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Title: The Elements of Perspective

Author: John Ruskin

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Most recently updated: October 17, 2024

Language: English

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i
Library Edition

THE COMPLETE WORKS
OF

JOHN RUSKIN

ELEMENTS OF DRAWING AND
PERSPECTIVE
THE TWO PATHS
UNTO THIS LAST
MUNERA PULVERIS
SESAME AND LILIES
ETHICS OF THE DUST

N AT I O N A L L I B R A RY A S S O C I AT I O N
NEW YORK CHICAGO

iii
THE ELEMENTS OF PERSPECTIVE

ARRANGED FOR THE USE OF SCHOOLS

AND INTENDED TO BE READ IN CONNECTION WITH THE

FIRST THREE BOOKS OF EUCLID.

v
CONTENTS.
PAGE

Preface
. . . . . . . . . . . . . . . . . . . . . . ix
Introduction
. . . . . . . . . . . . . . . . . . . . . . 1

PROBLEM I.
To fix. the
. P.osition
. . of. a .given
. .Point
. . . . . . . . . . . . . 10

PROBLEM II.
To draw
. .a R. ight
. L. ine. between
. . . two
. given
. . P.oints
. . . . . . . . . 13

PROBLEM III.
To find
. the
. .Vanishing
. . . -P.oint
. of. a .given
. .Horizontal
. . . .Line
. . . . . . 17

PROBLEM IV.
To find
. the
. .Dividing
. . -P . oints
. . of. a given
. . H . orizontal
. . . L. ine. . . . . . 23

PROBLEM V.
To draw a Horizontal Line, given in Position and Magnitude, by
. .of .its .Sight
means . -M
. agnitude
. . . and. . Dividing
. . .-Points
. . . . . . . . 24

PROBLEM VI.
To draw any Triangle, given in Position and Magnitude, in a
Horizontal
. . . P. lane
. . . . . . . . . . . . . . . . . . 27

PROBLEM VII.
To draw any Rectilinear Quadrilateral Figure, given in Position and
Magnitude
. . . , .in a. Horizontal
. . . . Plane
. . . . . . . . . . . . . 29

PROBLEM VIII.
To draw a Square, given in Position and Magnitude, in a Horizontal
Plane
. . . . . . . . . . . . . . . . . . . . . . 31

vi
PROBLEM IX.
To draw a Square Pillar, given in Position and Magnitude, its Base
and. Top
. being
. . in. H.orizontal
. . . Planes
. . . . . . . . . . . . . 34

PROBLEM X.
To draw a Pyramid, given in Position and Magnitude, on a Square
Base
. in. a.Horizontal
. . . .Plane
. . . . . . . . . . . . . . . 36

PROBLEM XI.
To draw
. .any. C.urve
. in. a.Horizontal
. . . .or .Vertical
. . .Plane
. . . . . . . 38

PROBLEM XII.
To divide a Circle drawn in Perspective into any given Number of
Equal
. . Parts
. . . . . . . . . . . . . . . . . . . . 42

PROBLEM XIII.
To draw a Square, given in Magnitude, within a larger Square given
in Position and Magnitude; the Sides of the two Squares being
Parallel
. . . . . . . . . . . . . . . . . . . . . . 45

PROBLEM XIV.
To draw a Truncated Circular Cone, given in Position and
Magnitude, the Truncations being in Horizontal Planes, and the
Axis
. of. the
. C. one
. vertical
. . . . . . . . . . . . . . . . . 47

PROBLEM XV.
To draw
. .an .Inclined
. . .Line
. , given
. . in. P.osition
. . and
. M. agnitude
. . . . . . . 50

PROBLEM XVI.
To find
. the
. .Vanishing
. . . -P.oint
. of. a .given
. .Inclined
. . L . ine
. . . . . . . 53

PROBLEM XVII.
To find
. the
. .Dividing
. . -P . oints
. . of. a given
. . I.nclined
. . L. ine. . . . . . . 55

PROBLEM XVIII.
To find the Sight-Line of an Inclined Plane in which Two Lines are
. in
given . P.osition
. . . . . . . . . . . . . . . . . . . 57

vii
PROBLEM XIX.
To find the Vanishing-Point of Steepest Lines in an Inclined Plane
. .Sight
whose . -L
. ine
. is. given
. . . . . . . . . . . . . . . . 59

PROBLEM XX.
To find the Vanishing-Point of Lines perpendicular to the Surface of
. . Inclined
a given . . . Plane
. . . . . . . . . . . . . . . . . 61

APPENDIX.

I.
Practice
. . and
. O. bservations
. . . .on .the. preceding
. . . P.roblems
. . . . . . . . 69

II.
Demonstrations which could not conveniently be included in
the. Text
. . . . . . . . . . . . . . . . . . . . . 99

ix
PREFACE.
For some time back I have felt the want, among Students of Drawing, of a
written code of accurate Perspective Law; the modes of construction in
common use being various, and, for some problems, insufficient. It would
have been desirable to draw up such a code in popular language, so as to do
away with the most repulsive difficulties of the subject; but finding this
popularization would be impossible, without elaborate figures and long
explanations, such as I had no leisure to prepare, I have arranged the
necessary rules in a short mathematical form, which any schoolboy may
read through in a few days, after he has mastered the first three and the
sixth books of Euclid.
Some awkward compromises have been admitted between the first-
attempted popular explanation, and the severer arrangement, involving
irregular lettering and redundant phraseology; but I cannot for the present
do more, and leave the book therefore to its trial, hoping that, if it be found
by masters of schools to answer its purpose, I may hereafter bring it into
better form.1
An account of practical methods, sufficient for general purposes of
x
sketching, might indeed have been set down in much less space: but if the
student reads the following pages carefully, he will not only find himself
able, on occasion, to solve perspective problems of a complexity greater
than the ordinary rules will reach, but obtain a clue to many important laws
of pictorial effect, no less than of outline. The subject thus examined
becomes, at least to my mind, very curious and interesting; but, for students
who are unable or unwilling to take it up in this abstract form, I believe
good help will be soon furnished, in a series of illustrations of practical
perspective now in preparation by Mr. Le Vengeur. I have not seen this
essay in an advanced state, but the illustrations shown to me were very clear
and good; and, as the author has devoted much thought to their
arrangement, I hope that his work will be precisely what is wanted by the
general learner.

Students wishing to pursue the subject into its more extended branches will
find, I believe, Cloquet’s treatise the best hitherto published.2

1 Some irregularities of arrangement have been admitted merely for the sake of convenient
reference; the eighth problem, for instance, ought to have been given as a case of the seventh,
but is separately enunciated on account of its importance.
Several constructions, which ought to have been given as problems, are on the contrary given
as corollaries, in order to keep the more directly connected problems in closer sequence; thus
the construction of rectangles and polygons in vertical planes would appear by the Table of
Contents to have been omitted, being given in the corollary to Problem IX.
Return to text
2 Nouveau Traité Élémentaire de Perspective. Bachelier, 1823. Return to text

1
THE
ELEMENTS OF PERSPECTIVE.

INTRODUCTION.

When you begin to read this book, sit down very near the window, and shut
the window. I hope the view out of it is pretty; but, whatever the view may
be, we shall find enough in it for an illustration of the first principles of
perspective (or, literally, of “looking through”).
Every pane of your window may be considered, if you choose, as a glass
picture; and what you see through it, as painted on its surface.
And if, holding your head still, you extend your hand to the glass, you may,
with a brush full of any thick color, trace, roughly, the lines of the landscape
on the glass.
But, to do this, you must hold your head very still. Not only you must not
move it sideways, nor up and down, but it must not even move backwards
or forwards; for, if you move your head forwards, you will see more of the
landscape through the pane; and, if you move it backwards, you will see
less: or considering the pane of glass as a picture, when you hold your head
near it, the objects are painted small, and a great many of them go into a
little space; but, when you hold your head some distance back, the objects
are painted larger upon the pane, and fewer of them go into the field of it.
But, besides holding your head still, you must, when you try to trace the
picture on the glass, shut one of your eyes. If you do not, the point of the
2
brush appears double; and, on farther experiment, you will observe that

each of your eyes sees the object in a different place on the glass, so that the
tracing which is true to the sight of the right eye is a couple of inches (or
more, according to your distance from the pane,) to the left of that which is
true to the sight of the left.
Thus, it is only possible to draw what you see through the window rightly
on the surface of the glass, by fixing one eye at a given point, and neither
moving it to the right nor left, nor up nor down, nor backwards nor
forwards. Every picture drawn in true perspective may be considered as an
upright piece of glass,3 on which the objects seen through it have been thus
drawn. Perspective can, therefore, only be quite right, by being calculated
for one fixed position of the eye of the observer; nor will it ever appear
deceptively right unless seen precisely from the point it is calculated for.
Custom, however, enables us to feel the rightness of the work on using both
our eyes, and to be satisfied with it, even when we stand at some distance
from the point it is designed for.
Supposing that, instead of a window, an unbroken plate of crystal extended
itself to the right and left of you, and high in front, and that you had a brush
as long as you wanted (a mile long, suppose), and could paint with such a
brush, then the clouds high up, nearly over your head, and the landscape far
away to the right and left, might be traced, and painted, on this enormous
crystal field.4 But if the field were so vast (suppose a mile high and a mile
wide), certainly, after the picture was done, you would not stand as near to
it, to see it, as you are now sitting near to your window. In order to trace the
upper clouds through your great glass, you would have had to stretch your
neck
3
quite back, and nobody likes to bend their neck back to see the top of a
picture. So you would walk a long way back to see the great picture—a
quarter of a mile, perhaps,—and then all the perspective would be wrong,
and would look quite distorted, and you would discover that you ought to
have painted it from the greater distance, if you meant to look at it from that
distance. Thus, the distance at which you intend the observer to stand from
a picture, and for which you calculate the perspective, ought to regulate to a
certain degree the size of the picture. If you place the point of observation
near the canvas, you should not make the picture very large: vice versâ, if
you place the point of observation far from the canvas, you should not make

it very small; the fixing, therefore, of this point of observation determines,
as a matter of convenience, within certain limits, the size of your picture.
But it does not determine this size by any perspective law; and it is a
mistake made by many writers on perspective, to connect some of their
rules definitely with the size of the picture. For, suppose that you had what
you now see through your window painted actually upon its surface, it
would be quite optional to cut out any piece you chose, with the piece of the
landscape that was painted on it. You might have only half a pane, with a
single tree; or a whole pane, with two trees and a cottage; or two panes,
with the whole farmyard and pond; or four panes, with farmyard, pond, and
foreground. And any of these pieces, if the landscape upon them were, as a
scene, pleasantly composed, would be agreeable pictures, though of quite
different sizes; and yet they would be all calculated for the same distance of
observation.
In the following treatise, therefore, I keep the size of the picture entirely
undetermined. I consider the field of canvas as wholly unlimited, and on
that condition determine the perspective laws. After we know how to apply
those laws without limitation, we shall see what limitations of the size of
the picture their results may render advisable.
But
4
although the size of the picture is thus independent of the observer’s
distance, the size of the object represented in the picture is not. On the
contrary, that size is fixed by absolute mathematical law; that is to say,
supposing you have to draw a tower a hundred feet high, and a quarter of a
mile distant from you, the height which you ought to give that tower on
your paper depends, with mathematical precision, on the distance at which
you intend your paper to be placed. So, also, do all the rules for drawing the
form of the tower, whatever it may be.
Hence, the first thing to be done in beginning a drawing is to fix, at your
choice, this distance of observation, or the distance at which you mean to
stand from your paper. After that is determined, all is determined, except
only the ultimate size of your picture, which you may make greater, or less,
not by altering the size of the things represented, but by taking in more, or
fewer of them. So, then, before proceeding to apply any practical

perspective rule, we must always have our distance of observation marked,
and the most convenient way of marking it is the following:

PLACING OF THE SIGHT-POINT, SIGHT-LINE, STATION-POINT, AND STATION-LINE.

Fig. 1.

5

I. The Sight-Point.—Let a b c d, Fig. 1., be your sheet of paper, the larger
the better, though perhaps we may cut out of it at last only a small piece for
our picture, such as the dotted circle n o p q. This circle is not intended to
limit either the size or shape of our picture: you may ultimately have it
round or oval, horizontal or upright, small or large, as you choose. I only
dot the line to give you an idea of whereabouts you will probably like to
have it; and, as the operations of perspective are more conveniently
performed upon paper underneath the picture than above it, I put this
conjectural circle at the top of the paper, about the middle of it, leaving
plenty of paper on both sides and at the bottom. Now, as an observer
generally stands near the middle of a picture to look at it, we had better at
first, and for simplicity’s sake, fix the point of observation opposite the
middle of our conjectural picture. So take the point s, the center of the circle
n o p q;—or, which will be simpler for you in your own work, take the point
s at random near the top of your paper, and strike the circle n o p q round it,

any size you like. Then the point s is to represent the point opposite which
you wish the observer of your picture to place his eye, in looking at it. Call
this point the “Sight-Point.”

