In Fig. 323, let F E D be the semicircle around which a pipe, of which A C B is a section, is to be carried by means of any suitable number of cross joints, in this instance ten. Divide the semicircle F E D into the same number of equal parts as there are to be joints, which, as just stated, is to be ten, all as shown by D, O, P, R, S, E, etc., and draw lines from each point to Z. As there are to be ten joints there must necessarily be eleven pieces, therefore, according to the directions given in the previous problem, the semicircle must be divided into such a number of equal parts that the number of lines radiating from Z shall be eleven, all as shown, each line serving as the center line of a piece. From D toward the center Z set off the diameter of the pipe A B, as shown by the point A1. From Z as center, with the radius Z A1, draw the dotted line representing the inner line of tin-pipe, and cutting the radial lines previously drawn in the points O1. P1, etc. Through O and O1 draw lines at right angles to O Z and continue them in either direction till they intersect with the lines drawn through P and P1 on the one side and through D and A1 on the other. Each pair of lines is to be drawn at right angles to its respective radial or center line. Through the points of intersection draw the lines T T1, U Ul, etc., which will represent the lines of the joints or miters.

Fig. 323.   A Pipe Carried Around a Semicircle by Means of Cross Joints.

Fig. 323. - A Pipe Carried Around a Semicircle by Means of Cross Joints.

It will appear by inspection that the point U is equidistant from P and O, and that U1 is also equidistant from P1 and O1, and that therefore the lines U U1, T T1, etc., if continued inward must arrive at the center Z. Thus the joint lines, like the center lines, must radiate from the center of the semicircle.

Draw the profile of the pipe A C B directly below and in line with one end of the pipe, all as shown in the engraving. As may be seen by inspection of the diagram, two patterns are required, one corresponding to the half section occurring at the end, and the other corresponding to the full sections composing the body of the pipe. For the pattern of the end section pro-ceed as follows: Divide the profile AC B in the usual manner into any convenient number of equal parts, and from the points thus obtained carry lines upward at right angles to Z D. cutting T1 T. Prolong the line Z D. and upon it place a stretchout from the profile A C B, perpendicular to which draw measuring lines in the usual manner. With the T-square placed parallel to Z D, and brought successively against the points in T1 T, cut the measuring lines of corresponding numbers. Then a line traced through the points of intersection thus obtained, as shown by I K L, will be the shape of the miter cut, and G I K L H will be the complete pattern for one of the end pieces. For the pattern of one of the large pieces, as U V V1 U1, lay off a stretchout of A C B upon its center line extended, as shown by M N, and through, the points in it draw measuring lines in the usual manner. Place the T-square parallel to U V and, bringing it against the points in U U1, cut the line V V. Next place the T-square parallel to the stretchout line, and, bringing it against the several points in the miter lines U1 U and V1 V, cut the corresponding measuring lines, all as shown, thus completing the pattern.

PROBLEM 44. The Patterns for an Elbow at Any Angle.

Let D F H K L I G E in Fig. 324 be the elevation of a pipe in which elbows are required at special angles. In convenient proximity to and in line with one end of the pipe draw a profile, as shown by A B C, which divide in the usual manner. Placing the T-square parallel to the first section of the pipe, and, bringing it against the several points in the profile, drop corresponding points upon F G. Shift the T-square, placing it parallel to the second section, and, bringing it against the several points in F G, drop them upon H I. At right angles to the first section lay off a stretchout of A B C, as shown by T U, through the points in which draw the customary measuring lines. Placing the T-square at right angles to this section of the pipe, and bringing it against the several points in F G, cut the corresponding measuring lines. Then the line RLS traced through these points will, with the line T U, be the pattern sought. The pattern for the opposite end is to be obtained in like manner, all as shown by M N O P, and therefore need not be described in detail. For the pattern of the middle section lay off a stretchout, W V, at right angles to it, with the customary, measuring lines. Placing the T-square at ri g h t angles to the section, bring it successively against the points in G F and I H, and cut the corresponding measuring lines, as shown. Then lines traced through these points, as shown by Y X Z Q, will be the pattern sought. The positions of the longitudinal joints in the several sections of this elbow, as well as those of all others, are determined by the order in which the measuring lines drawn through the stretchout are numbered. In the present instance the joints are allowed to come on the back of the pipe, or, in other words, upon D F H K, which corresponds to the point 1 in the profile. Hence, in numbering the measuring lines in the several stretchouts, point 1 is placed at the commencement and ending, while if it were desired to have the joint in either piece come on the opposite side, or at a point corresponding to 9 of the profile, the stretchout would have commenced and ended with that figure, the figure 1 in that case coming, in regular order, where 9 now occurs. The effect of such a change upon any of the patterns here given would be the same as if they were cut in two upon the line 9 and the two halves were transposed.

Fig. 324   An Elbow at Any Angle.

Fig. 324 - An Elbow at Any Angle.