which,
however, cannot be in the same straidit line.
Fig. 77.
112
DRAWING.
Let A, -B, C be the given points; join them to form a tri-
angle; with centre A, and distance greater than half of AC,
describe an arc; with centre C and same distance, describe an
arc to cut former; join points of section of arcs; this line will
pass through the centre of the circle; draw similar arcs on
AB, and join points of section to meet in D, the centre; then
a circle drawn with centre D, and distance DA, will pass
through A, B and C respectively. (Fig. 77.)
No. 10. — To find the centre of a whole or part of a circle.
Let BACD be a circle or arc; draw any two chords AB,
CD, and draw arcs EF and G-H, bisecting the chords, respec-
tively; join EF and HG and produce them to meet in K;
then K will be the centre, and if K and D be joined, KD
will be a radius, and a circle may be thus described with it
with the centre thus found. (Fig. 78.)
GEOMETRICAL DRAWING.
113
No. 11. — To draw a tangent to a given circle («) from a
point in the circumference.
Fig. 79.
Let C be a given point in the circumference; let A be the
centre; join CA; at C erect the perpendicular CE, which will
be tangent required. Also if H be the given point, join AH
and produce it, making HG equal to part produced ; bisect
this line by KL, which will also be a tangent at point H. If
AC be produced, the line FC, perpendicular to the tangent at
the point of contact C, is called a "normal." (Fig. 79.)
(6) To draw a tangent to a circle from an external point.
Join point with centre, and at point where it cuts the circum-
ference draw a perpendicular upon it ; thus, if D be given
point, HL will be a tangent.
114
DRAWING.
No. 12. — To construct an isosceles triangle of a given
altitude.
Let AD be the given altitude ; at A and D draw perpen-
diculars, and with A as centre and any distance describe an
arc EF : \vith D as centre and any distance less than DA
describe an arc to cut arc in E and F ; join AE and AF, and
produce them to base, forming isosceles triangle ABC. (Fig.
80.)
No. 13. — To construct an equilateral triangle of a given
altitude.
E A F
Let AD be the given altitude ; through A, draw EAF per-
GEOMETRICAL DRAWING. 115
pendicular to AD ; with centre A describe any semicircle
EGHF, and with centre F and distance FA, describe arc to
cut arc GH in H ; also with centre E and same distance de-
scribe arc to cut GH in G. Join AG and AH, and produce
them to meet BC in B and C : then ABC will be the equi-
lateral triangle required. (Fig. 81.)
No. IJf.. — {a) To draw, from a given point in a straight line,
an angle equal to a given angle.
Let BAG be given angle and let the given point be at D ;
with centre A and any distance less than a side describe an
arc CB ; with centre D and distance equal to AB describe arc
EF ; with centre E and distance equal to BC describe an arc
to cut EF in F ; join DF : then the angle EDF will be equal
to the angle BAG. (Fig. 82.)
(6) Within a given circle to construct a triangle similar to
another triangle. (Triangles are similar when the angles in
one are equal to the angles in the other, each to each. They
are similarly situated when the sides of one are parallel to the
sides of the other, each to each. These are called homologous
sides.)
Let ABC be the given circle and LHK, the given triangle.
At any point C, draw a tangent DE, and describe a semicircle
DE ; with centre H, and any distance HM, describe an arc,
and with centre K and same distance describe an arc. Make
EG = NO and DF = MO ; join CF and CG, and produce them
116
DRAWING.
to the circumference in A and B ; join AB, completing the
triangle required. (Fig. 83.)
Fig. 83.
No. 15. — To construct an equilateral triangle about a given
circle.
Fig. 84.
GEOMETRICAL DRAWING.
117
Let BCD be given circle ; at any point A in the circumfer-
ence, with distance equal to radius, describe an arc, cutting
circle in B and C ; with centre C, and same distance describe
an arc, cutting circle in D; describe similar arcs with centres
B and J); these arcs intersect in points E, F and G; join
these, completing the triangle required. (Fig. 84.)
iVo. 16. — About a given .circle to construct a triangle simi-
lar to a given triangle.
Fig. 86.
