The Theory and Practice of Perspective

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).

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).

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.

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.

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.

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.

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.

Fig. 240.

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.

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.

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.

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.

Fig. 246.

CXXXVII

To Divide an Inclined Plane into Equal Parts—such as a Ladder Placed against a Wall

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.

Fig. 247.

CXXXVIII

Steps and the Inclined Plane

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.

Fig. 248.

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).

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.

Fig. 250.

Fig. 251.

CXL

A Step Ladder at an Angle

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.

Fig. 252.

CXLI

Square Steps Placed over each Other

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

Fig. 253.

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.

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.

Fig. 255.

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.

Fig. 256.

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.

Fig. 257.

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.

Fig. 258.

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

Fig. 259. Sketch of Courtyard in Toledo.

CXLVI

Of the Cylindrical Picture or Diorama

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.

Fig. 260.

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.

Fig. 263.

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.

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.

CXLIX

Shadows Cast by the Sun

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

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

CL

The Sun in the Same Plane as the Picture

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

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.

Fig. 268.

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

Fig. 269.

CLI

The Sun Behind the Picture

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

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.

From point of sight S (Fig. 270) drop perpendicular to T, the station-point. From T draw TF at 45° to meet horizon at F. With radius FT make FO equal to it. Then O is the position of the spectator. From F raise vertical FL, and from O draw a line at 45° to meet FL at L, which is the luminary at an altitude of 45°, and at an angle of 45° to the picture.

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

Fig. 272.

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

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

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.

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.

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

Fig. 274.