394 PRACTICAL MATHEMATICS 



Then, resolving vertically, 



A 2 = e {(2/i + 2/7 - 2/4 - 2/io + 2/2 + 2/8 - 2/5 ~ 2/ii) sin 60 } 



or A 2 = -{(2/i + 2/2 + 2/7 + 2/8 - 2/4 - 2/5 - 2/io - 2/ii) sin 60 } 

 and resolving horizontally, 



B 2 = g {(2/o + 2/6 ~ 2/3 ~ 2/9) + (2/i + 2/7 - 2/4 - 2/io) cos 60 



- (2/2 + 2/8 - 2/5 - 2/ii) cos 60 } 

 1 



1 f 2 ' 1 r 2 ' 



(3) Now A 3 = I ?/ sin 30 d0, and B 3 = - 1 y cos 30 d0 

 Hence A 3 = - {sum of the ordinates of the 0, y sin 30 curve} 

 and B 3 = - {sum of the ordinates of the 0, y cos 30 curve} 



6 



or A 3 - \ 



and Bo = -\u n 



3 gltfO 



FIG. 130. 



If from the point O radial lines are drawn at intervals of 90 C 

 and the ordinates y , y v etc., are measured along these lines. 



