Plate 1.—Figure 1.—Let A BCD be the plan. Draw E F parallel to the sides, A n and n e, in the middle of the distance between them. On D c, as a diameter, describe the semi circle D F c: draw F D and F C, then the angle n F e is a right angle. Draw G F ft perpendicular to E F. cutting the sides A D and B c in G and H ; from F E cut off F I equal to the height or pitch of the roof, and join G I ; from F C cut off r K equal to F I, and join K D ; then G t is the length of a common rafter, and D K that of the hip ; for if the triangles G F I and D F K be turned round their seats, o F and D F, until their planes become perpendicular to the triangle G F D, the per pendicular F I will coincide with F K, and the point r will coincide with the point a ; the lines Gland D K, representing the rafters, will then be in their true position.
The same by calculation. = + (Euclid i. 47.) therefore, G I = G + F I 2 ) the length of the common rafter, ID = G + G the square of the seat of the hip. = = therefore D K = G + G In the same manner the other hip-rafter e L is found, as also the hip-rafters A M and B a.
Let it be required to find the backing of the hip-rafter whose seat is C F.
Geometrically.-1magine the triangle e F L to be raised upon its seat C F, until its plane becomes perpendicular to the plane of the wall-plate A B e D, then there will be two right-angled solid angles ; the three sides of the one are the plane angles of F c D, F C L, and the hypothenusal plane angle e L. In each of these solid angles the two sides, containing the right angle, viz., the plane angles F e n, F e n, and the perpendicular plane angle e F L, which is common to both, being given to find the two opposite inclinations to the sides F C n and F e D, and the remaining third sides.
Now the angles G D c and n e D are besected by the seats F D and F c of the hip-rafters ; for if E F is produced to meet c in u, u will be the centre of the circle n F e ; and u e, II F, u D, are equal to each other ; and because u F is equal to u c, the angle c F u is equal to F C II; but C F c is equal to the alternate angle F c 11 ; therefore, the angle r c u is equal to F c 11 ; that is, the angle u c n is besected by the seat F C of the hip-rafter. In the same manner may be slims]) that G is besected by the seat n F of the other hip-rafter. From any point, o, in F C, draw o v perpendicular to L e, cutting it in r, and o w perpendicular to FC, eutting n c in w; from o e cut ofF o o equal to o P. Join Q w, then o Q w will be the inclination opposite the plane angle F C u, and this is the angle which the end of the roof makes with the vertical triangle contained by the hip-rafter, its seat, and perpen dicular. Produce w o to meet B c in x, and join Q x, then
w Q x is the inclination of the two planes of a side and end of the roof, whose intersections area e and e D, on the plane of the wall-head. Now, the angle w Q x, which is double the angle w a o, is the backing of the hip. Make p v equal to Q w, and join e v, then will p c v be the angle contained by the two sides L e, c D, or that of the hypothenusal plane angle contained by the intersection B C, and the hip-rafter L e. This angle may be otherwise found thus :—Produce G n to K ; make C R equal to c L, then the angle n e R is equal to r e v. Now the angle n C R, or r e v, is the angle which the purlins (when one of their faces is in the side of the roof) make with the hip-rafter L e ; and the angle e v p, or c a it, is the angle which a jack-rafter makes with the same hip; in the same manner may the backings of the other hips be found. The other bevel of the jack-rafters is the angle nI F. To find the other bevel for cutting the shoulder of the purlin, proceed thus : on F, as a centre, with the distance F a, describe the arc G r; draw F Y perpendicular to G I, r z parallel to E F, cutting F D 111 z, and z & parallel to G H, cut ting A D in &. Join & F, then G & F is the angle which the other side of the shoulder makes with the length of the purlin.
At the other end of this diagram is shown the manner of finding the two bevels for cutting the shoulder of the purlin against the hip-rafter when the side of the purlin is not in the plane of the side of the roof.
To find the sante things by ca/culation.—The backing of the hip-rafter and hypothenusal side is obtained as follows:— It has been shown that the three plane angles, and the three inclinations of solid angles, consisting of three plane angles, are found exactly as the sides and angles of spheric triangles, any three parts being given ; the degrees of the plane angles being exactly the same as the sides of the spheric triangles, and the inclinations the proper measures of the spheric angles; therefore, if two of the plane angles should be perpendicular to each other, the spheric triangle representing this solid angle will have also two of its sides perpendicular to each other. Now, in this, there are given the two sides containing the right angle to find the hypothenuse and angles.