In Fig. 235, we show how all of the lengths, cuts and bevels may be obtained from the tri angle, bounded by A-B-C, formed by the runs of the common and hip rafters and the tangent, as shown at No. 1, as follows: From the run of the common rafter, erect the desired rise as at A-D and connect D-B. This forms the second triangle and contains the length, seat and plumb cuts of the common rafter, as shown in No. 2. At right angles from the common rafter draw a line equal to the tangent as and connect D-C'. This forms the third triangle, as shown in No. 3. In this are shown the face cut of the roof boards to fit in the valley or over the hip. This angle also gives the cut across the back of the jack to fit against the hip or valley, commonly called side cut of the jack. At right angles from the com mon rafter draw a line equal to the rise as D-A' and connect This forms the fourth triangle, as shown in No. 4. In this is shown the edge or miter cut of the roof boards to fit in the valley or over the hip. In other words, this is the same as the miter for a hopper. Now then, we will work from the other side of triangle No. 1. From the run of the hip draw a line at right angles from A-C equal to the rise, as at A-D' and connect C-D'. This forms triangle No. 5, and contains the length, seat and plumb cuts of the hip. From hip rafter and at right angles to A-C draw a line equal to as C-E and connect A-E. This forms triangle No. 6 and in it is contained the top or, commonly called, the side cut of the hip. The illustration. is for the 3-8 pitch or 9 inches rise to one foot run of the common rafter. For an octagon roof the angle at No. 1 would be 221 degrees. For a hexa gon roof it would be at 30 degrees.

In Fig. 236 are shown all of the above angles formed by as many steel squares, with the cor responding numbers placed on same that help to form two of the sides of each angle, and, by refer ing to the preceding illustration, the reader can readily see how the cuts are obtained on the steel square.


In Fig. 237 are shown the triangles for a hex agon (six sided) roof and by comparing with the preceding illustrations, the reader can see wherein they differ. The application is the same except in the sixth angle. Angle No. 1 represents the plan and governs the layout of the diagram. In this, the angle between the runs of the common rafter and the hip is at 30 degrees. Now, by referring to Fig. 238, we show the angle bounded by A-B-C as shown at No. 1 applied to the plan of the roof as follows : A-B run of the common rafter. A-C run of the hip. B-C tangent of the common rafter. Now by extending A-B intersecting a line at right angles from A-C as at E, then E-C will represent the length of the tangent from the hip. Note : See the difference in length when compared with the run of the hip.


In Fig. 239 is shown how these angles may be
obtained with the aid of two steel squares. It may be worked to a scale of one inch to the foot, or full scale for a one-foot run of the common rafter, as shown in the illustration. The figure on the tongue that gives the hexagon miter (6 11-12) represents the tangent for the common rafter as shown on square No. 1, and by placing square No. 2 with its heel resting at 6 11-12 and with the blade intersecting at 12 on square No. 1, then a line continued from the heel of square No. 1 and in line with the blade intersecting the tongue of square No. 2, which in this case is at 8 and represents the tangent for the hip. Now by erecting the rise from the intersection of the blades to D and D', then B-D represents the length of the common rafter and D'-C that of the hip and these lengths, taken on the blade of the respective squares, will give the figures to use for the side cuts of the rafters. Thus-6 11-12 and 15 1-8 as shown, will give the side cut of the jack, or of the common rafter, to fit in the angle between the hips at the peak (see Fig. 238) and 8 and 16 3-4 will give the side cut of the unbacked hip to fit in the peak, or if it is first backed, then the same figures as shown on square No. 1 applied to the backing plane will give the same result. But, of course, this is not practical because the ends of the hip would run to a feather edge, as shown. A better way is to insert a hexagon block with sides equal the width of the hip, then the cut of th - hip would simply be the plumb cut. This would afford a better nailing space and each hip would hav, a direct bearing against the one on the opposite side.
