Let a h, Fig. 23., be the square pillar.

Then, as it is given in position and magnitude, the position and magnitude of the square it stands upon must be given (that is, the line A B or A c in position), and the height of its side

Find the sight-magnitudes of a b and a E. Draw the two sides a b, a c, of the square of the base, by Problem VIII., as in Fig. 24. From the points a, b, and c, raise vertical lines, a e, c f, b g. Make a e equal to the sight-magnitude of A e. Now because the top and base of the pillar are in horizontal planes, the square of its top, f g, is parallel to the square of its base, b c.

Therefore the line E F is parallel to A c, and e g to A B.

Therefore E f has the same vanishing-point as A c, and E G the same vanishing-point as A B.

From e draw e f to the vanishing-point of a c, cutting c f in f.

Similarly draw e g to the vanishing-point of a b, cutting b g in g.

Complete the square in h, by drawing g h to

Problem IX To Draw A Square Pillar Given In Positi Perspective Elements 79

Fig. 23.

Problem IX To Draw A Square Pillar Given In Positi Perspective Elements 80

Fig. 24.

Problem IX - Corollary the vanishing-point of e f, and f h to the vanishing-point of e g, cutting each other in h. Then a g h f is the square pillar required.

It is obvious that if a E is equal to A c, the whole figure will be a cube, and each side, a e f c and a c g b, will be a square in a given vertical plane. And by making A B or A c longer or shorter in any given proportion, any form of rectangle may be given to either of the sides of the pillar. No other rule is therefore needed for drawing squares or rectangles in vertical planes.

Also any triangle may be thus drawn in a vertical plane, by enclosing it in a rectangle and determining, in perspective ratio, on the sides of the rectangle, the points of their contact with the angles of the triangle.

And if any triangle, then any polygon.

A less complicated construction will, however, be given hereafter.1

1 See page 299. (note), after you have read Problem XVI.

Problem IX

The drawing of most buildings occurring in ordinary practice will resolve itself into applications of this problem. In general, any house, or block of houses, presents itself under the main conditions assumed here in Fig. 54. There will be an angle or corner somewhere near the spectator, as A B; and the level of the eye will usually be above the base of the building, of which, therefore, the horizontal upper lines will slope down to the vanishing-points, and the base lines rise to them. The following practical directions will, however, meet nearly all cases: Let A B, Fig. 54.. be any important vertical line in the block of buildings; if it is the side of a street, you may fix upon such a line at the division between two houses. If its real height, distance, c, are given, you will proceed with the accurate construction of the problem; but usually you will neither know, nor care, exactly how high the building is,or how far off. In such case draw the line A B, as nearly as you can guess, about the part of the picture it ought to occupy, and on such a scale as you choose. Divide it into any convenient number of equal parts, according to the height you presume it to be. If you suppose it to be twenty feet high, you may divide it into twenty parts, and let each part stand for a foot; if thirty feet high, you may divide it into ten parts, and let each part stand for three feet; if seventy feet high, into fourteen parts, and let each part stand for five feet; and

Problem IX Perspective Elements 110

Fig. 54.

so on, avoiding thus very minute divisions till you come to details. Then observe how high your eye reaches upon this vertical line; suppose, for instance, that it is thirty feet high and divided into ten parts, and you are standing so as to raise your head to about six feet above its base, then the sight-line may be drawn, as in the figure, through the second division from the ground If you are standing above the house, draw the sight-line above B; if below the house, below A; at such height or depth as you suppose may be accurate (a yard or two more or less matters little at ordinary distances, while at great distances perspective rules become nearly useless, the eye serving you better than the necessarily imperfect calculation). Then fix your sight-point and station-point, the latter with proper reference to the scale of the line A B. As you cannot, in all probability, ascertain the exact direction of the line A V or B V, draw the slope B v as it appears to you, cutting the sight-line in v. Thus having fixed one vanishing-point, the other, and the dividing-points, must be accurately found by rule; for, as before stated, whether your entire group of points (vanishing and dividing) falls a little more or less to the right or left of S does not signify, but the relation of the points to each other does signify. Then draw the measuring-line B G, either through A or B, choosing always the steeper slope of the two; divide the measuring-line into parts of the

Problem IX Perspective Elements 111

Fig. 55.

same length as those used on A B, and let them stand for the same magnitudes. Thus, suppose there are two rows of windows in the house front, each window six feet high by three wide, and separated by intervals of three feet, both between window and window and between tier and tier: each of the divisions here standing for three feet, the lines drawn from B G to the dividing-point D fix the lateral dimensions, and the divisions on A B the vertical ones. For other magnitudes it would be necessary to subdivide the parts on the measuring-line, or on A B, as required. The lines which regulate the inner sides or returns of the windows (a, b, c, c.) of course are drawn to the vanishing-point of B F (the other side of the house), if F B v represents a right angle; if not, their own vanishing-point must be found separately for these returns. But see Practice on Problem XI

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