II. The Sight-Line.—Through the Sight-point, s, draw a horizontal line, g h,
right across your paper from side to side, and call this line the “Sight-Line.”
This line is of great practical use, representing the level of the eye of the
observer all through the picture. You will find hereafter that if there is a
horizon to be represented in your picture, as of distant sea or plain, this line
defines it.

III. The Station-Line.—From s let fall a perpendicular line, s r, to the
bottom of the paper, and call this line the “Station-Line.”
This represents the line on which the observer stands, at a greater or less
distance from the picture; and it ought to be imagined as drawn right out
from the paper at the point s. Hold your paper upright in front of you, and
hold
6
your pencil horizontally, with its point against the point s, as if you
wanted to run it through the paper there, and the pencil will represent the
direction in which the line s r ought to be drawn. But as all the
measurements which we have to set upon this line, and operations which we
have to perform with it, are just the same when it is drawn on the paper
itself, below s, as they would be if it were represented by a wire in the
position of the leveled pencil, and as they are much more easily performed
when it is drawn on the paper, it is always in practice, so drawn.

IV. The Station-Point.—On this line, mark the distance s t at your pleasure,
for the distance at which you wish your picture to be seen, and call the point
t the “Station-Point.”

Fig. 2.

In practice, it is generally advisable to make the distance s t about as great
as the diameter of your intended picture; and it should, for the most part, be
more rather than less; but, as I have just stated, this is quite arbitrary.
However, in this figure, as an approximation to a generally advisable
distance, I make the distance s t equal to the diameter of the circle n o p q.
Now, having fixed this distance, s t, all the dimensions of the objects in our
picture are fixed likewise, and for this reason:—
Let the upright line a b, Fig. 2., represent a pane of glass placed where our
picture
7
is to be placed; but seen at the side of it, edgeways; let s be the
Sight-point; s t the Station-line, which, in this figure, observe, is in its true
position, drawn out from the paper, not down upon it; and t the Station-
point.
Suppose the Station-line s t to be continued, or in mathematical language
“produced,” through s, far beyond the pane of glass, and let p q be a tower or
other upright object situated on or above this line.
Now the apparent height of the tower p q is measured by the angle q t p,
between the rays of light which come from the top and bottom of it to the
eye of the observer. But the actual height of the image of the tower on the

pane of glass a b, between us and it, is the distance p′ q′ between the points
where the rays traverse the glass.
Evidently, the farther from the point t we place the glass, making s t longer,
the larger will be the image; and the nearer we place it to t, the smaller the
image, and that in a fixed ratio. Let the distance d t be the direct distance
from the Station-point to the foot of the object. Then, if we place the glass
a b at one-third of that whole distance, p′ q′ will be one-third of the real
height of the object; if we place the glass at two-thirds of the distance, as at
e f, p″ q″ (the height of the image at that point) will be two-thirds the height5
of the object, and so on. Therefore the mathematical law is that p′ q′ will be
to p q as s t to d t. I put this ratio clearly by itself that you may remember it:

p′ q′ ∶ p q ∷ s t ∶ d t
or in words:
p dash q dash is to p q as s t to d t

In which formula, recollect that p′ q′ is the height of the appearance of the
object on the picture; p q the height of the object itself; s the Sight-point; t
the Station-point; d a point at the direct distance of the object; though the
object
8
is seldom placed actually on the line t s produced, and may be far to
the right or left of it, the formula is still the same.
For let s, Fig. 3., be the Sight-point, and a b the glass—here seen looking
down on its upper edge, not sideways;—then if the tower (represented now,
as on a map, by the dark square), instead of being at d on the line s t
produced, be at e, to the right (or left) of the spectator, still the apparent
height of the tower on a b will be as s′ t to e t, which is the same ratio as that
of s t to d t.

Now in many perspective problems, the position of an object is more
conveniently expressed by the two measurements d t and d e, than by the
single oblique measurement e t.
I shall call d t the “direct distance” of the object at e, and d e its “lateral
distance.” It is rather a license to call d t its “direct” distance, for e t is the

more direct of the two; but there is no other
term which would not cause confusion.
Lastly, in order to complete our knowledge
of the position of an object, the vertical
height of some point in it, above or below
the eye, must be given; that is to say, either
d p or d q in Fig. 2.6: this I shall call the
“vertical distance” of the point given. In all
perspective problems these three distances,
and the dimensions of the object, must be
stated, otherwise the problem is imperfectly
given. It ought not to be required of us
merely to draw a room or a church in
Fig. 3.
perspective; but to draw this room from this
corner, and that church on that spot, in
9
perspective. For want of knowing how to
base their drawings on the measurement and place of the object, I have
known practiced students represent a parish church, certainly in true
perspective, but with a nave about two miles and a half long.
It is true that in drawing landscapes from nature the sizes and distances of
the objects cannot be accurately known. When, however, we know how to
draw them rightly, if their size were given, we have only to assume a
rational approximation to their size, and the resulting drawing will be true
enough for all intents and purposes. It does not in the least matter that we
represent a distant cottage as eighteen feet long, when it is in reality only
seventeen; but it matters much that we do not represent it as eighty feet
long, as we easily might if we had not been accustomed to draw from
measurement. Therefore, in all the following problems the measurement of
the object is given.
The student must observe, however, that in order to bring the diagrams into
convenient compass, the measurements assumed are generally very
different from any likely to occur in practice. Thus, in Fig. 3., the distance
d s would be probably in practice half a mile or a mile, and the distance t s,
from the eye of the observer to the paper, only two or three feet. The

mathematical law is however precisely the same, whatever the proportions;
and I use such proportions as are best calculated to make the diagram clear.
Now, therefore, the conditions of a perspective problem are the following:

The Sight-line g h given, Fig. 1.;
The Sight-point s given;
The Station-point t given; and
The three distances of the object,7 direct, lateral, and vertical, with its
dimensions, given.

The size of the picture, conjecturally limited by the dotted circle, is to be
determined afterwards at our pleasure. On these conditions I proceed at
once to construction.

3 If the glass were not upright, but sloping, the objects might still be drawn through it, but their
perspective would then be different. Perspective, as commonly taught, is always calculated
for a vertical plane of picture. Return to text
4 Supposing it to have no thickness; otherwise the images would be distorted by refraction.
Return to text
5 I say “height” instead of “magnitude,” for a reason stated in Appendix I., to which you will
soon be referred. Read on here at present. Return to text
6 p and q being points indicative of the place of the tower’s base and top. In this figure both are
above the sight-line; if the tower were below the spectator both would be below it, and
therefore measured below d. Return to text
7 More accurately, “the three distances of any point, either in the object itself, or indicative of
its distance.” Return to text

10
PROBLEM I.

TO FIX THE POSITION OF A GIVEN POINT.8

Let p, Fig. 4., be the given point.

Fig. 4.

Let its direct distance be d t; its lateral distance to the left, d c; and vertical
distance beneath the eye of the observer, c p.
[Let g h be the Sight-line, s the Sight-point, and t the Station-point.]9

It is required to fix on the plane of the picture the position of the point p.
Arrange
11
the three distances of the object on your paper, as in Fig. 4.10
Join c t, cutting g h in q.

From q let fall the vertical line q p′.
Join p t, cutting q p in p′.
p′ is the point required.

If the point p is above the eye of the observer instead of below it, c p is to be
measured upwards from c, and q p′ drawn upwards from q. The construction
will be as in Fig. 5.

Fig. 5.

And
12
if the point p is to the right instead of the left of the observer, d c is to
be measured to the right instead of the left.
The figures 4. and 5., looked at in a mirror, will show the construction of
each, on that supposition.

Now read very carefully the examples and notes to this problem in
Appendix I. (page 69). I have put them in the Appendix in order to keep the
sequence of following problems more clearly traceable here in the text; but
you must read the first Appendix before going on.

8 More accurately, “To fix on the plane of the picture the apparent position of a point given in
actual position.” In the headings of all the following problems the words “on the plane of the
picture” are to be understood after the words “to draw.” The plane of the picture means a
surface extended indefinitely in the direction of the picture. Return to text
9 The sentence within brackets will not be repeated in succeeding statements of problems. It is
always to be understood. Return to text
10 In order to be able to do this, you must assume the distances to be small; as in the case of
some object on the table: how large distances are to be treated you will see presently; the
mathematical principle, being the same for all, is best illustrated first on a small scale.
Suppose, for instance, p to be the corner of a book on the table, seven inches below the eye,
five inches to the left of it, and a foot and a half in advance of it, and that you mean to hold
your finished drawing at six inches from the eye; then t s will be six inches, t d a foot and a
half, d c five inches, and c p seven. Return to text

13
PROBLEM II.

TO DRAW A RIGHT LINE BETWEEN TWO GIVEN POINTS.

Fig. 6.

Let a b, Fig. 6., be the given right line, joining the given points a and b.

Let the direct, lateral, and vertical distances of the point a be t d, d c, and c a.

Let the direct, lateral, and vertical distances of the point b be t d′, d c′, and
c′ b.

Then, by Problem I., the position of the point a on the plane of the picture is
a.
And similarly, the position of the point b on the plane of the picture is b.
Join a b.
Then a b is the line required.

14
COROLLARY I.
If the line a b is in a plane parallel to that of the picture, one end of the line
a b must be at the same direct distance from the eye of the observer as the
other.
Therefore, in that case, d t is equal to d′ t.
Then the construction will be as in Fig. 7.; and the student will find
experimentally that a b is now parallel to a b.11

Fig. 7.

And that a b is to a b as t s is to t d.
Therefore, to draw any line in a plane parallel to that of the picture, we have
only to fix the position of one of its extremities, a or b, and then to draw
from a or b a line parallel to the given line, bearing the proportion to it that
t s bears to t d.

15
COROLLARY II.
If the line a b is in a horizontal plane, the vertical distance of one of its
extremities must be the same as that of the other.
Therefore, in that case, a c equals b c′ (Fig. 6.).

And the construction is as in Fig. 8.

Fig. 8.

In Fig. 8. produce a b to the sight-line, cutting the sight-line in v; the point
v, thus determined, is called the Vanishing-Point of the line a b.

Join t v. Then the student will find experimentally that t v is parallel to a b.12

16
COROLLARY III.
If the line a b produced would pass through some point beneath or above the
station-point, c d is to d t as c′ d′ is to d′ t; in which case the point c
coincides with the point c′, and the line a b is vertical.
Therefore every vertical line in a picture is, or may be, the perspective
representation of a horizontal one which, produced, would pass beneath the
feet or above the head of the spectator.13

11 For by the construction a t ∶ a t ∷ b t ∶ b t; and therefore the two triangles a b t, a b t, (having
a common angle a t b,) are similar. Return to text
12 The demonstration is in Appendix II. Article I. Return to text
13 The reflection in water of any luminous point or isolated object (such as the sun or moon) is
therefore, in perspective, a vertical line; since such reflection, if produced, would pass under
the feet of the spectator. Many artists (Claude among the rest) knowing something of optics,
but nothing of perspective, have been led occasionally to draw such reflections towards a
point at the center of the base of the picture. Return to text

17
PROBLEM III.

TO FIND THE VANISHING-POINT OF A GIVEN HORIZONTAL LINE.

Fig. 9.

Let a b, Fig. 9., be the given line.

From t, the station-point, draw t v parallel to a b, cutting the sight-line in v.
v is the Vanishing-point required.14

18
COROLLARY I.
As, if the point b is first found, v may be determined by it, so, if the point v
is first found, b may be determined by it. For let a b, Fig. 10., be the given
line, constructed upon the paper as in Fig. 8.; and let it be required to draw
the line a b without using the point c′.

Fig. 10.

Find the position of the point a in a. (Problem I.)
Find
19
the vanishing-point of a b in v. (Problem III.)

Join a v.

Join b t, cutting a v in b.
Then a b is the line required.15

COROLLARY II.
We have hitherto proceeded on the supposition that the given line was small
enough, and near enough, to be actually drawn on our paper of its real size;
as in the example given in Appendix I. We may, however, now deduce a

construction available under all circumstances, whatever may be the
distance and length of the line given.

Fig. 11.

From
20
Fig. 8. remove, for the sake of clearness, the lines c′ d′, b v, and t v;
and, taking the figure as here in Fig. 11., draw from a, the line a r parallel to
a b, cutting b t in r.

Then a r is to a b as a t is to a t.
— — as c t is to c t.

— — as t s is to t d.
That is to say, a r is the sight-magnitude of a b.16

Fig. 12.

