In Fig. 689, B C F G represents the side elevation of the offset, A B G H a portion of a round pipe joining it below, and C D E F a portion of the oblong pipe joining it above. In the plan immediately below, J K L M shows the plan of the round pipe and NOP Q R S that, of the oblong pipe, while the distance L T shows the amount of the offset.

The piece forming the offset is similar in shape to that shown in Problem 189, the difference being that its liases B G and C F are neither horizontal nor parallel to each other and that sections on the lines of the bases are not given. Since the article required consists of symmetrical halves when divided on the line J T of the plan, the plane surface A B C D E F G H lying as it were back of the half shown by the elevation may, as in Problem 207, be regarded as a base from which to measure all hights, or projections, in obtaining the required profiles and sections necessary in developing the pattern. The first steps necessary will be to obtain true sections upon the lines C F and B G of the elevation. In Fig. 690, C D E F represents a duplicate of the part bearing the same letters in the elevation. Upon D E as a base line construct a duplicate of the half section of oblong pipe N O P T of Fig. 689, as shown by D N P E.

Divide the semicircle N P into any convenient number of equal parts, as shown by the small figures. With the blade of the T-square placed at right angles to D E, drop lines cutting C F. With the T-square placed at right angles to C F, and brought against the points in C F, draw lines, extending them indefinitely, as shown. Measuring in each instance from C F, set off on the lines just drawn the same length as similar lines in D N O P E, and through the points thus ohtained trace a line, as shown by n O p. Then C F p O n is the half shape of cut on line C F. In Fig. 691, A B G H is a duplicate of the elevation of the round pipe, below which is drawn a half profile of same, A M H. To obtain the shape of cut on line B G, divide the half profile A M H into the same number of parts as was N P, and, with the T-square placed parallel with A B, and brought successively against the points in A M H, carry lines rutting BG. With the T-square placed at right angles to B G, and brought against the points therein contained, erect perpendiculars, as shown. Measuring in each instance from B G, set off on the lines just drawn the same length as similar lines in A

Fig. 689 - Plan and Elevation of Offset.

Fig. 690. - Development of Section on Line C F of Elevation.

Fig. 691. - Development of Section on Line B G of Elevation.

M H, and through the points thus obtained trace a line, as shown by B m G. Then B m G is the half shape of cut on line B G.

In order to avoid a confusion of lines a duplicate of B C FG of the elevation is presented in Fig. 692, upon C F and B G of which, as base lines, arc drawn duplicates of the sections obtained in Figs, 690 and 691, all as shown. From points in n Op drop lines at right angles to C F, cutting the same, and from points in B m G drop lines at right angles to B G, cutting it. Connect points in these lines in consecutive order by solid lines, as shown, and subdivide the four sided figures thus obtained by dotted lines representing their shorter diagonals. The surface of the offset or transition piece is thus divided into a series of very tapering triangles, the lengths of whose bases or shortest sides are given in the two sections C n O p F and B m G. In order to obtain the correct lengths of their longer sides two diagrams or series of sections must be constructed for that purpose, which are shown in Figs. 693 and 694.

Fig. 692. - Middle Piece of Offset, Showing Method of Triangulation.

To obtain the various sections on the solid lines of the elevation proceed as follows: Draw the right angle U V W, Fig. 693. From V, on V U, set off the length of lines in B m G, Fig. 692. From V, on V W set off the length of solid lines in B C F G, and from the points thus obtained erect perpendiculars, in length equal to lines of similar number in C n G p F, Fig. 692. Thus make line W 1' of Fig. 693 equal to line C n or C 1, and draw V 1', which gives the distance from point B to point n in Fig. 692 as if measured on the finished article. Connect the perpendiculars drawn from V W with the points in V U, as shown, and corresponding with the figures in Fig. 692. Thus connect V and 8, 2' and 9, 3' and 10, etc. Then will the lengths of the oblique lines in Fig. 693 be the true lengths of the solid lines crossing the elevation. The diagram of sections shown in Fig. 694 is constructed in the same manner, using the dotted lines of the elevation as the basis of measurements.

Fig. 693. - Diagram of Sections on Solid Lines of Fig. 692.

Draw the right angle X Y Z, and from Y set off on Y Z the length of lines in C n O p F. From Y, on Y X, set off the length of dotted lines in B C F G, and from the points thus obtained erect perpendiculars, in length equal to lines of similar number in B m G, as indicated by the small figures. Connect the perpendiculars drawn from X Y with the points in Y Z, as shown, and corresponding with the figures in B C F G. Thus connect 1 and 9', 2 and 10', 3 and 11', etc.

Fig. 694. - Diagram of Sections on Dotted Lines of Fig. 692.

An inspection of the plan and elevation, Fig. 689, will show that the curved surface of the offset or transition piece B C F G, which has been divided into triangles, is shown by J M L Q R S of the plan, and that this piece is connected with its mate or equivalent in the opposite half of the article by a large plain triangular surface, S J N, on the upper side, and by another, Q L P, on its lower side, which must be added to the pattern of the curved portion after it has been developed. It will also be seen that V W 1' of Fig. 693 is one-half of J N S. Therefore to develop the pattern, first draw any line, as j x in Fig. 695, in length equal to B C of Fig. 689, or V W of Fig. 693. With j as center, and 1' V of Fig. 693 as radius, describe a small arc (near s), which intersect with another small arc struck from x as center, and with a radius equal to W 1' of Fig. 693, thus duplicating the triangle V W 1'. From s, or point 1, as center, with a radius equal to 1 9' of Fig. 694, describe a small arc (near 9), which intersect with another small arc struck from j or 8 of pattern as center, with a radius equal to 8 9 of the profile B m G, Fig. 692, thus establishing the position of point 9 of pattern. From 9 as a center, with a radius equal to 9 2' of Fig. 693, describe a small arc, which intersect with another small arc struck from point 1 of pattern as center, and a radius equal to 1 2 of the pro tile C a O p Fof Fig. 692. thus establishing the position of point 2 of pattern. Proceed in this manner, using the dotted oblique lines in Fig. 694, the lengths of the spaces in B m G in Fig. 692, the lengths of the solid oblique lines in Fig. 693 and the lengths of teh spaces in C n O p F of Fig. 692 in the order named until the line 7 14 is reached. Lines traced through the points of intersection from j to l and from s to t will give the shape of the curved portion of the pattern. From 14 of pattella as center, with a. radius equal to C F of Fig. 692, or V 7 of Fig. 693, describe a small are. which intersect with another small arc struck from point 7 of pattern as center, and a radius equal to 7 7' of Fig. 698, or T P of the plan. Draw 7 t and t l; then will l j x s g t be one-half the pattern required. The other half can be obtained by any means of duplication most convenient.

Fig. 695. - Pattern of Offset.