Publisher's Synopsis
This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1868 edition. Excerpt: ... The General Problem of the Sphere. Problem To find the perspective of the sphere, and of its curve of shade, when its centre is at the centre of the picture. 1.--To define the preliminaries, and find the perspective of the sphere. Let S, PI. XII., Fig. 83, be the centre, and Ss the radius of the given sphere. Then the great circle, with this centre and radius, is the intersection of the sphere with the perspective plane, and is its own perspective. Let SD be the horizon; D, the vanishing point of diagonals; S, the vertical projection of the point of sight, and VR the vertical trace of a plane of rays, perpendicular to the perspective, or vertical plane, and which we will call the perpendicular plane. Also ME, perpendicular to VR, is the vertical trace of a plane of rays perpendicular to the former one, and also of the plane of the curve of shade. Now, revolving D to E, with the centre S, gives E as the point of sight, after revolution, about VR, into the perspective plane (51) and makes VR represent the perspective plane, made perpendicular to the paper. Then EP.s', tangent to the vertical projection S--spt of the sphere, is an element of the visual cone, in the plane of the paper; and Ss' is found as the radius of the perspective, s'T'T, of the Bphere. 2.--To find the axes of the perspective of the curve of shade. In order to render more manifest the independence of this construction of that of the other points of this curve, it is given before the construction of the latter points. First: To find the conjugate axis. Let GL, Fig. 84, be the ground line, and RL and R'L, a ray of light. Revolve this ray about its own vertical projection, R'L, and into the vertical plane, when R'R" will be equal to GR, and R"L will be the revolved position...
