ANIMAL MECHANICS. 379 



Constants of Hyperboloid. 

 2C = 4.06 in. 

 20 = 91 0 1$'. 



The angle between the bones is so nearly 90 0 that we may 

 assume, as before, that A = c is a root of equation (92). 

 To determine I and I', we have 



P2 2 ~p 2 = 2a a + 2a' a' 4- {a % + a /2 ), 



which gives us 



5.36 a + 6.90 a' - 10. 17 =-- o, 

 a = - 0.78 a + 1.47. 



Substituting this value of a 7 in the expression for p 2 } we 

 obtain 



1. 61 a 2 - 2.29 a - 9.87 = o, 



and finally, 



( + c.qo in. , ( - 1. 10 in. 



a = „ a' = 



On inspection of the bones it is evident that the latter pair 

 of roots are those required in the problem. Hence we find — 



a = 



- 1.87 in. 



a' 



= + 2.93 in. 



«i = 



- °-53 » 



a/ 



= + 4-66 „ 



ct 2 = 



+ 0.81 „ 



a-2 / 



= + 6.39 „ 



a 



P ~ 



- °-35° 



a' 

 P 



= + 0.549 



«1 _ 



Pi ~ 



- 0.083 



a.i 



= + 0.728 



a 2 _ 



+ 0.107 



a 2 ' 



= + 0.841 





- 0.326 



<^ 



\P/ 



= + 2*118 



