Fig. 36. Laying Out Timbers of Roof with Two Unequal Pitches.

Fig. 36. Laying Out Timbers of Roof with Two Unequal Pitches.

With this knowledge of what figures to use, and why they are used, it will be an easy matter for anyone to lay out all rafters for equal-pitch roofs.

In Fig. 36 is shown a plan of a roof with two unequal pitches. The main roof is shown to have a rise of 12 inches to the foot run. The front wing is shown to have a run of 6 feet and to rise 12 feet; it has thus a pitch of 24 inches to the foot run. Therefore 12 on blade of the square and 12 on tongue will give the plumb and heel cuts for the main roof, and by stepping 12 times along the rafter timber the length of the rafter is found. The figures on the square to find the heel and plumb cuts for the rafter in the front wing, will be 12 run and 24 rise, and by stepping 6 times (the number of feet in the run of the rafter), the length will be found over the run of 6 feet, and it will measure 13 feet 6 inches.

If, in place of stepping along the timber, the diagonal of 12 and 24 is multiplied by 6, the number of feet in the run, the length may be found even to a greater exactitude.

Many carpenters use this method of framing; and to those who have confidence in their ability to figure correctly, it is a saving of time, and, as before said will result in a more accurate measurement; but the better and more scientific method of framing is to work to a scale of one inch, as has already been explained.

According to that method, the diagonal of a foot of run, and the number of inches to the foot run the roof is rising, measured to a scale, will give the exact length. For example, the main roof in Fig. 36 is rising 12 inches to a foot of run. The diagonal of 12 and 12 is 17 inches, which, considered as a scale of one inch to a foot, will give 17 feet, and this will be the exact length of the rafter for a roof rising 12 inches to the foot run and having a run of 12 feet.

Fig. 37. Finding; Length of Rafter for Front Wing in Roof Shown in Fig. 36.

Fig. 37. Finding; Length of Rafter for Front Wing in Roof Shown in Fig. 36.

Fig. 38. Laying Out Timbers of Roof Shown in Fig. 36, by Projecting Slope of Roof into Horizontal Plane.

Fig. 38. Laying Out Timbers of Roof Shown in Fig. 36, by Projecting Slope of Roof into Horizontal Plane.

The length of the rafter for the front wing, which has a run of 6 feet and a rise of 12 feet, may be obtained by placing the rule as shown in Fig. 37 from 6 on blade to 12 on tongue, which will give a length of 13 1/2 inches. If the scale be considered as one inch to a foot, this will equal 13 feet 6 inches, which will be the exact length of a common rafter rising 24 inches to the foot run and having a run of 6 feet.

It will be observed that the plan lines of the valleys in this figure in respect to one another deviate from forming a right angle. In equal-pitch roofs the plan lines are always at right angles to each other, and therefore the diagonal of 12 and 12, which is 17 inches, will be the relative foot run of valleys and hips in equal-pitch roofs.

In Fig. 36 is shown how to find the figures to use on the square for valleys and hips when deviating from the right angle. A line is drawn at a distance of 12 inches from the plate and parallel to it, cutting the valley in m as shown. The part of the valley from m to the plate will measure 13 1/2 inches, which is the figure that is to be used on the square to obtain the length and cuts of the valleys.

It will be observed that this equals the length of the common rafter as found by the square and rule in Fig. 37. In that figure is shown 12 on tongue and 6 on blade. The 12 here represents the rise, and the 6 the run of the front roof. If the 12 be taken to represent the run of the main roof, and the 6 to represent the run of the front roof, then, the diagonal 13 1/2 will indicate the length of the seat of the valley for 12 feet of run, and therefore for one foot it will be 13 1/2 inches. Now, by taking 13 1/2 on the blade for run, and 12 inches on the tongue for rise, and stepping along the valley rafter timber 12 times, the length of the valley will be found. The blade will give the heel cut, and the tongue the plumb cut.

In Fig. 38 is shown the slope of the roof projected into the horizontal plane. By drawing a figure based on a scale of one inch to one foot, all the timbers on the slope of the roof can be measured. Bevel 2, shown in this figure, is to fit the valleys against the ridge. By drawing a line from w square to the seat of the valley to m, making w 2 equal in length to the length of the valley, as shown, and by connecting 2 and m, the bevel at 2 is found, which will fit the valleys against the ridge, as shown at 3 and 3 in Fig. 36.

Fig. 39. Method of Finding Length and Cuts of Octagon Hips Intersecting a Roof.

Fig. 39. Method of Finding Length and Cuts of Octagon Hips Intersecting a Roof.

Fig. 40. Showing How Cornice Affects Valleys and Plates in Roof with Unequal Pitches.

Fig. 40. Showing How Cornice Affects Valleys and Plates in Roof with Unequal Pitches.

In Fig. 39, is shown how to find the length and cuts of octagon hips intersecting a roof. In Fig. 3G, half the plan of the octagon is shown to be inside of the plate, and the hips o, z, o intersect the slope of the roof. In Fig. 39, the lines below x y are the plan lines; and those above, the elevation. From z, o: o, in the plan, draw lines to x y, as shown from o to m and from z to m; from m and m, draw the elevation lines to the apex o", intersecting the line of the roof in d" and c". From d" and c", draw the lines d" v" and c" a" parallel to x y; from c", drop a line to intersect the plan line a o in c. Make a w equal in length to a" o" of the elevation, and connect w c; measure from w to n the full height of the octagon as shown from xy to the apex o"; and connect c n. The length from w to c is that of the two hips shown at o o in Fig. 36, both being equal hips intersecting the roof at an equal distance from the plate. The bevel at w is the top bevel, and the bevel at c will fit the roof.

Fig. 41. Showing Relative Position of Plates in Roof with Two Unequal Pitches.

Fig. 41. Showing Relative Position of Plates in Roof with Two Unequal Pitches.

Again, drop a line from d" to intersect the plan line a z in d. Make a 2 equal to v" o" in the elevation, and connect 2 d. Measure from 2 to b the full height of the tower as shown from x y to the apex o" in the elevation, and connect d b. The length 2 d represents the length of the hip z shown in Fig. 36; the bevel at 2 is that of the top; and the bevel at d, the one that will fit the foot of the hip to the intersecting roof.

When a cornice of any considerable width runs around a roof of this kind, it affects the plates and the angle of the valleys as shown in Fig. 40. In this figure are shown the same valleys as in Fig. 36; but, owing to the width of the cornice, the foot of each has been moved the distance a b along the plate of the main roof. Why this is done is shown in the drawing to be caused by the necessity for the valleys to intersect the corners c c of the cornice.

The plates are also affected as shown in Fig. 41, where the plate of the narrow roof is shown to be much higher than the plate of the main roof.

The bevels shown at 3, Fig. 40, are to fit the valleys against the ridge.

In Fig. 42 is shown a very simple method of finding the bevels for purlins in equal-pitch roofs. Draw the plan of the corner as shown, and a line from m to o; measure from o the length x y, representing the common rafter, to w; from w draw a line to m; the bevel shown at 2 will fit the top face of the purlin. Again, from o, describe an arc to cut the seat of the valley, and continue same around to S; connect S m; the bevel at 3 will be the side bevel.

Fig. 43. Method of Finding Bevels for Purlins in Equal Pitch Roofs.

Fig. 43. Method of Finding Bevels for Purlins in Equal-Pitch Roofs.