184 Mr. Tarn on the Stability of Domes. [May 31, 



If ff is the centre of gravity of the portion of the rib whose weight is P, 

 and G its projection on OA ; then if OG=z, we find the value of z from 



JJJr^ . sin^ . cos . dr . dd . d(f) 



Jjjr^ , sin 6. dr. d^ . d(j> 

 c^r, 1^ [[r' i (I -cos 26) dr. dd 

 1^ * ^^r\smd.dr.dd ' ^ ^ 

 And since 6 is small, we may consider ^^^-l^^j, nearly ; in which case 



^3 R"-y"0-jsin 20 

 ■^~8K'—r' 1-COS0 ' 

 Now X is the perpendicular distance of from E ; therefore 



sin d—z. 



rr.. . ^ . nxR'-' ^ f ' n 3R^-r*0-isin20l 



Therefore P.=a . ^ (1 -cos 0)-^ X | . sm 0-- -^^—^ | , 



or 



P^=?.l-? I Sr (R^-r^) sin (1 -cos 0)-3 (R'-r') (0-| sin 20) | . 

 And since y=R-rcos0, and N = P^, 



N-i^ 8r (n^-r') sin (1 -cos 0)~3 (U'-r') (d-j sin 20) ^ 

 24 R— r cos * 



I will take ^ as the circular measure of an angle of 2°, or (/)=^= '0349066, 



in which case we have 



N= -001454 a 8KR^-^0 sin (1 -cos 0)-3 (R^-rQ (0-^ sin 20) 



R — rcos0 



= •001454 ax N', say. 

 We have now to find what value of will make N' a maximum, and as 

 the ordinary rules for maxima will not apply to this expression, we must 

 find it by calculating N' for different values of 0. I have done this for 

 the case when r=10, R=ll, and find that N is greatest when 0=70°; 

 thus 



0=69° N'=506-034 



0=70° N'=506-241 



0==71° N'=506-0i8. 



We have now to substitute in N the values of 0, sin 0, &c. when the angle 



is 70°, or when 0=35 ^=1*22173, sin 0=-93969, cos 0=-342O2, sin 20 



= •64279. 



This reduces the expression for N to 



N-^ -007192 r (R^-r^)--0039273 (U'-r') . 



R--34202r • • • V ; 



We can now transpose N and P to the point E ; and in order to find the 