Therefore, when the position of the point a is fixed in a, as in Fig. 12., and
a v is drawn to the vanishing-point; if we draw a line a r from a, parallel to
a b, and make a r equal to the sight-magnitude of a b, and then join r t, the
line r t will cut a v in b.

So that, in order to determine the length of a b, we need not draw the long
and distant line a b, but only a r parallel to it, and of its sight-magnitude;
which is a great gain, for the line a b may be two miles long, and the line a r
perhaps only two inches.

21
COROLLARY III.
In Fig. 12., altering its proportions a little for the sake of clearness, and
putting it as here in Fig. 13., draw a horizontal line a r′ and make a r′ equal

to a r.

Through the points r and b draw r′ m, cutting the sight-line in m. Join t v.
Now the reader will find experimentally that v m is equal to v t.17

Fig. 13.

Hence it follows that, if from the vanishing-point v we lay off on the sight-
line a distance, v m, equal to v t; then draw through a a horizontal line a r′,
make a r′ equal to the sight-magnitude of a b, and join r′ m; the line r′ m will
cut a v in b. And this is in practice generally the most convenient way of
obtaining the length of a b.

22
COROLLARY IV.
Removing from the preceding figure the unnecessary lines, and retaining
only r′ m and a v, as in Fig. 14., produce the line a r′ to the other side of a,
and make a x equal to a r′.

Join x b, and produce x b to cut the line of sight in n.

Fig. 14.

Then as x r′ is parallel to m n, and a r′ is equal to a x, v n must, by similar
triangles, be equal to v m (equal to v t in Fig. 13.).
Therefore, on whichever side of v we measure the distance v t, so as to
obtain either the point m, or the point n, if we measure the sight-magnitude
a r′ or a x on the opposite side of the line a v, the line joining r′ m or x n will
equally cut a v in b.

The points m and n are called the “Dividing-Points” of the original line a b
(Fig. 12.), and we resume the results of these corollaries in the following
three problems.

14 The student will observe, in practice, that, his paper lying flat on the table, he has only to
draw the line t v on its horizontal surface, parallel to the given horizontal line a b. In theory,
the paper should be vertical, but the station-line s t horizontal (see its definition above, page
5); in which case t v, being drawn parallel to a b, will be horizontal also, and still cut the
sight-line in v.

The construction will be seen to be founded on the second Corollary of the preceding
problem.
It is evident that if any other line, as m n in Fig. 9., parallel to a b, occurs in the picture, the
line t v, drawn from t, parallel to m n, to find the vanishing-point of m n, will coincide with
the line drawn from t, parallel to a b, to find the vanishing-point of a b.
Therefore a b and m n will have the same vanishing-point.
Therefore all parallel horizontal lines have the same vanishing-point.

It will be shown hereafter that all parallel inclined lines also have the same vanishing-point;
the student may here accept the general conclusion—“All parallel lines have the same
vanishing-point.”

It is also evident that if a b is parallel to the plane of the picture, t v must be drawn parallel to
g h, and will therefore never cut g h. The line a b has in that case no vanishing-point: it is to be
drawn by the construction given in Fig. 7.
It is also evident that if a b is at right angles with the plane of the picture, t v will coincide
with t s, and the vanishing-point of a b will be the sight-point. Return to text
15 I spare the student the formality of the reductio ad absurdum, which would be necessary to
prove this. Return to text
16 For definition of Sight-Magnitude, see Appendix I. It ought to have been read before the
student comes to this problem; but I refer to it in case it has not. Return to text
17 The demonstration is in Appendix II. Article II. p. 101. Return to text

23
PROBLEM IV.

TO FIND THE DIVIDING-POINTS OF A GIVEN HORIZONTAL LINE.

Fig. 15.

Let the horizontal line a b (Fig. 15.) be given in position and magnitude. It
is required to find its dividing-points.
Find the vanishing-point v of the line a b.
With center v and distance v t, describe circle cutting the sight-line in m and
n.

Then m and n are the dividing-points required.
In general, only one dividing-point is needed for use with any vanishing-
point, namely, the one nearest s (in this case the point m). But its opposite n,
or both, may be needed under certain circumstances.

24
PROBLEM V.

TO DRAW A HORIZONTAL LINE, GIVEN IN POSITION AND MAGNITUDE, BY
MEANS OF ITS SIGHT-MAGNITUDE AND DIVIDING-POINTS.

Fig. 16.

Let a b (Fig. 16.) be the given line.
Find the position of the point a in a.

Find the vanishing-point v, and most convenient dividing-point m, of the
line a b.
Join a v.

Through a draw a horizontal line a b′ and make a b′ equal to the sight-
magnitude of a b. Join b′ m, cutting a v in b.

Then a b is the line required.

25
COROLLARY I.

Fig. 17.

Supposing it were now required to draw a line a c (Fig. 17.) twice as long as
a b, it is evident that the sight-magnitude a c′ must be twice as long as the
sight-magnitude a b′; we have, therefore, merely to continue the horizontal
line a b′, make b′ c′ equal to a b′, join c m′, cutting a v in c, and a c will be
the line required. Similarly, if we have to draw a line a d, three times the
length of a b, a d′ must be three times the length of a b′, and, joining d′ m,
a d will be the line required.
The student will observe that the nearer the portions cut off, b c, c d, etc.,
approach the point v, the smaller they become; and, whatever lengths may
be added to the line a d, and successively cut off from a v, the line a v will
never be cut off entirely, but the portions cut off will become infinitely

small, and apparently “vanish” as they approach the point v; hence this
point is called the “vanishing” point.

26
COROLLARY II.
It is evident that if the line a d had been given originally, and we had been
required to draw it, and divide it into three equal parts, we should have had
only to divide its sight-magnitude, a d′, into the three equal parts, a b′, b′ c′,
and c′ d′, and then, drawing to m from b′ and c′, the line a d would have
been divided as required in b and c. And supposing the original line a d be
divided irregularly into any number of parts, if the line a d′ be divided into
a similar number in the same proportions (by the construction given in
Appendix I.), and, from these points of division, lines are drawn to m, they
will divide the line a d in true perspective into a similar number of
proportionate parts.
The horizontal line drawn through a, on which the sight-magnitudes are
measured, is called the “Measuring-line.”
And the line a d, when properly divided in b and c, or any other required
points, is said to be divided “in perspective ratio” to the divisions of the
original line a d.
If the line a v is above the sight-line instead of beneath it, the measuring-
line is to be drawn above also: and the lines b′ m, c′ m, etc., drawn down to
the dividing-point. Turn Fig. 17. upside down, and it will show the
construction.

27
PROBLEM VI.

TO DRAW ANY TRIANGLE, GIVEN IN POSITION AND MAGNITUDE, IN A
HORIZONTAL PLANE.

Fig. 18.

Let a b c (Fig. 18.) be the triangle.
As it is given in position and magnitude, one of its sides, at least, must be
given in position and magnitude, and the directions of the two other sides.
Let a b be the side given in position and magnitude.

Then a b is a horizontal line, in a given position, and of a given length.

Draw the line a b. (Problem V.)

Let a b be the line so drawn.
Find v and v′, the vanishing-points respectively of the lines a c and b c.
(Problem III.)
From
28
a draw a v, and from b, draw b v′, cutting each other in c.
Then a b c is the triangle required.
If a c is the line originally given, a c is the line which must be first drawn,
and the line v′ b must be drawn from v′ to c and produced to cut a b in b.
Similarly, if b c is given, v c must be drawn to c and produced, and a b from
its vanishing-point to b, and produced to cut a c in a.

29
PROBLEM VII.

TO DRAW ANY RECTILINEAR QUADRILATERAL FIGURE, GIVEN IN POSITION
AND MAGNITUDE, IN A HORIZONTAL PLANE.

Fig. 19.

Let a b c d (Fig. 19.) be the given figure.

Join any two of its opposite angles by the line b c.
Draw first the triangle a b c. (Problem VI.)

And then, from the base b c, the two lines b d, c d, to their vanishing-points,
which will complete the figure. It is unnecessary to give a diagram of the
construction, which is merely that of Fig. 18. duplicated; another triangle
being drawn on the line a c or b c.

COROLLARY.
It is evident that by this application of Problem VI. any given rectilinear
figure whatever in a horizontal plane may be drawn, since any such figure

may be divided into a number of triangles, and the triangles then drawn in
succession.
More
30
convenient methods may, however, be generally found, according to
the form of the figure required, by the use of succeeding problems; and for
the quadrilateral figure which occurs most frequently in practice, namely,
the square, the following construction is more convenient than that used in
the present problem.

31
PROBLEM VIII.

TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN A
HORIZONTAL PLANE.

Fig. 20.

Let a b c d, Fig. 20., be the square.

As it is given in position and magnitude, the position and magnitude of all
its sides are given.
Fix the position of the point a in a.

Find v, the vanishing-point of a b; and m, the dividing-point of a b, nearest s.

Find v′, the vanishing-point of a c; and n, the dividing-point of a c, nearest s.
Draw
32
the measuring-line through a, and make a b′, a c′, each equal to the
sight-magnitude of a b.
(For since a b c d is a square, a c is equal to a b.)

Draw a v′ and c′ n, cutting each other in c.

Draw a v, and b′ m, cutting each other in b.
Then a c, a b, are the two nearest sides of the square.
Now, clearing the figure of superfluous lines, we have a b, a c, drawn in
position, as in Fig. 21.

Fig. 21.

And because a b c d is a square, c d (Fig. 20.) is parallel to a b.

And all parallel lines have the same vanishing-point. (Note to Problem III.)
Therefore, v is the vanishing-point of c d.
Similarly, v′ is the vanishing-point of b d.

Therefore, from b and c (Fig. 22.) draw b v′, c v, cutting each other in d.

Then a b c d is the square required.

COROLLARY I.
It is obvious that any rectangle in a horizontal plane may be drawn by this
problem, merely making a b′, on the measuring-line, Fig. 20., equal to the
sight-magnitude of one of its sides, and a c′ the sight-magnitude of the
other.

33
COROLLARY II.
Let a b c d, Fig. 22., be any square drawn in perspective. Draw the
diagonals a d and b c, cutting each other in c. Then c is the center of the
square. Through c, draw e f to the vanishing-point of a b, and g h to the
vanishing-point of a c, and these lines will bisect the sides of the square, so
that a g is the perspective representation of half the side a b; a e is half a c;
c h is half c d; and b f is half b d.

Fig. 22.

COROLLARY III.
Since a b c d, Fig. 20., is a square, b a c is a right angle; and as t v is parallel
to a b, and t v′ to a c, v′ t v must be a right angle also.
As the ground plan of most buildings is rectangular, it constantly happens in
practice that their angles (as the corners of ordinary houses) throw the lines

to the vanishing-points thus at right angles; and so that this law is observed,
and v t v′ is kept a right angle, it does not matter in general practice whether
the vanishing-points are thrown a little more or a little less to the right or
left of s: but it matters much that the relation of the vanishing-points should
be accurate. Their position with respect to s merely causes the spectator to
see a little more or less on one side or other of the house, which may be a
matter of chance or choice; but their rectangular relation determines the
rectangular shape of the building, which is an essential point.

34
PROBLEM IX.

TO DRAW A SQUARE PILLAR, GIVEN IN POSITION AND MAGNITUDE, ITS
BASE AND TOP BEING IN HORIZONTAL PLANES.

Let a h, Fig. 23., be the square pillar.
Then, as it is given in position and magnitude, the position and magnitude
of the square it stands upon must be given (that is, the line a b or a c in
position), and the height of its side a e.

Find the sight-magnitudes of a b and a e.
Draw the two sides a b, a c, of the square
of the base, by Problem VIII., as in
Fig. 24. From the points a, b, and c, raise
vertical lines a e, c f, b g.
Make a e equal to the sight-magnitude of
a e.

Now because the top and base of the
Fig. 23. Fig. 24. pillar are in horizontal planes, the square
of its top, f g, is parallel to the square of
its base, b c.

Therefore the line e f is parallel to a c, and e g to a b.
Therefore e f has the same vanishing-point as a c, and e g the same
vanishing-point as a b.
From e draw e f to the vanishing-point of a c, cutting c f in f.
Similarly draw e g to the vanishing-point of a b, cutting b g in g.

Complete the square g f in h, by drawing g h to the vanishing-point of e f,
and f h to the vanishing-point of e g, cutting each other in h. Then a g h f is
the square pillar required.

