COR

218

COR

I

i

That is, the weights on any two angles are as the sinesof these angles directly, and reciprocally as the product ofthe sines of the two parts of these angles formed by thevertical lines.

Corollary 1. Hence the weights on any two angles are asthe sines of the angles directly, and as the produce of theco-sines of the angles of elevation reciprocally. For, drawb h perpendicular to b i, and produce i b and A b to I and K,then will the angle k b i, equal the angle h b i, he the com-plement of the angle hbi, viz., the complement of the angleof elevation of the rafter A b above the horizon ; and becausec b i is the supplement of c b i, the angles c b i and CBi havethe same sine, and the angle CBiis the complement of theangle hbc, viz., the angle of elevation of the rafter b C.

Corollary 2. Hence also the weights on any two anglesare as the sines of the angles directly, and as the productsof the secants of elevation jointly; for the secants of any twoangles are reciprocally as the co-sines of these angles.

Corollary 3. The force which any rafter makes in itsown direction, is as the secant of its elevation. For, makea p equal to b h, draw the lines p n, A h, » l, Sec., parallelto the vertical lines b i, C m, &c., and draw A n, b n, cl, &c.parallel to the horizon’; then, because the angles nap,hb1,I. C n, Sec., are the angles of elevation, and A n, b h, c l, &c.,are all equal, if A n, bh,c l, &c. be considered as radii, a p,b k, c n, Sec. are the secants of elevation, and also representthe forces on the rafters.

Corollary 4. Hence the horizontal pressures at A and care equal; for all the perpendiculars drawn from the oppositeangles of each parallelogram to meet the vertical diagonal,are all equal.

Corollary 5. Hence, if the position of any two rafters,and the proportion of the weights be given, the position ofthe remaining rafters may be determined.

Corollary 6. If tire vertical line SDvbe drawn, the hori-zontal line a v G and the lines a s, a r, a q, a t, &c. be drawnparallel to the rafters a b, b c, C d, d e, &c. meeting thevertical line in s, R, Q, t ; then will a s, ar, a q, at, repre-sent the forces, and s r, rq, QT, the forces upon the angles ;for A s, AR, aq, A t, &c. are the secants of the elevation, andthe triangles a s r, a r Q, A Q t, are all similar to the trianglesh b i, lent, o v>p, Sec.

Corollary 7. In every roof kept in equilibrio by the weightof the rafters, if u, v, W, &c. be the centres of gravity of therafters, and also represent their weights, then the weightpressing vertically on B, will be

ad X u

AB

+

VC XVBC ’

and the weight on

vb X v

c =-

bc

wd X w

-I-, and

CD

so on.

Hence

AD X PAB

VC XVBC

VB X V , WDXir _ _ S.ABC S.BCD

bc *" CD ' ' S./abj X S.iBK ’ s.lcm X S.mcnCorollary 8. Hence, if the rafters be prismatic figures,the weights on the angles b, c, d, &c. will be respectivelyas A b+b C, b C+C d, C d+d e, and so on.

Proposition II. Figure 2. Given, the vertical angle of aroof, and the proportion of the rafters on each side; to describethe roof to a given width, so that it shall be in equilibrio.

Let the proportion of the rafters from the bottom, upwards,be as 4, 3, 2 :

Then 4 + 3 = 7 represents the weight on g.

3+2 = 5 represents the weight on h.and 2+2 = 4 represents the weight on i.

Now let f b e be half the given angle, produce e b throughc to d, draw f a m perpendicular to e d cutting e d at a ;

make AE equal to ab, and join ef; let be re P resen ot lierweight on the vertical angle: make e b, b c, c d, to one an sas 4, 5, 7 ; join fd, fc, fe ; from any scale of e< l uamake fg = 7 ; draw GH parallel to f c, equal to 5 p :ir ’^ e sIt i parallel to F b, equal to 4 parts, and the contiguous ^f G, G it, h I, will be similar to the half roof. The ot \j na te swill be found by drawing the vertical line IN, and or<^ #1)( jperpendicular thereto, from the points G H, to l an® ^ ff[lT emaking the distances On both sides of imay he reduced or enlarged to any given width, bya similar figure upon a given line.

Proposition III. Figure 3. The angular points ating of every two rafters of a roof in equilibrio, by equal' ^gr,hung at the angles, in directions equidistant from eacare in the curve of a parabola. . ff eigl lf ’ 3

Let abode, Sec. be kept in equilibrio, by equal ' e ° u j-suspended at the angular points b, c, d, e, Sec ., i° 1 ^c, vdistant directions bf, c g, d h, e i, &c. the point 3 A > ’e, &c., are in the curve of a parabola. at f , G ’

For, let the lines bf, c g, dii, and ei, meet AN A >ih, i: draw a k parallel toparallel to B C, cutting

D E, ALF B at K, L,

parallel to c p ’ v -

m. Draw B Q,parallel to an, cutting the middle line ie at Q. r >

Then, because that the weights on the angles ai er ef° r6f k, k l, l m, m b, are as the numbers 1 , 2 , 2, 2, ^ . but

F K, f l, F m, f b, are as the odd numbers 1> 3, ’ j-Jlsbbecause of the equidistant lines b f, c g, d h> and P t ;vel/d o, c p, b q, the triangles afi,afl,af M, are re3 ”. tb ete \equal and similar to the triangles doe, CS», !jjeq u#fore f K is equal to o e, f l equal to s d equal to o i’> j iere fo^toRC equal to p q, and lastly, bf is equal to Q 1 ’ . 3,5,"E o, o P, p Q, Q i, are to one another as the numbei 3 ^ ^ fo>

and e o, e p, e q, e i, are as the square numbers U ^ g, 4 <

but the lines od,pc,qb, are to one another a 3 ’ ,’ u -e3 0therefore the abscissas e o, e p, e q, e i, are as the a ,. e ithe ordinates od,pc,qb, and the points A, b, c , v> ’the curve of a parabola. . . g foe cl ' s ’

In the same manner it may be shown that this 1whatever be the number of ordinates.

i vi a ciili \jl vi uuiaiCo< , IIJ £ v »i

Corollary. —Hence a roof of this constructs ^ gif* 3described to any given height and vertical angle, 01 jjgide* fwidth and height with any number of rafters on ea oJD bef ^Example. —To describe a roof with any g> v ® n g0 rSrafters on each side, of a given width and height, ^the weights suspended from the angular points o eqin vertical equidistant lines, may keep the rattelibrio.

side,

let i 1

Figure 3. —Let there be four rafters on each sl “’ T pai'y^half the width, and i e the height. Draw N T anto I E and I N ; divide N T into four equal parts, ^ intod t, and draw dg e, e h E,/i e : likewise divide .equal parts, i c, c b, b a, a N, and draw c g, b ", *? e ’foe l " M foeie. Join E g,gh, h i, i N, and these lines will ge eof half the roof required. For the demonstiaarticle Conic Sections. r eq u ' r6 ° ef

Proposition IY. Figure 4. —Suppose it ' verC t0 th 6 11 »

construct a curb roof to have the bottom ra ^ el j i ie top,rafter as 2 to 3, to a given vertical angle al t

given width ab. _ 4hp w e '°

Now the weight on the upper angle is i0 ^ fo tthe lower angle, as is to H 1 that is, 2

jj 2i

3 + 2 „#• a to 5, ° T

to —-— = 2 this is in the proportion ol o

the

h»“

weight at n is to the bottom weight at i,Bisect A b by the perpendicular c d, and

as 3 to* foe ^