ROOF. 



in equilibrio by the weight of the rafters ; of a given 

 width A E, Plate XLII1. fig. 5, and height F C. 



Join E C, and bifeft it in H, by a perpendicular D H G, 

 meeting A E in G ; on G, as a centre, with the diftance 

 G E or G C, defcribe the circle CEO. Draw K H I 

 parallel to F E, meeting the vertical line O C in K, and the 

 circle in I. Draw I D C, and join DE; then make the 

 fide C B A fimilar to C D E, and AB, BC, CD.DE, 

 will be the rafters of the roof required. 



For complete the parallelogram C D Q B, and join B D, 

 I F, and draw C L perpendicular to C F, and equal to 

 F G. On L, with the diftance G E, defcribe the circle 

 N I F, meeting the vertical line at N and F ; produce E D 

 to meet it alfo in M, and B C to P. 



Then becaufe K F is equal to K C, and R C equal to 

 RQ, the triangles CIF and C D Cj are fimilar ; there- 

 fore I F is parallel to D Q : and becaufe the two fegment< 

 N I F and CEO are equal to one another, the angle N I F 

 is equal to the angle CEO, equal to twice the angle C E F, 

 or twice the alternate angle E C L equal toECD + DCL, 

 but E C D is equal to half the external angle M D C, and 

 D C L is half the angle D C P equal to CDQ. There- 

 fore the angle N I F is equal to the angles MDC f CDQ 

 equal to the angle MDQ, and C F : C N :: C Q : C M ; 

 but C F and C N are equal, therefore C Q and C M are 

 equal ; but C Q is to CM as the weight on C is to the 

 weight on B ; therefore the weights on C and B are equal, 

 and the rafters A B, B C, CD, D E, are in equilibrio. 



Prop. VII. 



Suppofe it were required to conftruft a curb roof, the 

 bottom rafter being in proportion to the upper rafter, as 



2 to 3, and to a given vertical angle at the top, and to be 

 of a given width A B, Plate XLIII. fig. 6. 



Now the weight on the upper angle is to the weight on 



2HI. HI + I A , . 



the lower angle, as is to , that is, as 



2 2 



3 + 3 3 + 

 J J = 3 is to 2 — 



2 J 2 



of 6 to 5, or the half weight at II is to the bottom weight 

 at I, as 3 is to 5. 



Bifeft A B by the perpendicular C D, and make the 

 angle A E C equal to half the vertical angle, or the angle 

 E A C equal to its complement. Make ED to EC as 

 5 to 3. Join D A and D B. Take A G of any length ; 

 draw F G parallel to A E, and make A G to G F as 

 2 to 3. Draw A F H, meeting CD at H ; and H I 

 parallel to F G or E A, cutting A D at I ; make B K 

 equal to A I, and join K H ; then AIHKB is the eon- 

 tour of the roof required. This is fo evident from its con- 

 ftru&ion, that it does not require demon ftration. 



Prop. VIII. 



In any roof confirudled with two equal rafters only ; as 

 the height of the roof is to half the breadth of the building, 

 fo is half the weight of the roof to the horizontal thrufl, or 

 lateral prefTure. 



Let ABC, Plate XLIV. fig. 1, be a roof, having the 

 two equal rafters A B, BC; join the bottom of the 

 rafters A C ; draw B D perpendicular to it ; complete the 

 parallelogram B E F G, and draw E G, cutting B D in H. 

 Then, becaufe the triangles B H E and B D A are fimilar, 

 BD : DA :: BH : HE. 



Cor. 1 — Hence, in a roof with two rafters and a tie. 

 Vol. XXX- 



= 2^ ; this is in the proportion 



beam at the bottom, the tenfion, H E, of the tie-beam is 

 _ D A x B H 

 BD — ' 

 Cor. 2 — Hence, alfo, BD: B A:: BH : BE, that 

 is, as the height of the roof is to the length of the rafter, 

 fo is half the weight, reprefented by B H, of the roof to 



B A y B H 



the comprcflion of the rafters =x . 



BD 

 Cor. 3. — Half the weight of the roof, the tenfion of the 

 tie-beam, and the comprefiion of the rafters, are to one 

 another as the height of the roof, half the breadth of the 

 fpan, and the length of the rafters ; for the triangle B H E 

 is fimilar to the triangle B D A. 



Prop. IX. 



If a rafter bear any weight, or have a weight uniformly 

 diffufed over it, the force tending to break it is equal to the 

 cofine of elevation, multiplied into the weight, divided by 

 radius. 



Let A B, B C, fig. 2, be two equal rafters ; join A C, 

 draw B G perpendicular to it, meeting it in G ; and let the 

 weight W be fufpended by the firing D E. Draw D F 

 perpendicular to A B, and E F parallel to it ; then if D E 

 reprefent the weight, D F will reprefent the force tending 

 to break the rafter ; F E its tendency to pufh it from B 

 towards A. 



Now, becaufe E F is parallel to A B, the alternate 

 angles A D E and D E F are equal, and the angles D F E 

 and A G B are right angles ; the triangles EDF and 

 BAG are iimilar ; therefore AB:AG:: DE:DF 



AG x DE . . 

 = -r-p » tnat ls > as radius is to the cofine of ele- 

 vation, fo is the force employed to its tendency to break the 



r .__ ., . r, r iat- -O.T. cof. elevation x DE 



rafter, that is, as R : cof. : : D E : D F = . 



R 



Cor. 1.— Hence the weight employed, the prefTure in a 



direction of the length of the rafter at A, the tendency to 



break it, are as radius, the fine, and cofine of elevation. 



r -a r nr cof - elevation X D E . 



Cor. 2. — Became Dr= 5 , and be- 



R 



caufe the flrefs is as the length when the weight is given, 



the flrefs is as the cofine of elevation multiplied into the 



weight, and this produft multiplied into the length of the 



rafter, the radius being a conflant quantity. 



Prop. X. 



To prevent the rafters of a roof, with a tie-beam, from 

 bending in the middle, and to remove lateral preffiire from 

 the walls, when there is no beam. 



A variety of methods may be ufed for this purpofe ; but 

 the befl are thofe where the fhortell and lcafl quantity of 

 timber arc employed without producing a tranfverfe flrain 

 upon any part. When a roof confifts of two rafters only, 

 no part of the rafters can be loaded between their extre- 

 mities, nor indeed will they bear their own weight without 

 producing a concavity on the upper fide, which will be 

 greater as the length of the rafter and weight applied to it 

 arc greater. Now becaufe the fhortcr the rafters arc at the 

 fame elevation, the greater the weight they will bear, and 

 be more able to fupport their own weight ; the thing to be 

 done is to fupport them by a fuflicicnt number of fixed 

 3 R points 



