of the studio with long strings, &c., which is a very clumsy and
unscientific way of setting to work. The architects in such cases make
use of the centrolinead, a clever mechanical contrivance for getting
over the difficulty of the far-off vanishing point, but by the method I
have shown you, and shall further illustrate, you will find that you can
dispense with all this trouble, and do all your perspective either
inside the picture or on a very small margin outside it.

Perhaps another drawback to this method is that it is not self-evident,
as in the former one, and being rather difficult to explain, the student
is apt to take it on trust, and not to trouble about the reasons for its
construction: but to show that it is equally correct, I will draw the
two methods in one figure.

LXIII

TWO METHODS OF ANGULAR PERSPECTIVE IN ONE FIGURE

  [Illustration: Fig. 125.]

It matters little whether the station-point is placed above or below the
horizon, as the result is the same. In Fig. 125 it is placed above, as
the lower part of the figure is occupied with the geometrical plan of
the other method.

In each case we make the square _K_ the same size and at the same angle,
its near corner being at _A_. It must be seen that by whichever method
we work out this perspective, the result is the same, so that both are
correct: the great advantage of the first or geometrical system being,
that we can place the square at any angle, as it is drawn without
reference to vanishing points.

We will, however, work out a few figures by the second method.

LXIV

TO DRAW A CUBE, THE POINTS BEING GIVEN

As in a previous figure (124) we found the various working points of
angular perspective, we need now merely transfer them to the horizontal
line in this figure, as in this case they will answer our purpose
perfectly well.

  [Illustration: Fig. 126.]

Let _A_ be the nearest angle touching the base. Draw AV1, AV2. From
_A_, raise vertical _Ae_, the height of the cube. From _e_ draw eV1,
eV2, from the other angles raise verticals _bf_, _dh_, _cg_, to meet
eV1, eV2, fV2, &c., and the cube is complete.

LXV

AMPLIFICATION OF THE CUBE APPLIED TO DRAWING A COTTAGE

  [Illustration: Fig. 127.]

Note that we have started this figure with the cube _Adhefb_. We have
taken three times _AB_, its width, for the front of our house, and twice
_AB_ for the side, and have made it two cubes high, not counting the
roof. Note also the use of the measuring-points in connexion with the
measurements on the base line, and the upper measuring line _TPK_.

LXVI

HOW TO DRAW AN INTERIOR AT AN ANGLE

Here we make use of the same points as in a previous figure, with the
addition of the point _G_, which is the vanishing point of the diagonals
of the squares on the floor.

  [Illustration: Fig. 128.]

From _A_ draw square _Abcd_, and produce its sides in all directions;
again from _A_, through the opposite angle of the square _C_, draw a
diagonal till it cuts the horizon at _G_. From _G_ draw diagonals
through _b_ and _d_, cutting the base at _o_, _o_, make spaces _o_, _o_,
equal to _Ao_ all along the base, and from them draw diagonals to _G_;
through the points where these diagonals intersect the vanishing lines
drawn in the direction of _Ab_, _dc_ and _Ad_, _bc_, draw lines to the
other vanishing point V1, thus completing the squares, and so cover
the floor with them; they will then serve to measure width of door,
windows, &c. Of course horizontal lines on wall 1 are drawn to V1, and
those on wall 2 to V2.

In order to see this drawing properly, the eye should be placed about
3 inches from it, and opposite the point of sight; it will then stand
out like a stereoscopic picture, and appear as actual space, but
otherwise the perspective seems deformed, and the angles exaggerated.
To make this drawing look right from a reasonable distance, the point of
distance should be at least twice as far off as it is here, and this
would mean altering all the other points and sending them a long way out
of the picture; this is why artists use those long strings referred to
above. I would however, advise them to make their perspective drawing on
a small scale, and then square it up to the size of the canvas.

LXVII

HOW TO CORRECT DISTORTED PERSPECTIVE BY DOUBLING THE LINE OF DISTANCE

Here we have the same interior as the foregoing, but drawn with double
the distance, so that the perspective is not so violent and the objects
are truer in proportion to each other.

  [Illustration: Fig. 129.]

To redraw the whole figure double the size, including the station-point,
would require a very large diagram, that we could not get into this book
without a folding plate, but it comes to the same thing if we double the
distances between the various points. Thus, if from _S_ to _G_ in the
small diagram is 1 inch, in the larger one make it 2 inches. If from _S_
to M2 is 2 inches, in the larger make it 4, and so on.

Or this form may be used: make _AB_ twice the length of _AC_ (Fig. 130),
or in any other proportion required. On _AC_ mark the points as in the
drawing you wish to enlarge. Make _AB_ the length that you wish to
enlarge to, draw _CB_, and then from each division on _AC_ draw lines
parallel to _CB_, and _AB_ will be divided in the same proportions, as I
have already shown (Fig. 117).

There is no doubt that it is easier to work direct from the vanishing
points themselves, especially in complicated architectural work, but at
the same time I will now show you how we can dispense with, at all
events, one of them, and that the farthest away.

  [Illustration: Fig. 130.]

LXVIII

HOW TO DRAW A CUBE ON A GIVEN SQUARE, USING ONLY ONE VANISHING POINT

_ABCD_ is the given square (Fig. 131). At _A_ raise vertical _Aa_ equal
to side of square _AB·_, from _a_ draw _ab_ to the vanishing point.
Raise _Bb_. Produce _VD_ to _E_ to touch the base line. From _E_ raise
vertical _EF_, making it equal to _Aa_. From _F_ draw _FV_. Raise _Dd_
and _Cc_, their heights being determined by the line _FV_. Join _da_ and
the cube is complete. It will be seen that the verticals raised at each
corner of the square are equal perspectively, as they are drawn between
parallels which start from equal heights, namely, from _EF_ and _Aa_ to
the same point _V_, the vanishing point. Any other line, such as _OO·_,
can be directed to the inaccessible vanishing point in the same way as
_ad_, &c.

_Note._ This is only one of many original figures and problems in this
book which have been called up by the wish to facilitate the work of the
artist, and as it were by necessity.

  [Illustration: Fig. 131.]

LXIX

A COURTYARD OR CLOISTER DRAWN WITH ONE VANISHING POINT

  [Illustration: Fig. 132.]

In this figure I have first drawn the pavement by means of the diagonals
_GA_, _Go_, _Go_, &c., and the vanishing point _V_, the square at _A_
being given. From _A_ draw diagonal through opposite corner till it cuts
the horizon at _G_. From this same point _G_ draw lines through the
other corners of the square till they cut the ground line at _o_, _o_.
Take this measurement _Ao_ and mark it along the base right and left of
_A_, and the lines drawn from these points _o_ to point _G_ will give
the diagonals of all the squares on the pavement. Produce sides of
square _A_, and where these lines are intersected by the diagonals _Go_
draw lines from the vanishing point _V_ to base. These will give us the
outlines of the squares lying between them and also guiding points that
will enable us to draw as many more as we please. These again will give
us our measurements for the widths of the arches, &c., or between the
columns. Having fixed the height of wall or dado, we make use of _V_
point to draw the sides of the building, and by means of proportionate
measurement complete the rest, as in Fig. 128.

LXX

HOW TO DRAW LINES WHICH SHALL MEET AT A DISTANT POINT,
BY MEANS OF DIAGONALS

This is in a great measure a repetition of the foregoing figure, and
therefore needs no further explanation.

  [Illustration: Fig. 133.]

I must, however, point out the importance of the point _G_. In angular
perspective it in a measure takes the place of the point of distance in
parallel perspective, since it is the vanishing point of diagonals at
45° drawn between parallels such as _AV_, _DV_, drawn to a vanishing
point _V_. The method of dividing line _AV_ into a number of parts equal
to _AB_, the side of the square, is also shown in a previous figure
(Fig. 120).

LXXI

HOW TO DIVIDE A SQUARE PLACED AT AN ANGLE INTO A GIVEN NUMBER
OF SMALL SQUARES

_ABCD_ is the given square, and only one vanishing point is accessible.
Let us divide it into sixteen small squares. Produce side _CD_ to base
at _E_. Divide _EA_ into four equal parts. From each division draw lines
to vanishing point _V_. Draw diagonals _BD_ and _AC_, and produce the
latter till it cuts the horizon in _G_. Draw the three cross-lines
through the intersections made by the diagonals and the lines drawn to
_V_, and thus divide the square into sixteen.

  [Illustration: Fig. 134.]

This is to some extent the reverse of the previous problem. It also
shows how the long vanishing point can be dispensed with, and the
perspective drawing brought within the picture.

LXXII

FURTHER EXAMPLE OF HOW TO DIVIDE A GIVEN OBLIQUE SQUARE INTO
A GIVEN NUMBER OF EQUAL SQUARES, SAY TWENTY-FIVE

Having drawn the square _ABCD_, which is enclosed, as will be seen, in a
dotted square in parallel perspective, I divide the line _EA_ into five
equal parts instead of four (Fig. 135), and have made use of the device
for that purpose by measuring off the required number on line _EF_, &c.
Fig. 136 is introduced here simply to show that the square can be
divided into any number of smaller squares. Nor need the figure be
necessarily a square; it is just as easy to make it an oblong, as _ABEF_
(Fig. 136); for although we begin with a square we can extend it in any
direction we please, as here shown.

  [Illustration: Fig. 135.]

  [Illustration: Fig. 136.]

LXXIII

OF PARALLELS AND DIAGONALS

  [Illustration: Fig. 137 A.]

  [Illustration: Fig. 137 B.]

  [Illustration: Fig. 137 C.]

To find the centre of a square or other rectangular figure we have but
to draw its two diagonals, and their intersection will give us the
centre of the figure (see 137 A). We do the same with perspective
figures, as at B. In Fig. C is shown how a diagonal, drawn from one
angle of a square _B_ through the centre _O_ of the opposite side of the
square, will enable us to find a second square lying between the same
parallels, then a third, a fourth, and so on. At figure _K_ lying on
the ground, I have divided the farther side of the square _mn_ into ¼,
1/3, ½. If I draw a diagonal from _G_ (at the base) through the half
of this line I cut off on _FS_ the lengths or sides of two squares;
if through the quarter I cut off the length of four squares on the
vanishing line _FS_, and so on. In Fig. 137 D is shown how easily any
number of objects at any equal distances apart, such as posts, trees,
columns, &c., can be drawn by means of diagonals between parallels,
guided by a central line _GS_.

  [Illustration: Fig. 137 D.]

LXXIV

THE SQUARE, THE OBLONG, AND THEIR DIAGONALS

  [Illustration: Fig. 138.]

  [Illustration: Fig. 139.]

Having found the centre of a square or oblong, such as Figs. 138 and
139, if we draw a third line through that centre at a given angle and
then at each of its extremities draw perpendiculars _AB_, _DC_, we
divide that square or oblong into three parts, the two outer portions
being equal to each other, and the centre one either larger or smaller
as desired; as, for instance, in the triumphal arch we make the centre
portion larger than the two outer sides. When certain architectural
details and spaces are to be put into perspective, a scale such as that
in Fig. 123 will be found of great convenience; but if only a ready
division of the principal proportions is required, then these diagonals
will be found of the greatest use.

LXXV

SHOWING THE USE OF THE SQUARE AND DIAGONALS IN DRAWING DOORWAYS,
WINDOWS, AND OTHER ARCHITECTURAL FEATURES

This example is from Serlio's _Architecture_ (1663), showing what
excellent proportion can be obtained by the square and diagonals. The
width of the door is one-third of the base of square, the height
two-thirds. As a further illustration we have drawn the same figure in
perspective.

  [Illustration: Fig. 140.]

  [Illustration: Fig. 141.]

LXXVI

HOW TO MEASURE DEPTHS BY DIAGONALS

If we take any length on the base of a square, say from _A_ to _g_, and
from _g_ raise a perpendicular till it cuts the diagonal _AB_ in _O_,
then from _O_ draw horizontal _Og·_, we form a square AgOg·, and thus
measure on one side of the square the distance or depth _Ag·_. So can we
measure any other length, such as _fg_, in like manner.

  [Illustration: Fig. 142.]

  [Illustration: Fig. 143.]

To do this in perspective we pursue precisely the same method, as shown
in this figure (143).

To measure a length _Ag_ on the side of square _AC_, we draw a line from
_g_ to the point of sight _S_, and where it crosses diagonal _AB_ at _O_
we draw horizontal _Og_, and thus find the required depth _Ag_ in the
picture.

LXXVII

HOW TO MEASURE DISTANCES BY THE SQUARE AND DIAGONAL

It may sometimes be convenient to have a ready method by which to
measure the width and length of objects standing against the wall of a
gallery, without referring to distance-points, &c.

  [Illustration: Fig. 144.]

In Fig. 144 the floor is divided into two large squares with their
diagonals. Suppose we wish to draw a fireplace or a piece of furniture
_K_, we measure its base _ef_ on _AB_, as far from _B_ as we wish it to
be in the picture; draw _eo_ and _fo_ to point of sight, and proceed as
in the previous figure by drawing parallels from _Oo_, &c.

Let it be observed that the great advantage of this method is, that we
can use it to measure such distant objects as _XY_ just as easily as
those near to us.

There is, however, a still further advantage arising from it, and that
is that it introduces us to a new and simpler method of perspective, to
which I have already referred, and it will, I hope, be found of infinite
use to the artist.

_Note._--As we have founded many of these figures on a given square in
angular perspective, it is as well to have a ready and certain means of
drawing that square without the elaborate setting out of a geometrical
plan, as in the first method, or the more cumbersome and extended system
of the second method. I shall therefore show you another method equally
correct, but much simpler than either, which I have invented for our
use, and which indeed forms one of the chief features of this book.

LXXVIII

HOW BY MEANS OF THE SQUARE AND DIAGONAL WE CAN DETERMINE
THE POSITION OF POINTS IN SPACE

Apart from the aid that perspective affords the draughtsman, there is a
further value in it, in that it teaches us almost a new science, which
we might call the mystery of aspect, and how it is that the objects
around us take so many different forms, or rather appearances, although
they themselves remain the same. And also that it enables us, with,
I think, great pleasure to ourselves, to fathom space, to work out
difficult problems by simple reasoning, and to exercise those inventive
and critical faculties which give strength and enjoyment to mental life.

And now, after this brief excursion into philosophy, let us come down to
the simple question of the perspective of a point.

  [Illustration: Fig. 145.]

  [Illustration: Fig. 146.]

