30 
MR. G. T. WALKER ON BOOMERANGS. 
= - s 
T 
Due to the twisting we have 
Zo = — y {%{?■ + + 2 (ccv — yu) coo + {x^ + y^) coo^] [{u — yoj^) cos (f) 
— (v + sin (f>] 
+ (42 + 4') ^ 3 } + ( 3 / 4 ' + 243 + 4 s) C. 32 ,, + 2 (4V + 4V) coo 
+ (4^ + 4'^ + 4'^ + 4o^) ^^3 
The resultanl. couple about the first axis will be the integral over the surface of 
— Kt\ {lu^ + Qm\ + — {y — s sin <^) X + i\^) {hi^ + mi\ + iCj). 
As before this may be divided into three portions, of which the first, on a plane 
boomerang, is 
^0 = “ 4^ “ ^^2 + yo)^) [k (n d- xo)^) 4- X^ {ti^ d- 4 - 2 {xv — yii) Wg 
in which terms iu .s® have been omitted as before. 
Therefore, 
F(j = ( — KV 2\k^^O}.^u) -j- SwoWo {kk2~ — 2Xk/2<) 
— XSw, (?f2 -I- ^,2) -j_ 2k^^(O^V d- (k-o^ d- K~^) 0^3^] • 
Y, = - b/s 
while 
F.= 
[K-ojg COS (f> ^ (22/W3 {u sin ^ -f cos <f)) -f- 2^3% [x cos (f) — y sin 
— sin {lY d- 4" 2a)2vx — + x'^ojf -fi 
X [(n — 2/W3) cos (/» — ('?’ 4 " ^(^3) sin 0 ] 
- - [K’niyaojgd- 2Xa;3{2??22^a^4- 4-^b<®d-'?’^) (4>^4^d“'>^2^)‘^3d-wVv} 
-f- X-yaj 3 ^( 3 wf3“^ — 2??i ) 4 4* 34 " 3 »2y^)], 
y [o^S [ — iiy cos cf) {Xy [v'^ d~ 'V' + 2 (o^vx -\- x'^oj^- -}" y^<^3^)] 
+ y ("3?/ cos (fi -\- V sin cf) -j- W3X sin cf)) \ k {v -\- oj^x) — 2\oj^uy^}J 
kS 
= , + (z'Z + V) + (V + 4’) <"3‘i 
xs« 
^ (4^ 0^' + H“ 2 (4^ d" 4'") ‘^ 3 '^’ + (4'^ + 24 ^ + 34o^) ^ 3 ^} 
