603 



2. Calculation of relief pressure . 



In order to simplify (18) we take polar coordinates X, a with origin P (Figure 5). 

 Then using (21) we obtain 



dS' = X dx cH (22) 



so that after integration Dy part^ with respect to X and use of geometry of Figure 5 we find 



2 77 



Jo J 



" (T - -^j (1 - Xj - x^ + 2 xxj cos a) dX d a (23) 



.77 (1- Xj') J- (T) - 32 TT-?! (T) 



2 7r 



i ( T - i.) COS ^ d a 



,2 77 



(2«) 



i^ {T-il) 6 a 



= 4 77 (1 - Xj ) ■!• (T) - 32Tr-ij (T) + 



2 -^jd -i^) COS 5 + i{T -^) cos^ 



(25) 



the integral being taken round the boundary. 

 ^. Contribution of relief pressure to equation of motion . 



From equations (20) and (25) we have 

 ,1 



I = STT^ -i- (T) 



(1 - Xj^) Xj dXj - 6U 77^ ^j (T) 



(1 - x^^) Xj dXj 



(1 - Xj') II 



■Z (T - ^) COS^ ff + 2 -Jj (T - i^) cos 5 



(26) 



the first two terms givin'j a contribution 



±2^. i' (T) - 16 77^ 



(T) - 16 Tr''-?^ (T) 



(27) 



Interchanging the line and surface integrals and using syrametry the last term of (26) 

 gives a contribution 



4 (X - ^) cos^ e + 2 -2, (T - i^) cos e , 

 L_, , i 2 (1 _ X, 2) dS- 



(28) 



where this integral is to Be evaluated for any given element dS of the circumference and will 

 by symmetry be the same for any element on the circumference. 



Taking polar coordinates (X', 6) as in Figure 6 then 

 /ri'' 2 /2 cos 9 



16 77- 



{ -i (T - ^) cos^ 5 + 2 -ij (T - i^) cos 5 } (1 - x^^) dX' d e 



(29) 



using the fact that 



Xj^ = X'^ + 1 - 2 X' cos i9 (30) 



we finally have after some integration by parts and thence using (27) 



I, 



