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For the projections of the " cone " the first problem is Problem 50 (Fig. 149). Given the plan of a cone, having its axis perpendicular to the HP, and an altitude equal to twice the diame- ter of its base, to find its elevation. Let the circle No. 1, Fig. 149, be the given plan of the cone ; its centre a as in the cylinder will be the plan of its axis. Find by pro- jection the elevation of this axis, and on it set off from the IL the altitude of the cone in point a. Through a, the centre of the circle in No. 1, draw AB parallel to the IL. Find by projection the elevation of points A and B in the base of the cone on the IL, and join these by right lines with point a on the axis a' a. The triangle A a B will be the required elevation. Next, let the base of the cone be inclined to the HP at an angle of 45, keeping its axis parallel to the VP, and its plan be required. Here, an elevation of the cone at the given angle must first be drawn in, as in No. 3 ; then find the plan of its axis, which, being parallel to the YP, will be a line parallel to the IL, as in No. 4. Obtain by projection the plan of the base of the cone, which is an ellipse, its major axis being CD arid its minor AB, No. 4 ; then find the plan of the apex a, No. 3, of the cone in a, No. 4, and join it by right lines to points C and D ; the required plan of the cone in the position stated will thus have been obtained. Again, let the axis of the cone be parallel to the HP, but inclined to the VP, at an angle of 4^ V an d its elevation be required. To obtain this, draw-in the plan of the cone in the position stated, then find the elevation of its axis, and on this, by projection from the plan, draw-in the base, which will be an ellipse ; join the two