35
COROLLARY.
It is obvious that if a e is equal to a c, the whole figure will be a cube, and
each side, a e f c and a e g b, will be a square in a given vertical plane. And
by making a b or a c longer or shorter in any given proportion, any form of
rectangle may be given to either of the sides of the pillar. No other rule is
therefore needed for drawing squares or rectangles in vertical planes.
Also any triangle may be thus drawn in a vertical plane, by inclosing it in a
rectangle and determining, in perspective ratio, on the sides of the
rectangle, the points of their contact with the angles of the triangle.
And if any triangle, then any polygon.
A less complicated construction will, however, be given hereafter.18

18 See page 96 (note), after you have read Problem XVI. Return to text

36
PROBLEM X.

TO DRAW A PYRAMID, GIVEN IN POSITION AND MAGNITUDE, ON A
SQUARE BASE IN A HORIZONTAL PLANE.

Fig. 25.

Let a b, Fig. 25., be the four-sided pyramid. As it is given in position and
magnitude, the square base on which it stands must be given in position and
magnitude, and its vertical height, c d.19

Fig. 26.

Draw a square pillar, a b g e, Fig. 26., on the square base of the pyramid, and
make
36
the height of the pillar a f equal to the vertical height of the pyramid
c d (Problem IX.). Draw the diagonals g f, h i, on the top of the square pillar,
cutting each other in c. Therefore c is the center of the square f g h i.
(Prob. VIII. Cor. II.)

Fig. 27.

Join c e, c a, c b.

Then a b c e is the pyramid required. If the base of the pyramid is above the
eye, as when a square spire is seen on the top of a church-tower, the
construction will be as in Fig. 27.

19 If, instead of the vertical height, the length of a d is given, the vertical must be deduced from
it. See the Exercises on this Problem in the Appendix, p. 79. Return to text

38
PROBLEM XI.

TO DRAW ANY CURVE IN A HORIZONTAL OR VERTICAL PLANE.

Fig. 28.

Let a b, Fig. 28., be the curve.

Inclose it in a rectangle, c d e f.
Fix the position of the point c or d, and draw the rectangle. (Problem VIII.
Coroll. I.)20
Let c d e f, Fig. 29., be the rectangle so drawn.
If an extremity of the curve, as a, is in a side of the rectangle, divide the side
c e, Fig. 29., so that a c shall be (in perspective ratio) to a e as a c is to a e in
Fig. 28. (Prob. V. Cor. II.)
Similarly determine the points of contact of the curve and rectangle e, f, g.

If an extremity of the curve, as b, is not in a
39
side of the rectangle, let fall the
perpendiculars b a, b b on the rectangle
sides. Determine the correspondent points a
and b in Fig. 29., as you have already
determined a, b, e, and f.

From b, Fig. 29., draw b b parallel to c d,21
Fig. 29. and from a draw a b to the vanishing-point
of d f, cutting each other in b. Then b is the
extremity of the curve.
Determine any other important point in the curve, as p, in the same way, by
letting fall p q and p r on the rectangle’s sides.
Any number of points in the curve may be thus determined, and the curve
drawn through the series; in most cases, three or four will be enough.
Practically, complicated curves may be better drawn in perspective by an
experienced eye than by rule, as the fixing of the various points in haste
involves too many chances of error; but it is well to draw a good many by
rule first, in order to give the eye its experience.22

COROLLARY.
If the curve required be a circle, Fig. 30., the rectangle which incloses it
will become a square, and the curve will have four points of contact, a b c d,
in the middle of the sides of the square.
Draw the square, and as a square may be drawn about a circle in any
position, draw it with its nearest side, e g, parallel to the sight-line.
Let e f, Fig. 31., be the square so drawn.

Draw
40
its diagonals e f, g h; and through the center of the square (determined
by their intersection) draw a b to the vanishing-point of g f, and c d parallel
to e g. Then the points a b c d are the four points of the circle’s contact.

Fig. 30.

Fig. 31.

On e g describe a half square, e l; draw the semicircle k a l; and from its
center, r, the diagonals r e, r g, cutting the circle in x, y.

From the points x y, where the circle cuts the diagonals, raise
perpendiculars, p x, q y, to e g.

From p and q draw p p′, q q′, to the vanishing-point of g f, cutting the
diagonals in m, n, and o, p.
Then m, n, o, p are four other points in the circle.

Through these eight points the circle may be drawn by the hand accurately
enough for general purposes; but any number of points required may, of
course, be determined, as in Problem XI.
The distance e p is approximately one-seventh of e g, and may be assumed to
be so in quick practice, as the error involved is not greater than would be
incurred in the hasty operation of drawing the circle and diagonals.
It41 may frequently happen that, in consequence of associated constructions,
it may be inconvenient to draw e g parallel to the sight-line, the square being
perhaps first constructed in some oblique direction. In such cases, q g and e p
must be determined in perspective ratio by the dividing-point, the line e g
being used as a measuring-line.

[Obs. In drawing Fig. 31. the station-point has been taken much nearer the paper than is usually
advisable, in order to show the character of the curve in a very distinct form.
If the student turns the book so that e g may be vertical, Fig. 31. will represent the construction for
drawing a circle in a vertical plane, the sight-line being then of course parallel to g l; and the
semicircles a d b, a c b, on each side of the diameter a b, will represent ordinary semicircular arches
seen in perspective. In that case, if the book be held so that the line e h is the top of the square, the
upper semicircle will represent a semicircular arch, above the eye, drawn in perspective. But if the
book be held so that the line g f is the top of the square, the upper semicircle will represent a
semicircular arch, below the eye, drawn in perspective.

If the book be turned upside down, the figure will represent a circle drawn on the ceiling, or any
other horizontal plane above the eye; and the construction is, of course, accurate in every case.]

20 Or if the curve is in a vertical plane, Coroll. to Problem IX. As a rectangle may be drawn in
any position round any given curve, its position with respect to the curve will in either case be
regulated by convenience. See the Exercises on this Problem, in the Appendix, p. 85.
Return to text
21 Or to its vanishing-point, if c d has one. Return to text
22 Of course, by dividing the original rectangle into any number of equal rectangles, and
dividing the perspective rectangle similarly, the curve may be approximately drawn without
any trouble; but, when accuracy is required, the points should be fixed, as in the problem.
Return to text

42
PROBLEM XII.
TO DIVIDE A CIRCLE DRAWN IN PERSPECTIVE INTO ANY GIVEN NUMBER
OF EQUAL PARTS.

Let a b, Fig. 32., be the circle drawn in perspective. It is required to divide it
into a given number of equal parts; in this case, 20.
Let k a l be the semicircle used in the construction. Divide the semicircle
k a l into half the number of parts required; in this case, 10.

Produce the line e g laterally, as far as may be necessary.

From o, the center of the semicircle k a l, draw radii through the points of
division of the semicircle, p, q, r, etc., and produce them to cut the line e g
in p, q, r, etc.

From the points p q r draw the lines p p′, q q′, r r′, etc., through the center of
the circle a b, each cutting the circle in two points of its circumference.

Then these points divide the perspective circle as required.
If from each of the points p, q, r, a vertical were raised to the line e g, as in
Fig. 31., and from the point where it cut e g a line were drawn to the
vanishing-point, as q q′ in Fig. 31., this line would also determine two of the
points of division.

43

Fig. 32.

If it is required to divide a circle into any number of given unequal parts (as
in the points a, b, and c, Fig. 33.), the shortest way is thus to raise vertical
lines from a and b to the side of the perspective square x y, and then draw to
the vanishing-point, cutting the perspective circle in a and b, the points
required. Only notice that if any point, as a, is on the nearer side of the
circle a b c, its representative point, a, must be on the nearer side of the
circle a b c; and if the point b is on the farther side of the circle a b c, b must
be
44
on the farther side of a b c. If any point, as c, is so much in the lateral arc
of the circle as not to be easily determinable by the vertical line, draw the
horizontal c p, find the correspondent p in the side of the perspective square,
and draw p c parallel to x y, cutting the perspective circle in c.

Fig. 33.

COROLLARY.
It is obvious that if the points p′, q′, r, etc., by which the circle is divided in
Fig. 32., be joined by right lines, the resulting figure will be a regular
equilateral figure of twenty sides inscribed in the circle. And if the circle be
divided into given unequal parts, and the points of division joined by right
lines, the resulting figure will be an irregular polygon inscribed in the circle
with sides of given length.
Thus any polygon, regular or irregular, inscribed in a circle, may be
inscribed in position in a perspective circle.

45
PROBLEM XIII.
TO DRAW A SQUARE, GIVEN IN MAGNITUDE, WITHIN A LARGER SQUARE
GIVEN IN POSITION AND MAGNITUDE; THE SIDES OF THE TWO SQUARES
BEING PARALLEL.

Fig. 34.

Let a b, Fig. 34., be the sight-magnitude of the side of the smaller square,
and a c that of the side of the larger square.
Draw the larger square. Let d e f g be the square so drawn.
Join e g and d f.

On either d e or d g set off, in perspective ratio, d h equal to one half of b c.
Through h draw h k to the vanishing-point of d e, cutting d f in i and e g in k.
Through i and k draw i m, k l, to vanishing-point of d g, cutting d f in l and
e g in m. Join l m.

Then i k l m is the smaller square, inscribed as required.23

46
COROLLARY.
If, instead of one square within
another, it be required to draw one
circle within another, the
dimensions of both being given,
inclose each circle in a square.
Fig. 36.
Draw the squares first, and then the
circles within, as in Fig. 36.

23 If either of the sides of the greater square is
parallel to the plane of the picture, as d g in
Fig. 35., d g of course must be equal to a c,
and d h equal to b c/2, and the construction is
as in Fig. 35.

Fig. 35.

Return to text

47
PROBLEM XIV.

TO DRAW A TRUNCATED CIRCULAR CONE, GIVEN IN POSITION AND
MAGNITUDE, THE TRUNCATIONS BEING IN HORIZONTAL PLANES, AND THE
AXIS OF THE CONE VERTICAL.

Let a b c d, Fig. 37., be the portion of the cone required.

Fig. 37.

As it is given in magnitude, its diameters must be given at the base and
summit, a b and c d; and its vertical height, c e.24
And as it is given in position, the center of its base must be given.
Draw
48
in position, about this center,25 the square pillar a f d, Fig. 38.,
making its height, b g, equal to c e; and its side, a b, equal to a b.

In the square of its base, a b c d, inscribe a circle, which therefore is of the
diameter of the base of the cone, a b.

In the square of its top, e f g h, inscribe concentrically a circle whose
diameter shall equal c d. (Coroll. Prob. XIII.)
Join the extremities of the circles by the right lines k l, n m. Then k l n m is
the portion of cone required.

Fig. 38.

COROLLARY I.
If similar polygons be inscribed in similar positions in the circles k n and
l m (Coroll. Prob. XII.), and the corresponding angles of the polygons
joined by right lines, the resulting figure will be a portion of a polygonal
pyramid. (The dotted lines in Fig. 38., connecting the extremities of two
diameters and one diagonal in the respective circles, occupy the position of
the three nearest angles of a regular octagonal pyramid, having its angles set
on the diagonals and diameters of the square a d, inclosing its base.)
If the cone or polygonal pyramid is not truncated, its apex will be the center
of the upper square, as in Fig. 26.

COROLLARY II.
If equal circles, or equal and similar polygons, be inscribed in the upper and
lower squares in Fig. 38., the resulting figure will be a vertical cylinder, or a
vertical
49
polygonal pillar, of given height and diameter, drawn in position.

COROLLARY III.
If the circles in Fig. 38., instead of being inscribed in the squares b c and
f g, be inscribed in the sides of the solid figure b e and d f, those sides being
made square, and the line b d of any given length, the resulting figure will
be, according to the constructions employed, a cone, polygonal pyramid,
cylinder, or polygonal pillar, drawn in position about a horizontal axis
parallel to b d.
Similarly, if the circles are drawn in the sides g d and e c, the resulting
figures will be described about a horizontal axis parallel to a b.

24 Or if the length of its side, a c, is given instead, take a e, Fig. 37., equal to half the excess of
a b over c d; from the point e raise the perpendicular c e. With center a, and distance a c,
describe a circle cutting c e in c. Then c e is the vertical height of the portion of cone required,
or c e. Return to text
25 The direction of the side of the square will of course be regulated by convenience.
Return to text

50
PROBLEM XV.

TO DRAW AN INCLINED LINE, GIVEN IN POSITION AND MAGNITUDE.

We have hitherto been examining the conditions of horizontal and vertical
lines only, or of curves inclosed in rectangles.

Fig. 39. Fig. 40.

We must, in conclusion, investigate the perspective of inclined lines,
beginning with a single one given in position. For the sake of completeness
of system, I give in Appendix II. Article III. the development of this
problem from the second. But, in practice, the position of an inclined line
may be most conveniently defined by considering it as the diagonal of a
rectangle, as a b in Fig. 39., and I shall therefore, though at some sacrifice of
system, examine it here under that condition.
If the sides of the rectangle a c and a d are given, the slope of the line a b is
determined; and then its position will depend on that of the rectangle. If, as
in Fig. 39., the rectangle is parallel to the picture plane, the line a b must be
so
51
also. If, as in Fig. 40., the rectangle is inclined to the picture plane, the
line a b will be so also. So that, to fix the position of a b, the line a c must be
given in position and magnitude, and the height a d.

