Let A B D C of Fig. 578 represent the side elevation of the pipe and C D E F the side view of the collar, fitting against the pitch of the roof shown by G H.

Construct a plan below the elevation, as shown, making J K M L the plan view of the pipe and N O P R the plan view of the collar on a horizontal line, giving the collar an equal projection at the bottom on the four sides, as shown. Through the center point X in plan draw a line parallel to N R, intersecting the circle at K and L; likewise through the center X, and parallel to O N, intersect the circle at J and M. From J and K draw lines to the corner N; likewise from J and L, L and M, and M and K, draw lines to the corners R, P and O. It will be seen that by this operation the collar has been divided in such a manner that the four corner pieces are portions of oblique cones whose apices lay at the corners of the collar, while the side pieces between are simply flat triangular pieces of metal. The dotted lines connecting the plan with the elevation show corresponding points in the two views. Divide the quarter circles K J and J L into any convenient number of equal spaces, as shown by the small figures, and from points on each draw lines to the corners N and R. Then will these lines represent the bases of a series of right angled triangles, whose hypothenuses will give the correct distances across the pattern of the collar. To construct these triangles proceed as follows: Upon C Y extended assume any point, as S, at which erect the perpendicular S T, equal in hight to the cone C Y F, as shown by the dotted line from F. From S on S C set off the Lengths of the several lines in K N J of the plan, as shown by 1', 2',

Fig. 578. - Plan and Side Elevation of a Collar to Fit Around a Pipe Pa sing throu h an Inclined Roof.

3', etc., and from these points draw lines to T. In the same manner construct the diagram of triangles shown at the right. At U upon the line Y D extended erect the perpendicular U V, equal in hight„to the cone Y D E, as shown by the dotted line E V. From U on U D set off the lengths of the lines in J R L and draw the hypothenuses, as shown.

To develop the pattern, first draw any horizontal line, as A A1 of Fig. 579, equal in length to P R in plan of Fig. 578. With A and A1 as centers, and the hypothenuse V 9' of Fig. 578 as radios, describe arcs intersecting each other at 9. Now, with 9 of the pattern as center, and 9 8 of the plan as radius, describe the arc 8; then with V 8' of Fig. 578 as radius, and A of the pattern as center, describe an arc, intersecting the arc previously drawn, thus establishing the point 8. Proceed in this manner, using alternately first the divisions on the quarter circle L J in plan, then the hypothenuses of the triangles whose bases are shown by the lines in J L R, until the point 5 in pattern has been obtained. Draw a line from 5 to A in Fig. 579. Then with A as center, and E F in side elevation, Fig. 578, as radius, describe an arc, shown at C of Fig. 579, and with 5 of the pattern as center, and the hypothenuse T 5' of Fig. 578 as radius, describe an arc intersecting the previous arc at C. Draw a line from 5 to C. Now proceed as above described, using alternately first the spaces on the quarter circle in J K in plan, then the hypothenuses of the triangles whose bases are shown in J K N in plan, until the point 1 in pattern has been obtained. Then with C of the pattern as center, and W N of the plan as radius, describe an arc, as shown at D, and with F C in side elevation as radius, and 1 of the pattern as center, describe an arc, intersecting the arc previously described at D. Draw the lines 1 D, D C, C A, and through the intersections of the arcs trace a line, shown from 1 to 9 on pattern. This will complete one-half the pattern. The entire pattern may be completed by duplicating the part 1 5 9 A C D and adding the same to that already obtained in such a manner that the side 9 A will coincide with 9 A1, as shown by 9 5' 1' D1 C1 A1.

Fig. 579. - Pattern of Collar Shown in Fig. 578.