Problem IX
To Draw a Square Pillar, Given in Position and Magnitude, Its Base and Top Being in Horizontal Planes
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 A E.
[Geometric diagram]
Fig. 23.[Geometric diagram]
Fig. 24.
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 g f in h, by drawing g h to 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.
Corollary
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 e 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 inclosing 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.18
Notes
18 See page 96 (note), after you have read Problem XVI.