ANIMAL MECHANICS. 



243 



rotation round the socket S; and let the point a move to 

 the point b; and let OX be the bisector of the vertical angle 

 AOB. Also, let 



OSa = w, 

 SOX = 0, OS = 

 XOa = 0, £a=A 



Draw and let fall 5p perpendicular upon act' We now 

 have 



d (I- I) = ap ■-= ab x cos = ab x sin ( SaO), 



but 

 and 



sin (SaO) = sin (w + 0 + 0) ; 

 therefore, 



3 (7 - /') = A Si*) sin 4- 0 + 0) ; 



but 



sin (o> + 0 + 0) 

 Therefore, finally, 



§ (/ - = ^gft, sin (0 4 0). 



The work done by the muscle during the rotation through dw, 

 is, therefore, 



S(l-J)d9 = k^ 



sin (0 + 0) e?0. 



J- 



The integration of this expression gives us for the work done 



jg (I - I') dO = 2kSto sin 0 sin 0 ( 44 ) 



This result will be a maximum when 0 = 90 0 . Hence we ob- 

 tain the following solution of the problem proposed : 



Produce the extreme fibres A A' and BB, to meet in the point 

 0, and draw OX, the bisector of the vertical angle of the triangle 

 R 2 



