Chapter VIII: Book IV (4)

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.

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.

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.

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.

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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.

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.

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.

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.

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).

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_.

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.

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.

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.

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.

CXVII

FURTHER ILLUSTRATION OF THE HEXAGON

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.

CXVIII

ANOTHER VIEW OF THE HEXAGON IN ANGULAR PERSPECTIVE

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.

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

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

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.

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

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.

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.

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.

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.

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.

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.

CXXVI

OF ARCHES, ARCADES, BRIDGES, &C.

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.

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).

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).

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 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.

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.

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.

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 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.

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_.

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_.