If these are given, and it is only required to draw the single line
a b in perspective, the construction is entirely simple; thus:—

Draw the line a c by Problem I.
Let a c, Fig. 41., be the line so drawn. From a and c raise the
vertical lines a d, c b. Make a d equal to the sight-magnitude of Fig. 41.
a d. From d draw d b to the vanishing-point of a c, cutting b c
in b.
Join a b. Then a b is the inclined line required.
If the line is inclined in the opposite direction, as d c in
Fig. 42., we have only to join d c instead of a b in
Fig. 41., and d c will be the line required.
I shall hereafter call the line a c, when used to define the
position of an inclined line a b (Fig. 40.), the “relative
horizontal” of the line a b. Fig. 42.

Observation.

In general, inclined lines are most needed
for gable roofs, in which, when the
conditions are properly stated, the vertical
height of the gable, x y, Fig. 43., is given,
and the base line, a c, in position. When
these are given, draw a c; raise vertical
a d; make a d equal to sight-magnitude of
x y; complete the perspective-rectangle
Fig. 43.
a d b c; join a b and d c (as by dotted lines
in figure); and through the intersection of
the dotted lines draw vertical x y, cutting d b in y. Join a y, c y; and these lines
are
52
the sides of the gable. If the length of the roof a a′ is also given, draw in
perspective the complete parallelopiped a′ d′ b c, and from y draw y y′ to the

vanishing-point of a a′, cutting d′ b′ in y′. Join a′ y, and you have the slope of
the farther side of the roof.

Fig. 44.

The construction above the eye is as in Fig. 44.; the roof is reversed in
direction merely to familiarize the student with the different aspects of its
lines.

53
PROBLEM XVI.
TO FIND THE VANISHING-POINT OF A GIVEN INCLINED LINE.

If, in Fig. 43. or Fig. 44., the lines a y and a′ y′ be produced, the student will
find that they meet.
Let p, Fig. 45., be the point at which they meet.
From p let fall the vertical p v on the sight-line, cutting the sight-line in v.
Then the student will find experimentally that v is the vanishing-point of the
line a c.26
Complete the rectangle of the base a c′, by drawing a′ c′ to v, and c c′ to the
vanishing-point of a a′.
Join y′ c′.

Now if y c and y′ c′ be produced downwards, the student will find that they
meet.
Let them be produced, and meet in p′.
Produce p v, and it will be found to pass through the point p′.
Therefore if a y (or c y), Fig. 45., be any inclined line drawn in perspective
by Problem XV., and a c the relative horizontal (a c in Figs. 39, 40.), also
drawn in perspective.
Through v, the vanishing-point of a v, draw the vertical p p′ upwards and
downwards.
Produce a y (or c y), cutting p p′ in p (or p′).

Then p is the vanishing-point of a y (or p′ of c y).

Fig. 45.

The student will observe that, in order to find the point p by this method, it
is necessary first to draw a portion of the given inclined line by
Problem XV. Practically, it is always necessary to do so, and, therefore, I
give the problem in this form.

Theoretically,
54
as will be shown in the analysis of the problem, the point p
should be found by drawing a line from the station-point parallel to the
given inclined line: but there is no practical means of drawing such a line;
so that in whatever terms the problem may be given, a portion of the
inclined line (a y or c y) must always be drawn in perspective before p can be
found.

26 The demonstration is in Appendix II. Article III. Return to text

55
PROBLEM XVII.
TO FIND THE DIVIDING-POINTS OF A GIVEN INCLINED LINE.

Fig. 46.

Let p, Fig. 46., be the vanishing-point of the inclined line, and v the
vanishing-point of the relative horizontal.
Find the dividing-points of the relative horizontal, d and d′.
Through p draw the horizontal line x y.
With center p and distance d p describe the two arcs d x and d′ y, cutting the
line x y in x and y.

Then x and y are the dividing-points of the inclined line.27
Obs. The dividing-points found by the above rule, used with the ordinary
measuring-line, will lay off distances on the retiring inclined line, as the
ordinary dividing-points lay them off on the retiring horizontal line.
Another dividing-point, peculiar in its application, is sometimes useful, and
is to be found as follows:—

56

Fig. 47.

Let a b, Fig. 47., be the given inclined line drawn in perspective, and a c the
relative horizontal.
Find the vanishing-points, v and e, of a c and a b; d, the dividing-point of a c;
and the sight-magnitude of a c on the measuring-line, or a c.
From d erect the perpendicular d f.

Join c b, and produce it to cut d e in f. Join e f.
Then, by similar triangles, d f is equal to e v, and e f is parallel to d v.
Hence it follows that if from d, the dividing-point of a c, we raise a
perpendicular and make d f equal to e v, a line c f, drawn from any point c on
the measuring-line to f, will mark the distance a b on the inclined line, a b
being the portion of the given inclined line which forms the diagonal of the
vertical rectangle of which a c is the base.

27 The demonstration is in Appendix II., p. 104. Return to text

57
PROBLEM XVIII.

TO FIND THE SIGHT-LINE OF AN INCLINED PLANE IN WHICH TWO LINES
ARE GIVEN IN POSITION.28

As in order to fix the position of a line two points in it must be given, so in
order to fix the position of a plane, two lines in it must be given.

Fig. 48.

Let the two lines be a b and c d, Fig. 48.

As
58
they are given in position, the relative horizontals a e and c f must be
given.
Then by Problem XVI. the vanishing-point of a b is v, and of c d, v′.
Join v v′ and produce it to cut the sight-line in x.

Then v x is the sight-line of the inclined plane.

Like the horizontal sight-line, it is of indefinite length; and may be
produced in either direction as occasion requires, crossing the horizontal
line of sight, if the plane continues downward in that direction.
x is the vanishing-point of all horizontal lines in the inclined plane.

28 Read the Article on this problem in the Appendix, p. 97, before investigating the problem
itself. Return to text

59
PROBLEM XIX.
TO FIND THE VANISHING-POINT OF STEEPEST LINES IN AN INCLINED
PLANE WHOSE SIGHT-LINE IS GIVEN.

Fig. 49.

Let v x, Fig. 49., be the given sight-line.

Produce it to cut the horizontal sight-line in x.

Therefore x is the vanishing-point of horizontal lines in the given inclined
plane. (Problem XVIII.)
Join t x, and draw t y at right angles to t x.

Therefore y is the rectangular vanishing-point corresponding to x.29

From y erect the vertical y p, cutting the sight-line of the inclined plane in p.

Then
60
p is the vanishing-point of steepest lines in the plane.

All lines drawn to it, as q p, r p, n p, etc., are the steepest possible in the
plane; and all lines drawn to x, as q x, o x, etc., are horizontal, and at right
angles to the lines p q, p r, etc.

29 That is to say, the vanishing-point of horizontal lines drawn at right angles to the lines whose
vanishing-point is x. Return to text

61
PROBLEM XX.
TO FIND THE VANISHING-POINT OF LINES PERPENDICULAR TO THE
SURFACE OF A GIVEN INCLINED PLANE.

Fig. 50.

As the inclined plane is given, one of its steepest lines must be given, or
may be ascertained.
Let
62
a b, Fig. 50., be a portion of a steepest line in the given plane, and v the
vanishing-point of its relative horizontal.
Through v draw the vertical g f upwards and downwards.

From a set off any portion of the relative horizontal a c, and on a c describe a
semicircle in a vertical plane, a d c, cutting a b in e.

Join e c, and produce it to cut g f in f.

Then f is the vanishing-point required.
For, because a e c is an angle in a semicircle, it is a right angle; and therefore
the line e f is at right angles to the line a b; and similarly all lines drawn to f,
and therefore parallel to e f, are at right angles with any line which cuts
them, drawn to the vanishing-point of a b.
And because the semicircle a d c is in a vertical plane, and its diameter a c is
at right angles to the horizontal lines traversing the surface of the inclined
plane, the line e c, being in this semicircle, is also at right angles to such
traversing lines. And therefore the line e c, being at right angles to the
steepest lines in the plane, and to the horizontal lines in it, is perpendicular
to its surface.

T63he preceding series of constructions, with the examples in the first Article
of the Appendix, put it in the power of the student to draw any form,
however complicated,30 which does not involve intersection of curved
surfaces. I shall not proceed to the analysis of any of these more complex
problems, as they are entirely useless in the ordinary practice of artists. For
a few words only I must ask the reader’s further patience, respecting the
general placing and scale of the picture.
As the horizontal sight-line is drawn through the sight-point, and the sight-
point is opposite the eye, the sight-line is always on a level with the eye.
Above and below the sight-line, the eye comprehends, as it is raised or
depressed while the head is held upright, about an equal space; and, on each
side of the sight-point, about the same space is easily seen without turning
the head; so that if a picture represented the true field of easy vision, it
ought to be circular, and have the sight-point in its center. But because some
parts of any given view are usually more interesting than others, either the
uninteresting parts are left out, or somewhat more than would generally be
seen of the interesting parts is included, by moving the field of the picture a
little upwards or downwards, so as to throw the sight-point low or high. The
operation will be understood in a moment by cutting an aperture in a piece
of pasteboard, and moving it up and down in front of the eye, without
moving the eye. It will be seen to embrace sometimes the low, sometimes
the
64
high objects, without altering their perspective, only the eye will be
opposite the lower part of the aperture when it sees the higher objects, and
vice versâ.
There is no reason, in the laws of perspective, why the picture should not be
moved to the right or left of the sight-point, as well as up or down. But
there is this practical reason. The moment the spectator sees the horizon in a
picture high, he tries to hold his head high, that is, in its right place. When
he sees the horizon in a picture low, he similarly tries to put his head low.
But, if the sight-point is thrown to the left hand or right hand, he does not
understand that he is to step a little to the right or left; and if he places

himself, as usual, in the middle, all the perspective is distorted. Hence it is
generally unadvisable to remove the sight-point laterally, from the center of
the picture. The Dutch painters, however, fearlessly take the license of
placing it to the right or left; and often with good effect.
The rectilinear limitation of the sides, top, and base of the picture is of
course quite arbitrary, as the space of a landscape would be which was seen
through a window; less or more being seen at the spectator’s pleasure, as he
retires or advances.
The distance of the station-point is not so arbitrary. In ordinary cases it
should not be less than the intended greatest dimension (height or breadth)
of the picture. In most works by the great masters it is more; they not only
calculate on their pictures being seen at considerable distances, but they like
breadth of mass in buildings, and dislike the sharp angles which always
result from station-points at short distances.31
Whenever perspective, done by true rule, looks wrong, it is always because
the
65
station-point is too near. Determine, in the outset, at what distance the
spectator is likely to examine the work, and never use a station-point within
a less distance.
There is yet another and a very important reason, not only for care in
placing the station-point, but for that accurate calculation of distance and
observance of measurement which have been insisted on throughout this
work. All drawings of objects on a reduced scale are, if rightly executed,
drawings of the appearance of the object at the distance which in true
perspective reduces it to that scale. They are not small drawings of the
object seen near, but drawings the real size of the object seen far off. Thus if
you draw a mountain in a landscape, three inches high, you do not reduce
all the features of the near mountain so as to come into three inches of
paper. You could not do that. All that you can do is to give the appearance
of the mountain, when it is so far off that three inches of paper would really
hide it from you. It is precisely the same in drawing any other object. A face
can no more be reduced in scale than a mountain can. It is infinitely delicate
already; it can only be quite rightly rendered on its own scale, or at least on
the slightly diminished scale which would be fixed by placing the plate of

glass, supposed to represent the field of the picture, close to the figures.
Correggio and Raphael were both fond of this slightly subdued magnitude
of figure. Colossal painting, in which Correggio excelled all others, is
usually the enlargement of a small picture (as a colossal sculpture is of a
small statue), in order to permit the subject of it to be discerned at a
distance. The treatment of colossal (as distinguished from ordinary)
paintings will depend therefore, in general, on the principles of optics more
than on those of perspective, though, occasionally, portions may be
represented as if they were the projection of near objects on a plane behind
them. In all points the subject is one of great difficulty and subtlety; and its
examination does not fall within the compass of this essay.
Lastly,
66
it will follow from these considerations, and the conclusion is one of
great practical importance, that, though pictures may be enlarged, they
cannot be reduced, in copying them. All attempts to engrave pictures
completely on a reduced scale are, for this reason, nugatory. The best that
can be done is to give the aspect of the picture at the distance which reduces
it in perspective to the size required; or, in other words, to make a drawing
of the distant effect of the picture. Good painting, like nature’s own work, is
infinite, and unreduceable.
I wish this book had less tendency towards the infinite and unreduceable. It
has so far exceeded the limits I hoped to give it, that I doubt not the reader
will pardon an abruptness of conclusion, and be thankful, as I am myself, to
get to an end on any terms.

