Problem 25
RECTANGULAR PIPE OFFSET
47. The Rectangular Pipe Offset.—As was pointed out in the preceding problem, the elevation of a fitting for rectangular piping presents much the same appearance as an elevation for a round pipe fitting. For this reason the description given below will answer equally as well for an offset in round piping, the only difference being the shape of the profile.
The Profile.—A profile, as shown in Fig. 138, using the dimensions given, should be drawn. Each vertex (corner) of the profile should be numbered. Extension lines are carried upwards from points 2 and 3.
The Elevation.—A base line AH in Fig. 137 should be drawn equal in length to line 2–3 of the profile. A perpendicular 3 inches high should be erected at point A. The upper point of this fine should be lettered B. The line BC is drawn according to the dimensions given in Fig. 137. The 3-inch dimension gives the fitting its name, it being the amount that the third piece sets off to one side of the first piece of the elbow. The line CD is drawn at right angles to the base line. This completes the outline of one side of the elevation. The miter line BG must next be drawn. Every miter line of an elbow bisects the angle formed by the adjacent sides of the pipe. Therefore, in order to get the miter line BG the angle ABC must be bisected.
The procedure for bisecting an angle is as follows: With B as a center and any radius, set off equal distances on each side of the point B, on lines AB and BC. Letter these points P and R. With P and R as centers and any radius greater than RB, draw intersecting arcs (to the right of the figure). Letter this intersection J.
The straight line BJ will bisect the angle and can be used as the first miter line of the fitting. A perpendicular is erected at the point H until it cuts the miter line at the point G, Fig. 137. The next line to be drawn is FG.
In order to draw the line FG the angle HGB must be copied. This is done as follows: With G as a center and any radius, draw an arc cutting line HG at the point K and also cutting the miter line at the point M. With M as a center and a radius equal to
