CONSTRUCTION FOR TORQUE 193 



or 



T v QB 1 n _ 



la = -V2/-\ ~ T^HT* *J/j 



OB sin QPB 



QB 



where K< is a constant whose value is K4 = K 2 ,^ Tl 'f^^ . t 



' ' I > sin 



Hence 



I 8 a = K4 3 . OZ a 



Again, by (12), s = K 5 . OS, where K 5 = ^ T a np ,. Substituting 

 these values for I a 2 and s in (13), we find 



KIT 2 O72 

 ... _ 8-^4 UZ| _ Tf 



Kg ' OS ~ ^OS 



K K 2 



where K 6 is the constant *L . Let ZT be drawn || SO. Then since 



v = OZ x 7 1 = OZ cos ZOS = OZ cos OZT = ZT, we have 



Oo Oo 



Tf VT 1 /1 A\ 



J\Q . /-I ( L4 ) 



i.e. the torque is pi'oportional to the length of the perpendicular let fall 

 from Z on OK. 



It is evident that the maximum torque which the motor is capable 

 of exerting is given by the length XY of the perpendicular erected 

 at the middle point X of OK. 



Since (Fig. 128) FB = ,, =, and AF = AB - FB 



sin 



r M 2 \ 



= p I LI y- ) , we see that 



m 



104 (15) 



La 



AF 

 FB 

 and 



AF 



So long, therefore, as LI, La, and M remain unaltered, AF and FB, 

 and the circle described on FB as diameter, remain unchanged. We 



* The straight line joining O and Z is, for the sake of clearness, not shown in the 

 diagram. 



t The angle QPB remains constant, because QB is a fixed chord of the circle 

 described by 1'. 