30 As in algebraic science, much depends, in complicated perspective, on the student’s ready
invention of expedients, and on his quick sight of the shortest way in which the solution may
be accomplished, when there are several ways. Return to text
31 The greatest masters are also fond of parallel perspective, that is to say, of having one side of
their buildings fronting them full, and therefore parallel to the picture plane, while the other
side vanishes to the sight-point. This is almost always done in figure backgrounds, securing
simple and balanced lines. Return to text

67
APPENDIX.

I.
PRACTICE AND OBSERVATIONS.
II.
DEMONSTRATIONS.

69
I.

PRACTICE AND OBSERVATIONS ON THE
PRECEDING PROBLEMS.

Problem I.
An example will be necessary to make this problem clear to the general
student.
The nearest corner of a piece of pattern on the carpet is 4½ feet beneath the
eye, 2 feet to our right and 3½ feet in direct distance from us. We intend to
make a drawing of the pattern which shall be seen properly when held
1½ foot from the eye. It is required to fix the position of the corner of the
piece of pattern.
Let a b, Fig. 51., be our sheet of paper, some
3 feet wide. Make s t equal to 1½ foot. Draw
the line of sight through s. Produce t s, and
make d s equal to 2 feet, therefore t d equal
to 3½ feet. Draw d c, equal to 2 feet; c p,
equal to 4 feet. Join t c (cutting the sight-line
in q) and t p.

Let fall the vertical q p′, then p′ is the point
required.
If the lines, as in the figure, fall outside of
your sheet of paper, in order to draw them, it Fig. 51.
is necessary to attach other sheets of paper to

its
70
edges. This is inconvenient, but must be done at first that you may see
your way clearly; and sometimes afterwards, though there are expedients
for doing without such extension in fast sketching.
It is evident, however, that no extension of surface could be of any use to
us, if the distance t d, instead of being 3½ feet, were 100 feet, or a mile, as
it might easily be in a landscape.
It is necessary, therefore, to obtain some other means of construction; to do
which we must examine the principle of the problem.

In the analysis of Fig. 2., in the introductory remarks, I used the word
“height” only of the tower, q p, because it was only to its vertical height that
the law deduced from the figure could be applied. For suppose it had been a
pyramid, as o q p, Fig. 52., then the image of its side, q p, being, like every
other magnitude, limited on the glass a b by the lines coming from its
extremities, would appear only of the length q′ s; and it is not true that q′ s is
to q p as t s is to t p. But if we let fall a vertical q d from q, so as to get the
vertical height of the pyramid, then it is true that q′ s is to q d as t s is to t d.

Fig. 52.

Supposing this figure represented, not a pyramid, but a triangle on the
ground, and that q d and q p are horizontal lines, expressing lateral distance
from the line t d, still the rule would be false for q p and true for q d. And,
similarly,
71
it is true for all lines which are parallel, like q d, to the plane of the
picture a b, and false for all lines which are inclined to it at an angle.

Hence generally. Let p q (Fig. 2. in Introduction, p. 6) be any magnitude
parallel to the plane of the picture; and p′ q′ its image on the picture.
Then always the formula is true which you learned in the Introduction: p′ q′
is to p q as s t is to d t.
Now the magnitude p dash q dash in this formula I call the “sight-
magnitude” of the line p q. The student must fix this term, and the meaning
of it, well in his mind. The “sight-magnitude” of a line is the magnitude
which bears to the real line the same proportion that the distance of the
picture bears to the distance of the object. Thus, if a tower be a hundred feet
high, and a hundred yards off; and the picture, or piece of glass, is one yard
from the spectator, between him and the tower; the distance of picture being
then to distance of tower as 1 to 100, the sight-magnitude of the tower’s
height will be as 1 to 100; that is to say, one foot. If the tower is two
hundred yards distant, the sight-magnitude of its height will be half a foot,
and so on.
But farther. It is constantly necessary, in perspective operations, to measure
the other dimensions of objects by the sight-magnitude of their vertical
lines. Thus, if the tower, which is a hundred feet high, is square, and
twenty-five feet broad on each side; if the sight-magnitude of the height is
one foot, the measurement of the side, reduced to the same scale, will be the
hundredth part of twenty-five feet, or three inches: and, accordingly, I use in
this treatise the term “sight-magnitude” indiscriminately for all lines
reduced in the same proportion as the vertical lines of the object. If I tell
you to find the “sight-magnitude” of any line, I mean, always, find the
magnitude which bears to that line the proportion of s t to d t; or, in simpler
terms, reduce the line to the scale which you have fixed by the first
determination of the length s t.

Therefore, you must learn to draw quickly to scale before you do anything
else;
72
for all the measurements of your object must be reduced to the scale
fixed by s t before you can use them in your diagram. If the object is fifty
feet from you, and your paper one foot, all the lines of the object must be
reduced to a scale of one fiftieth before you can use them; if the object is
two thousand feet from you, and your paper one foot, all your lines must be

reduced to the scale of one two-thousandth before you can use them, and so
on. Only in ultimate practice, the reduction never need be tiresome, for, in
the case of large distances, accuracy is never required. If a building is three
or four miles distant, a hairbreadth of accidental variation in a touch makes
a difference of ten or twenty feet in height or breadth, if estimated by
accurate perspective law. Hence it is never attempted to apply
measurements with precision at such distances. Measurements are only
required within distances of, at the most, two or three hundred feet. Thus it
may be necessary to represent a cathedral nave precisely as seen from a spot
seventy feet in front of a given pillar; but we shall hardly be required to
draw a cathedral three miles distant precisely as seen from seventy feet in
advance of a given milestone. Of course, if such a thing be required, it can
be done; only the reductions are somewhat long and complicated: in
ordinary cases it is easy to assume the distance s t so as to get at the reduced
dimensions in a moment. Thus, let the pillar of the nave, in the case
supposed, be 42 feet high, and we are required to stand 70 feet from it:
assume s t to be equal to 5 feet. Then, as 5 is to 70 so will the sight-
magnitude required be to 42; that is to say, the sight-magnitude of the
pillar’s height will be 3 feet. If we make s t equal to 2½ feet, the pillar’s
height will be 1½ foot, and so on.
And for fine divisions into irregular parts which cannot be measured, the
ninth and tenth problems of the sixth book of Euclid will serve you: the
following construction is, however, I think, more practically convenient:—
The line a b (Fig. 53.) is divided by given points, a, b, c, into a given
number of irregularly unequal parts; it is required to divide any other line,
c d, into an equal number of parts, bearing to each other the same
73

proportions as the parts of a b, and arranged in the same order.

Draw the two lines parallel to each other, as in the figure.
Join a c and b d, and produce the lines a c, b d, till they meet in p.
Join a p, b p, c p, cutting c d in f, g, h.
Then the line c d is divided as required, in f, g, h.

In the figure the lines a b and c d are accidentally perpendicular to a p. There
is no need for their being so.

Fig. 53.

Now, to return to our first problem.
The construction given in the figure is only the quickest mathematical way
of obtaining, on the picture, the sight-magnitudes of d c and p c, which are
both magnitudes parallel with the picture plane. But if these magnitudes are
too great to be thus put on the paper, you have only to obtain the reduction
by scale. Thus, if t s be one foot, t d eighty feet, d c forty feet, and c p ninety
feet, the distance q s must be made equal to one eightieth of d c, or half a
foot; and the distance q p′, one eightieth of c p, or one eightieth of ninety
feet; that is to say, nine eighths of a foot, or thirteen and a half inches. The
lines c t and p t are thus practically useless, it being only necessary to
measure
74
q s and q p, on your paper, of the due sight-magnitudes. But the
mathematical construction, given in Problem I., is the basis of all
succeeding problems, and, if it is once thoroughly understood and practiced
(it can only be thoroughly understood by practice), all the other problems
will follow easily.

Lastly. Observe that any perspective operation whatever may be performed
with reduced dimensions of every line employed, so as to bring it
conveniently within the limits of your paper. When the required figure is
thus constructed on a small scale, you have only to enlarge it accurately in
the same proportion in which you reduced the lines of construction, and you
will have the figure constructed in perspective on the scale required for use.

75
PROBLEM IX.
The drawing of most buildings occurring in ordinary practice will resolve
itself into applications of this problem. In general, any house, or block of
houses, presents itself under the main conditions assumed here in Fig. 54.
There will be an angle or corner somewhere near the spectator, as a b; and
the level of the eye will usually be above the base of the building, of which,
therefore, the horizontal upper lines will slope down to the vanishing-
points, and the base lines rise to them. The following practical directions
will, however, meet nearly all cases:—

Fig. 54.

Let a b, Fig. 54., be any important vertical line in the block of buildings; if it
is the side of a street, you may fix upon such a line at the division between
two houses. If its real height, distance, etc., are given, you will proceed with
76
the accurate construction of the problem; but usually you will neither know,

nor care, exactly how high the building is, or how far off. In such case draw
the line a b, as nearly as you can guess, about the part of the picture it ought
to occupy, and on such a scale as you choose. Divide it into any convenient
number of equal parts, according to the height you presume it to be. If you
suppose it to be twenty feet high, you may divide it into twenty parts, and
let each part stand for a foot; if thirty feet high, you may divide it into ten
parts, and let each part stand for three feet; if seventy feet high, into
fourteen parts, and let each part stand for five feet; and so on, avoiding thus
very minute divisions till you come to details. Then observe how high your
eye reaches upon this vertical line; suppose, for instance, that it is thirty feet
high and divided into ten parts, and you are standing so as to raise your
head to about six feet above its base, then the sight-line may be drawn, as in
the figure, through the second division from the ground. If you are standing
above the house, draw the sight-line above b; if below the house, below a; at
such height or depth as you suppose may be accurate (a yard or two more or
less matters little at ordinary distances, while at great distances perspective
rules become nearly useless, the eye serving you better than the necessarily
imperfect calculation). Then fix your sight-point and station-point, the latter
with proper reference to the scale of the line a b. As you cannot, in all
probability, ascertain the exact direction of the line a v or b v, draw the slope
b v as it appears to you, cutting the sight-line in v. Thus having fixed one
vanishing-point, the other, and the dividing-points, must be accurately
found by rule; for, as before stated, whether your entire group of points
(vanishing and dividing) falls a little more or less to the right or left of s
does not signify, but the relation of the points to each other does signify.
Then draw the measuring-line b g, either through a or b, choosing always the
steeper slope of the two; divide the measuring-line into parts of the same
length
77
as those used on a b, and let them stand for the same magnitudes.
Thus, suppose there are two rows of windows in the house front, each
window six feet high by three wide, and separated by intervals of three feet,
both between window and window and between tier and tier; each of the
divisions here standing for three feet, the lines drawn from b g to the
dividing-point d fix the lateral dimensions, and the divisions on a b the
vertical ones. For other magnitudes it would be necessary to subdivide the
parts on the measuring-line, or on a b, as required. The lines which regulate

the inner sides or returns of the windows (a, b, c, etc.) of course are drawn
to the vanishing-point of b f (the other side of the house), if f b v represents a
right angle; if not, their own vanishing-point must be found separately for
these returns. But see Practice on Problem XI.

Fig. 55.

Interior angles, such as e b c, Fig. 55. (suppose the corner of a room), are to
be treated in the same way, each side of the room having its measurements
separately carried to it from the measuring-line. It may sometimes happen
in such cases that we have to carry the measurement up from the corner b,
and that the sight-magnitudes are given us from the length of the line a b.
For instance, suppose the room is eighteen feet high, and therefore a b is
eighteen feet; and we have to lay off lengths of six feet on the top of the
room
78
wall, b c. Find d, the dividing-point of b c. Draw a measuring-line, b f,
from b; and another, g c, anywhere above. On b f lay off b g equal to one
third of a b, or six feet; and draw from d, through g and b, the lines g g, b b,
to the upper measuring-line. Then g b is six feet on that measuring-line.
Make b c, c h, etc., equal to b g; and draw c e, h f, etc., to d, cutting b c in e
and f, which mark the required lengths of six feet each at the top of the wall.