Here, for instance, are two aspects of the same thing: the geometrical
square _A_, which is facing us, and the perspective square _B_, which we
suppose to lie flat on the table, or rather on the perspective plane.
Line _A·C·_ is the perspective of line _AC_. On the geometrical square
we can make what measurements we please with the compasses, but on the
perspective square _B·_ the only line we can actually measure is the
base line. In both figures this base line is the same length. Suppose we
want to find the perspective of point _P_ (Fig. 146), we make use of the
diagonal _CA_. From _P_ in the geometrical square draw _PO_ to meet the
diagonal in _O_; through _O_ draw perpendicular _fe_; transfer length
_fB_, so found, to the base of the perspective square; from _f_ draw
_fS_ to point of sight; where it cuts the diagonal in _O_, draw
horizontal _OP·_, which gives us the point required. In the same way we
can find the perspective of any number of points on any side of the
square.

LXXIX

PERSPECTIVE OF A POINT PLACED IN ANY POSITION WITHIN THE SQUARE

Let the point _P_ be the one we wish to put into perspective. We have
but to repeat the process of the previous problem, making use of our
measurements on the base, the diagonals, &c.

  [Illustration: Fig. 147.]

Indeed these figures are so plain and evident that further description
of them is hardly necessary, so I will here give two drawings of
triangles which explain themselves. To put a triangle into perspective
we have but to find three points, such as _fEP_, Fig. 148 A, and then
transfer these points to the perspective square 148 B, as there shown,
and form the perspective triangle; but these figures explain themselves.
Any other triangle or rectilineal figure can be worked out in the same
way, which is not only the simplest method, but it carries its
mathematical proof with it.

  [Illustration: Fig. 148 A.]

  [Illustration: Fig. 148 B.]

  [Illustration: Fig. 149 A.]

  [Illustration: Fig. 149 B.]

LXXX

PERSPECTIVE OF A SQUARE PLACED AT AN ANGLE NEW METHOD

As we have drawn a triangle in a square so can we draw an oblique square
in a parallel square. In Figure 150 A we have drawn the oblique square
_GEPn_. We find the points on the base _Am_, as in the previous figures,
which enable us to construct the oblique perspective square _n·G·E·P·_
in the parallel perspective square Fig. 150 B. But it is not necessary
to construct the geometrical figure, as I will show presently. It is
here introduced to explain the method.

  [Illustration: Fig. 150 A.]

  [Illustration: Fig. 150 B.]

Fig. 150 B. To test the accuracy of the above, produce sides _G·E·_ and
_n·P·_ of perspective square till they touch the horizon, where they
will meet at _V_, their vanishing point, and again produce the other
sides _n·G·_ and _P·E·_ till they meet on the horizon at the other
vanishing point, which they must do if the figure is correctly drawn.

In any parallel square construct an oblique square from a given
point--given the parallel square at Fig. 150 B, and given point _n·_ on
base. Make _A·f·_ equal to _n·m·_, draw _f·S_ and _n·S_ to point of
sight. Where these lines cut the diagonal _AC_ draw horizontals to _P·_
and _G·_, and so find the four points _G·E·P·n·_ through which to draw
the square.

LXXXI

ON A GIVEN LINE PLACED AT AN ANGLE TO THE BASE DRAW A SQUARE IN ANGULAR
PERSPECTIVE, THE POINT OF SIGHT, AND DISTANCE, BEING GIVEN.

  [Illustration: Fig. 151.]

Let _AB_ be the given line, _S_ the point of sight, and _D_ the distance
(Fig. 151, 1). Through _A_ draw _SC_ from point of sight to base (Fig.
151, 2 and 3). From _C_ draw _CD_ to point of distance. Draw _Ao_
parallel to base till it cuts _CD_ at _o_, through _O_ draw _SP_, from
_B_ mark off _BE_ equal to _CP_. From _E_ draw _ES_ intersecting _CD_ at
_K_, from _K_ draw _KM_, thus completing the outer parallel square.
Through _F_, where _PS_ intersects _MK_, draw _AV_ till it cuts the
horizon in _V_, its vanishing point. From _V_ draw _VB_ cutting side
_KE_ of outer square in _G_, and we have the four points _AFGB_, which
are the four angles of the square required. Join _FG_, and the figure is
complete.

Any other side of the square might be given, such as _AF_. First through
_A_ and _F_ draw _SC_, _SP_, then draw _Ao_, then through _o_ draw _CD_.
From _C_ draw base of parallel square _CE_, and at _M_ through _F_ draw
_MK_ cutting diagonal at _K_, which gives top of square. Now through _K_
draw _SE_, giving _KE_ the remaining side thereof, produce _AF_ to _V_,
from _V_ draw _VB_. Join _FG_, _GB_, and _BA_, and the square required
is complete.

The student can try the remaining two sides, and he will find they work
out in a similar way.

LXXXII

HOW TO DRAW SOLID FIGURES AT ANY ANGLE BY THE NEW METHOD

As we can draw planes by this method so can we draw solids, as shown in
these figures. The heights of the corners of the triangles are obtained
by means of the vanishing scales _AS_, _OS_, which have already been
explained.

  [Illustration: Fig. 152.]

  [Illustration: Fig. 153.]

In the same manner we can draw a cubic figure (Fig. 154)--a box, for
instance--at any required angle. In this case, besides the scale _AS_,
_OS_, we have made use of the vanishing lines _DV_, _BV_, to corroborate
the scale, but they can be dispensed with in these simple objects, or we
can use a scale on each side of the figure as _a·o·S_, should both
vanishing points be inaccessible. Let it be noted that in the scale
_AOS_, _AO_ is made equal to _BC_, the height of the box.

  [Illustration: Fig. 154.]

By a similar process we draw these two figures, one on the square, the
other on the circle.

  [Illustration: Fig. 155.]

  [Illustration: Fig. 156.]

LXXXIII

POINTS IN SPACE

The chief use of these figures is to show how by means of diagonals,
horizontals, and perpendiculars almost any figure in space can be set
down. Lines at any slope and at any angle can be drawn by this
descriptive geometry.

The student can examine these figures for himself, and will understand
their working from what has gone before. Here (Fig. 157) in the
geometrical square we have a vertical plane _AabB_ standing on its base
_AB_. We wish to place a projection of this figure at a certain distance
and at a given angle in space. First of all we transfer it to the side
of the cube, where it is seen in perspective, whilst at its side is
another perspective square lying flat, on which we have to stand our
figure. By means of the diagonal of this flat square, horizontals from
figure on side of cube, and lines drawn from point of sight (as already
explained), we obtain the direction of base line _AB_, and also by means
of lines _aa·_ and _bb·_ we obtain the two points in space _a·b·_. Join
_Aa·_, _a·b·_ and _Bb·_, and we have the projection required, and which
may be said to possess the third dimension.

  [Illustration: Fig. 157.]

In this other case (Fig. 158) we have a wedge-shaped figure standing on
a triangle placed on the ground, as in the previous figure, its three
corners being the same height. In the vertical geometrical square we
have a ground-plan of the figure, from which we draw lines to diagonal
and to base, and notify by numerals 1, 3, 2, 1, 3; these we transfer to
base of the horizontal perspective square, and then construct shaded
triangle 1, 2, 3, and raise to the height required as shown at
1·, 2·, 3·. Although we may not want to make use of these special
figures, they show us how we could work out almost any form or object
suspended in space.

  [Illustration: Fig. 158.]

LXXXIV

THE SQUARE AND DIAGONAL APPLIED TO CUBES AND SOLIDS DRAWN THEREIN

  [Illustration: Fig. 159.]

As we have made use of the square and diagonal to draw figures at
various angles so can we make use of cubes either in parallel or angular
perspective to draw other solid figures within them, as shown in these
drawings, for this is simply an amplification of that method. Indeed we
might invent many more such things. But subjects for perspective
treatment will constantly present themselves to the artist or
draughtsman in the course of his experience, and while I endeavour to
show him how to grapple with any new difficulty or subject that may
arise, it is impossible to set down all of them in this book.

  [Illustration: Fig. 160.]

LXXXV

TO DRAW AN OBLIQUE SQUARE IN ANOTHER OBLIQUE SQUARE
WITHOUT USING VANISHING POINTS

It is not often that both vanishing points are inaccessible, still it is
well to know how to proceed when this is the case. We first draw the
square _ABCD_ inside the parallel square, as in previous figures. To
draw the smaller square _K_ we simply draw a smaller parallel square _h
h h h_, and within that, guided by the intersections of the diagonals
therewith, we obtain the four points through which to draw square _K_.
To raise a solid figure on these squares we can make use of the
vanishing scales as shown on each side of the figure, thus obtaining the
upper square 1 2 3 4, then by means of the diagonal 1 3 and 2 4 and
verticals raised from each corner of square _K_ to meet them we obtain
the smaller upper square corresponding to _K_.

It might be said that all this can be done by using the two vanishing
points in the usual way. In the first place, if they were as far off as
required for this figure we could not get them into a page unless it
were three or four times the width of this one, and to use shorter
distances results in distortion, so that the real use of this system is
that we can make our figures look quite natural and with much less
trouble than by the other method.

  [Illustration: Fig. 161.]

LXXXVI

SHOWING HOW A PEDESTAL CAN BE DRAWN BY THE NEW METHOD

This is a repetition of the previous problem, or rather the application
of it to architecture, although when there are many details it may be
more convenient to use vanishing points or the centrolinead.

  [Illustration: Fig. 162.]

  [Illustration: Fig. 163. Honfleur.]

LXXXVII

SCALE ON EACH SIDE OF THE PICTURE

As one of my objects in writing this book is to facilitate the working
of our perspective, partly for the comfort of the artist, and partly
that he may have no excuse for neglecting it, I will here show you how
you may, by a very simple means, secure the general correctness of your
perspective when sketching or painting out of doors.

Let us take this example from a sketch made at Honfleur (Fig. 163), and
in which my eye was my only guide, but it stands the test of the rule.
First of all note that line _HH_, drawn from one side of the picture to
the other, is the horizontal line; below that is a wall and a pavement
marked _aV_, also going from one side of the picture to the other, and
being lower down at _a_ than at _V_ it runs up as it were to meet the
horizon at some distant point. In order to form our scale I take first
the length of _Ha_, and measure it above and below the horizon, along
the side to our left as many times as required, in this case four or
five. I now take the length _HV_ on the right side of the picture and
measure it above and below the horizon, as in the other case; and then
from these divisions obtain dotted lines crossing the picture from one
side to the other which must all meet at some distant point on the
horizon. These act as guiding lines, and are sufficient to give us the
direction of any vanishing lines going to the same point. For those that
go in the opposite direction we proceed in the same way, as from _b_ on
the right to _V·_ on the left. They are here put in faintly, so as not
to interfere with the drawing. In the sketch of Toledo (Fig. 164) the
same thing is shown by double lines on each side to separate the two
sets of lines, and to make the principle more evident.

  [Illustration: Fig. 164. Toledo.]

LXXXVIII

THE CIRCLE

If we inscribe a circle in a square we find that it touches that square
at four points which are in the middle of each side, as at _a b c d_. It
will also intersect the two diagonals at the four points _o_ (Fig. 165).
If, then, we put this square and its diagonals, &c., into perspective we
shall have eight guiding points through which to trace the required
circle, as shown in Fig. 166, which has the same base as Fig. 165.

  [Illustration: Fig. 165.]

  [Illustration: Fig. 166.]

LXXXIX

THE CIRCLE IN PERSPECTIVE A TRUE ELLIPSE

Although the circle drawn through certain points must be a freehand
drawing, which requires a little practice to make it true, it is
sufficient for ordinary purposes and on a small scale, but to be
mathematically true it must be an ellipse. We will first draw an ellipse
(Fig. 167). Let _ee_ be its long, or transverse, diameter, and _db_ its
short or conjugate diameter. Now take half of the long diameter _eE_,
and from point _d_ with _cE_ for radius mark on _ee_ the two points
_ff_, which are the foci of the ellipse. At each focus fix a pin, then
make a loop of fine string that does not stretch and of such a length
that when drawn out the double thread will reach from _f_ to _e_. Now
place this double thread round the two pins at the foci _ff·_ and
distend it with the pencil point until it forms triangle _fdf·_, then
push the pencil along and right round the two foci, which being guided
by the thread will draw the curve, which is a true ellipse, and will
pass through the eight points indicated in our first figure. This will
be a sufficient proof that the circle in perspective and the ellipse are
identical curves. We must also remember that the ellipse is an oblique
projection of a circle, or an oblique section of a cone. The difference
between the two figures consists in their centres not being in the same
place, that of the perspective circle being at _c_, higher up than _e_
the centre of the ellipse. The latter being a geometrical figure, its
long diameter is exactly in the centre of the figure, whereas the centre
_c_ and the diameter of the perspective are at the intersection of the
diagonals of the perspective square in which it is inscribed.

  [Illustration: Fig. 167.]

XC

FURTHER ILLUSTRATION OF THE ELLIPSE

In order to show that the ellipse drawn by a loop as in the previous
figure is also a circle in perspective we must reconstruct around it the
square and its eight points by means of which it was drawn in the first
instance. We start with nothing but the ellipse itself. We have to find
the points of sight and distance, the base, &c. Let us start with base
_AB_, a horizontal tangent to the curve extending beyond it on either
side. From _A_ and _B_ draw two other tangents so that they shall touch
the curve at points such as _TT·_ a little above the transverse diameter
and on a level with each other. Produce these tangents till they meet at
point _S_, which will be the point of sight. Through this point draw
horizontal line _H_. Now draw tangent _CD_ parallel to _AB_. Draw
diagonal _AD_ till it cuts the horizon at the point of distance, this
will cut through diameter of circle at its centre, and so proceed to
find the eight points through which the perspective circle passes, when
it will be found that they all lie on the ellipse we have drawn with the
loop, showing that the two curves are identical although their centres
are distinct.

  [Illustration: Fig. 168.]

XCI

HOW TO DRAW A CIRCLE IN PERSPECTIVE WITHOUT A GEOMETRICAL PLAN

Divide base _AB_ into four equal parts. At _B_ drop perpendicular _Bn_,
making _Bn_ equal to _Bm_, or one-fourth of base. Join _mn_ and transfer
this measurement to each side of _d_ on base line; that is, make _df_
and _df·_ equal to _mn_. Draw _fS_ and _f·S_, and the intersections of
these lines with the diagonals of square will give us the four points _o
o o o_.

  [Illustration: Fig. 169.]

The reason of this is that _ff·_ is the measurement on the base _AB_ of
another square _o o o o_ which is exactly half of the outer square. For
if we inscribe a circle in a square and then inscribe a second square in
that circle, this second square will be exactly half the area of the
larger one; for its side will be equal to half the diagonal of the
larger square, as can be seen by studying the following figures. In Fig.
170, for instance, the side of small square _K_ is half the diagonal of
large square _o_.

  [Illustration: Fig. 170.]

  [Illustration: Fig. 171.]