Let DEF be the given circle and TON the given triangle
find the centre G, and draw any radius GD ; at D, draw i
118
DRAWING.
tangent to the circle. With centre N and any distance MN,
describe an arc MS, and with centre O, and same distance
describe an arc RP ; with centre G, and distance equal to
MN or OP, describe a circle cutting GD in H. Make arc
HK = RP and HL = MS ; join GK and GL, and produce them
to the circumference in E and F respectively. Draw tangents
at E and F to meet the other tangent in A and C ; then
triangle ABC will be similar to TON. (Figs. 85 and 86.)
No. 17. — Within a circle to draw any number of equal
smaller circles, each touching two others and the outer circle.
Let KME be the given circle, and divide it (in this case)
into six equal parts. Take centre O, and join any two, as OA,
OB ; bisect the angle AOB by OE, and at E draw a tangent
CD ; produce OA and OB to meet the tangent in C and D.
GEOMETRICAL DRAWING. 119
Bisect the angles at C and D by CF and DF, meeting at F ;
then with centre O and distance OF describe a circle ; also
with centre F and distance FE describe a circle : this will be
one, and the remaining iive may be similarly drawn. (Fig. 87.)
No. 18. — To construct a regular polygon, (r^) on a given line.
D A B C
Fig. 88.
Let AB be the given line, produce it both ways; then with,
centre A and distance AB, describe the semicircle DEB,
describe also a similar semicircle AFC. Divide the circumfer-
ence DEB into as many equal parts as the polygon is to have
sides (in tliis case live), and join A with the second point of
division; make the arc rC = DE; join BF and with centres
E and F and distance EA and FB describe arcs to intersect
at G ; join GE and GF, 'completing the polygon.
Note. — This method will be clear if it be remembered that,
if from a point within a polygon straight lines be drawn to
the angles, the figure will be divided into as many triangles
as it has sides, and each triangle will contain two right
angles, but the angles around the common point within, to-
gether make four right angles. Then if N represent the
number of sides, the number of degrees in the angle of a
, . .... 90(2n-4) . ^ . 180 (n- 2) ^,
regular polygon will be - — ^^ ^, that is, ^ '. Now,
n n
120
DRAWING.
in the above figure the line DB may be called 180"^, or two
right angles. Then the angle EAB will be represented by
5-2
— —"j or f of 180"; hence it is ahcays necessary to draw
through the second point of division.
(5) In a given circle.
Let ABE be the given circle; draw any diameter FC, and
di%dde it into as many equal parts as the figure is to have
sides (in this case five) ; with centre C and distance CF
describe arc FG, and describe similar arc CG, intersecting at
G. Draw GA from A, through second point of division ; join
FA, and continue this around the circumference, completing
the polygon. (Fig. 89.)
No. 19. — To construct a regular pentagon on a given line
by a special method.
Let AB be the given line ; describe arcs CAD and CBD,
with radius AB ; join CD : with centre D and distance same
as AB describe arc EABF, cutting former arc in E and F.
GEOMETRICAL DRAWING.
121
Join FG and EG, and produce them to meet arcs in H, K;
join AH and BK ; with centre H and distance HA describe
arc; and with centre K and distance KB, describe arc cutting
former arc in L. Join LH and LK, completing the penta-
gon. (Fig. 90.)
No. 20. — To construct a regular hexagon on a given straight
line.
Fig. 91.
122
DRAWING.
Let AB be the given line ; with centre A and distance AB
describe arc ; with centre B and distance BA, describe arc
cutting at C; join CA, CB. With centre C and distance CA,
describe circle cutting in D, B and A ; join DC and produce
to E ; produce AC to G and BC to F ; join AE, EF, FG,
GD and DB, completing the hexagon.
N'o. 21. — To construct a regular octagon («) on a given
straight line.
5
Hy
E
^
\
\
/
/
h
F
^
\
\
i
/
M
A B
Fig. 92.
Let AB be the given line ; produce it both ways, and
describe the semicircles CKB and AMD. Erect a perpen-
dicular at A and at B ; bisect the right angles CAG and DBH
by AK and BM, respectively, and erect perpendiculars KE
and ]MF, at K and ^I, respectively. With centre K and dis-
tance KA describe arc EA cutting KE in E; draw similar
arc BF cutting MF in F; with centre E and distance EK
describe arc to cut AG in G ; similarly find H ; join EG, GH
and HF, completing the octagon. (Fig. 92.)
(b) In a given square.
