300 Proceedings of the Royal Society of Edinburgh. [Sess. 
The components of these forces along the axis of 0 and y and the 
moment about O are for the general case stated above as follows : — 
X = P b 2 sin a/2 - pga 2 cos (a - /3) sin 2a cos f $/ 2 cos (a + /3) , 
Y = - P b 2 cos a/2 - pga 2 cos (a — /3) sin 2a sin /?/2 cos (a 4- /3) 
M = - P# 3 /6 - pga 3 cos 2 (a - ft) sin 2a tan (3 / 3 cos (a + /3) . 
The general solution in plane polars is as follows : — 
rr = P?’[(cos 2 a - sin 2 #) sin 0 - tan 3 a(cos 2 # - sin 2 a) cos #]/4 sin 3 a 
- [s(cos 2 a - sin 2 #) sin # - c( cos 2 # - sin 2 a) cos # + 2 cos (/ 3 - #)] 
Lj 
+ P#[sin 2a - sin 2# - 2 cos 2a(a + #)]/2(sin 2a - 2a cos 2a) 
- 2Xr _1 cos #/( 2a + sin 2a) - 2 Y r~ x sin #/( 2a - sin 2a) 
+ 2M>~ 2 sin 2#/(sin 2a - 2a cos 2a) , 
00= — Pr[(3 sin 2 a - sin 2 #) sin # + tan 3 a(3 cos 2 a — cos 2 #) cos #]/4 sin 3 a 
+ [s(3 sin 2 a - sin 2 #) 4- c ( 3 cos 2 a - cos 2 #) cos # - 2 cos ((3 - #)] 
4- P#[sin 2a 4- sin 2# - 2 cos 2a(a + #)]/2(sin 2a - 2a cos 2a) , 
rO — Pr[sin 2 a - sin 2 #) cos # 4- tan 3 a(cos 2 # - cos 2 a) sin #]/4 sin 3 a 
- [s(sin 2 a - sin 2 #) cos # 4- c(cos 2 # - cos 2 a) sin #] 
4- P6[cos 2a - cos 2#j/2(sin 2a - 2a cos 2a) 
+ Mr _2 [cos 2# - cos 2a]/(sin 2a - 2a cos 2a) , 
where 
s = sin (3 j t>in 2 a ? and c = cos /3/cos 2 a . 
The solution in rectangular co-ordinates is : — 
xx = - -^-[(3 sin 2 a - l)2//sin 3 a + aj/cos a] - PjL [sin /3.cot 2 a y + cos /3.x] 
— [sin 2a - 2a cos 2a — 2 cos 2a(# 4- ^z//?‘ 2 )]/(sin 2a - 2a cos 2a) 
2 
- 2Xz 3 /?‘ 4 (2a 4- sin 2a) - 2Yx 2 y/r 4 (2a - sin 2a) 
- 2M[cos 2a xy(x 2 4- y 2 ) — 3a: 3 ?/ 4- a??/ 3 ]/r 6 (sin 2a — 2a cos 2a) , 
yy = - — [2// sin a 4- (3 cos 2 a - l)a?/cos 3 a] - ^ [cos /3 tan 2 a x + sin f3.y] 
4 2 
P# 
(32) 
(33) 
[sin 2a - 2a cos 2a - 2 cos 2a(# - xy/r 2 )~\/( sin 2a - 2a cos 2a) \ . (34) 
- 2 Xxy 2 /r 4 (2a + sin 2a) - 2Y?/ 3 /r 4 (2a - sin 2a) 
4- 2M[cos 2 a xy{x 2 4 - y 2 ) 4 - 3 a?// 3 - x B y] r 6 (sin 2 a - 2 a cos 2 a) , 
xy = -^-[sc/ sin a 4 - y/ cos a ]—E£.[x sin (3 4- y cos f3] 
4- ^ [cos 2a(a? 2 - y 2 ) /r 2 - 1]/ (sin 2a - 2a cos 2a) 
- 2Xx 2 y/r 4 (2a + sin 2 a) - 2Yxy 2 / ? ,4 ( 2 a - sin 2 a) 
4- M[cos 2a(x 4 - y 4 ) 4- 6 x 2 y 2 — x 4 - ;?/ 4 ] / r 6 (sin 2 a - 2 a cos 2 a) . 