In Fig. 171, _CB_ represents half of diagonal _EB_ of the outer square
in which the circle is inscribed. By taking a fourth of the base _mB_
and drawing perpendicular _mh_ we cut _CB_ at _h_ in two equal parts,
_Ch_, _hB_. It will be seen that _hB_ is equal to _mn_, one-quarter of
the diagonal, so if we measure _mn_ on each side of _D_ we get _ff·_
equal to _CB_, or half the diagonal. By drawing _ff_, _f·f_ passing
through the diagonals we get the four points _o o o o_ through which to
draw the smaller square. Without referring to geometry we can see at a
glance by Fig. 172, where we have simply turned the square _o o o o_ on
its centre so that its angles touch the sides of the outer square, that
it is exactly half of square _ABEF_, since each quarter of it, such as
EoCo, is bisected by its diagonal _oo_.

  [Illustration: Fig. 172.]

  [Illustration: Fig. 173.]

XCII

HOW TO DRAW A CIRCLE IN ANGULAR PERSPECTIVE

Let _ABCD_ be the oblique square. Produce _VA_ till it cuts the base
line at _G_.

  [Illustration: Fig. 174.]

Take _mD_, the fourth of the base. Find _mn_ as in Fig. 171, measure it
on each side of _E_, and so obtain _Ef_ and _Ef·_, and proceed to draw
_fV_, _EV_, _f·V_ and the diagonals, whose intersections with these
lines will give us the eight points through which to draw the circle. In
fact the process is the same as in parallel perspective, only instead of
making our divisions on the actual base _AD_ of the square, we make them
on _GD_, the base line.

To obtain the central line _hh_ passing through _O_, we can make use of
diagonals of the half squares; that is, if the other vanishing point is
inaccessible, as in this case.

XCIII

HOW TO DRAW A CIRCLE IN PERSPECTIVE MORE CORRECTLY,
BY USING SIXTEEN GUIDING POINTS

First draw square _ABCD_. From _O_, the middle of the base, draw
semicircle _AKB_, and divide it into eight equal parts. From each
division raise perpendiculars to the base, such as _2 O_, _3 O_, _5 O_,
&c., and from divisions _O_, _O_, _O_ draw lines to point of sight,
and where these lines cut the diagonals _AC_, _DB_, draw horizontals
parallel to base _AB_. Then through the points thus obtained draw the
circle as shown in this figure, which also shows us how the
circumference of a circle in perspective may be divided into any
number of equal parts.

  [Illustration: Fig. 175.]

XCIV

HOW TO DIVIDE A PERSPECTIVE CIRCLE INTO ANY NUMBER OF EQUAL PARTS

This is simply a repetition of the previous figure as far as its
construction is concerned, only in this case we have divided the
semicircle into twelve parts and the perspective into twenty-four.

  [Illustration: Fig. 176.]

  [Illustration: Fig. 177.] We have raised perpendiculars from the
divisions on the semicircle, and proceeded as before to draw lines to
the point of sight, and have thus by their intersections with the
circumference already drawn in perspective divided it into the required
number of equal parts, to which from the centre we have drawn the radii.
This will show us how to draw traceries in Gothic windows, columns in a
circle, cart-wheels, &c.

The geometrical figure (177) will explain the construction of the
perspective one by showing how the divisions are obtained on the line
_AB_, which represents base of square, from the divisions on the
semicircle _AKB_.

XCV

HOW TO DRAW CONCENTRIC CIRCLES

  [Illustration: Fig. 178.]

First draw a square with its diagonals (Fig. 178), and from its centre
_O_ inscribe a circle; in this circle inscribe a square, and in this
again inscribe a second circle, and so on. Through their intersections
with the diagonals draw lines to base, and number them 1, 2, 3, 4, &c.;
transfer these measurements to the base of the perspective square (Fig.
179), and proceed to construct the circles as before, drawing lines from
each point on the base to the point of sight, and drawing the curves
through the inter-sections of these lines with the diagonals.

  [Illustration: Fig. 179.]

Should it be required to make the circles at equal distances, as for
steps for instance, then the geometrical plan should be made
accordingly.

Or we may adopt the method shown at Fig. 180, by taking quarter base of
both outer and inner square, and finding the measurement _mn_ on each
side of _C_, &c.

  [Illustration: Fig. 180.]

XCVI

THE ANGLE OF THE DIAMETER OF THE CIRCLE IN ANGULAR
AND PARALLEL PERSPECTIVE

The circle, whether in angular or parallel perspective, is always an
ellipse. In angular perspective the angle of the circle's diameter
varies in accordance with the angle of the square in which it is placed,
as in Fig. 181, _cc_ is the diameter of the circle and _ee_ the diameter
of the ellipse. In parallel perspective the diameter of the circle
always remains horizontal, although the long diameter of the ellipse
varies in inclination according to the distance it is from the point of
sight, as shown in Fig. 182, in which the third circle is much elongated
and distorted, owing to its being outside the angle of vision.

  [Illustration: Fig. 181.]

  [Illustration: Fig. 182.]

XCVII

HOW TO CORRECT DISPROPORTION IN THE WIDTH OF COLUMNS

  [Transcriber's Note:
  The column referred to as "1" in the text is marked "S" in both
  Figures.]

The disproportion in the width of columns in Fig. 183 arises from the
point of distance being too near the point of sight, or, in other words,
taking too wide an angle of vision. It will be seen that column 3 is
much wider than column 1.

  [Illustration: Fig. 183.]

  [Illustration: Fig. 184.]

In our second figure (184) is shown how this defect is remedied, by
doubling the distance, or by counting the same distance as half, which
is easily effected by drawing the diagonal from _O_ to ½D, instead of
from _A_, as in the other figure, _O_ being at half base. Here the
squares lie much more level, and the columns are nearly the same width,
showing the advantage of a long distance.

XCVIII

HOW TO DRAW A CIRCLE OVER A CIRCLE OR A CYLINDER

First construct square and circle _ABE_, then draw square _CDF_ with its
diagonals. Then find the various points _O_, and from these raise
perpendiculars to meet the diagonals of the upper square at points _P_,
which, with the other points will be sufficient guides to draw the
circle required. This can be applied to towers, columns, &c. The size of
the circles can be varied so that the upper portion of a cylinder or
column shall be smaller than the lower.

  [Illustration: Fig. 185.]

XCIX

TO DRAW A CIRCLE BELOW A GIVEN CIRCLE

Construct the upper square and circle as before, then by means of the
vanishing scale _POV_, which should be made the depth required, drop
perpendiculars from the various points marked _O_, obtained by the
diagonals, making them the right depth by referring them to the
vanishing scale, as shown in this figure. This can be used for drawing
garden fountains, basins, and various architectural objects.

  [Illustration: Fig. 186.]

C

APPLICATION OF PREVIOUS PROBLEM

That is, to draw a circle above a circle. In Fig. 187 can be seen how by
means of the vanishing scale at the side we obtain the height of the
verticals 1, 2, 3, 4, &c., which determine the direction of the upper
circle; and in this second figure, how we resort to the same means to
draw circular steps.

  [Illustration: Fig. 187.]

  [Illustration: Fig. 188.]

CI

DORIC COLUMNS

It is as well for the art student to study the different orders of
architecture, whether architect or not, as he frequently has to
introduce them into his pictures, and at least must know their
proportions, and how columns diminish from base to capital, as shown in
this illustration.

  [Illustration: Fig. 189.]

CII

TO DRAW SEMICIRCLES STANDING UPON A CIRCLE AT ANY ANGLE

  [Illustration: Fig. 190.]

Given the circle _ACBH_, on diagonal _AB_ draw semicircle _AKB_, and on
the same line _AB_ draw rectangle _AEFB_, its height being determined by
radius _OK_ of semicircle. From centre _O_ draw _OF_ to corner of
rectangle. Through _f·_, where that line intersects the semicircle, draw
_mn_ parallel to _AB_. This will give intersection _O·_ on the vertical
_OK_, through which all such horizontals as _m·n·_, level with _mn_,
must pass. Now take any other diameter, such as _GH_, and thereon raise
rectangle _GghH_, the same height as the other. The manner of doing this
is to produce diameter _GH_ to the horizon till it finds its vanishing
point at _V_. From _V_ through _K_ draw _hg_, and through _O·_ draw
_n·m·_. From _O_ draw the two diagonals _og_ and _oh_, intersecting
_m·n·_ at _O_, _O_, and thus we have the five points _GOKOH_ through
which to draw the required semicircle.

CIII

A DOME STANDING ON A CYLINDER

  [Illustration: Fig. 191.]

This figure is a combination of the two preceding it. A cylinder is
first raised on the circle, and on the top of that we draw semicircles
from the different divisions on the circumference of the upper circle.
This, however, only represents a small half-globular object. To draw the
dome of a cathedral, or other building high above us, is another matter.
From outside, where we can get to a distance, it is not difficult, but
from within it will tax all our knowledge of perspective to give it
effect.

We shall go more into this subject when we come to archways and vaulted
roofs, &c.

CIV

SECTION OF A DOME OR NICHE

  [Illustration: Fig. 192.]

First draw outline of the niche _GFDBA_ (Fig. 193), then at its base
draw square and circle _GOA_, _S_ being the point of sight, and divide
the circumference of the circle into the required number of parts. Then
draw semicircle _FOB_, and over that another semicircle _EOC_. The
manner of drawing them is shown in Fig. 192. From the divisions on the
circle _GOA_ raise verticals to semicircle _FOB_, which will divide it
in the same way. Divide the smaller semicircle _EOC_ into the same
number of parts as the others, which divisions will serve as guiding
points in drawing the curves of the dome that are drawn towards _D_, but
the shading must assist greatly in giving the effect of the recess.

  [Illustration: Fig. 193.]

In Fig. 192 will be seen how to draw semicircles in perspective.
We first draw the half squares by drawing from centres _O_ of their
diameters diagonals to distance-point, as _OD_, which cuts the vanishing
line BS at _m_, and gives us the depth of the square, and in this we
draw the semicircle in the usual way.

  [Illustration: Fig. 194. A Dome.]

CV

A DOME

First draw a section of the dome ACEDB (Fig. 194) the shape required.
Draw _AB_ at its base and _CD_ at some distance above it. Keeping these
as central lines, form squares thereon by drawing _SA_, _SB_, _SC_,
_SD_, &c., from point of sight, and determining their lengths by
diagonals _fh_, _f·h·_ from point of distance, passing through _O_.
Having formed the two squares, draw perspective circles in each, and
divide their circumferences into twelve or whatever number of parts are
needed. To complete the figure draw from each division in the lower
circle curves passing through the corresponding divisions in the upper
one, to the apex. But as these are freehand lines, it requires some
taste and knowledge to draw them properly, and of course in a large
drawing several more squares and circles might be added to aid the
draughtsman. The interior of the dome can be drawn in the same way.

  [Illustration: Fig. 195.]

CVI

HOW TO DRAW COLUMNS STANDING IN A CIRCLE

In Fig. 195 are sixteen cylinders or columns standing in a circle. First
draw the circle on the ground, then divide it into sixteen equal parts,
and let each division be the centre of the circle on which to raise the
column. The question is how to make each one the right width in
accordance with its position, for it is evident that a near column must
appear wider than the opposite one. On the right of the figure is the
vertical scale _A_, which gives the heights of the columns, and at its
foot is a horizontal scale, or a scale of widths _B_. Now, according to
the line on which the column stands, we find its apparent width marked
on the scale. Thus take the small square and circle at 15, without its
column, or the broken column at 16; and note that on each side of its
centre _O_ I have measured _oa_, _ob_, equal to spaces marked 3 on the
same horizontal in the scale _B_. Through these points _a_ and _b_ I
have drawn lines towards point of sight _S_. Through their intersections
with diagonal _e_, which is directed to point of distance, draw the
farther and nearer sides of the square in which to describe the circle
and the cylinder or column thereon. I have made all the squares thus
obtained in parallel perspective, but they do not represent the bases of
columns arranged in circles, which should converge towards the centre,
and I believe in some cases are modified in form to suit that design.

CVII

COLUMNS AND CAPITALS

This figure shows the application of the square and diagonal in drawing
and placing columns in angular perspective.

  [Illustration: Fig. 196.]

CVIII

METHOD OF PERSPECTIVE EMPLOYED BY ARCHITECTS

The architects first draw a plan and elevation of the building to be put
into perspective. Having placed the plan at the required angle to the
picture plane, they fix upon the point of sight, and the distance from
which the drawing is to be viewed. They then draw a line _SP_ at right
angles to the picture plane _VV·_, which represents that distance so
that _P_ is the station-point. The eye is generally considered to be
the station-point, but when lines are drawn to that point from the
ground-plan, the station-point is placed on the ground, and is in fact
the trace or projection exactly under the point at which the eye is
placed. From this station-point _P_, draw lines _PV_ and _PV·_ parallel
to the two sides of the plan _ba_ and _ad_ (which will be at right
angles to each other), and produce them to the horizon, which they will
touch at points _V_ and _V·_. These points thus obtained will be the
two vanishing points.

  [Illustration: Fig. 197.
  A method of angular Perspective employed by architects.
  [_To face p. 171_] ]

The next operation is to draw lines from the principal points of the
plan to the station-point _P_, such as _bP_, _cP_, _dP_, &c., and where
these lines intersect the picture plane (_VV·_ here represents it as
well as the horizon), drop perpendiculars _b·B_, _aA_, _d·D_, &c., to
meet the vanishing lines _AV_, _AV·_, which will determine the points
_A_, _B_, _C_, _D_, 1, 2, 3, &c., and also the perspective lengths of
the sides of the figure _AB_, _AD_, and the divisions _B_, 1, 2, &c.
Taking the height of the figure _AE_ from the elevation, we measure it
on _Aa_; as in this instance _A_ touches the ground line, it may be used
as a line of heights.

I have here placed the perspective drawing under the ground plan to show
the relation between the two, and how the perspective is worked out, but
the general practice is to find the required measurements as here shown,
to mark them on a straight edge of card or paper, and transfer them to
the paper on which the drawing is to be made.

This of course is the simplest form of a plan and elevation. It is easy
to see, however, that we could set out an elaborate building in the same
way as this figure, but in that case we should not place the drawing
underneath the ground-plan, but transfer the measurements to another
sheet of paper as mentioned above.

CIX

THE OCTAGON

To draw the geometrical figure of an octagon contained in a square, take
half of the diagonal of that square as radius, and from each corner
describe a quarter circle. At the eight points where they touch the
sides of the square, draw the eight sides of the octagon.

  [Illustration: Fig. 198.]

  [Illustration: Fig. 199.]