Let BCDE be the given square; draw diagonals intersecting
at A. With centre B, and distance. B A, describe arc cutting
sides of square in H and P ; similarly find points K, iS", F, M
GEOMETRICAL DRAWING.
123
and G, L; join FG, PN, ML and HK, completing the octagon.
(Fig. 93.)
N M
Fig. 93.
No. 22. — To draw a perfect ellipse by means of the foci
and intersecting arcs, axes being given.
Let the axes, AB and CD, be placed centrally, at right
angles to each other ; then measure from C the distance from
A to D, and describe arc cutting AB in F and F : these are
the foci. Between O and F, take any number of points, 1, 2,
etc. — the more the better ; then with centre F (left) and dis-
124
DRAWING.
tance equal to distance from 1 to B, describe arcs E, E ; and
with centre F (right) and same distance describe arcs E, E,
Then with centre F (left) and distance equal to that from 1
to A, describe arcs cutting the former, so also describe arcs
from centre F (right). Thus for each point between O and F,
we get four points. Having thus found a number of points,
join them, or rather draw a curve through them : this curve
will be an ellipse. (Fig. 9-i.)
No. 23. — To draw an ellipse (a) by means of concentric
circles and intersecting perpendiculars.
Let the concentric circles EF and AB be drawn; draw
diameters AB and CD at right angles to each other, divide
each quadrant into the same number of equal parts, and join
opposite points ; draw perpendiculars from the outer points
and horizontals from the inner points to meet them ; thus
draw perpendicular from H and horizontal from a to meet
in 1 ; similarly tind 2, 3, 4, etc., all around the circle. Draw
a curve through the points of intersection thus found, which
will form an ellipse. (Fig. 95.)
GEOMETRICAL DRAWING. 125
(b) When the major axis (transverse diameter) only is given.
£■
Let AB be the given diameter; dix-ide it into four equal
parts in G, M, D ; with centre D and distance DG describe
arc EF, and with centre G and same distance describe arcs to
intersect in E and F. Join FG, FD, EG and ED, and produce
them to the circumference in C, H, K and L respectively.
Then with centre F and distance FC describe arc CH, and
with centre E and distance EK describe arc KL, completing
the elliptical curve. (Fig. 96.)
Note. — No part of a true ellipse is an arc of a circle.
iVo. 24- — --^n ellipse being given, to find axis and foci.
Draw any two parallel chords AB and CD; bisect each
and join points of section FE, and produce each way to
meet the circumference in G and H; bisect GH in K, and
with centre K describe a circle to cut the ellipse in four
points X, O, R and P; join these to form a rectangular paral-
lelogram; bisect each side and join the opposite points of sec-
tion, and produce both ways to meet circumference in L, M,
T and V ; then LM and TV vriW be the axes ; and if the dis-
126
DRAWING.
tance TK be taken with centre L or M, the arc will cut TV
in S, S, which will be required foci. (Fig. 97.)
Fig. 97.
No. 25. — To draw a tangent to an ellipse {a) from a point
in the circumference.
Fig. 98.
GEOMETRICAL DRAWING.
127
Let B, C be the foci and A, given point on the curve; join
BA and CA and produce them to D and F; bisect the angle
BAD by AK, and the angle DAF by AG. Then will KA be
a tangent and GA a perpendicular or normal to it, at the
point of contact, A. (Fig. 98.)
(b) From an external point.
Fig. 99.
Let A be an external point ; draw major axis BC, and on
it describe the semi-circle BFC; draw a tangent AF to the
circle at F; draw FE perpendicular to BC, cutting curve in
G; join AG, which will be a tangent, to the curve. (Fig. 99.)
A^o. 26. — To draw an oval of a given width.
Let AB be the given width ; bisect it in C, and on it describe
the circle ADB; draw CD at right angles to AB; with centre
B and distance BA describe curve AE; similarly describe
128
DRAWING.
curve BF; join BD and AD, and produce them to meet curve
in E and F; with centre 1) and distance DE describe curve
EF, completing the oval. (Fig. 100.)
No. 27, — To construct the involute of the circle.
Divide the circle into any number of equal parts and draw
the radii, numbering them 1, 2, etc. Draw the tangents, mak-
ing the jfirst the length from 1 to 2, the second twice this
length, the third three times, and so on. When all the tan-
gents have been drawn thus, begin again at 1 by producing
it, and so get a second series of points. Then draw a curve
through the points, commencing with 8, or last, and joining
it with 1, then 2, etc. (Fig. 101.)
