192 ALTERNATING CURRENTS 



- SpLi2 



tan t/2 = ~ 

 or 



5 = -5-.tan0 2 . (11) 



so that the slip is proportional to tan 0%. 



In Fig. 129, join OF, and draw OS_LOF. Join BP, and produce 

 it until it intersects OS at S. Since the angle FPS is a right angle, 

 both the triangles FSP and FSO will be right-angled, and hence a 

 circle described on FS as diameter will pass through P and 0. Since 

 the angle OFS stands on the same arc of this circle as OPS, it follows 



OS 

 that /OFS = 2 , hence tan 2 = ~,-, and by (11) 



but OF being constant, it follows that the slip is proportional to OS. 



We may notice that the point K where OF intersects the circle 

 corresponds to an infinite value of the slip, BK being parallel to OS, 

 i.e. intersecting it at an infinite distance. 



We shall next explain the graphical construction for the torque. 



On OB as diameter, describe a semicircle intersecting BS in Z, 

 and from Z let fall the perpendicular ZT on OK. 



Equation (3), 107, given above may be written 



T = K 3y ........ (13) 



K 3 being a constant whose value (if the torque is expressed in 

 Ib.-feet) is given by 



K 3 = 



P 

 Now in the triangle QPB 



QP _ sin QBP 

 QB ~ sin QPB 



so that 



