measured by the degrees of the angle formed by straight lines drawn from its extremities to the center of the elbow, Fig. 77. Since all backsets in the same elbow are equal, the value of the backset can be expressed in degrees. A five-piece elbow has 8 backsets. A five-piece 90° elbow would have each backset equal to 90°÷8 or 11¼°. The first piece of any elbow contains one backset, the last piece one, and every other piece contains two. Therefore, the rise of the miter lines for a five-piece 90° elbow would be:
| Rise of | 1st | miter | line would be | 11¼°×1 | =11¼°, having but one backset. |
| " | 2nd | " | " | 11¼×(1+2) | =33¾°, by adding two backsets. |
| " | 3rd | " | " | 11¼×(3+2) | =56¼°, by adding two backsets. |
| " | 4th | " | " | 11¼×(5+2) | =78¾°, by adding two backsets. |
Problem 13B.—Give the rise of each miter line in a four-inch, four-piece, 75° elbow.
Problem 13C.—Give the rise of each miter line in a three-piece, 90° elbow.
Problem 13D.—Give the value in degrees of the backset of a two-piece, 12° elbow.
Problem 13E.—Give the value in degrees of the backset of a three-piece, 24° elbow.
Problem 13F.—Give the value in degrees of the backset of a four-piece, 36° elbow.
Problem 13G.—Give the value in degrees of the backset of a five-piece, 48° elbow.
Problem 13H.—Give the value in degrees of the backset of a six-piece, 60° elbow.
Problem 13I.—Give the value in degrees of the backset of a seven-piece, 72° elbow.
Problem 13J.—Give the value in degrees of the backset of an eight-piece, 84° elbow.
Problem 13K.—For the same big end diameter, why would one pattern answer for all of the elbows mentioned in problems 13D to 13J inclusive?
PROBLEMS ON THE "CUTS OF AN ELBOW"
Problem 13L.—What would be the "big end cuts" of the following sizes of elbows? Add the standard stovepipe lock of one inch.