To put this into perspective take the base of the square _AB_ and
thereon form the perspective square _ABCD_. From either extremity of
that base (say _B_) drop perpendicular _BF_, draw diagonal _AF_, and
then from _B_ with radius _BO_, half that diagonal, describe arc _EOE_.
This will give us the measurement _AE_. Make _GB_ equal to _AE_. Then
draw lines from _G_ and _E_ towards _S_, and by means of the diagonals
find the transverse lines _KK_, _hh_, which will give us the eight
points through which to draw the octagon.

CX

HOW TO DRAW THE OCTAGON IN ANGULAR PERSPECTIVE

Form square _ABCD_ (new method), produce sides _BC_ and _AD_ to the
horizon at _V_, and produce _VA_ to _a·_ on base. Drop perpendicular
from _B_ to _F_ the same length as _a·B_, and proceed as in the previous
figure to find the eight points on the oblique square through which to
draw the octagon.

  [Illustration: Fig. 200.]

It will be seen that this operation is very much the same as in parallel
perspective, only we make our measurements on the base line _a·B_ as we
cannot measure the vanishing line _BA_ otherwise.

CXI

HOW TO DRAW AN OCTAGONAL FIGURE IN ANGULAR PERSPECTIVE

In this figure in angular perspective we do precisely the same thing as
in the previous problem, taking our measurements on the base line _EB_
instead of on the vanishing line _BA_. If we wish to raise a figure on
this octagon the height of _EG_ we form the vanishing scale _EGO_, and
from the eight points on the ground draw horizontals to _EO_ and thus
find all the points that give us the perspective height of each angle of
the octagonal figure.

  [Illustration: Fig. 201.]

CXII

HOW TO DRAW CONCENTRIC OCTAGONS, WITH ILLUSTRATION OF A WELL

The geometrical figure 202 A shows how by means of diagonals _AC_ and
_BD_ and the radii 1 2 3, &c., we can obtain smaller octagons inside the
larger ones. Note how these are carried out in the second figure
(202 B), and their application to this drawing of an octagonal well on
an octagonal base.

  [Illustration: Fig. 202 A.]

  [Illustration: Fig. 202 B.]

  [Illustration: Fig. 203.]

CXIII

A PAVEMENT COMPOSED OF OCTAGONS AND SMALL SQUARES

To draw a pavement with octagonal tiles we will begin with an octagon
contained in a square _abcd_. Produce diagonal _ac_ to _V_. This will be
the vanishing point for the sides of the small squares directed towards
it. The other sides are directed to an inaccessible point out of the
picture, but their directions are determined by the lines drawn from
divisions on base to V2 (see back, Fig. 133).

  [Illustration: Fig. 204.]

  [Illustration: Fig. 205.]

I have drawn the lower figure to show how the squares which contain the
octagons are obtained by means of the diagonals, _BD_, _AC_, and the
central line OV2. Given the square _ABCD_. From _D_ draw diagonal to
_G_, then from _C_ through centre _o_ draw _CE_, and so on all the way
up the floor until sufficient are obtained. It is easy to see how other
squares on each side of these can be produced.

CXIV

THE HEXAGON

The hexagon is a six-sided figure which, if inscribed in a circle, will
have each of its sides equal to the radius of that circle (Fig. 206). If
inscribed in a rectangle _ABCD_, that rectangle will be equal in length
to two sides of the hexagon or two radii of the circle, as _EF_, and its
width will be twice the height of an equilateral triangle _mon_.

  [Illustration: Fig. 206.]

To put the hexagon into perspective, draw base of quadrilateral _AD_,
divide it into four equal parts, and from each division draw lines to
point of sight. From _h_ drop perpendicular _ho_, and form equilateral
triangle _mno_. Take the height _ho_ and measure it twice along the base
from _A_ to 2. From 2 draw line to point of distance, or from 1 to ½
distance, and so find length of side _AB_ equal to A2. Draw _BC_,
and _EF_ through centre _o·_, and thus we have the six points through
which to draw the hexagon.

  [Illustration: Fig. 207.]

CXV

A PAVEMENT COMPOSED OF HEXAGONAL TILES

In drawing pavements, except in the cases of square tiles, it is
necessary to make a plan of the required design, as in this figure
composed of hexagons. First set out the hexagon as at _A_, then draw
parallels 1 1, 2 2, &c., to mark the horizontal ends of the tiles
and the intermediate lines _oo_. Divide the base into the required
number of parts, each equal to one side of the hexagon, as 1, 2, 3, 4,
&c.; from these draw perpendiculars as shown in the figure, and also the
diagonals passing through their intersections. Then mark with a strong
line the outlines of the hexagonals, shading some of them; but the
figure explains itself.

It is easy to put all these parallels, perpendiculars, and diagonals
into perspective, and then to draw the hexagons.

First draw the hexagon on _AD_ as in the previous figure, dividing _AD_
into four, &c., set off right and left spaces equal to these fourths,
and from each division draw lines to point of sight. Produce sides _me_,
_nf_ till they touch the horizon in points _V_, _V·_; these will be the
two vanishing points for all the sides of the tiles that are receding
from us. From each division on base draw lines to each of these
vanishing points, then draw parallels through their intersections as
shown on the figure. Having all these guiding lines it will not be
difficult to draw as many hexagons as you please.

  [Illustration: Fig. 208.]

Note that the vanishing points should be at equal distances from _S_,
also that the parallelogram in which each tile is contained is oblong,
and not square, as already pointed out.

We have also made use of the triangle _omn_ to ascertain the length and
width of that oblong. Another thing to note is that we have made use of
the half distance, which enables us to make our pavement look flat
without spreading our lines outside the picture.

  [Illustration: Fig. 209.]

CXVI

A PAVEMENT OF HEXAGONAL TILES IN ANGULAR PERSPECTIVE

This is more difficult than the previous figure, as we only make use of
one vanishing point; but it shows how much can be done by diagonals, as
nearly all this pavement is drawn by their aid. First make a geometrical
plan _A_ at the angle required. Then draw its perspective _K_. Divide
line 4b into four equal parts, and continue these measurements all
along the base: from each division draw lines to _V_, and draw the
hexagon _K_. Having this one to start with we produce its sides right
and left, but first to the left to find point _G_, the vanishing point
of the diagonals. Those to the right, if produced far enough, would meet
at a distant vanishing point not in the picture. But the student should
study this figure for himself, and refer back to Figs. 204 and 205.

  [Illustration: Fig. 210.]

CXVII

FURTHER ILLUSTRATION OF THE HEXAGON

  [Illustration: Fig. 211 A.]

  [Illustration: Fig. 211 B.]

To draw the hexagon in perspective we must first find the rectangle in
which it is inscribed, according to the view we take of it. That at _A_
we have already drawn. We will now work out that at _B_. Divide the base
_AD_ into four equal parts and transfer those measurements to the
perspective figure _C_, as at _AD_, measuring other equal spaces along
the base. To find the depth _An_ of the rectangle, make _DK_ equal to
base of square. Draw _KO_ to distance-point, cutting _DO_ at _O_, and
thus find line _LO_. Draw diagonal _Dn_, and through its intersections
with the lines 1, 2, 3, 4 draw lines parallel to the base, and we shall
thus have the framework, as it were, by which to draw the pavement.

  [Illustration: Fig. 212.]

CXVIII

ANOTHER VIEW OF THE HEXAGON IN ANGULAR PERSPECTIVE

  [Illustration: Fig. 213.]

Given the rectangle _ABCD_ in angular perspective, produce side _DA_ to
_E_ on base line. Divide _EB_ into four equal parts, and from each
division draw lines to vanishing point, then by means of diagonals, &c.,
draw the hexagon.

In Fig. 214 we have first drawn a geometrical plan, _G_, for the sake of
clearness, but the one above shows that this is not necessary.

  [Illustration: Fig. 214.]

To raise the hexagonal figure _K_ we have made use of the vanishing
scale _O_ and the vanishing point _V_. Another method could be used by
drawing two hexagons one over the other at the required height.

CXIX

APPLICATION OF THE HEXAGON TO DRAWING A KIOSK

  [Illustration: Fig. 215.]

This figure is built up from the hexagon standing on a rectangular base,
from which we have raised verticals, &c. Note how the jutting portions
of the roof are drawn from _o·_. But the figure explains itself, so
there is no necessity to repeat descriptions already given in the
foregoing problems.

CXX

THE PENTAGON

  [Illustration: Fig. 216.]

The pentagon is a figure with five equal sides, and if inscribed in a
circle will touch its circumference at five equidistant points. With any
convenient radius describe circle. From half this radius, marked 1, draw
a line to apex, marked 2. Again, with 1 as centre and 1 2 as radius,
describe arc 2 3. Now with 2 as centre and 2 3 as radius describe arc
3 4, which will cut the circumference at point 4. Then line 2 4 will be
one of the sides of the pentagon, which we can measure round the circle
and so produce the required figure.

To put this pentagon into parallel perspective inscribe the circle in
which it is drawn in a square, and from its five angles 4, 2, 4, &c.,
drop perpendiculars to base and number them as in the figure. Then draw
the perspective square (Fig. 217) and transfer these measurements to its
base. From these draw lines to point of sight, then by their aid and the
two diagonals proceed to construct the pentagon in the same way that we
did the triangles and other figures. Should it be required to place this
pentagon in the opposite position, then we can transfer our measurements
to the far side of the square, as in Fig. 218.

  [Illustration: Fig. 217.]

  [Illustration: Fig. 218.]

Or if we wish to put it into angular perspective we adopt the same
method as with the hexagon, as shown at Fig. 219.

  [Illustration: Fig. 219.]

Another way of drawing a pentagon (Fig. 220) is to draw an isosceles
triangle with an angle of 36° at its apex, and from centre of each side
of the triangle draw perpendiculars to meet at _o_, which will be the
centre of the circle in which it is inscribed. From this centre and
with radius _OA_ describe circle A 3 2, &c. Take base of triangle 1 2,
measure it round the circle, and so find the five points through which
to draw the pentagon. The angles at 1 2 will each be 72°, double that at
_A_, which is 36°.

  [Illustration: Fig. 220.]

CXXI

THE PYRAMID

Nothing can be more simple than to put a pyramid into perspective. Given
the base (_abc_), raise from its centre a perpendicular (_OP_) of the
required height, then draw lines from the corners of that base to a
point _P_ on the vertical line, and the thing is done. These pyramids
can be used in drawing roofs, steeples, &c. The cone is drawn in the
same way, so also is any other figure, whether octagonal, hexangular,
triangular, &c.

  [Illustration: Fig. 221.]

  [Illustration: Fig. 222.]

  [Illustration: Fig. 223.]

  [Illustration: Fig. 224.]

CXXII

THE GREAT PYRAMID

This enormous structure stands on a square base of over thirteen acres,
each side of which measures, or did measure, 764 feet. Its original
height was 480 feet, each side being an equilateral triangle. Let us see
how we can draw this gigantic mass on our little sheet of paper.

In the first place, to take it all in at one view we must put it very
far back, and in the second the horizon must be so low down that we
cannot draw the square base of thirteen acres on the perspective plane,
that is on the ground, so we must draw it in the air, and also to a very
small scale.

Divide the base _AB_ into ten equal parts, and suppose each of these
parts to measure 10 feet, _S_, the point of sight, is placed on the left
of the picture near the side, in order that we may get a long line of
distance, _S ½ D_; but even this line is only half the distance we
require. Let us therefore take the 16th distance, as shown in our
previous illustration of the lighthouse (Fig. 92), which enables us to
measure sixteen times the length of base _AB_, or 1,600 feet. The base
_ef_ of the pyramid is 1,600 feet from the base line of the picture, and
is, according to our 10-foot scale, 764 feet long.

The next thing to consider is the height of the pyramid. We make a scale
to the right of the picture measuring 50 feet from _B_ to 50 at point
where _BP_ intersects base of pyramid, raise perpendicular _CG_ and
thereon measure 480 feet. As we cannot obtain a palpable square on the
ground, let us draw one 480 feet above the ground. From _e_ and _f_
raise verticals _eM_ and _fN_, making them equal to perpendicular _G_,
and draw line _MN_, which will be the same length as base, or 764 feet.
On this line form square _MNK_ parallel to the perspective plane, find
its centre _O·_ by means of diagonals, and _O·_ will be the central
height of the pyramid and exactly over the centre of the base. From this
point _O·_ draw sloping lines _O·f_, _O·e_, _O·Y_, &c., and the figure
is complete.

Note the way in which we find the measurements on base of pyramid and on
line _MN_. By drawing _AS_ and _BS_ to point of sight we find _Te_,
which measures 100 feet at a distance of 1,600 feet. We mark off seven
of these lengths, and an additional 64 feet by the scale, and so obtain
the required length. The position of the third corner of the base is
found by dropping a perpendicular from _K_, till it meets the line _eS_.

Another thing to note is that the side of the pyramid that faces us,
although an equilateral triangle, does not appear so, as its top angle
is 382 feet farther off than its base owing to its leaning position.

CXXIII

THE PYRAMID IN ANGULAR PERSPECTIVE

In order to show the working of this proposition I have taken a much
higher horizon, which immediately detracts from the impression of the
bigness of the pyramid.

  [Illustration: Fig. 225.]

We proceed to make our ground-plan _abcd_ high above the horizon instead
of below it, drawing first the parallel square and then the oblique one.
From all the principal points drop perpendiculars to the ground and thus
find the points through which to draw the base of the pyramid. Find
centres _OO·_ and decide upon the height _OP_. Draw the sloping lines
from _P_ to the corners of the base, and the figure is complete.

CXXIV

TO DIVIDE THE SIDES OF THE PYRAMID HORIZONTALLY

Having raised the pyramid on a given oblique square, divide the vertical
line OP into the required number of parts. From _A_ through _C_ draw
_AG_ to horizon, which gives us _G_, the vanishing point of all the
diagonals of squares parallel to and at the same angle as _ABCD_. From
_G_ draw lines through the divisions 2, 3, &c., on _OP_ cutting the
lines _PA_ and _PC_, thus dividing them into the required parts. Through
the points thus found draw from _V_ all those sides of the squares that
have _V_ for their vanishing point, as _ab_, _cd_, &c. Then join _bd_,
_ac_, and the rest, and thus make the horizontal divisions required.

  [Illustration: Fig. 226.]

  [Illustration: Fig. 227.]

The same method will apply to drawing steps, square blocks, &c., as
shown in Fig. 227, which is at the same angle as the above.

CXXV

OF ROOFS

The pyramidal roof (Fig. 228) is so simple that it explains itself. The
chief thing to be noted is the way in which the diagonals are produced
beyond the square of the walls, to give the width of the eaves,
according to their position.

  [Illustration: Fig. 228.]