79
PROBLEM X.
This is one of the most important foundational problems in perspective, and
it is necessary that the student should entirely familiarize himself with its
conditions.
In order to do so, he must first observe these general relations of magnitude
in any pyramid on a square base.
Let a g h′, Fig. 56., be any pyramid on a square base.
The best terms in which its magnitude can be given, are
the length of one side of its base, a h, and its vertical
altitude (c d in Fig. 25.); for, knowing these, we know all
the other magnitudes. But these are not the terms in
which its size will be usually ascertainable. Generally, we
shall have given us, and be able to ascertain by
Fig. 56. measurement, one side of its base a h, and either a g the
length of one of the lines of its angles, or b g (or b′ g) the
length of a line drawn from its vertex, g, to the middle of the side of its
base. In measuring a real pyramid, a g will usually be the line most easily
found; but in many architectural problems b g is given, or is most easily
ascertainable.
Observe therefore this general construction.
Let a b d e, Fig. 57., be the square base of any pyramid.
Draw its diagonals, a e, b d, cutting each other in its center, c.

Bisect any side, a b, in f.
From f erect vertical f g.
Produce f b to h, and make f h equal to a c.

Now if the vertical altitude of the pyramid (c d in
Fig. 25.) be given, make f g equal to this vertical
altitude.
Join
80
g b and g h.

Then g b and g h are the true magnitudes of g b and
g h in Fig. 56.

If g b is given, and not the vertical altitude, with
center b, and distance g b, describe circle cutting f g
in g, and f g is the vertical altitude.
Fig. 57.
If g h is given, describe the circle from h, with
distance g h, and it will similarly cut f g in g.

It is especially necessary for the student to examine this construction
thoroughly, because in many complicated forms of ornaments, capitals of
columns, etc., the lines b g and g h become the limits or bases of curves,
which are elongated on the longer (or angle) profile g h, and shortened on
the shorter (or lateral) profile b g. We will take a simple instance, but must
previously note another construction.
It is often necessary, when pyramids are the roots of some ornamental form,
to divide them horizontally at a given vertical height. The shortest way of
doing so is in general the following.

Fig. 58.

Let a e c, Fig. 58., be any pyramid on a square base a b c, and a d c the square
pillar used in its construction.
Then
81
by construction (Problem X.) b d and a f are both of the vertical height
of the pyramid.
Of the diagonals, f e, d e, choose the shortest (in this case d e), and produce
it to cut the sight-line in v.
Therefore v is the vanishing-point of d e.
Divide d b, as may be required, into the sight-magnitudes of the given
vertical heights at which the pyramid is to be divided.

Fig. 59. Fig. 60.

From the points of division, 1, 2, 3, etc., draw to the vanishing-point v. The
lines so drawn cut the angle line of the pyramid, b e, at the required
elevations. Thus, in the figure, it is required to draw a horizontal black band
on the pyramid at three fifths of its height, and in breadth one twentieth of
its height. The line b d is divided into five parts, of which three are counted
from b upwards. Then the line drawn to v marks the base of the black band.
Then one fourth of one of the five parts is measured, which similarly gives
the breadth of the band. The terminal lines of the band are then drawn on
the sides of the pyramid parallel to a b (or to its vanishing-point if it has
one), and to the vanishing-point of b c.
If82 it happens that the vanishing-points of the diagonals are awkwardly
placed for use, bisect the nearest base line of the pyramid in b, as in Fig. 59.

Erect the vertical d b and join g b and d g (g being the apex of pyramid).
Find the vanishing-point of d g, and use d b for division, carrying the
measurements to the line g b.

In Fig. 59., if we join a d and d c, a d c is the vertical profile of the whole
pyramid, and b d c of the half pyramid, corresponding to f g b in Fig. 57.

Fig. 61.

We may now proceed to an architectural example.
Let a h, Fig. 60., be the vertical profile of the capital of a pillar, a b the semi-
diameter of its head or abacus, and f d the semi-diameter of its shaft.

Let the shaft be circular, and the abacus square, down to the level e.
Join b d, e f, and produce them to meet in g.
Therefore e c g is the semi-profile of a reversed pyramid containing the
capital.
Construct
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this pyramid, with the square of the abacus, in the required
perspective, as in Fig. 61.; making a e equal to a e in Fig. 60., and a k, the
side of the square, equal to twice a b in Fig. 60. Make e g equal to c g, and e d

equal to c d. Draw d f to the vanishing-point of the diagonal d v (the figure is
too small to include this vanishing-point), and f is the level of the point f in
Fig. 60., on the side of the pyramid.
Draw f m, f n, to the vanishing-points of a h and a k. Then f n and f m are
horizontal lines across the pyramid at the level f, forming at that level two
sides of a square.

Fig. 62.

Complete the square, and within it inscribe a circle, as in Fig. 62., which is
left unlettered that its construction may be clear. At the extremities of this
draw vertical lines, which will be the sides of the shaft in its right place. It
will be found to be somewhat smaller in diameter than the entire shaft in
Fig. 60., because at the center of the square it is more distant than the
nearest edge of the square abacus. The curves of the capital may then be
drawn approximately by the eye. They are not quite accurate in Fig. 62.,
there
84
being a subtlety in their junction with the shaft which could not be
shown on so small a scale without confusing the student; the curve on the
left springing from a point a little way round the circle behind the shaft, and

that on the right from a point on this side of the circle a little way within the
edge of the shaft. But for their more accurate construction see Notes on
Problem XIV.

85
PROBLEM XI.
It is seldom that any complicated curve, except occasionally a spiral, needs
to be drawn in perspective; but the student will do well to practice for some
time any fantastic shapes which he can find drawn on flat surfaces, as on
wall-papers, carpets, etc., in order to accustom himself to the strange and
great changes which perspective causes in them.

Fig. 63.

The curves most required in architectural drawing, after the circle, are those
of pointed arches; in which, however, all that will be generally needed is to
fix the apex, and two points in the sides. Thus if we have to draw a range of
pointed arches, such as a p b, Fig. 63., draw the measured arch to its sight-
magnitude first neatly in a rectangle, a b c d; then draw the diagonals a d and
b c; where they cut the curve draw a horizontal line (as at the level e in the
figure), and carry it along the range to the vanishing-point, fixing the points
where the arches cut their diagonals all along. If the arch is cusped, a line
should be drawn, at f to mark the height of the cusps, and verticals raised at
86
g and h, to determine the interval between them. Any other points may be
similarly determined, but these will usually be enough. Figure 63. shows

the perspective construction of a square niche of good Veronese Gothic,
with an uncusped arch of similar size and curve beyond.

Fig. 64.

In Fig. 64. the more distant arch only is lettered, as the construction of the
nearest explains itself more clearly to the eye without letters. The more
distant arch shows the general construction for all arches seen underneath,
as of bridges, cathedral aisles, etc. The rectangle a b c d is first drawn to
contain the outside arch; then the depth of the arch, a a, is determined by the
measuring-line, and the rectangle, a b c d, drawn for the inner arch.
a a, b b, etc., go to one vanishing-point; a b, a b, etc., to the opposite one.

In the nearer arch another narrow rectangle is drawn to determine the cusp.
The parts which would actually come into sight are slightly shaded.

87
PROBLEM XIV.
Several exercises will be required on this important problem.
I. It is required to draw a circular flat-bottomed dish narrower at the bottom
than the top; the vertical depth being given, and the diameter at the top and
bottom.

Fig. 65.

Let a b, Fig. 65., be the diameter of the bottom, a c the diameter of the top,
and a d its vertical depth.
Take a d in position equal to a c.

On a d draw the square a b c d, and inscribe in it a circle.
Therefore, the circle so inscribed has the diameter of the top of the dish.
From a and d let fall verticals, a e, d h, each equal to a d.

Join e h, and describe square e f g h, which accordingly will be equal to the
square a b c d, and be at the depth a d beneath it.
Within the square e f g h describe a square i k, whose diameter shall be equal
to a b.
Describe a circle within the square i k. Therefore the circle so inscribed has
its
88
diameter equal to a b; and it is in the center of the square e f g h, which is
vertically beneath the square a b c d.
Therefore the circle in the square i k represents the bottom of the dish.

Now the two circles thus drawn will either intersect one another, or they
will not.
If they intersect one another, as in the figure, and they are below the eye,
part of the bottom of the dish is seen within it.

Fig. 66.

To avoid confusion, let us take then two intersecting circles without the
inclosing squares, as in Fig. 66.
Draw right lines, a b, c d, touching both circles externally. Then the parts of
these lines which connect the circles are the sides of the dish. They are
drawn in Fig. 65. without any prolongations, but the best way to construct
them is as in Fig. 66.
If the circles do not intersect each other, the smaller must either be within
the larger or not within it.
If within the larger, the whole of the bottom of the dish is seen from above,
Fig. 67. a.

If the smaller circle is not within the larger, none of the
bottom is seen inside the dish, b.
If the circles are above instead of beneath the eye, the
bottom of the dish is seen beneath it, c.
If one circle is above and another beneath the eye, neither
the bottom nor top of the dish is seen, d. Unless the
object be very large, the circles in this case will have
little apparent curvature.
89
II. The preceding problem is simple, because the lines of
the profile of the object (a b and c d, Fig. 66.) are
straight. But if these lines of profile are curved, the
Fig. 67.
problem becomes much more complex: once mastered,
however, it leaves no farther difficulty in perspective.
Let it be required to draw a flattish circular cup or vase, with a given curve
of profile.
The basis of construction is given in Fig. 68., half of it only being drawn, in
order that the eye may seize its lines easily.

Fig. 68.

Two squares (of the required size) are first drawn, one above the other, with
a given vertical interval, a c, between them, and each is divided into eight
parts by its diameters and diagonals. In these squares two circles are drawn;
which are, therefore, of equal size, and one above the other. Two smaller

circles, also of equal size, are drawn within these larger circles in the
construction of the present problem; more may be necessary in some, none
at all in others.
It will be seen that the portions of the diagonals and diameters of squares
which are cut off between the circles represent radiating planes, occupying
the position of the spokes of a wheel.
Now let the line a e b, Fig. 69., be the profile of the vase or cup to be drawn.

Inclose it in the rectangle c d, and if any portion of it is not curved, as a e,
cut off the curved portion by the vertical line e f, so as to include it in the
smaller rectangle f d.

Draw
90
the rectangle a c b d in position, and upon it construct two squares, as
they are constructed on the rectangle a c d in Fig. 68.; and complete the
construction of Fig. 68., making the radius of its large outer circles equal to
a d, and of its small inner circles equal to a e.

The planes which occupy the position of the wheel spokes will then each
represent a rectangle of the size of f d. The construction is shown by the
dotted lines in Fig. 69.; c being the center of the uppermost circle.

Fig. 69.

Within each of the smaller rectangles between the circles, draw the curve e b
in perspective, as in Fig. 69.
Draw the curve x y, touching and inclosing the curves in the rectangles, and
meeting the upper circle at y.32
Then x y is the contour of the surface of the cup, and the upper circle is its
lip.
If the line x y is long, it may be necessary to draw other rectangles between
the eight principal ones; and, if the curve of profile a b is complex or
retorted, there may be several lines corresponding to x y, inclosing the
successive waves of the profile; and the outer curve will then be an
undulating or broken one.
91

Fig. 70.

III. All branched ornamentation, forms of flowers, capitals of columns,
machicolations of round towers, and other such arrangements of radiating
curve, are resolvable by this problem, using more or fewer interior circles
according to the conditions of the curves. Fig. 70. is an example of the
construction of a circular group of eight trefoils with curved stems. One
outer or limiting circle is drawn within the square e d c f, and the extremities
of
92
the trefoils touch it at the extremities of its diagonals and diameters. A
smaller circle is at the vertical distance b c below the larger, and a is the
angle of the square within which the smaller circle is drawn; but the square

is not given, to avoid confusion. The stems of the trefoils form drooping
curves, arranged on the diagonals and diameters of the smaller circle, which
are dotted. But no perspective laws will do work of this intricate kind so
well as the hand and eye of a painter.
IV. There is one common construction, however, in which, singularly, the
hand and eye of the painter almost always fail, and that is the fillet of any
ordinary capital or base of a circular pillar (or any similar form). It is rarely
necessary in practice to draw such minor details in perspective; yet the
perspective laws which regulate them should be understood, else the eye
does not see their contours rightly until it is very highly cultivated.

Fig. 71.

Fig. 71. will show the law with sufficient clearness; it represents the
perspective construction of a fillet whose profile is a semicircle, such as f h
in Fig. 60., seen above the eye. Only half the pillar with half the fillet is
drawn, to avoid confusion.

q is the center of the shaft.
93

p q the thickness of the fillet, sight-magnitude at the shaft’s center.

Round p a horizontal semicircle is drawn on the diameter of the shaft a b.