129
Fig. 101.
No. 28. — To find {a) a mean proportional between two given
lines.
Fig. 102.
Let AC and BC be the given lines ; place them in a
straight line AB ; bisect AB in D, and on AB describe
130
DRAWING.
semicircle AEB; through C draw CE at riglit angles to AB;
then will AC : CE : : CE : CB. (Fig. 102.)
(b) To draw a third proportional (greater).
Fiff 103
Let AB and BC be the given lines ; draw AE greater than
AC, making any angle with AC ; make AD = BC ; join BD,
and through C draw CE parallel to BD ; then will AB : BC : :
BC:DE. (Fig. 103.)
(c) (Less). Make AE greater than BC and less than AC.
Fi^. 104.
GEOMETRICAL DRAWING.
131
No. 29. — To draw a circle of given radius which shall touch
both lines of a given angle.
Let BAG be the given angle and ST the given radius ; bisect
angle by line AE: erect on either line a perpendicular DE
equal to ST; through E draw EG parallel to AD, and cutting
AF in G; then a circle drawn with centre G and radius equal
to DE or ST, will touch the sides AB and AD. (Fig. 104.)
No. SO. — To draw a circle of given radius which shall touch
another given circle and a given straight line.
Fig. 105.
Let FG be a given line, BX a given circle, and DE a given
radius; draw a line GH perpendicular to FG and equal to
DE; through H draw HK parallel to FG; draw any radius
AB and produce it, making the part produced equal to DE;
then with centre A and distance AC, describe an arc cutting
HK in K: the circle drawn with centre K, and at a distance
equal to GH or DE will touch the circle BX and the line
FG. (Fig. 105.)
132 DRAWING.
Graded EiKercises.
1. Construct a square whose side is 2".
2. Construct a square whose diagonal is 4".
3. Construct an oblong whose sides are IJ" and 2|" respec-
tively.
4. A rectangular field is 900 yds. long and 400 yds. wide;
divide it into four equal fields each 400 vds. long. Scale
100 yds. = 1".
5. Divide a line 3J" long into two parts in the ratio 3 : 4.
6. Draw a line parallel to and between two other parallel
lines 2^" apart, the line to be twice as near one as the other.
7. Construct an equilateral triangle whose side is 2".
8. Centrally within the triangle in Xo. 7 construct a tri-
angle whose side is 1:^^".
9. Construct a triangle whose sides are 2^" 3^ and 3|^"
respectively.
10. Inscribe a circle in a triangle whose sides are same as
Xo. 9.
11. The diagonal of a parallelogram is 4' and one side is 2'.
Draw it;
12. Find the extent of an angle of 22J',of 37 J ^ of 41^^
13. Construct an isosceles triangle whose base is 2 and
vertical angle 37^^^.
14. Construct an isosceles triangle whose base is 2" and
altitude 3 J".
15. Construct an equilateral triangle about a circle whose
diameter is 4".
16. Construct a triangle whose sides are in the ratio 3:4:5
about a circle whose diameter is 4".
17. Describe a circle about a square whose side is 3 ".
18. Within a triangle whose sides are 3 ", 4" and 5" respec-
tively, inscribe a circle.
19. Construct a regular pentagon whose side is IJ".
20. Construct a regular heptagon in a circle 4' diameter.
GRADED EXERCISES. 133
21. Draw a hexagon whose side is 2".
22. Draw a hexagon within an equilateral triangle of 3".
23. Construct a regular octagon whose side is ly.
24. Construct a regular octagon in a square, side 3".
25. Construct a regular octagon in a circle of 3" diameter.
26. Inscribe seven equal circles in a circle of 3" diameter.
27. The diameters of an ellipse being 3h" and2 J" respec-
tively, draw it.
28. Draw an elliptical curve on a transverse axis of 3 ".
29. Draw an elliptical figure from two squares, diagonal of
each 3".
30. A circle and an ellipse touch the angles of an oblong
3" X 5", find axes and foci of the ellipse.
31. Draw an oval whose shorter axis is 2^".
32. Construct an involute to a circle whose diameter is J".
33. Find a mean proportional between two lines 2V and
3J" respectively.
3-1. Draw a third proportional (greater) to two lines 2V
and 3J" respectively.