Another form of the pyramidal roof is here given (Fig. 229). First draw
the cube _edcba_ at the required height, and on the side facing us,
_adcb_, draw triangle _K_, which represents the end of a gable roof.
Then draw similar triangles on the other sides of the cube (see Fig.
159, LXXXIV). Join the opposite triangles at the apex, and thus form two
gable roofs crossing each other at right angles. From _o_, centre of
base of cube, raise vertical _OP_, and then from _P_ draw sloping lines
to each corner of base _a_, _b_, &c., and by means of central lines
drawn from _P_ to half base, find the points where the gable roofs
intersect the central spire or pyramid. Any other proportions can be
obtained by adding to or altering the cube.

  [Illustration: Fig. 229.]

To draw a sloping or hip-roof which falls back at each end we must first
draw its base, _CBDA_ (Fig. 230). Having found the centre _O_ and
central line _SP_, and how far the roof is to fall back at each end,
namely the distance _Pm_, draw horizontal line _RB_ through _m_. Then
from _B_ through _O_ draw diagonal _BA_, and from _A_ draw horizontal
_AD_, which gives us point _n_. From these two points _m_ and _n_ raise
perpendiculars the height required for the roof, and from these draw
sloping lines to the corners of the base. Join _ef_, that is, draw the
top line of the roof, which completes it. Fig. 231 shows a plan or
bird's-eye view of the roof and the diagonal _AB_ passing through centre
_O_. But there are so many varieties of roofs they would take almost a
book to themselves to illustrate them, especially the cottages and
farm-buildings, barns, &c., besides churches, old mansions, and others.
There is also such irregularity about some of them that perspective
rules, beyond those few here given, are of very little use. So that the
best thing for an artist to do is to sketch them from the real whenever
he has an opportunity.

  [Illustration: Fig. 230.]

  [Illustration: Fig. 231.]

CXXVI

OF ARCHES, ARCADES, BRIDGES, &C.

  [Illustration: Fig. 232.]

For an arcade or cloister (Fig. 232) first set up the outer frame _ABCD_
according to the proportions required. For round arches the height may
be twice that of the base, varying to one and a half. In Gothic arches
the height may be about three times the width, all of which proportions
are chosen to suit the different purposes and effects required. Divide
the base _AB_ into the desired number of parts, 8, 10, 12, &c., each
part representing 1 foot. (In this case the base is 10 feet and the
horizon 5 feet.) Set out floor by means of ¼ distance. Divide it into
squares of 1 foot, so that there will be 8 feet between each column or
pilaster, supposing we make them to stand on a square foot. Draw the
first archway _EKF_ facing us, and its inner semicircle _gh_, with also
its thickness or depth of 1 foot. Draw the span of the archway _EF_,
then central line _PO_ to point of sight. Proceed to raise as many other
arches as required at the given distances. The intersections of the
central line with the chords _mn_, &c., will give the centres from which
to describe the semicircles.

CXXVII

OUTLINE OF AN ARCADE WITH SEMICIRCULAR ARCHES

This is to show the method of drawing a long passage, corridor, or
cloister with arches and columns at equal distances, and is worked in
the same way as the previous figure, using ¼ distance and ¼ base.
The floor consists of five squares; the semicircles of the arches are
described from the numbered points on the central line _OS_, where it
intersects the chords of the arches.

  [Illustration: Fig. 233.]

CXXVIII

SEMICIRCULAR ARCHES ON A RETREATING PLANE

First draw perspective square _abcd_. Let _ae·_ be the height of the
figure. Draw _ae·f·b_ and proceed with the rest of the outline. To draw
the arches begin with the one facing us, _Eo·F_ enclosed in the
quadrangle _Ee·f·F_. With centre _O_ describe the semicircle and across
it draw the diagonals _e·F_, _Ef·_, and through _nn_, where these lines
intersect the semicircle, draw horizontal _KK_ and also _KS_ to point of
sight. It will be seen that the half-squares at the side are the same
size in perspective as the one facing us, and we carry out in them much
the same operation; that is, we draw the diagonals, find the point _O_,
and the points _nn_, &c., through which to draw our arches. See
perspective of the circle (Fig. 165).

  [Illustration: Fig. 234.]

If more points are required an additional diagonal from _O_ to _K_ may
be used, as shown in the figure, which perhaps explains itself. The
method is very old and very simple, and of course can be applied to any
kind of arch, pointed or stunted, as in this drawing of a pointed arch
(Fig. 235).

  [Illustration: Fig. 235.]

CXXIX

AN ARCADE IN ANGULAR PERSPECTIVE

First draw the perspective square _ABCD_ at the angle required, by new
method. Produce sides _AD_ and _BC_ to _V_. Draw diagonal _BD_ and
produce to point _G_, from whence we draw the other diagonals to _cfh_.
Make spaces 1, 2, 3, &c., on base line equal to _B 1_ to obtain sides of
squares. Raise vertical _BM_ the height required. Produce _DA_ to _O_ on
base line, and from _O_ raise vertical _OP_ equal to _BM_. This line
enables us to dispense with the long vanishing point to the left; its
working has been explained at Fig. 131. From _P_ draw _PRV_ to vanishing
point _V_, which will intersect vertical _AR_ at _R_. Join _MR_, and
this line, if produced, would meet the horizon at the other vanishing
point. In like manner make O2 equal to B2·. From 2 draw line to _V_, and
at 2, its intersection with _AR_, draw line 2 2, which will also meet
the horizon at the other vanishing point. By means of the quarter-circle
_A_ we can obtain the points through which to draw the semicircular
arches in the same way as in the previous figure.

  [Illustration: Fig. 236.]

CXXX

A VAULTED CEILING

From the square ceiling _ABCD_ we have, as it were, suspended two arches
from the two diagonals _DB_, _AC_, which spring from the four corners of
the square _EFGH_, just underneath it. The curves of these arches, which
are not semicircular but elongated, are obtained by means of the
vanishing scales _mS_, _nS_. Take any two convenient points _P_, _R_, on
each side of the semicircle, and raise verticals _Pm_, _Rn_ to _AB_, and
on these verticals form the scales. Where _mS_ and _nS_ cut the diagonal
_AC_ drop perpendiculars to meet the lower line of the scale at points
1, 2. On the other side, using the other scales, we have dropped
perpendiculars in the same way from the diagonal to 3, 4. These points,
together with _EOG_, enable us to trace the curve _E 1 2 O 3 4 G_. We
draw the arch under the other diagonal in precisely the same way.

  [Illustration: Fig. 237.]

The reason for thus proceeding is that the cross arches, although
elongated, hang from their diagonals just as the semicircular arch _EKF_
hangs from _AB_, and the lines _mn_, touching the circle at _PR_, are
represented by 1, 2, hanging from the diagonal _AC_.

  [Illustration: Fig. 238.]

Figure 238, which is practically the same as the preceding only
differently shaded, is drawn in the following manner. Draw arch _EGF_
facing us, and proceed with the rest of the corridor, but first finding
the flat ceiling above the square on the ground _ABcd_. Draw diagonals
_ac_, _bd_, and the curves pending from them. But we no longer see the
clear arch as in the other drawing, for the spaces between the curves
are filled in and arched across.

CXXXI

A CLOISTER, FROM A PHOTOGRAPH

This drawing of a cloister from a photograph shows the correctness of
our perspective, and the manner of applying it to practical work.

  [Illustration: Fig. 239.]

CXXXII

THE LOW OR ELLIPTICAL ARCH

Let _AB_ be the span of the arch and _Oh_ its height. From centre _O_,
with _OA_, or half the span, for radius, describe outer semicircle. From
same centre and _oh_ for radius describe the inner semicircle. Divide
outer circle into a convenient number of parts, 1, 2, 3, &c., to which
draw radii from centre _O_. From each division drop perpendiculars.
Where the radii intersect the inner circle, as at _gkmo_, draw
horizontals _op_, _mn_, _kj_, &c., and through their intersections with
the perpendiculars _f_, _j_, _n_, _p_, draw the curve of the flattened
arch. Transfer this to the lower figure, and proceed to draw the tunnel.
Note how the vanishing scale is formed on either side by horizontals
_ba_, _fe_, &c., which enable us to make the distant arches similar to
the near ones.

  [Illustration: Fig. 240.]

  [Illustration: Fig. 241.]

CXXXIII

OPENING OR ARCHED WINDOW IN A VAULT

First draw the vault _AEB_. To introduce the window _K_, the upper part
of which follows the form of the vault, we first decide on its width,
which is _mn_, and its height from floor _Ba_. On line _Ba_ at the side
of the arch form scales _aa·S_, _bb·S_, &c. Raise the semicircular arch
_K_, shown by a dotted line. The scale at the side will give the lengths
_aa·_, _bb·_, &c., from different parts of this dotted arch to
corresponding points in the curved archway or window required.

  [Illustration: Fig. 242.]

Note that to obtain the width of the window _K_ we have used the
diagonals on the floor and width _m n_ on base. This method of
measurement is explained at Fig. 144, and is of ready application in a
case of this kind.

CXXXIV

STAIRS, STEPS, &C.

Having decided upon the incline or angle, such as _CBA_, at which the
steps are to be placed, and the height _Bm_ of each step, draw _mn_ to
_CB_, which will give the width. Then measure along base _AB_ this width
equal to _DB_, which will give that for all the other steps. Obtain
length _BF_ of steps, and draw _EF_ parallel to _CB_. These lines will
aid in securing the exactness of the figure.

  [Illustration: Fig. 243.]

  [Illustration: Fig. 244.]

CXXXV

STEPS, FRONT VIEW

In this figure the height of each step is measured on the vertical line
_AB_ (this line is sometimes called the line of heights), and their
depth is found by diagonals drawn to the point of distance _D_. The rest
of the figure explains itself.

  [Illustration: Fig. 245.]

CXXXVI

SQUARE STEPS

Draw first step _ABEF_ and its two diagonals. Raise vertical _AH_, and
measure thereon the required height of each step, and thus form scale.
Let the second step _CD_ be less all round than the first by _Ao_ or
_Bo_. Draw _oC_ till it cuts the diagonal, and proceed to draw the
second step, guided by the diagonals and taking its height from the
scale as shown. Draw the third step in the same way.

  [Illustration: Fig. 246.]

CXXXVII

TO DIVIDE AN INCLINED PLANE INTO EQUAL PARTS--SUCH AS A LADDER PLACED
AGAINST A WALL

  [Illustration: Fig. 247.]

Divide the vertical _EC_ into the required number of parts, and draw
lines from point of sight _S_ through these divisions 1, 2, 3, &c.,
cutting the line _AC_ at 1, 2, 3, &c. Draw parallels to _AB_, such as
_mn_, from _AC_ to _BD_, which will represent the steps of the ladder.

CXXXVIII

STEPS AND THE INCLINED PLANE

  [Illustration: Fig. 248.]

In Fig. 248 we treat a flight of steps as if it were an inclined plane.
Draw the first and second steps as in Fig. 245. Then through 1, 2, draw
1V, _AV_ to _V_, the vanishing point on the vertical line _SV_. These
two lines and the corresponding ones at _BV_ will form a kind of
vanishing scale, giving the height of each step as we ascend. It is
especially useful when we pass the horizontal line and we no longer see
the upper surface of the step, the scale on the right showing us how to
proceed in that case.

In Fig. 249 we have an example of steps ascending and descending. First
set out the ground-plan, and find its vanishing point _S_ (point of
sight). Through _S_ draw vertical _BA_, and make _SA_ equal to _SB_. Set
out the first step _CD_. Draw _EA_, _CA_, _DA_, and _GA_, for the
ascending guiding lines. Complete the steps facing us, at central line
_OO_. Then draw guiding line _FB_ for the descending steps (see Rule 8).

  [Illustration: Fig. 249.]

CXXXIX

STEPS IN ANGULAR PERSPECTIVE

First draw the base _ABCD_ (Fig. 251) at the required angle by the new
method (Fig. 250). Produce _BC_ to the horizon, and thus find vanishing
point _V_. At this point raise vertical _VV·_. Construct first step
_AB_, refer its height at _B_ to line of heights hI on left, and thus
obtain height of step at _A_. Draw lines from _A_ and _F_ to _V·_. From
_n_ draw diagonal through _O_ to _G_. Raise vertical at _O_ to represent
the height of the next step, its height being determined by the scale of
heights at the side. From _A_ and _F_ draw lines to _V·_, and also
similar lines from _B_, which will serve as guiding lines to determine
the height of the steps at either end as we raise them to the required
number.

  [Illustration: Fig. 250.]

  [Illustration: Fig. 251.]

CXL

A STEP LADDER AT AN ANGLE

  [Illustration: Fig. 252.]

First draw the ground-plan _G_ at the required angle, using vanishing
and measuring points. Find the height _hH_, and width at top _HH·_, and
draw the sides _HA_ and _H·E_. Note that _AE_ is wider than _HH·_, and
also that the back legs are not at the same angle as the front ones, and
that they overlap them. From _E_ raise vertical _EF_, and divide into as
many parts as you require rounds to the ladder. From these divisions
draw lines 1 1, 2 2, &c., towards the other vanishing point (not in the
picture), but having obtained their direction from the ground-plan in
perspective at line _Ee_, you may set up a second vertical _ef_ at any
point on _Ee_ and divide it into the same number of parts, which will be
in proportion to those on _EF_, and you will obtain the same result by
drawing lines from the divisions on _EF_ to those on _ef_ as in drawing
them to the vanishing point.

CXLI

SQUARE STEPS PLACED OVER EACH OTHER

  [Illustration: Fig. 253.]

This figure shows the other method of drawing steps, which is simple
enough if we have sufficient room for our vanishing points.

The manner of working it is shown at Fig. 124.

CXLII

STEPS AND A DOUBLE CROSS DRAWN BY MEANS OF DIAGONALS
AND ONE VANISHING POINT

Although in this figure we have taken a longer distance-point than in
the previous one, we are able to draw it all within the page.

  [Illustration: Fig. 254.]

Begin by setting out the square base at the angle required. Find point
_G_ by means of diagonals, and produce _AB_ to _V_, &c. Mark height of
step _Ao_, and proceed to draw the steps as already shown. Then by the
diagonals and measurements on base draw the second step and the square
inside it on which to stand the foot of the cross. To draw the cross,
raise verticals from the four corners of its base, and a line _K_ from
its centre. Through any point on this central line, if we draw a
diagonal from point _G_ we cut the two opposite verticals of the shaft
at _mn_ (see Fig. 255), and by means of the vanishing point _V_ we cut
the other two verticals at the opposite corners and thus obtain the four
points through which to draw the other sides of the square, which go to
the distant or inaccessible vanishing point. It will be seen by
carefully examining the figure that by this means we are enabled to draw
the double cross standing on its steps.

  [Illustration: Fig. 255.]

  [Illustration: Fig. 256.]

