264? Archdeacon Pratt on the Stability of Arches, 



I have kept a 2 and a 3 general that I might obtain these equa- 

 tions by differentiation. Now consider them equal, and obtain 

 Bac 3 and 8a 4 in terms of Ba x and Sa 2 by these equations. Mul- 

 tiply the first by cos a 2 or cos a 3 , and the second by sin a 2 or 

 sin a 3 , and subtract, 



.\ a 1 sin(a 1 — ot 2 )Bu l + a 4 ^in (a 3 + a 4 )S« 4 = 0, 



. . OCt 4 — : - -0«,. 



a 4 sin(« 8 + a 4 ) l 

 Again, multiply the two equations by cos a 4 , sin a 4 , and add; then 

 a x sin [u l + a 4 )5« l + (« 2 S# 2 + ^3 ^ a a) sm ( a a + a <0 = ^ 

 ^ fl 2 j, a x sin (a! + « 4 ) * 



c 3 a 3 sm(« 3 4-a 4 ) 



Now differentiate the expression for V, observing that /3 { . . . are 

 constant, and substitute the values of Sa 4 and 8a, 3 ; also substi- 

 tute h l £L l . . . which have been explained above. Hence 



(W 1 + W 2 + W 3 + W 4 )5V 



=W 1 A 1 S^H- W^Hj^H- V« 2 ) + W 8 (H 4 S« 4 -^8« 8 ) + W 4 /^ 4 



= ^{w A + w iHl - (w A+ w 3 h 4 ) J "gzsj 



Suppose that AB and ED are produced to meet in the point 

 F (not drawn in the diagram), and that the angles of the triangle 

 FBD are called F, B, D. Then the above formula becomes 



(W 1 + W 2 +AV 3 +W 4 )SV 



Ba x and Ba 2 are independent and arbitrary variations of a l and 

 et 2 : Bix l cannot be positive and Bu 2 cannot be negative, owing to the 

 structure of the arch and the way in which the openings take place. 

 Take the case where 8« 2 =0, that is, suppose the point C slides 

 slightly to the right, along CD. Then BV, or the variation of 

 height of the centre of gravity of the four weights (measured 

 upwards), will be positive or negative according as 



■xtt ^1 Txr Hi , nr ho sin F . . . /, Tr h. ^ T H A sin B 



W 2 -i + W 2 — > +TV 3 -^ -v-=r is < or > ( W 4 -^ + W 3 — % )-— fv 



!:«! 2 «! d a 3 smD \ 4 « 4 d « 4 /sinD 



