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in No. 1 in the figure, from this obtain by projection an elevation of them as if entire. Then draw in, parallel to the IL and through the front halves of the sphere and cylinder, as many lines as it is intended to use section planes ; find by projection elevations of the circular and rectangular sections produced by them, and through the points of intersection of these, draw in the lines of penetration as shown in No. 2, Fig. 187. The next problem is a combination of a sphere with a cons. It is- Problem 84 (Fig. 188). A sphere is penetrated by a cone having its axis vertical ; required the lines of intersection of the solids when their axes are not coincident, but parallel, and lie in a plane which is inclined to the VP. It may not have occurred to the student, in dealing with the previous problems in which the sphere is involved, that it is quite possible for its axis to lie in a vertical plane and yet be inclined at an angle to either or both of the planes of projection. In the problem for solution the axis of the cone being vertical, and that of the sphere parallel to it, both will be vertical, and be represented by points in plan. Therefore in No. 1 (Fig. 188) let the points c and s be the plans of the axes of the two solids, c being that of the cone and s that of the sphere, both being in a plane of which the line a b is the plan at an angle of 60 with the VP. Then as the sphere is the solid penetrated, with s as centre, and half the sphere's diameter as radius, describe the circle to represent its plan, and with c as a centre and half the diameter of the base of the intended cone as radius, describe a second circle, dotting in that part of it covered by the sphere, as shown