162 SYNCHRONOUS MOTORS 



will be observed that there is a limit to the possible increase of OBo, 

 namely, when is 90 and OBo = DB, beyond which the tangential 

 crank effort OBo decreases, to zero when 6 =180, to a negative 

 maximum when 0= 270, and_back to zero again when 6 = 360 or o. 

 Thus although the tension OB goes on increasing after has passed 

 the 90 point, the lever arm decreases more rapidly than the tension 

 increases, i.e., the tangential component (OBo) of the tension decreases. 



Assume now that with constant load-torque the driven crank pin 

 radius is altered. The tangential crank effort, OBo, must remain the 

 same, and the driven crank pin B must lie somewhere on the line 

 BoBoo. Thus for a given driver crank radius, the angle and the 

 tension OB, depend upon the torque demanded by the load (which 

 is proportional to OBo), and upon the driven crank pin radius DB. 

 For each value of the load torque and of OBo, the tension will be a 

 minimum for that value of the driven crank pin radius DB, which 

 causes B to fall at .Bo- It is obvious that this particular value of 

 DB varies with the load. 



If now the driver crank pin radius DO be increased, the value of 

 OBo for a given load will be decreased and vice versa, since their 

 product must be constant. 



It has been assumed thus far that a fixed relation exists between 

 the crank pin displacement OB and the coupling tension; but while 

 this is true for a given case, it is evident that a different elastic band 

 or spring could be substituted which would have a different elastic 

 coefficient, and with which the angle would be quite different for 

 the same load. 



The angle will be hereafter referred to as the coupling angle. 

 It obviously depends upon the load torque, the two crank pin radii, 

 and the elastic coefficient of the coupling. 



Let p= tension in pounds = kXOB, where k is the coefficient of the 

 elastic band. Then if DO and DB are measured in feet, the torque 

 in pound-feet will be 



T=kXOBoXDO=kXOB cos tyXDO=kXDB sin QXDO. 



Thus for any given value of the crank pin radii, the torque varies 

 as the sin 0. 



A comparison of the diagrams of Figs. 81 and 82 will show their 

 exact mechanical equivalence. In both cases the torque is propor- 

 tional to OBo for a given value of OD; therefore the relation of the 

 torque to and to DB is exactly the same in the two cases. In 



