ROOF. 



equal ; for all the perpendiculars drawn from the oppofite an- 

 gles of each parallelogram to meet the vertical diagonal, are 

 all equal. 



Cor. 5. — Hence, if the pofition of any two rafters, and 

 the proportion of the weights, be given, the pofition of the 

 remaining rafters may he determined. 



Cor. 6 If the vertical line S D V be drawn, the hori- 

 zontal line A V G, and the lines A S, A R, A Q, A T, &c. 

 be drawn parallel to the rafters A B, BC, CD, D E, &c. 

 meeting the vertical line in S, R, Q, T ; then will A S, 

 A R, AQ, AT, reprefent the forces, and S R, R Q, 

 Q T, the forces upon the angles ; for A S, A R, A Q, A T, 

 &c. are the fecants of the elevation, and the triangles A S R, 

 A R Q, AQT, are all fimilar to the triangles h B i, I C m, 

 oDp, &c. 



Cor. 7. — In every roof kept in equilibrio by the weights 

 of the rafters, if U, V, W, &c. be the centres of gravity 

 of the rafters, and alfo reprefent their weights ; then the 



A U x U 



weight prefling 

 VC x V 



BC 



vertically on B, will be 

 , and the weight on C = 



AB 



W D x W 



and fo on ; hence 

 x W 



BC 

 BC D 



WD 



CD "S.iBr S.iBi ' S.lCmX S.mCn' 



Cor. 8. — Hence, if the rafters be prifmatic figures, the 



weights on the angles B. C, D, &c. will be refpeftively as 



AB + BCBC+CDCD + DE Jr 

 , , , and lo on. 



2 2 2 



PftOP. IV. 



Given the vertical angle of a roof, and the proportion of 

 the rafters on each fide, to defcribe the roof to a given width, 

 fo that it fhall be in equilibrio. 



Let the proportion of the rafters from the top down- 

 wards be as 2, 3, 4, that is, as 4, 6, 8 ; then the weight 



4 "I - 4 

 on the vertical angle is = 4 ; on the next fucceeding 



angle 



4 + 



5 ; and on the bottom angle 



6 + 8 



2 2 



( Plate XLIII. Jig. 2. ) Now let A B C be the given angle. 

 Make B A equal to B C ; join A C, and draw D B E per- 

 pendicular to A C ; then make D B to reprefent half the 

 weight of the vertical rafter : let B D be divided into two 

 equal parts ; make B M five of thefe parts, and M E feven ; 

 join M A, M C, E A, E C ; then from any fcale of equal 

 parts, make B F and B G each two parts ; draw F H and 

 G I parallel to M A and M C, and make F H and G I 

 each equal to three equal parts of the fame fcale. Laitly, 

 draw H K and I L parallel to E A and E C, and make 

 H K, I L, each equal to four equal parts ; then K H, H F, 

 F B, B G, G I, I L, are the rafters, in the proportion re- 

 quired. 



How to reduce this proportion of figure to a given 

 width is obvious ; it is only drawing a figure, having a given 

 fide in the fame proportion as another. 



Prop. V. 



The angular points at the meeting of every two rafters of 

 a roof in equilibrio, by equal weights hung at the angles in 

 vertical directions, equidillant from each other, are in the 

 curve of a parabola. 



Let AB C D E, &c. (Plate XLIII. fg. 3.) be kept 



in equilibrio by equal weights, fufpended at the angular 

 points B, C, D, £, &c. in the equidiftant direftions B F, 

 C G, D H, E I, &c. the points A, B, C, D, E, &c. are in 

 the curve of a parabola. 



For let B F meet A N at F. Draw A K parallel to 

 D E, A L parallel to C D, and A M parallel to B C, 

 cutting F B at K, L, M. Draw B Q, C P, DO, parallel 

 to A N, cutting the middle line I E at Q, P, O. 



Then, becaufe the weights on the angles are equal, F K, 

 K L, L M, MB, are alfo equal, the firft excepted, being 

 half, or as the numbers I, 2, 2, 2 ; therefore F K, F L, 

 F M, F B, are as the odd numbers 1, 3, 5, 7 : but be- 

 caufe of the equidiftant lines B F, C G, D H, E I, &c. and 

 the parallels DO, CP, BO, AI, the triangles A F K, 

 A F L, A F M, are refpe&ively equal and fimilar to the 

 triangles DOE, C S D, BRC; therefore K F is equal 

 to E O, L F equal to D S,andD S equal to O P, M F equal to 

 C R, and C R equal to P Q : and laftly, B F is equal to Q I ; 

 therefore E O, O P, P Q, Q I, are to one another as the num. 

 bers i, 3, 5, 7 5 and E 0, E P, E Q, E I, are as the 

 fquare numbers 1, 4, 9, 16 ; but the lines O D, P C, Q B, 

 are to one another as 1, 2, 3, 4 ; therefore the abfcifies E O, 

 E P, E Q, EI, are as the fquares of the ordinates O D, 

 PC, Q B, I A, and the points A, B, C, D, E, are placed 

 in the curve of a parabola. In the fame manner it may be 

 (hewn, that this is the cafe, whatever be the number of or- 

 dinates. 



Corol. — Hence a roof of this conftruftion may be defcribed 

 to any given height and vertical angle, or to a given width 

 and height, with any number of rafters on each fide. 



Example. — To defcribe a roof with any given number of 

 rafters on each fide to a given width and height, fo that 

 all the vertical lines palling through the angular points of 

 the rafters ihall be equidiftant, and the rafters in equi- 

 librio. 



Let there be four rafters on each fide, (Plate XLIII. 

 jig. 3. ) Let I N be half the width, and I E the 

 height. Draw N T and E T parallel to I E and I N ; di- 

 vide N T into four equal parts N/, f e, e d, dT, and draw 

 d g E, e h E,y « E ; likewife divide I N into four equal 

 parts I c, c b, b a, a N, and draw eg, b k, a i, parallel to I E. 

 Join E g, g h, h i, i N, and thefe lines will be the rafters of 

 half the roof required. 



For let C B (Plate XLIII. fg. 4.) be the height, 

 and A B half the bafe. Draw C D parallel to A B, and 

 A D parallel to B C. In A B take any point a, and divide 

 A D, fo that b D may be to A D as B a to B A. Join 

 b C, draw a c parallel to B C, meeting b C at c ; and c d and 

 b e parallel to A B, cutting BC at; and d, then the tri- 

 angles C dc and C e b are fimilar. 

 Therefore Cd: C e :: c d : e b, or A B 



And by conftruaion, C<: CB::nBorc(/: ABor{( 

 Therefore C d: C B :: c d> . :AB" 



Therefore the point C is in the curve of a parabola. 



Another method may be as follows: Let BE (Plate 

 XLIII. Jig. 4.) be half the width, B C the height. Pro- 

 duce B C to F, making C F equal to C B. Divide C F into 

 four equal parts, CJ, J g, gh, hY, and ]o\n f TL,g E, h E. 

 Divide B E into four equal parts, B i, i i, i I, / E„ and draw 

 the lines i m, in, lo, parallel to B C, cutting the former 

 lines _/E, g E, h E, at m,n,o; the points m, n, 0, are in the 

 curve of a parabola. For the principles of this conftruftion, 

 the reader may confult Theorem VII. p. 123. vol. ii. Hut- 

 ton's Mathematics. 



Prop. VI. 

 To defcribe a roof with four equal rafters, that ftiall be 



