In the solution of this problem two conditions may arise; in the first, the two pieces of the elbow have the same flare or taper, while in the second case one of the pieces may have more flare than the other. It has been shown in the chapter on geometrical problems that an oblique section through the opposite sides of a cone is a perfect ellipse. Keeping this in mind, it is evident that if the cone shown by A B C in Fig. 507 were made of some solid material and cut obliquely by the plane D E and the severed parts placed side by side, both would present to view two perfect ellipses of exactly the same proportions and dimensions; and, therefore, that if the two parts be placed together again, turning the upper piece half way around, as shown by D E A1, the edges of the two pieces from D to E would exactly coincide.
Taking advantage of this fact, then, it only becomes necessary to ascertain the angle of the line D E, necessary to produce the required angle between the two pieces of an elbow, both of which have equal flare.
Therefore, at any convenient point upon the axis A H, as I, draw I J at the angle which the axis of the upper piece is required to make with that of the lower, then bisect the angle J I H. as shown by the line I K. Draw D E parallel to I K at the required hight of the lower piece, which will be the miter line sought.
Before completing the elevation of the elbow it will be necessary to notice a peculiarity of the oblique section of a cone - viz., that although the line A H bisects the cone and its base it does not bisect the oblique line D E, as by measurement the center of D E is found to bee at x. Therefore, through the point b, which is as far to the right of x as point a is to the left of it, draw any line, as b A1, parallel to I J and make b A1 equal in length to a A. and draw A1 D and A1 E, Next draw G1 E1 at righl angles to A1 A. representing the upper end of the elbow. .Make D F equal to E F1. and E G equal to D G1. Then B F G C will he the elevation of a frustum of a cone, which, when cut in two upon the line D E, will, when the upper section is turned half-way around upon the lower part, form the elbow B D G1 F1 E C.
At any convenient distance below the base of cone B C draw half the plan, as shown by L H M, which divide into any convenient number of equal spaces. From the points of division erect lines vertically, cutting the base of the cone B C, and thence carry them toward the point A. cutting the miter lino D E. Placing the T-square parallel to the base line B C bring it successively against the points in D E, cutting the side of the cone, as shown below D.
From A as center, with radii A B and A F, draw ares, as shown. Upon the arc drawn from B, beginning at any convenient point, as N, step off a stretchout of L H M, as shown by the small figures. From each of the points thus obtained draw measuring lines toward the point A, and from the last point O one cutting the are drawn from F at Q. Placing one point of the compasses at the point A, bring the pencil point in turn to each of the points in the side of the cone below D and cut measuring lines of corresponding number. Then a line traced through the points of intersection, as shown from S to R, will be the miter cut between the two parts of the pattern of the frustum O N P Q necessary to form the patterns of the required elbow.
As but half the plan of the cone was used in obtaining a stretchout, the drawing shows but halves of the patterns. In duplicating the halves to form the complete patterns the upper piece can be doubled upon the line Q S and the lower part upon the line R N, thus bringing the joints on the short sides.
If, according to the second condition stated at the beginning of this problem, the upper section of this elbow is required to have more or less Hare than the lower section, thereby placing the apex A1 nearer to or farther away from the line D E, a different course will have to be pursued in obtaining the pattern. If, for instance, the hight of the cone A H be reduced, the base BC remaining the same, the proportions - that is, the comparative length and width - of the ellipse derived from the cut D E would be different from those derived from the same cut were the proportions of the cone to remain unchanged. Therefore, since the shape of the lower piece at the line D E is a fixed factor, if the circle at G1 F1 be shifted up or down the axis, or, remaining where it is, its diameter be changed, the piece D G1 P1 E becomes an irregular tapering article, in which case its pattern can most easily be obtained by triangulation. Patterns for pieces embodying those conditions can be found in Section 3 of this chapter, to which the reader is referred.