CXLIII

A STAIRCASE LEADING TO A GALLERY

In this figure we have made use of the devices already set forth in the
foregoing figures of steps, &c., such as the side scale on the left of
the figure to ascertain the height of the steps, the double lines drawn
to the high vanishing point of the inclined plane, and so on; but the
principal use of this diagram is to show on the perspective plane, which
as it were runs under the stairs, the trace or projection of the flights
of steps, the landings and positions of other objects, which will be
found very useful in placing figures in a composition of this kind.
It will be seen that these underneath measurements, so to speak, are
obtained by the half-distance.

CXLIV

WINDING STAIRS IN A SQUARE SHAFT

Draw square _ABCD_ in parallel perspective. Divide each side into four,
and raise verticals from each division. These verticals will mark the
positions of the steps on each wall, four in number. From centre _O_
raise vertical _OP_, around which the steps are to wind. Let _AF_ be the
height of each step. Form scale _AB_, which will give the height of each
step according to its position. Thus at _mn_ we find the height at the
centre of the square, so if we transfer this measurement to the central
line _OP_ and repeat it upwards, say to fourteen, then we have the
height of each step on the line where they all meet. Starting then with
the first on the right, draw the rectangle _gD1f_, the height of _AF_,
then draw to the central line _go_, f1, and 1 1, and thus complete the
first step. On _DE_, measure heights equal to _D 1_. Draw 2 2 towards
central line, and 2n towards point of sight till it meets the second
vertical _nK_. Then draw n2 to centre, and so complete the second
step. From 3 draw 3a to third vertical, from 4 to fourth, and so on,
thus obtaining the height of each ascending step on the wall to the
right, completing them in the same way as numbers 1 and 2, when we come
to the sixth step, the other end of which is against the wall opposite
to us. Steps 6, 7, 8, 9 are all on this wall, and are therefore equal in
height all along, as they are equally distant. Step 10 is turned towards
us, and abuts on the wall to our left; its measurement is taken on the
scale _AB_ just underneath it, and on the same line to which it is
drawn. Step 11 is just over the centre of base _mo_, and is therefore
parallel to it, and its height is _mn_. The widths of steps 12 and 13
seem gradually to increase as they come towards us, and as they rise
above the horizon we begin to see underneath them. Steps 13, 14, 15, 16
are against the wall on this side of the picture, which we may suppose
has been removed to show the working of the drawing, or they might be an
open flight as we sometimes see in shops and galleries, although in that
case they are generally enclosed in a cylindrical shaft.

  [Illustration: Fig. 257.]

  [Illustration: Fig. 258.]

CXLV

WINDING STAIRS IN A CYLINDRICAL SHAFT

First draw the circular base _CD_. Divide the circumference into equal
parts, according to the number of steps in a complete round, say twelve.
Form scale _ASF_ and the larger scale _ASB_, on which is shown the
perspective measurements of the steps according to their positions;
raise verticals such as _ef_, _Gh_, &c. From divisions on circumference
measure out the central line _OP_, as in the other figure, and find the
heights of the steps 1, 2, 3, 4, &c., by the corresponding numbers in
the large scale to the left; then proceed in much the same way as in the
previous figure. Note the central column _OP_ cuts off a small portion
of the steps at that end.

In ordinary cases only a small portion of a winding staircase is
actually seen, as in this sketch.

  [Illustration: Fig. 259. Sketch of Courtyard in Toledo.]

CXLVI

OF THE CYLINDRICAL PICTURE OR DIORAMA

  [Illustration: Fig. 260.]

Although illusion is by no means the highest form of art, there is no
picture painted on a flat surface that gives such a wonderful appearance
of truth as that painted on a cylindrical canvas, such as those
panoramas of 'Paris during the Siege', exhibited some years ago; 'The
Battle of Trafalgar', only lately shown at Earl's Court; and many
others. In these pictures the spectator is in the centre of a cylinder,
and although he turns round to look at the scene the point of sight is
always in front of him, or nearly so. I believe on the canvas these
points are from 12 to 16 feet apart.

The reason of this look of truth may be explained thus. If we place
three globes of equal size in a straight line, and trace their apparent
widths on to a straight transparent plane, those at the sides, as _a_
and _b_, will appear much wider than the centre one at _c_. Whereas, if
we trace them on a semicircular glass they will appear very nearly equal
and, of the three, the central one _c_ will be rather the largest, as
may be seen by this figure.

We must remember that, in the first case, when we are looking at a globe
or a circle, the visual rays form a cone, with a globe at its base. If
these three cones are intersected by a straight glass _GG_, and looked
at from point _S_, the intersection of _C_ will be a circle, as the cone
is cut straight across. The other two being intersected at an angle,
will each be an ellipse. At the same time, if we look at them from the
station point, with one eye only, then the three globes (or tracings of
them) will appear equal and perfectly round.

Of course the cylindrical canvas is necessary for panoramas; but we
have, as a rule, to paint our pictures and wall-decorations on flat
surfaces, and therefore must adapt our work to these conditions.

In all cases the artist must exercise his own judgement both in the
arrangement of his design and the execution of the work, for there is
perspective even in the touch--a painting to be looked at from a
distance requires a bold and broad handling; in small cabinet pictures
that we live with in our own rooms we look for the exquisite workmanship
of the best masters.

BOOK FOURTH

CXLVII

THE PERSPECTIVE OF CAST SHADOWS

There is a pretty story of two lovers which is sometimes told as the
origin of art; at all events, I may tell it here as the origin of
sciagraphy. A young shepherd was in love with the daughter of a potter,
but it so happened that they had to part, and were passing their last
evening together, when the girl, seeing the shadow of her lover's
profile cast from a lamp on to some wet plaster or on the wall, took a
metal point, perhaps some sort of iron needle, and traced the outline of
the face she loved on to the plaster, following carefully the outline of
the features, being naturally anxious to make it as like as possible.
The old potter, the father of the girl, was so struck with it that he
began to ornament his wares by similar devices, which gave them
increased value by the novelty and beauty thus imparted to them.

Here then we have a very good illustration of our present subject and
its three elements. First, the light shining on the wall; second, the
wall or the plane of projection, or plane of shade; and third, the
intervening object, which receives as much light on itself as it
deprives the wall of. So that the dark portion thus caused on the plane
of shade is the cast shadow of the intervening object.

We have to consider two sorts of shadows: those cast by a luminary a
long way off, such as the sun; and those cast by artificial light, such
as a lamp or candle, which is more or less close to the object. In the
first case there is no perceptible divergence of rays, and the outlines
of the sides of the shadows of regular objects, as cubes, posts, &c.,
will be parallel. In the second case, the rays diverge according to the
nearness of the light, and consequently the lines of the shadows,
instead of being parallel, are spread out.

CXLVIII

THE TWO KINDS OF SHADOWS

In Figs. 261 and 262 is seen the shadow cast by the sun by parallel
rays.

Fig. 263 shows the shadows cast by a candle or lamp, where the rays
diverge from the point of light to meet corresponding diverging lines
which start from the foot of the luminary on the ground.

  [Illustration: Fig. 261.]

  [Illustration: Fig. 262.]

The simple principle of cast shadows is that the rays coming from the
point of light or luminary pass over the top of the intervening object
which casts the shadow on to the plane of shade to meet the horizontal
trace of those rays on that plane, or the lines of light proceed from
the point of light, and the lines of the shadow are drawn from the foot
or trace of the point of light.

  [Illustration: Fig. 263.]

  [Illustration: Fig. 264.]

Fig. 264 shows this in profile. Here the sun is on the same plane as the
picture, and the shadow is cast sideways.

Fig. 265 shows the same thing, but the sun being behind the object,
casts its shadow forwards. Although the lines of light are parallel,
they are subject to the laws of perspective, and are therefore drawn
from their respective vanishing points.

  [Illustration: Fig. 265.]

CXLIX

SHADOWS CAST BY THE SUN

Owing to the great distance of the sun, we have to consider the rays of
light proceeding from it as parallel, and therefore subject to the same
laws as other parallel lines in perspective, as already noted. And for
the same reason we have to place the foot of the luminary on the
horizon. It is important to remember this, as these two things make the
difference between shadows cast by the sun and those cast by artificial
light.

The sun has three principal positions in relation to the picture. In the
first case it is supposed to be in the same plane either to the right or
to the left, and in that case the shadows will be parallel with the base
of the picture. In the second position it is on the other side of it,
or facing the spectator, when the shadows of objects will be thrown
forwards or towards him. In the third, the sun is in front of the
picture, and behind the spectator, so that the shadows are thrown in the
opposite direction, or towards the horizon, the objects themselves being
in full light.

CL

THE SUN IN THE SAME PLANE AS THE PICTURE

Besides being in the same plane, the sun in this figure is at an angle
of 45° to the horizon, consequently the shadows will be the same length
as the figures that cast them are high. Note that the shadow of step
No. 1 is cast upon step No. 2, and that of No. 2 on No. 3, the top of
each of these becoming a plane of shade.

  [Illustration: Fig. 266.]

  [Illustration: Fig. 267.]

  [Illustration: Fig. 268.]

When the shadow of an object such as _A_, Fig. 268, which would fall
upon the plane, is interrupted by another object _B_, then the outline
of the shadow is still drawn on the plane, but being interrupted by the
surface _B_ at _C_, the shadow runs up that plane till it meets the rays
1, 2, which define the shadow on plane _B_. This is an important point,
but is quite explained by the figure.

Although we have said that the rays pass over the top of the object
casting the shadow, in the case of an archway or similar figure they
pass underneath it; but the same principle holds good, that is, we draw
lines from the guiding points in the arch, 1, 2, 3, &c., at the same
angle of 45° to meet the traces of those rays on the plane of shade, and
so get the shadow of the archway, as here shown.

  [Illustration: Fig. 269.]

CLI

THE SUN BEHIND THE PICTURE

We have seen that when the sun's altitude is at an angle of 45° the
shadows on the horizontal plane are the same length as the height of the
objects that cast them. Here (Fig. 270), the sun still being at 45°
altitude, although behind the picture, and consequently throwing the
shadow of _B_ forwards, that shadow must be the same length as the
height of cube _B_, which will be seen is the case, for the shadow _C_
is a square in perspective.

  [Illustration: Fig. 270.]

To find the angle of altitude and the angle of the sun to the picture,
we must first find the distance of the spectator from the foot of the
luminary.

  [Illustration: Fig. 271.]

From point of sight _S_ (Fig. 270) drop perpendicular to _T_, the
station-point. From _T_ draw _TF_ at 45° to meet horizon at _F_. With
radius _FT_ make _FO_ equal to it. Then _O_ is the position of the
spectator. From _F_ raise vertical _FL_, and from _O_ draw a line at 45°
to meet _FL_ at _L_, which is the luminary at an altitude of 45°, and at
an angle of 45° to the picture.

Fig. 272 is similar to the foregoing, only the angles of altitude and of
the sun to the picture are altered.

_Note._--The sun being at 50° to the picture instead of 45°, is nearer
the point of sight; at 90° it would be exactly opposite the spectator,
and so on. Again, the elevation being less (40° instead of 45°) the
shadow is longer. Owing to the changed position of the sun two sides of
the cube throw a shadow. Note also that the outlines of the shadow, 1 2,
2 3, are drawn to the same vanishing points as the cube itself.

It will not be necessary to mark the angles each time we make a drawing,
as it must be seen we can place the luminary in any position that suits
our convenience.

  [Illustration: Fig. 272.]

CLII

SUN BEHIND THE PICTURE, SHADOWS THROWN ON A WALL

As here we change the conditions we must also change our procedure. An
upright wall now becomes the plane of shade, therefore as the principle
of shadows must always remain the same we have to change the relative
positions of the luminary and the foot thereof.

At _S_ (point of sight) raise vertical _SF·_, making it equal to _fL_.
_F·_ becomes the foot of the luminary, whilst the luminary itself still
remains at _L_.

  [Illustration: Fig. 273.]

We have but to turn this page half round and look at it from the right,
and we shall see that _SF·_ becomes as it were the horizontal line. The
luminary _L_ is at the right side of point _S_ instead of the left, and
the foot thereof is, as before, the trace of the luminary, as it is just
underneath it. We shall also see that by proceeding as in previous
figures we obtain the same results on the wall as we did on the
horizontal plane. Fig. B being on the horizontal plane is treated as
already shown. The steps have their shadows partly on the wall and
partly on the horizontal plane, so that the shadows on the wall are
outlined from _F·_ and those on the ground from _f_. Note shadow of roof
_A_, and how the line drawn from _F·_ through _A_ is met by the line
drawn from the luminary _L_, at the point _P_, and how the lower line of
the shadow is directed to point of sight _S_.

  [Illustration: Fig. 274.]

Fig. 274 is a larger drawing of the steps, &c., in further illustration
of the above.

CLIII

SUN BEHIND THE PICTURE THROWING SHADOW ON AN INCLINED PLANE

  [Illustration: Fig. 275.]

The vanishing point of the shadows on an inclined plane is on a vertical
dropped from the luminary to a point (_F_) on a level with the vanishing
point (_P_) of that inclined plane. Thus _P_ is the vanishing point of
the inclined plane _K_. Draw horizontal _PF_ to meet _fL_ (the line
drawn from the luminary to the horizon). Then _F_ will be the vanishing
point of the shadows on the inclined plane. To find the shadow of _M_
draw lines from _F_ through the base _eg_ to _cd_. From luminary _L_
draw lines through _ab_, also to _cd_, where they will meet those drawn
from _F_. Draw _CD_, which determines the length of the shadow _egcd_.

CLIV

THE SUN IN FRONT OF THE PICTURE

  [Illustration: Fig. 276.]

When the sun is in front of the picture we have exactly the opposite
effect to that we have just been studying. The shadows, instead of
coming towards us, are retreating from us, and the objects throwing them
are in full light, consequently we have to reverse our treatment. Let us
suppose the sun to be placed above the horizon at _L·_, on the right of
the picture and behind the spectator (Fig. 276). If we transport the
length _L·f·_ to the opposite side and draw the vertical downwards from
the horizon, as at _FL_, we can then suppose point _L_ to be exactly
opposite the sun, and if we make that the vanishing point for the sun's
rays we shall find that we obtain precisely the same result. As in Fig.
277, if we wish to find the length of _C_, which we may suppose to be
the shadow of _P_, we can either draw a line from _A_ through _O_ to
_B_, or from _B_ through _O_ to _A_, for the result is the same. And as
we cannot make use of a point that is behind us and out of the picture,
we have to resort to this very ingenious device.

  [Illustration: Fig. 277.]

In Fig. 276 we draw lines L1, L2, L3 from the luminary to the top of the
object to meet those drawn from the foot _F_, namely F1, F2, F3, in the
same way as in the figures we have already drawn.

