In Fig. 657, let I D E F H represent the front elevation of the article, showing the circular opening DEF6 forming its upper perimeter or profile. The triangle A B C shows the shape of the article as it appears when viewed from the side, below which is drawn the plan, showing its circular base, J K L M. The line N P shows the plan of the opening D E F G, which opening is shown in the side view by that portion of the line A B from A to Q. Opposite the front side of the plan N P is drawn a duplicate of the profile D E F G, as shown by E' F' G', so placed that its vertical center line E' G' shall coincide with the center line of the plan, as shown. As the article consists of two symmetrical halves it will only be necessary to develop one-half the complete pattern. Therefore divide one-half of Loth profiles E F G and E' F' G' into the same number of equal parts, numbering each in the same order, as shown by the small figures; also divide the plan of the base into the same number of equal parts as the profile, numbering the points to correspond with the same. Drop lines from the points on the profile E' F' G' on to the line N P, at right ancles to the same, as shown, and connect these points with those of similar number upon the plan of base by solid lines, as shown. Also connect points upon the base with those of the next higher number upon the line N P by dotted lines.

Fig.657 . - Elevations and Plan of an Irregular Shape Forming a Trunsition from a Round Horizontal Base to a Round Top Placed Vertically.

It will be noticed that the point J of the plan

Pattern Problems 377 represents at once the point 1 of the base and the points 1 and 7 of the profile, shown by B, Q and A of the side elevation. The lines drawn across the plan represent the horizontal distances between the points which they connect and will form the bases of a series of right angled triangles, whose altitudes can be derived from the elevations, as will be shown, and whose hypothenuses, when drawn, will give the true distances between points of corresponding number across the finished article or its pattern. To obtain the altitudes of the triangles carry lines from the points in the half profile E F G horizontally across, cutting the line A Q, as shown; then the distances of the points in A Q from B will constitute the respective altitudes of the triangles. Therefore, to construct a diagram of all the triangles, draw any horizontal line, as 1) C, Fig. 658, near the center of which erect a perpendicular, B A. Upon B A set off from B the various distances from B to points in A Q of Fig. 657, numbering the same as shown by the small figures. From B set off on B C the lengths of the various solid lines drawn across the plan, Fig. 657, and connect points in B C with those of like number in B A. From B set off toward D the lengths of the various dotted lines drawn across the plan and connect them by dotted lines with points of the next higher number in the line B A, all as shown; then these various hypothenuses will constitute the true distances across the finished article between points of corresponding number indicated on the plan and elevations. The distances between points in the base line forming the larger or outer curve of the pattern can be measured from the base line in plan, while spaces forming the upper or shorter side of pattern can be measured from either of the profiles.

Fig. 658. - Diagram of Triangles.

To develop the pattern it is simply necessary to construct the various triangles whose dimensions have been obtained in the previous operations, beginning at either end most convenient and using the dimensions in the order in which they occur until all have been used and the pattern is complete.

Therefore, upon any straight line, as A C of Fig. 659, set off a distance equal to A C of Fig. 657 or the solid line 7 7 of Fig. 658. From C as a center, with a radius equal to 7 6 of the plan, Fig. 657, describe a small arc to the left, which intersect with another small arc struck from A as a center, and with a radius equal to the dotted line 7 6 of the diagram of triangles, Fig. 658, thus establishing the position of the point 6 of the pattern. From A of Fig. 659 as center, with a radius equal to 7 6 of the profile, Fig. 657, describe a small arc, which intersect with another struck from point 6 of pattern as center, with a radius equal to the solid line 6 6 of Fig. 2, thus locating the position of the point 6' of the pattern.

Fig. 659. - Pattern of Shape Shown in Fig. 657.

So continue to use the spaces of the plan, the lengths of the dotted lines of the diagram of triangles, the spaces in the profile and the lengths of the solid lines of the diagram of triangles in the order named until all have been used and the pattern is complete. Lines traced through the numbered points obtained, as shown from C to B and from A to Q, will form the outlines of the pattern for half the article. The other half,

A Q' B' C, can be obtained by any means of duplication most convenient.