While the ordinary principles of mensuration are all that are necessary to calculate any roof area, yet the modern house, with its numerous gables and irregular surfaces introduces complications which render some further explanation of roof measurement desirable. The most common error made in figuring roofs, and one which should be carefully guarded against, is that of using the apparent length of slopes, as shown by the plan or side elevation, instead of the true length, obtained from the end elevations.

The area of a plain gable roof, as shown in end and side elevations in (a), Fig. 1, is found by multiplying the length g j by the slope length b d, and further multiplying by 2, for both sides. The area of the gable is found by multiplying the width of the gable a d by the altitude c 6 and dividing by 2.

At (b) is shown the plan and the elevation of a hip roof, having a deck z. The pitch of the roof being the same on each side, the line c d shows the true length of the common rafter lm,ce being the height of the deck above ad. At (c) is shown the method of determining the true lengths of the hips, and the true size of one side of the roof. Let abcd represent the same lines as the corresponding ones in (6).

Fig. 1.

From the line a d, through 6 and c, draw perpendiculars, as g h and ef; lay off from g and e on these lines the length of the common rafter ab, in (b), and draw the lines ah and df; then the figure ahfd will show the true shape and size of that side of the roof shown in the elevation in (6). The triangle def equals in area the triangle agh or a similar triangle aih. Hence, the portion of the roof ahfd is equal in area to the rectangle aife, whose length is one-half the sum of the eaves and deck lengths, and whose breadth is the length of a common rafter.

A method of obtaining the lengths of valley - applicable also to hip rafters - is shown at (d), which is the plan of a hip-and-gable roof. To ascertain the length of the valley rafter ab, draw the line ac perpendicular to a b and equal in length to the altitude of the gable; then draw the line c b, Which will be the length required.

As an example of roof mensuration, the number of square feet of surface on the roof shown in Fig. 2 will be calculated. The area of the triangular portion acb is equal to one-half the base a 6, multiplied by the slope length of c d. The latter is found by making c c', perpendicular to d c, equal to the height of the ridge (10 ft.) above a b. and drawing c' d, which is the required slope length. Using dimensions, the area of a c b is(23x13.75) / 2 = 158.1 sq. ft.

Fig. 2.

The ana of the trapezoid gf ih is one-half the sum of f i and g h, multiplied by the true length of h i. which, by laying off i i'. 8 ft. along f i, and drawing i' h, is found to be 10 ft. 7 1/2 in., or, say, 10.6 ft. Then g f i h =[(14+5) / 2] x 10.6 = 100.7 sq. ft.; or, for the two slopes of the gable, the area is 201.4 sq. ft. As the opposite gable is the same size, the area of the two is

201.4 x2 = 402.8 sq. ft.

The area of q p n k is equal to the area q p w, minus the area k n w, which is covered by the intersecting gable roof.

The area q p w is equal to the area a c b, or 158.1 sq. ft. The area of k n w is equal to 1/2 the product of n w and the slope of s k (which, by laying off k k' equal to the height of the gable,

5.5 ft., and drawing s k', is found to be nearly 7.5 ft.). Then area k n w = (13x7.5) / 2 = 48.7 sq. ft., which, deducted from

158.1 sq. ft., shows the area of q p n k to be 109.4 sq. ft.

The area of a p q c is (ap+qc) / 2 multiplied by the true slope length of t v, which is t v', measuring 15.25 ft. Substituting dimensions, the area is found to be (6+24) / 2 x 15.25= 228.7 sq.ft.

From this deduct the area of y z u, which is the portion covered by the intersecting gable roof. The true length of t u along the slope is t u', which measures 12 ft.; hence, the area of y z u is (14x12) / 2 = 84 sq. ft. The net area of a p q c is, therefore, 228.7-84=144.7 sq. ft.; b c q w being equal to a p q c, its area is the same, making the area of both sides 289.4 sq. ft.

The area of k n m l is [(mn+lk) / 2]-----------x m I', the slope length of m I. Substituting dimensions, the area is [(11+16) / 2] x8.5=114.7 sq. ft. As k l x w is equal to k n m l, the area of both is 229.4 sq. ft. Adding the partial areas thus obtained, the sum is 158.1 + 402.8 + 109.4 + 289.4 + 229.4 = 1,189.1 sq. ft., or 11.9 squares.