  [Illustration: Fig. 278.]

Fig. 278 gives further illustration of this problem.

CLV

THE SHADOW OF AN INCLINED PLANE

The two portions of this inclined plane which cast the shadow are first
the side _fbd_, and second the farther end _abcd_. The points we have to
find are the shadows of _a_ and _b_. From luminary _L_ draw _La_, _Lb_,
and from _F_, the foot, draw _Fc_, _Fd_. The intersection of these lines
will be at _a·b·_. If we join _fb·_ and _db·_ we have the shadow of the
side _fbd_, and if we join _ca·_ and _a·b·_ we have the shadow of
_abcd_, which together form that of the figure.

  [Illustration: Fig. 279.]

CLVI

SHADOW ON A ROOF OR INCLINED PLANE

To draw the shadow of the figure _M_ on the inclined plane _K_ (or a
chimney on a roof). First find the vanishing point _P_ of the inclined
plane and draw horizontal _PF_ to meet vertical raised from _L_, the
luminary. Then _F_ will be the vanishing point of the shadow. From _L_
draw L1, L2, L3 to top of figure _M_, and from the base of _M_ draw
1F, 2F, 3F to _F_, the vanishing point of the shadow. The
intersections of these lines at 1, 2, 3 on _K_ will determine the
length and form of the shadow.

  [Illustration: Fig. 280.]

CLVII

TO FIND THE SHADOW OF A PROJECTION OR BALCONY ON A WALL

To find the shadow of the object _K_ on the wall _W_, drop verticals
_OO_ till they meet the base line _B·B·_ of the wall. Then from the
point of sight _S_ draw lines through _OO_, also drop verticals _Dd·_,
_Cc·_, to meet these lines in _d·c·_; draw _c·F_ and _d·F_ to foot of
luminary. From the points _xx_ where these lines cut the base _B_ raise
perpendiculars _xa·_, _xb·_. From _D_, _A_, and _B_ draw lines to the
luminary _L_. These lines or rays intersecting the verticals raised from
_xx_ at _a·b·_ will give the respective points of the shadow.

  [Illustration: Fig. 281.]

The shadow of the eave of a roof can be obtained in the same way. Take
any point thereon, mark its trace on the ground, and then proceed as
above.

CLVIII

SHADOW ON A RETREATING WALL, SUN IN FRONT

Let _L_ be the luminary. Raise vertical _LF_. _F_ will be the vanishing
point of the shadows on the ground. Draw _Lf·_ parallel to _FS_. Drop
_Sf·_ from point of sight; _f·_ (so found) is the vanishing point of the
shadows on the wall. For shadow of roof draw _LE_ and _f·B_, giving us
_e_, the shadow of _E_. Join _Be_, &c., and so draw shadow of eave of
roof.

  [Illustration: Fig. 282.]

For shadow of _K_ draw lines from luminary _L_ to meet those from _f·_
the foot, &c.

The shadow of _D_ over the door is found in a similar way to that of the
roof.

  [Illustration: Fig. 283.]

Figure 283 shows how the shadow of the old man in the preceding drawing
is found.

CLIX

SHADOW OF AN ARCH, SUN IN FRONT

Having drawn the arch, divide it into a certain number of parts, say
five. From these divisions drop perpendiculars to base line. From
divisions on _AB_ draw lines to _F_ the foot, and from those on the
semicircle draw lines to _L_ the luminary. Their intersections will give
the points through which to draw the shadow of the arch.

  [Illustration: Fig. 284.]

CLX

SHADOW IN A NICHE OR RECESS

In this figure a similar method to that just explained is adopted. Drop
perpendiculars from the divisions of the arch 1 2 3 to the base. From
the foot of each draw 1S, 2S, 3S to foot of luminary _S_, and
from the top of each, A 1 2 3 B, draw lines to _L_ as before. Where the
former intersect the curve on the floor of the niche raise verticals
to meet the latter at P 1 2 B, &c. These points will indicate about the
position of the shadow; but the niche being semicircular and domed at
the top the shadow gradually loses itself in a gradated and somewhat
serpentine half-tone.

  [Illustration: Fig. 285.]

CLXI

SHADOW IN AN ARCHED DOORWAY

  [Illustration: Fig. 286.]

This is so similar to the last figure in many respects that I need not
repeat a description of the manner in which it is done. And surely an
artist after making a few sketches from the actual thing will hardly
require all this machinery to draw a simple shadow.

CLXII

SHADOWS PRODUCED BY ARTIFICIAL LIGHT

  [Illustration: Fig. 287.]

Shadows thrown by artificial light, such as a candle or lamp, are found
by drawing lines from the seat of the luminary through the feet of the
objects to meet lines representing rays of light drawn from the luminary
itself over the tops or the corners of the objects; very much as in the
cases of sun-shadows, but with this difference, that whereas the foot of
the luminary in this latter case is supposed to be on the horizon an
infinite distance away, the foot in the case of a lamp or candle may be
on the floor or on a table close to us. First draw the table and chair,
&c. (Fig. 287), and let _L_ be the luminary. For objects on the table
such as _K_ the foot will be at _f_ on the table. For the shadows on the
floor, of the chair and table itself, we must find the foot of the
luminary on the floor. Draw _So_, find trace of the edge of the table,
drop vertical _oP_, draw _PS_ to point of sight, drop vertical from foot
of candlestick to meet _PS_ in _F_. Then _F_ is the foot of the luminary
on the floor. From this point draw lines through the feet or traces of
objects such as the corners of the table, &c., to meet other lines drawn
from the point of light, and so obtain the shadow.

CLXIII

SOME OBSERVATIONS ON REAL LIGHT AND SHADE

Although the figures we have been drawing show the principles on which
sun-shadows are shaped, still there are so many more laws to be
considered in the great art of light and shade that it is better to
observe them in Nature herself or under the teaching of the real sun. In
the study of a kitchen and scullery in an old house in Toledo (Fig. 288)
we have an example of the many things to be considered besides the mere
shapes of shadows of regular forms. It will be seen that the light is
dispersed in all directions, and although there is a good deal of
half-shade there are scarcely any cast shadows except on the floor; but
the light on the white walls in the outside gallery is so reflected into
the cast shadows that they are extremely faint. The luminosity of this
part of the sketch is greatly enhanced by the contrast of the dark legs
of the bench and the shadows in the roof. The warm glow of all this
portion is contrasted by the grey door and its frame.

  [Illustration: Fig. 288.]

Note that the door itself is quite luminous, and lighted up by the
reflection of the sun from the tiled floor, so that the bars in the
upper part throw distinct shadows, besides the mystery of colour thus
introduced. The little window to the left, though not admitting much
direct sunlight, is evidence of the brilliant glare outside; for the
reflected light is very conspicuous on the top and on the shutters on
each side; indeed they cast distinct shadows up and down, while some
clear daylight from the blue sky is reflected on the window-sill. As to
the sink, the table, the wash-tubs, &c., although they seem in strong
light and shade they really receive little or no direct light from a
single point; but from the strong reflected light re-reflected into them
from the wall of the doorway. There are many other things in such
effects as this which the artist will observe, and which can only be
studied from real light and shade. Such is the character of reflected
light, varying according to the angle and intensity of the luminary and
a hundred other things. When we come to study light in the open air we
get into another region, and have to deal with it accordingly, and yet
we shall find that our sciagraphy will be a help to us even in this
bewilderment; for it will explain in a manner the innumerable shapes of
sun-shadows that we observe out of doors among hills and dales, showing
up their forms and structure; its play in the woods and gardens, and its
value among buildings, showing all their juttings and abuttings,
recesses, doorways, and all the other architectural details. Nor must we
forget light's most glorious display of all on the sea and in the clouds
and in the sunrises and the sunsets down to the still and lovely
moonlight.

These sun-shadows are useful in showing us the principle of light and
shade, and so also are the shadows cast by artificial light; but they
are only the beginning of that beautiful study, that exquisite art of
tone or _chiaro-oscuro_, which is infinite in its variety, is full of
the deepest mystery, and is the true poetry of art. For this the student
must go to Nature herself, must study her in all her moods from early
dawn to sunset, in the twilight and when night sets in. No mathematical
rules can help him, but only the thoughtful contemplation, the silent
watching, and the mental notes that he can make and commit to memory,
combining them with the sentiments to which they in turn give rise. The
_plein air_, or broad daylight effects, are but one item of the great
range of this ever-changing and deepening mystery--from the hard reality
to the soft blending of evening when form almost disappears, even to the
merging of the whole landscape, nay, the whole world, into a
dream--which is felt rather than seen, but possesses a charm that almost
defies the pencil of the painter, and can only be expressed by the deep
and sweet notes of the poet and the musician. For love and reverence are
necessary to appreciate and to present it.

There is also much to learn about artificial light. For here, again, the
study is endless: from the glare of a hundred lights--electric and
otherwise--to the single lamp or candle. Indeed a whole volume could be
filled with illustrations of its effects. To those who aim at producing
intense brilliancy, refusing to acknowledge any limitations to their
capacity, a hundred or a thousand lights commend themselves; and even
though wild splashes of paint may sometimes be the result, still the
effort is praiseworthy. But those who prefer the mysterious lighting of
a Rembrandt will find, if they sit contemplating in a room lit with one
lamp only, that an endless depth of mystery surrounds them, full of dark
recesses peopled by fancy and sweet thought, whilst the most beautiful
gradations soften the forms without distorting them; and at the same
time he can detect the laws of this science of light and shade a
thousand times repeated and endless in its variety.

_Note._--Fig. 288 must be looked upon as a rough sketch which only gives
the general effect of the original drawing; to render all the delicate
tints, tones and reflections described in the text would require a
highly-finished reproduction in half-tone or in colour.

As many of the figures in this book had to be re-drawn, not a light
task, I must here thank Miss Margaret L. Williams, one of our Academy
students, for kindly coming to my assistance and volunteering her
careful co-operation.

CLXIV

REFLECTION

  [Transcriber's Note:
  In this chapter, [R] represents "R" printed upside-down.]

Reflections in still water can best be illustrated by placing some
simple object, such as a cube, on a looking-glass laid horizontally on a
table, or by studying plants, stones, banks, trees, &c., reflected in
some quiet pond. It will then be seen that the reflection is the
counterpart of the object reversed, and having the same vanishing points
as the object itself.

  [Illustration: Fig. 289.]

Let us suppose _R_ (Fig. 289) to be standing on the water or reflecting
plane. To find its reflection make square [R] equal to the original
square _R_. Complete the reversed cube by drawing its other sides, &c.
It is evident that this lower cube is the reflection of the one above
it, although it differs in one respect, for whereas in figure _R_ the
top of the cube is seen, in its reflection [R] it is hidden, &c. In
figure A of a semicircular arch we see the underneath portion of the
arch reflected in the water, but we do not see it in the actual object.
However, these things are obvious. Note that the reflected line must be
equal in length to the actual one, or the reflection of a square would
not be a square, nor that of a semicircle a semicircle. The apparent
lengthening of reflections in water is owing to the surface being broken
by wavelets, which, leaping up near to us, catch some of the image of
the tree, or whatever it is, that it is reflected.

  [Illustration: Fig. 290.]

In this view of an arch (Fig. 290) note that the reflection is obtained
by dropping perpendiculars from certain points on the arch, 1, 0, 2,
&c., to the surface of the reflecting plane, and then measuring the same
lengths downwards to corresponding points, 1, 0, 2, &c., in the
reflection.

CLXV

ANGLES OF REFLECTION

In Fig. 291 we take a side view of the reflected object in order to show
that at whatever angle the visual ray strikes the reflecting surface it
is reflected from it at the same angle.

  [Illustration: Fig. 291.]

We have seen that the reflected line must be equal to the original line,
therefore _mB_ must equal _Ma_. They are also at right angles to _MN_,
the plane of reflection. We will now draw the visual ray passing from
_E_, the eye, to _B_, which is the reflection of _A_; and just
underneath it passes through _MN_ at _O_, which is the point where the
visual ray strikes the reflecting surface. Draw _OA_. This line
represents the ray reflected from it. We have now two triangles, _OAm_
and _OmB_, which are right-angled triangles and equal, therefore angle
_a_ equals angle _b_. But angle _b_ equals angle _c_. Therefore angle
_EcM_ equals angle _Aam_, and the angle at which the ray strikes the
reflecting plane is equal to the angle at which it is reflected from it.

CLXVI

REFLECTIONS OF OBJECTS AT DIFFERENT DISTANCES

In this sketch the four posts and other objects are represented standing
on a plane level or almost level with the water, in order to show the
working of our problem more clearly. It will be seen that the post _A_
is on the brink of the reflecting plane, and therefore is entirely
reflected; _B_ and _C_ being farther back are only partially seen,
whereas the reflection of _D_ is not seen at all. I have made all the
posts the same height, but with regard to the houses, where the length
of the vertical lines varies, we obtain their reflections by measuring
from the points _oo_ upwards and downwards as in the previous figure.

  [Illustration: Fig. 292.]

Of course these reflections vary according to the position they are
viewed from; the lower we are down, the more do we see of the
reflections of distant objects, and vice versa. When the figures are on
a higher plane than the water, that is, above the plane of reflection,
we have to find their perspective position, and drop a perpendicular
_AO_ (Fig. 293) till it comes in contact with the plane of reflection,
which we suppose to run under the ground, then measure the same length
downwards, as in this figure of a girl on the top of the steps. Point
_o_ marks the point of contact with the plane, and by measuring
downwards to _a·_ we get the length of her reflection, or as much as is
seen of it. Note the reflection of the steps and the sloping bank, and
the application of the inclined plane ascending and descending.

  [Illustration: Fig. 293.]

CLXVII

REFLECTION IN A LOOKING-GLASS

I had noticed that some of the figures in Titian's pictures were only
half life-size, and yet they looked natural; and one day, thinking I
would trace myself in an upright mirror, I stood at arm's length from it
and with a brush and Chinese white, I made a rough outline of my face
and figure, and when I measured it I found that my drawing was exactly
half as long and half as wide as nature. I went closer to the glass, but
the same outline fitted me. Then I retreated several paces, and still
the same outline surrounded me. Although a little surprising at first,
the reason is obvious. The image in the glass retreats or advances
exactly in the same measure as the spectator.

  [Illustration: Fig. 294.]

Suppose him to represent one end of a parallelogram _e·s·_, and his
image _a·b·_ to represent the other. The mirror _AB_ is a perpendicular
half-way between them, the diagonal _e·b·_ is the visual ray passing
from the eye of the spectator to the foot of his image, and is the
diagonal of a rectangle, therefore it cuts _AB_ in the centre _o_, and
_AO_ represents _a·b·_ to the spectator. This is an experiment that any
one may try for himself. Perhaps the above fact may have something to do
with the remarks I made about Titian at the beginning of this chapter.

