306 SUPPLEMENT TO CHAP. XIV. [CHAP. V. 



From (57), since f or <j> = 0, A x = 0, we have, inserting the value of A 0, 

 above, 



M r 



A X = 



^[^'Jo^'jo 00 *'*'! 



HrT A A 1 



1 (1 cos <j>) I (1 cos 0) d $ I (1 cos 0)* d I 



E I L J J J 



Hr + Mo A 



H cos / a 6 r tt sin o. 



BA V^ 



For ^ = a, A = 0, hence from the first of these expressions 



r _ A- d +- r** 



* Jo >8 ^ + A J> 



If the distance at which the horizontal thrust H acts from the crown is 

 0) we have M = H , whence we see at once that e Q is the fraction in 

 (63) multiplied by r. For <j> = 0, A a? must also be zero, and -we thus 

 obtain another relation between M and H which does not contain A. If 

 we multiply the expression thus obtained by r cos a, and then subtract the 

 result from that previously obtained f or $ = 0, A <j> = 0, we have 



M r C* Hr 2 r /"" /"" T 



-= I (cos0)d$4--jj-j I / (1 cos0)cf0 / (1 cos0) 2 d0 j= ttsino 



.... (64) 

 Performing the integrations jndicated in (63), we have (Art. 7) 



(1 + K) a 



(66) 



where, as before, * = - . 

 Ar* 



From (64) we obtain 



M, r sin a iHr* (a 2 sin a + sin a cos a) = E 1 1 1 sin a. 

 Inserting the value of M above, we have 



= _ _ (66) 



r* [(1 + K) (a* + a sin a cos a) 2 sin* a] 

 and hence 



_ 8 E I g < sin a [(1 + K) a sin a] . 



r [(1 + ) (q* + q sin q cos q) 2 sin 2 q] 

 From these two we obtain for the point of application of H 



(68) 