Round q another horizontal semicircle is drawn on diameter c d.
These two semicircles are the upper and lower edges of the fillet.
Then diagonals and diameters are drawn as in Fig. 68., and, at their
extremities, semicircles in perspective, as in Fig. 69.
The letters a, b, c, d, and e, indicate the upper and exterior angles of the
rectangles in which these semicircles are to be drawn; but the inner vertical
line is not dotted in the rectangle at c, as it would have confused itself with
other lines.
Then the visible contour of the fillet is the line which incloses and touches33
all the semicircles. It disappears behind the shaft at the point h, but I have
drawn it through to the opposite extremity of the diameter at d.
Turned upside down the figure shows the construction of a basic fillet.
The capital of a Greek Doric pillar should be drawn frequently for exercise
on this fourteenth problem, the curve of its echinus being exquisitely subtle,
while the general contour is simple.

32 This point coincides in the figure with the extremity of the horizontal diameter, but only
accidentally. Return to text
33 The engraving is a little inaccurate; the inclosing line should touch the dotted semicircles at a
and b. The student should draw it on a large scale. Return to text

94
PROBLEM XVI.
It is often possible to shorten other perspective operations considerably, by
finding the vanishing-points of the inclined lines of the object. Thus, in
drawing the gabled roof in Fig. 43., if the gable a y c be drawn in
perspective, and the vanishing-point of a y determined, it is not necessary to
draw the two sides of the rectangle, a′ d′ and d′ b′, in order to determine the
point y′; but merely to draw y y′ to the vanishing-point of a a′ and a′ y′ to the
vanishing-point of a y, meeting in y′, the point required.

Again, if there be a series of gables, or other figures produced by parallel
inclined lines, and retiring to the point v, as in Fig. 72.,34 it is not necessary
to draw each separately, but merely to determine their breadths on the line
a v, and draw the slopes of each to their vanishing-points, as shown in
Fig. 72. Or if the gables are equal in height, and a line be drawn from y to v,
the construction resolves itself into a zigzag drawn alternately to p and q,
between the lines y v and a v.
The student must be very cautious, in finding the vanishing-points of
inclined lines, to notice their relations to the horizontals beneath them, else
he may easily mistake the horizontal to which they belong.
Thus, let a b c d, Fig. 73., be a rectangular inclined plane, and let it be
required to find the vanishing-point of its diagonal b d.
Find v, the vanishing-point of a d and b c.

Draw a e to the opposite vanishing-point, so that d a e may represent a right
angle.
Let fall from b the vertical b e, cutting a e in e.

Join e d, and produce it to cut the sight-line in v′.

95

Fig. 72.

Then,
96
since the point e is vertically under the point b, the horizontal line e d
is vertically under the inclined line b d.

Fig. 73.

So that if we now let fall the vertical v′ p from v′, and produce b d to cut v′ p
in p, the point p will be the vanishing-point of b d, and of all lines parallel to
it.35

34 The diagram is inaccurately cut. y v should be a right line. Return to text
35 The student may perhaps understand this construction better by completing the rectangle
a d f e, drawing d f to the vanishing-point of a e, and e f to v. The whole figure, b f, may then
be conceived as representing half the gable roof of a house, a f the rectangle of its base, and
a c the rectangle of its sloping side.

In nearly all picturesque buildings, especially on the Continent, the slopes of gables are much
varied (frequently unequal on the two sides), and the vanishing-points of their inclined lines
become very important, if accuracy is required in the intersections of tiling, sides of dormer
windows, etc.
Obviously, also, irregular triangles and polygons in vertical planes may be more easily
constructed by finding the vanishing-points of their sides, than by the construction given in
the corollary to Problem IX.; and if such triangles or polygons have others concentrically
inscribed within them, as often in Byzantine mosaics, etc., the use of the vanishing-points will
become essential. Return to text

97
PROBLEM XVIII.
Before examining the last three problems it is necessary that you should
understand accurately what is meant by the position of an inclined plane.
Cut a piece of strong white pasteboard into any irregular shape, and dip it in
a sloped position into water. However you hold it, the edge of the water, of
course, will always draw a horizontal line across its surface. The direction
of this horizontal line is the direction of the inclined plane. (In beds of rock
geologists call it their “strike.”)

Fig. 74.

Next, draw a semicircle on the piece of pasteboard; draw its diameter, a b,
Fig. 74., and a vertical line from its center, c d; and draw some other lines,
c e, c f, etc., from the center to any points in the circumference.

Now dip the piece of pasteboard again into water, and, holding it at any
inclination and in any direction you choose, bring the surface of the water
to the line a b. Then the line c d will be the most steeply inclined of all the
lines drawn to the circumference of the circle; g c and h c will be less steep;
and e c and f c less steep still. The nearer the lines to c d, the steeper they
will be; and the nearer to a b, the more nearly horizontal.
98
When, therefore, the line a b is horizontal (or marks the water surface), its
direction is the direction of the inclined plane, and the inclination of the line

d c is the inclination of the inclined plane. In beds of rock geologists call the
inclination of the line d c their “dip.”

To fix the position of an inclined plane, therefore, is to determine the
direction of any two lines in the plane, a b and c d, of which one shall be
horizontal and the other at right angles to it. Then any lines drawn in the
inclined plane, parallel to a b, will be horizontal; and lines drawn parallel to
c d will be as steep as c d, and are spoken of in the text as the “steepest
lines” in the plane.
But farther, whatever the direction of a plane may be, if it be extended
indefinitely, it will be terminated, to the eye of the observer, by a boundary
line, which, in a horizontal plane, is horizontal (coinciding nearly with the
visible horizon);—in a vertical plane, is vertical;—and, in an inclined plane,
is inclined.
This line is properly, in each case, called the “sight-line” of such plane; but
it is only properly called the “horizon” in the case of a horizontal plane: and
I have preferred using always the term “sight-line,” not only because more
comprehensive, but more accurate; for though the curvature of the earth’s
surface is so slight that practically its visible limit always coincides with the
sight-line of a horizontal plane, it does not mathematically coincide with it,
and the two lines ought not to be considered as theoretically identical,
though they are so in practice.
It is evident that all vanishing-points of lines in any plane must be found on
its sight-line, and, therefore, that the sight-line of any plane may be found
by joining any two of such vanishing-points. Hence the construction of
Problem XVIII.

99
II.

DEMONSTRATIONS WHICH COULD NOT
CONVENIENTLY BE INCLUDED IN THE TEXT.

I.

THE SECOND COROLLARY, PROBLEM II.

In Fig. 8. omit the lines c d, c′ d′, and d s; and, as here in Fig. 75., from a
draw a d parallel to a b, cutting b t in d; and from d draw d e parallel to b c′.

Fig. 75.

Now as a d is parallel to a b—

a c ∶ a c ∷ b c′ ∶ d e;
but a c is equal to b c′—
∴ a c = d e.

Now
100
because the triangles a c v, b c′ v, are similar—

a c ∶ b c′ ∷ a v ∶ b v;
and because the triangles d e t, b c′ t are similar—
d e ∶ b c′ ∷ d t ∶ b t.

But a c is equal to d e—

∴ a v ∶ b v ∷ d t ∶ b t;
∴ the two triangles a b d, b t v, are similar, and their angles are alternate;
∴ t v is parallel to a d.

But a d is parallel to a b—

∴ t v is parallel to a b.

101
II.

THE THIRD COROLLARY, PROBLEM III.

In Fig. 13., since a r is by construction parallel to a b in Fig. 12., and t v is
by construction in Problem III. also parallel to a b—

∴ a r is parallel to t v,
∴ a b r and t b v are alternate triangles,
∴ a r ∶ t v ∷ a b ∶ b v.

Again, by the construction of Fig. 13., a r′ is parallel to m v—

∴ a b r′ and m b v are alternate triangles,
∴ a r′ ∶ m v ∷ a b ∶ b v.

And it has just been shown that also
a r ∶ t v ∷ a b ∶ b v—
∴ a r′ ∶ m v ∷ a r ∶ t v.

But by construction, a r′ = a r—

∴ m v = t v.

102
III.
ANALYSIS OF PROBLEM XV.

We proceed to take up the general condition of the second problem, before
left unexamined, namely, that in which the vertical distances b c′ and a c
(Fig. 6. page 13), as well as the direct distances t d and t d′ are unequal.
In Fig. 6., here repeated (Fig. 76.), produce c′ b downwards, and make c′ e
equal to c a.

Fig. 76.

Join a e.

Then, by the second Corollary of Problem II., a e is a horizontal line.

Draw t v parallel to a e, cutting the sight-line in v.

∴ v is the vanishing-point of a e.

Complete
103
the constructions of Problem II. and its second Corollary.
Then by Problem II. a b is the line a b drawn in perspective; and by its
Corollary a e is the line a e drawn in perspective.

From v erect perpendicular v p, and produce a b to cut it in p.
Join t p, and from e draw e f parallel to a e, and cutting a t in f.

Now in triangles e b t and a e t, as e b is parallel to e b and e f to a e;—
e b ∶ e f ∷ e b ∶ a e.
But t v is also parallel to a e and p v to e b.

Therefore also in the triangles a p v and a v t,

e b ∶ e f ∷ p v ∶ v t.

Therefore p v ∶ v t ∷ e b ∶ a e.

And, by construction, angle t p v = ∠ a e b.
Therefore the triangles t v p, a e b, are similar; and t p is parallel to a b.

Now
104
the construction in this problem is entirely general for any inclined
line a b, and a horizontal line a e in the same vertical plane with it.
So that if we find the vanishing-point of a e in v, and from v erect a vertical
v p, and from t draw t p parallel to a b, cutting v p in p, p will be the
vanishing-point of a b, and (by the same proof as that given at page 17) of
all lines parallel to it.

Fig. 77.

Next, to find the dividing-point of the inclined line.
I remove some unnecessary lines from the last figure and repeat it here,
Fig. 77., adding the measuring-line a m, that the student may observe its
position with respect to the other lines before I remove any more of them.
Now if the line a b in this diagram represented the length of the line a b in
reality (as a b does in Figs. 10. and 11.), we should only have to proceed to
modify Corollary III. of Problem II. to this new construction. We shall see
presently that a b does not represent the actual length of the inclined line a b
in nature, nevertheless we shall first proceed as if it did, and modify our
result afterwards.
In
105
Fig. 77. draw a d parallel to a b, cutting b t in d.
Therefore a d is the sight-magnitude of a b, as a r is of a b in Fig. 11.

Fig. 78.

Remove again from the figure all lines except p v, v t, p t, a b, a d, and the
measuring-line.
Set off on the measuring-line a m equal to a d.
Draw p q parallel to a m, and through b draw m q, cutting p q in q.
Then, by the proof already given in page 20, p q = p t.

Therefore if p is the vanishing-point of an inclined line a b, and q p is a
horizontal line drawn through it, make p q equal to p t, and a m on the
measuring-line equal to the sight-magnitude of the line a b in the diagram,
and the line joining m q will cut a p in b.

We have now, therefore, to consider what relation the length of the line a b
in this diagram, Fig. 77., has to the length of the line a b in reality.

Now the line a e in Fig. 77. represents the length of a e in reality.

But the angle a e b, Fig. 77., and the corresponding angle in all the
constructions of the earlier problems, is in reality a right angle, though in
the diagram necessarily represented as obtuse.
Therefore, if from e we draw e c, as in Fig. 79., at right
angles to a e, make e c = e b, and join a c, a c will be the
real length of the line a b.
Now, therefore, if instead of a m in Fig. 78., we take
the real length of a b, that real length will be to a m as
a c to a b in Fig. 79.
Fig. 79.
And then, if the line drawn to the measuring-line p q is
still to cut a p in b, it is evident that the line p q must be shortened in the
same ratio that a m was shortened; and the true dividing-point will be q′ in
Fig. 80., fixed so that q′ p shall be to q p as a m′ is to a m; a m′ representing
the real length of a b.
But
106
a m′ is therefore to a m as a c is to a b in Fig. 79.

Therefore p q′ must be to p q as a c is to a b.
But p q equals p t (Fig. 78.); and p v is to v t (in Fig. 78.) as b e is to a e
(Fig. 79.).
Hence we have only to substitute p v for e c, and v t for a e, in Fig. 79., and
the resulting diagonal a c will be the required length of p q′.

Fig. 80.

It will be seen that the construction given in the text (Fig. 46.) is the
simplest means of obtaining this magnitude, for v d in Fig. 46. (or v m in
Fig. 15.) = v t by construction in Problem IV. It should, however, be
observed, that the distance p q′ or p x, in Fig. 46., may be laid on the sight-
line of the inclined plane itself, if the measuring-line be drawn parallel to
that sight-line. And thus any form may be drawn on an inclined plane as
conveniently as on a horizontal one, with the single exception of the
radiation of the verticals, which have a vanishing-point, as shown in
Problem XX.

the end.

Transcriber’s Note
A handful of unequivocal typographical errors has been corrected.
For increased clarity, a few diagrams have been shifted from their
original position in the text.

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