35. Draw same as No. 34, less.
36. Lay out a circular garden whose radius is 30 yds., which
shall just touch a fence on one side, and another garden whose
radius is 50 yds. on the other. Scale 20 yds. = I".
37. Lay out a circular garden, radius 30 yds., which shall
touch two fences not parallel. Scale 20 yds. = 1".
38. Lay out a circular garden, radius 30 yds., which shall
touch two fences not parallel, and whose edge shall just touch
a tree in a given position. Scale 20 yds. = 1".
39. The axes of an elliptical flower garden are 40 and 60
yds, respectively, and a point is taken 60 yds, to left of the
shorter (produced), and 40 yds, above the longer (produced).
Draw a path from this point to touch the elliptical garden.
Scale 20 yds. = 1". Longer axis horizontal.
40. Construct a triangle whose angles are 75°, 45° and 60°
respectively, on a line 2" in length.
41. Two circles, whose diameters are each 3|", intersect, the
intercepted arc of each being one-fifth of the whole circum-
ference.
42. The base of a right-angle triangle being 2", and the
134 DRAWING.
perpendicular on the hypothenuse from the right angle being
H", construct the triangle.
43. Two lines meet at a point. Find a point bet%veen them
that will be 2" from one and 3 " from the other.
44. Show that the number of degrees in the angle of any
regular polygon is represented by ^ — , where 7t = num-
ber of sides.
45. Show that a circle is but a particular form of an
ellipse.
46. If an ellipse be drawn with a string around two fixed
pins as foci, show that the sum of the distances from any
point in the curve to the foci is the same.
47. From the point found in Xo. 43 draw two equal
straight lines to the given line.
48. Draw a parallelogram 3J" x oh", and within this draw
one of half the size similar and similarly situated.
49. If one hexagon be inscribed in, and another inscribed
about, a circle, show that their areas are in ratio 3 : 4.
50. Three circular gardens, diameters 20 yds., 30 yds. and
40 yds. respectively, are to be placed with walks of 5 yds.
between them. Scale 20 yds. = 1 ".
51. Divide a triangle whose sides are 3", 4" and 5" respec-
tively, into four equal and similar triangles.
52. From the vertex of a scalene triangle draw a straight
line to the base which shall exceed the less side as much as it
is exceeded by the greater.
53. One of the acute angles of a right-angled triangle is
three times as great as the other; trisect the smaller.
54. One side of a right-angled triangle is 4", and the differ-
ence between the hypothenuse and the sum of the other two
sides is 2" ; construct the triangle.
55. The altitude of an equilateral triangle is 2"; construct it.
56. Place a straight line, 3" in length, between two straight
lines, each 2" in length, which meet, so that it shall be equally
inclined to each of them.
57. Describe an isosceles triangle upon a given base, having
each of the sides double of the base.
58. Draw a square equal in area to two unequal oblongs.
GRADED EXERCISES. 135
59. Given the base, the vertical angle and the perpendicular
of a plane triangle, construct it.
60. Cut off two-thirds of an isosceles triangle by a line
parallel to the base.
61. Describe two circles with given radii which shall cut
each other and have the line between the points of section
equal to a given, limited line.
62. Describe a circle with a given centre cutting a given
circle in the extremities of the diameter.
63. Describe a circle which shall pass through a given point
and which shall touch a given straight line in a given point.
64. Describe a circle which shall touch a given straiglit
line at a given point and bisect the circumference of a given
circle.
65. Two circles are described about the same centre; draw
a chord to the outer circle, which shall be divided into three
equal parts by the inner one. What are the limits of the
diameters 1
66. The perimeter of an oblong is 20", and the sides are in
ratio 3:2; construct it.
67. Construct an isosceles triangle of given vertical angle
and given altitude.
68. Draw a square equal in area to an oblong 3" x 2".
69. Describe a circle of given radius to touch two points.
What limits the position of the points 1
70. Show how to draw a similar triangle within another,
71. Two equal ellipses, axes 5" and 3\ cut each other at
right angles, and their centres are coincident ; draw them.
72. Trisect a given line.
73. Show how an angle may be trisected (mechanically).
74. Divide a square into three equal parts by lines drawn
from an angle.
75. Divide an oblong into three equal parts by lines drawn
from an angle.
76. Find the number of degrees in the angle of a regular
duodecagon.
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