[p 34 ] PROBLEM IX. TO DRAW A SQUARE PILLAR, GIVEN IN POSITION AND MAGNITUDE, ITS BASE AND TOP BEING IN HORIZONTAL PLANES .

Let AH, 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 AB or AC in position), and the height of its side AE.

[Geometric diagram]
Fig.23.[Geometric diagram]
Fig.24.

Find the sight-magnitudes of AB and AE. Draw the two sides ab, ac, of the square of the base, by ProblemVIII., as in Fig.24. From the points a, b, and c, raise vertical lines ae, cf, bg.

Make ae equal to the sight-magnitude of AE.

Now because the top and base of the pillar are in horizontal planes, the square of its top, FG, is parallel to the square of its base, BC.

Therefore the line EF is parallel to AC, and EG to AB.

Therefore EF has the same vanishing-point as AC, and EG the same vanishing-point as AB.

From e draw ef to the vanishing-point of ac, cutting cf in f.

Similarly draw eg to the vanishing-point of ab, cutting bg in g.

Complete the square gf in h, by drawing gh to the vanishing-point of ef, and fh to the vanishing-point of eg, cutting each other in h. Then aghf is the square pillar required.

[p35]
COROLLARY.

It is obvious that if AE is equal to AC, the whole figure will be a cube, and each side, aefc and aegb, will be a square in a given vertical plane. And by making AB or AC 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.[Footnote 18]