In Fig. 546, let G H F E be the elevation of one side of the article, LMUR the elevation of an end, E1 H1 F1 U1 the plan of the article at the base, and T V S P the plan at the top. Produce E G and F H of the side elevation until they meet in the point I, the apex of the cone.

Divide one-quarter of the plan E1 R1 into any convenient number of equal parts, as indicated by the small figures. From the points thus determined draw-lines to the center C. These lines will form the bases of a series of right angled triangles whose common altitude is the hight of the cone, and whose hypothenuses when drawn will give the true distances from the apex to the several points assumed in the base of the cone. Therefore at any convenient place draw the straight line D A of Fig. 547, in length equal to I E2. Make D B equal to I G1. From A and B of Fig. 547 draw perpendiculars to D A indefinitely, as shown by A O and B N. Take the distances C 5, C 4, C 3, etc., of the plan and set off corresponding distances from A on A O, as shown by A 5, A 4, A 3, etc. From these points in A O draw lines to D, cutting B N. These lines are also shown in the elevation, but are not necessary in that view in obtaining the pattern. From D as center describe arcs whose radii are equal to the lengths of the several lines just drawn from D to the points in A O.

Fig. 546. - Plan, Hide and End Elevations of the Frustum of an Elliptical Cone.

From any convenient point in the first arc draw a straight line to D, as shown by W D. This will form one side of the pattern. From W, as a starting point, lay off the stretchout of the plan E1 R1 F1, etc., using the same length of spaces as employed in dividing it, stepping from one arc to the next each time, as shown. A line traced through these points will be the outline of the base of the pattern, one-half of the entire envelope being shown in the pattern from W to Z.

Fig. 547. - Diagram of Triangles and Pattern of Frustum Shows in Fig. 546.

From the points in W Z draw lines to D, which intersect by arcs drawn with D as center and starting from points of corresponding number in B N. A line traced through the points of intersection will form the upper line of the pattern, as shown. Then W X Y Z will constitute the pattern of one-half of the envelope, to which add a duplicate of itself for the complete pattern.