  [Illustration: Fig. 295.]

  [Illustration: Fig. 296.]

CLXVIII

THE MIRROR AT AN ANGLE

If an object or line _AB_ is inclined at an angle of 45° to the mirror
_RR_, then the angle _BAC_ will be a right angle, and this angle is
exactly divided in two by the reflecting plane _RR_. And whatever the
angle of the object or line makes with its reflection that angle will
also be exactly divided.

  [Illustration: Fig. 297.]

  [Illustration: Fig. 298.]

Now suppose our mirror to be standing on a horizontal plane and on a
pivot, so that it can be inclined either way. Whatever angle the mirror
is to the plane the reflection of that plane in the mirror will be at
the same angle on the other side of it, so that if the mirror _OA_ (Fig.
298) is at 45° to the plane _RR_ then the reflection of that plane in
the mirror will be 45° on the other side of it, or at right angles, and
the reflected plane will appear perpendicular, as shown in Fig. 299,
where we have a front view of a mirror leaning forward at an angle of
45° and reflecting the square _aob_ with a cube standing upon it, only
in the reflection the cube appears to be projecting from an upright
plane or wall.

  [Illustration: Fig. 299.]

If we increase the angle from 45° to 60°, then the reflection of the
plane and cube will lean backwards as shown in Fig. 300. If we place it
on a level with the original plane, the cube will be standing upright
twice the distance away. If the mirror is still farther tilted till it
makes an angle of 135° as at _E_ (Fig. 298), or 45° on the other side of
the vertical _Oc_, then the plane and cube would disappear, and objects
exactly over that plane, such as the ceiling, would come into view.

In Fig. 300 the mirror is at 60° to the plane _mn_, and the plane itself
at about 15° to the plane _an_ (so that here we are using angular
perspective, _V_ being the accessible vanishing point). The reflection
of the plane and cube is seen leaning back at an angle of 60°. Note the
way the reflection of this cube is found by the dotted lines on the
plane, on the surface of the mirror, and also on the reflection.

  [Illustration: Fig. 300.]

CLXIX

THE UPRIGHT MIRROR AT AN ANGLE OF 45° TO THE WALL

In Fig. 301 the mirror is vertical and at an angle of 45° to the wall
opposite the spectator, so that it reflects a portion of that wall as
though it were receding from us at right angles; and the wall with the
pictures upon it, which appears to be facing us, in reality is on our
left.

  [Illustration: Fig. 301.]

An endless number of complicated problems could be invented of the
inclined mirror, but they would be mere puzzles calculated rather to
deter the student than to instruct him. What we chiefly have to bear in
mind is the simple principle of reflections. When a mirror is vertical
and placed at the end or side of a room it reflects that room and gives
the impression that we are in one double the size. If two mirrors are
placed opposite to each other at each end of a room they reflect and
reflect, so that we see an endless number of rooms.

Again, if we are sitting in a gallery of pictures with a hand mirror,
we can so turn and twist that mirror about that we can bring any picture
in front of us, whether it is behind us, at the side, or even on the
ceiling. Indeed, when one goes to those old palaces and churches where
pictures are painted on the ceiling, as in the Sistine Chapel or the
Louvre, or the palaces at Venice, it is not a bad plan to take a hand
mirror with us, so that we can see those elevated works of art in
comfort.

There are also many uses for the mirror in the studio, well known to the
artist. One is to look at one's own picture reversed, when faults become
more evident; and another, when the model is required to be at a longer
distance than the dimensions of the studio will admit, by drawing his
reflection in the glass we double the distance he is from us.

The reason the mirror shows the fault of a work to which the eye has
become accustomed is that it doubles it. Thus if a line that should be
vertical is leaning to one side, in the mirror it will lean to the
other; so that if it is out of the perpendicular to the left, its
reflection will be out of the perpendicular to the right, making a
double divergence from one to the other.

CLXX

MENTAL PERSPECTIVE

Before we part, I should like to say a word about mental perspective,
for we must remember that some see farther than others, and some will
endeavour to see even into the infinite. To see Nature in all her
vastness and magnificence, the thought must supplement and must surpass
the eye. It is this far-seeing that makes the great poet, the great
philosopher, and the great artist. Let the student bear this in mind,
for if he possesses this quality or even a share of it, it will give
immortality to his work.

To explain in detail the full meaning of this suggestion is beyond the
province of this book, but it may lead the student to think this
question out for himself in his solitary and imaginative moments, and
should, I think, give a charm and virtue to his work which he should
endeavour to make of value, not only to his own time but to the
generations that are to follow. Cultivate, therefore, this mental
perspective, without forgetting the solid foundation of the science I
have endeavoured to impart to you.

INDEX

  [Transcriber's Note:
  Index citations in the original book referred to page numbers.
  References to chapters (Roman numerals) or figures (Arabic numerals)
  have been added in brackets where possible. Note that the last two
  entries for "Toledo" are figure numbers rather than pages; these have
  not been corrected.]

A
Albert Dürer, 2, 9.
Angles of Reflection, 259 [CLXV].
Angular Perspective, 98 [XLIX] - 123 [LXXII], 133 [LXXX], 170.
   "         "       New Method, 133 [LXXX],
                       134 [LXXXI], 135 [LXXXII], 136 [LXXXIII].
Arches, Arcades, &c., 198 [CXXVI], 200 [CXXVII] - 208 [CXXIII].
Architect's Perspective, 170 [CVIII], 171 [197].
Art Schools Perspective, 112 [LXII] - 118 [LXVI], 217 [CXLI].
Atmosphere, 1, 74 [XXX].

B
Balcony, Shadow of, 246 [CLVII].
Base or groundline, 89 [XLI].

C
Campanile Florence, 5, 59.
Cast Shadows, 229 [CXLVII] - 253 [CLXII].
Centre of Vision, 15 [II].
Chessboard, 74 [XXXI].
Chinese Art, 11.
Circle, 145 [LXXXVIII], 151 [XCII] - 156 [XCVI], 159 [XCIX].
Columns, 157 [XCVII], 159 [XCIX], 161 [CI], 169 [CVI], 170 [CVII].
Conditions of Perspective, 24 [VII], 25.
Cottage in Angular Perspective, 116 [LXV].
Cube, 53 [XVII], 65 [XXIII], 115 [LXIV], 119 [LXVIII].
Cylinder, 158 [XCVIII], 159 [CXIX].
Cylindrical picture, 227 [CXLVI].

D
De Hoogh, 2, 62 [68], 73 [82].
Depths, How to measure by diagonals, 127 [LXXVI], 128 [LXXVII].
Descending plane, 92 [XLIV] - 95 [XLV].
Diagonals, 45, 124 [LXXIII], 125 [LXXIV], 126 [LXXV].
Disproportion, How to correct, 35, 118 [LXVII], 157 [XCVII].
Distance, 16 [III], 77 [XXXIII], 78 [XXXIV], 85 [XXXVII],
                            87 [XXXIX], 103 [LIV], 128 [LXXVII].
Distorted perspective, How to correct, 118 [LXVII].
Dome, 163 [CIII] - 167 [CV].
Double Cross, 218 [CXLII].

E
Ellipse, 145 [LXXXIX], 146 [XC], 147 [168].
Elliptical Arch, 207 [CXXXII].

F
Farningham, 95 [103].
figures on descending plane, 92 [XLIV], 93 [100],
                                             94 [102], 95 [XLV].
   "    "  an inclined plane, 88 [XL].
   "    "  a level plane, 70 [79], 71 [XXVIII], 72 [81],
                                   73 [82], 74 [XXX], 75 [XXXI].
   "    "  uneven ground, 90 [XLII], 91 [XLIII].

G
Geometrical and Perspective figures contrasted, 46 [XII] - 48.
     "      plane, 99 [L].
Giovanni da Pistoya, Sonnet to, by Michelangelo, 60.
Great Pyramid, 190 [CXXII].

H
Hexagon, 177 [CXIV], 183 [CXVII], 185 [CXIX].
Hogarth, 9.
Honfleur, 83 [92], 142 [163].
Horizon, 3, 4, 15 [II], 20, 59 [XX], 60 [66].
Horizontal line, 13 [I], 15 [II].
Horizontals, 30, 31, 36.

Inaccessible vanishing points, 77 [XXXII], 78 [XXXIII],
                                                 136, 140 - 144.
Inclined plane, 33, 118, 213 [CXXXVIII], 244 [XLV], 245 [XLVI].
Interiors, 62 [XXI], 117 [LXVI], 118 [LXVII], 128.

J
Japanese Art, 11.
Jesuit of Paris, Practice of Perspective by, 9.

K
Kiosk, Application of Hexagon, 185 [XCIX].
Kirby, Joshua, Perspective made Easy (?), 9.

L
Ladder, Step, 212 [CXXXVII], 216 [CXL].
Landscape Perspective, 74 [XXX].
Landseer, Sir Edwin, 1.
Leonardo da Vinci, 1, 61.
Light, Observations on, 253 [CLXIII].
Light-house, 84 [XXXVII].
Long distances, 85 [XXXVIII], 87 [XXXIX].

M
Measure distances by square and diagonal, 89 [XLI],
                                              128 [LXXVII], 129.
   "    vanishing lines, How to, 49 [XIV], 50 [XV].
Measuring points, 106 [LVII], 113.
    "     point O, 108, 109, 110 [LX].
Mental Perspective, 269 [CLXX].
Michelangelo, 5, 57, 58, 60.

N
Natural Perspective, 12, 82 [91], 95 [103], 142 [163], 144 [164].
New Method of Angular Perspective, 133 [LXXX], 134 [LXXXI],
                  135 [LXXXII], 141 [LXXXVI], 215 [CXXXIX], 219.
Niche, 164 [CIV], 165 [193], 250 [CLX].

O
Oblique Square, 139 [LXXXV].
Octagon, 172 [CIX] - 175 [202].
O, measuring point, 110 [LX].
Optic Cone, 20 [IV].

P
Parallels and Diagonals, 124 [LXXIII] - 128 [LXXVI].
Paul Potter, cattle, 19 [16].
Paul Veronese, 4.
Pavements, 64 [XXII], 66 [XXIV], 176 [CXIII], 178 [CXV],
                              180 [209],181 [CXVI], 183 [CXVII].
Pedestal, 141 [LXXXVI], 161 [CI].
Pentagon, 186 [CXX], 187 [217], 188 [219].
Perspective, Angular, 98 [XLIX] - 123 [LXXII].
     "       Definitions, 13 [I] - 23 [VI].
     "       Necessity of, 1.
     "       Parallel, 42 - 97 [XLVII].
     "       Rules and Conditions of, 24 [VII] - 41.
     "       Scientific definition of, 22 [VI].
     "       Theory of, 13 - 24 [VI].
     "       What is it? 6 - 12.
Pictures painted according to positions they are to occupy,
                                                        59 [XX].
Point of Distance, 16 [III] - 21 [IV].
  "  "   Sight, 12, 15 [II].
Points in Space, 129 [LXXVIII], 137 [LXXXIII].
Portico, 111 [122].
Projection, 21 [V], 137.
Pyramid, 189 [CXXI], 190 [224], 191 [CXXII],
                                      193 [CXXIII] - 196 [CXXV].

R
Raphael, 3.
Reduced distance, 77 [XXXIII], 78 [XXXIV], 79 [XXXV], 84 [90].
Reflection, 257 [CLXIV] - 268 [CLXIX].
Rembrandt, 59 [XX], 256.
Reynolds, Sir Joshua, 9, 60.
Rubens, 4.
Rules of Perspective, 24 - 41.

S
Scale on each side of Picture, 141 [LXXXVII],
                                          142 [163] - 144 [164].
  "   Vanishing, 69 [XXVI], 71 [XXVII], 81 [XXXVI], 84 [90].
Serlio, 5, 126 [LXXV].
Shadows cast by sun, 229 [CXLVII] - 252 [CLXI].
   "     "   "  artificial light, 252 [CLXII].
Sight, Point of, 12, 15 [II].
Sistine Chapel, 60.
Solid figures, 135 [LXXXII] - 140 [LXXXV].
Square in Angular Perspective, 105 [LVI], 106 [LVII], 109 [120],
                 112 [LXII], 114 [LXIII], 121 [LXX], 122 [LXXI],
              123 [LXXII], 133 [LXXX], 134 [LXXXI], 139 [LXXXV].
   "   and diagonals, 125 [LXXIV], 138 [LXXXIV], 139 [LXXXV],
                                                   141 [LXXXVI].
   "   of the hypotenuse (fig. 170), 149 [170].
   "   in Parallel Perspective, 42 [IX], 43 [X], 50 [XV],
                                            53 [XVII], 54 [XIX].
   "   at 45°, 64 [XXII] - 66 [XXIV].
Staircase leading to a Gallery, 221 [CXLIII].
Stairs, Winding, 222 [CXLIV], 225 [CXLV].
Station Point, 13 [I].
Steps, 209 [CXXXIV] - 218 [CXLII].

T
Taddeo Gaddi, 5.
Terms made use of, 48 [XIII].
Tiles, 176 [CXIII], 178 [CXV], 181 [CXVI].
Tintoretto, 4.
Titian, 59 [XX], 262 [CLXVII].
Toledo, 96 [104], 144 [164], 259 [259], 288 [288].
Trace and projection, 21 [V].
Transposed distance, 53 [XVIII].
Triangles, 104 [LV], 106 [LVII], 132 [148], 135 [151], 138 [158].
Turner, 2, 87 [95].

U
Ubaldus, Guidus, 9.

V
Vanishing lines, 49 [XIV].
    "     point, 119 [LXVIII].
    "     scale, 68 [XXV] - 72 [XXVIII], 74 [XXX], 77 [XXXII],
                                             79 [XXXV], 84 [90].
Vaulted Ceiling, 203 [CXXX].
Velasquez, 59 [XX].
Vertical plane, 13 [I].
Visual rays, 20 [IV].

W
Winding Stairs, 222 [CXLIV] - 225 [CXLV].
Water, Reflections in, 257 [CLXIV], 258 [CLXV], 260 [CLXVI],
                                                      261 [293].

Errors and Anomalies:

Missing punctuation in the Index has been silently supplied.

The name form "Albert Dürer" (for Albrecht) is used throughout.
In all references to Kirby, _Perspective made Easy_ (?), the question
  mark is in the original text.

Figure 66:
  _Caption missing, but number is given in text_
ground plan of the required design, as at Figs. 73 and 74
  _text reads "Figs. 74 and 75"_
CV [Chapter head]
  _"C" invisible_

_Index_
Dürer, Albert
  _umlaut missing_
Taddeo Gaddi
  _text reads "Tadeo"_
Titian
  _text reads Titien_