Fig. 632. -Plan and Elevation of Flaring Collar.

In Fig. 632 E F G H shows the side elevation of a flaring collar, the profile of the small end or top being shown at A B C D, and that of the bottom at K L M N of the plan. The conditions embodied in this problem are in no respect different from those of the problem immediately preceding. A slight difference in detail consists in the fact that in the former case the short side was at right angles to the larger end, while in the present case it is at right angles to the smaller end, but the pattern may be obtained by exactly the same method as that employed in the previous problem. However, as the elevation was there made use of to determine the triangulation, the plan will here be used upon which to determine the position of the triangles of which the pattern will subsequently be constructed.

Divide A B C of the profile into any convenient number of parts, and, with the T-square at right angles to E F of the elevation, carry lines from the points in ABC, cutting E F, as shown. Extend the base line G H to the left indefinitely, and through the center of plan of base draw O M, parallel with H G of elevation. Drop lines from the points in E F, extending them vertically through O M. With the dividers take the distance across the profile A B C D on each of the several lines drawn through it, and set the same distance off on corresponding lines drawn through O M. That is, taking A C as the base of measurement in the one case, and O M in the other, set off on the latter, on each side, the same length as the several lines measure on each side of A C. Through the points thus obtained trace a line, as shown by O P Q R, thus obtaining the shape of the upper outline as it would appear in the plan.

As both halves of the plan when divided by the line O M are exactly alike, it will only be necessary to use one-half in obtaining the pattern; therefore divide K N M of plan into the same number of parts as was the half of profile, in the present instance six, as shown by the small figures. Number the points in KNM to correspond with the points in O R Q, and connect corresponding points by solid lines, as shown by 1 1', 2 2', 3 3', etc. Also connect the points in O R Q with those of the next higher number in K N M, as shown by the dotted lines 1 2', 2 3', 3 4', etc. The solid and dotted lines thus drawn across the plan will represent the bases of a number of right angled triangles whose altitudes are equal to the vertical lines between E F and J Y of the elevation, and whose hvpothenuses, when obtained, will give the real distances across the sides of the finished article in the direction indicated by the lines across the plan.

Fig. 633. - Diagram of Triangles Based upon the Solid Lines of the Plan in Fig. 632.

Fig. 634. - Diagram of Triangles Based upon the Dotted Lines of the Plan in Fig. 632.

To construct these triangles proceed as follows: Draw any right angle, as S T U in Fig. 633. On T U, measuring from T, set off the lengths of solid lines in plan, making T U of diagram equal to Q M of plan, T 2 of diagram equal to 2 2' of plan, T 3 of diagram equal to 3 3' of plan, etc. From T on T S set off the length of lines in E F Y J of elevation, making T S of diagram equal to F Y of elevation, T 2 of diagram equal to a 2 of elevation, T 3 of diagram equal to b 3 of elevation, etc. Connect points in T S with those of similar number in T U, as shown by the solid lines. The hypothenuses of the triangles thus obtained will give the distance from points in plan of base to points of similar number of top as if measured on the finished article. The diagram of triangles in Fig. 634 is constructed in a similar manner. Draw the right angle

V W X and on W X set off the lengths of dotted lines in plan, and from W on W V the lengths of lines in E F Y J of elevation, excepting the line E J, or No. 7, which is not used. Connect the points in W V of diagram with those of the next higher number in AY X, as shown by the dotted lines in diagram and by similar lines in the plan. The resulting hypothenuses will give the correct distances from points in top of article to points of next higher number in the plan of base.

Fig. 635. - Pattern of Flaring Collar.

Having now obtained all the necessary measurements, the pattern may be developed as follows: Draw any line, as q m in Fig. 635, in length equal to S U of Fig. 633, or F G of elevation. With m of pattern as center, and M 2' of plan as radius, describe a small arc (2"), which intersect with one struck from point q of pattern as center, and V X of diagram of triangles in Fig. 634 as radius, thus establishing the point 2" of pattern. With point 2" of pattern as center, and 2 2 of Fig. 633 as radius, describe a small arc (2), which intersect with another struck from point q of pattern as center, and C 2 of profile as radius, thus establishing the point 2 of pattern. With 2 of pattern as center, and 2 3 of Fig. 634 as radius, describe a small arc (3"), which intersect with another struck from 2" of pattern as center, and 2' 3' of plan as radius, thus establishing point 3" of pattern. With 3" of pattern as center, and 3 3 of Fig. 633 as radius, describe a small arc (3), which intersect with one struck from point 2 of pattern as center, and the distance 2 3 of profile as radius, thus establishing point 3 of the pattern. Proceed in this manner, using the hypothenuses of the triangles in Figs. 633 and 634 for the distances across the pattern; the distances between the points in the plan of base for the stretchout of the bottom of pattern; and the distances between the points in the profile of top for the stretchout of the top of pattern. Lines drawn through the points of intersection, as shown by q o and m k, will, with q m and ok; constitute the pattern for half the article. The other half of the pattern q o' k' m can be obtained by any convenient means of duplication.