HIP

497

IIIP

also an isosceles triangle ; the other two opposite sides are theright-angled triangles contained by the two hips and the twoadjoining rafters on the side of the roof. This rectangularpyramid may be divided into three triangular pyramids bythe two vertical triangular planes, formed by the hip-rafters,their seats, and the common perpendicular from their vertex.

Two of these pyramids, when the plan of the building is arectangle, are equal and opposite. In each of these equal andopposite pyramids the base is a right-angled triangle, con-tained by the seat of the hip-rafter, the seat of the adjoiningcommon rafter, and the part of the wall-plate between thehip and the adjoining common rafter. One of the sides is aright-angled triangle contained by the adjoining commonrafter, its seat, and perpendicular ; a second side is a right-angled triangle contained by the common rafter, the hip-rafters, and the wall-plate, between them ; and the remainingthird side is the triangle contained by the hip-rafter, its seat,and perpendicular. With regard to the remaining pyramid,its base is a right-angled triangle contained by the seats ofthe two hips and the wall-plate between them, the right anglebeing that contained by the seats of the two hips ; two of itssides are the triangular planes passing the hip-rafter, whichare also common to the other two pyramids ; its third side isthe hipped end of the roof.

Given the plan of a building, or the form of a wall-plate ofa hip-roof, and the pitch of the roof, to find the various lengthsand angles concerned, ivhether the roof is square or bevel.

Example I.— To find the length of the rafters, the backingof the hips, and the shoulders of jack-rafters and purlins,geometrically.

Plate I. —Figure 1.—Let ab cd be the plan. Draw efparallel to the sides, A d and b c, in the middle of the distancebetween them. On DC, as a diameter, describe the semi-circle dfc; draw p d and f c, then the angle dfc is a rightangle. Draw G F n perpendicular to e f, cutting the sidesA d and b C in G and H ; from FE cut off FI equal to theheight or pitch of the roof, and join G I ; from F c cut off f itequal to f i, and join k d ; then GI is the length of a commonrafter, and d k that of the hip ; for if the triangles G F i andD f.k be turned round their seats, gf and df, until theirplanes become perpendicular to the triangle Gfd, the per-pendicular F I will coincide with f it, and the point i willcoincide with the point K; the lines G I and D it, representingthe rafters, will then be in their true position.

The same by calculation. Gl 2 = GF* + FI 2 (Euclidi.47.), therefore gi = (gf 2 + Fi a )J the length of thecommon rafter, df 2 = gf 2 + GD 2 the square of the seatof the hip. dk 2 = df 2 + fk 2 = Gf 2 + G d 2 + f i 2 ,

-- 1 .

therefore d k = gf’ + gd 1 + fi ,s

In the same manner the other liip-rafter C l is found, asalso the hip-rafters AM and BN.

Let it be required to find the backing of the hip-rafter,whose seat is c F.

Geometrically. —Imagine the triangle cfl to be raisedupon its seat c F, until its plane becomes perpendicular tothe plane of the wall-plate abcd, then there will be tworight-angled solid angles ; the three sides of the one are theplane angles of f c d, f c l, and the hypothenusal plane angleDCL. In each of these solid angles the two sides, containingthe right angle, viz., the plane angles fch, fcd, and theperpendicular plane angle cfl, which is common to both,being given, to find the two opposite inclinations to the sidesfch and fcd, and the remaining third sides.

Now the angles GDC and H C d are besected by the seatsV D and f c of the hip-rafters ; for if E F is produced to meetD c in u, u will be the centre of the circle dfc; and u C,

6 L

u f, u d, are equal to each other ; and because u f is equalto u c, the angle c f u is equal to f c u ; but c f u is equal tothe alternate angle fch; therefore, the angle pcuis equalto f c h ; that is, the angle u C n is besected by the seat f cof the hip-rafter. In the same manner may be shown thatitdg is besected by the seat df of the other liip-rafter.From any point, o, in FC, draw ov perpendicular to l c,cutting it in p, and o w perpendicular to f c, cutting d c in jtv ; from o C cut off O Q equal to O P. Join Q tv, then o Q tvwill be the inclination opposite the plane angle f cu, and thisis the angle which the end of the roof makes with the verticaltriangle contained by the hip-rafter, its seat, and perpen-dicular. Produce tvo to meet bc in x, and join q x, thenw Q x is the inclination of the two planes of a side and endof the roof, whose intersections are b c and c d, on the planeof the wall-head. Now, the angle w Q x, which is doublethe angle w Q o, is the backing of the hip. Make p v equal ito Q w, and join c v, then will pc v be the angle containedby the two sides l c, CD, or that of the hypothenusal planeangle contained by the intersection bc, and the liip-rafterl c. This angle may be otherwise found thus :—Produce ghto R; make cR equal to c l, then the angle hcr is equal to /p c v. Now the angle hcr, or pc 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 c ; and the angle cvp, orCRH, jis the angle which a jack-rafter makes with the same hip ; in jthe same manner may the backings of the other hips be found. !The other bevel of the jack-rafters is the angle h if. Tofind the other bevel for cutting the shoulder of the purlin,proceed thus : on f, as a centre, with the distance f g,describe the arc G y ; draw f y perpendicular to g i, y zparallel to e f, cutting f d in z, and z & parallel to G h, cut-ting a d in &. Join & f, then g & f is the angle which theother side of the shoulder makes with the length of the purlin.

At the other end of this diagram is shown the manner offinding the two bevels for cutting the shoulder of the purlinagainst the hip-rafter, when the side of the purlin is not inthe plane of the side of the roof.

To find the same things by calculation. —The backing ofthe liip-rafter and hypothenusal side is obtained as follows :—

It has been shown that the three plane angles, and the threeinclinations 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 anglesbeing 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 perpendicularto each other, the spheric triangle representing this solid anglewill have also two of its sides perpendicular to each other.Now, in this, there are given the two sides containing theright angle to find the hypothenuse and angles.

It is shown, by writers on spherical trigonometry, that inany right-angled spherical triangle, radius is to the cosine ofeither of the sides, as the cosine of the other side to the cosineof the hypothenuse. Suppose the plane angle fcl to be27°, and the angle v c h 52°, to find the hypothenuse andangles of a right-angled spherical triangle, one of whose legs

is 27° and the other 52°, it will therefore be—

As radius, sine of 90° . = 10.00000

Is to the cosine of f c l, 27°. = 9.94988

So is the cosine of f c h 52°. = 9.7893 4

19,73922

10.00000

To the cosine of the hypothenusal side 56° 44' 9.73922

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