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 1/2-D, 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]

[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

1/2 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 deg 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 deg, double that

at _A_, which is 36 deg.

[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 1/2 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 1/4 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 1/4 distance and 1/4 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 deg 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 deg 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 deg 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 deg

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 deg 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 deg to meet _FL_ at _L_, which is the luminary at an altitude of

45 deg, and at an angle of 45 deg 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 deg to the picture instead of 45 deg, is

nearer the point of sight; at 90 deg it would be exactly opposite the

spectator, and so on. Again, the elevation being less (40 deg instead of

45 deg) 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 deg 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 deg to the plane _RR_ then the reflection of that plane in

the mirror will be 45 deg 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 deg 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 deg to 60 deg, 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 deg as at _E_ (Fig. 298), or 45 deg 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 deg to the plane _mn_, and the plane

itself at about 15 deg 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 deg. 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 DEG. TO THE WALL

In Fig. 301 the mirror is vertical and at an angle of 45 deg 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 Durer, 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.

I

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 deg, 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 Duerer" (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_

Durer, Albert

_umlaut missing_

Taddeo Gaddi

_text reads "Tadeo"_

Titian

_text reads Titien_

The Theory and Practice of Perspective · 第 5 / 5 章