Translating Framing Problems From Protractor To Framing Square And Vice Versa. Frequently it is desirable to translate framing problems from degrees to numbers to be used upon the steel square, and vice versa. To change from degree framing to steel square framing it is only necessary to remember that the numbers to use on the square must be numbers such that their ratio one to the other shall give a tangent value equal to that given in the Table of Natural Functions, Appendix II, for the angle under consideration.

Fig. 81. Side Cut

Fig. 81. Side Cut.

Fig. 82. End Cut

Fig. 82. End Cut.

Example:

Given: Angle of inclination of common rafter or of roof = 3O degrees. Find the numbers to take on the square to frame seat and plumb cuts. Solution:

Tan 30° = .577 (by Table, Appendix II).

By agreement, run of common rafter takes 12" on one member of the square for constant of run. Therefore, rise must be 6.93", or 6 11/12", must be taken on the other member of the square. (The base of the table is 1 so that for 12" run we must have 12" x .5774.)

In a similar manner the number to be taken on the blade, when the inclination of any other common or jack rafter is given, may be determined. In the case of the hip or valley inclination, however, it must be remembered that 17" is to be taken on the tongue for the run in the square cornered house, with 13" in the octagon, which will necessitate multiplying the tangent value of the angle of inclination by 17 and 13 respectively to find the number to take on the blade when one or the other of these is taken on the tongue.

Example:

Given: Side cut of jack rafter, or hip or valley = 38 degrees.

Find the numbers to use on the framing square to lay off this same angle.

Solution:

Tan 38° - .7813

By agreement, we shall take 12" on the tongue.

The number to be taken on the blade must be, therefore, 12" x .7813 =

9°⅜"

(In the case of the side cut, any number other than 12" might have been assumed on the tongue.) Example:

Given: A hip rafter on a square roof which is framed with 17" on the tongue and 8" on the blade. To find the angle of inclination of the hip.

Solution:

8/17 = .4706

The angle whose tan = .4706 is 25° 19'