150 II IK DIRECT-rrUHKNT MOTOR (II. VII 



The acceleration a in f.p.s. per second produced by a force 

 of T pounds acting on a mass of W tons is given by 



32-2 T _ 1 T 

 2240 W 69-5 W 



If the centre of gravity of this mass rotate in a circle of d 

 inches diameter, and if the motion is derived from a motor 

 with a velocity ratio of v, we have 



(82). 



Inserting this value of T in Equation 81 we obtain the 

 following : 



rt = -405-10- 4 ?- 5" f.p.s. per second ...... (83.) 



d W 



or a=288-10- 4 V ~, ^ f.p.s. per second ...... (84). 



where t (l is the torque on the motor shaft available for 

 acceleration. 



Now one revolution of the motor corresponds to a 



motion of the rotating mass of - - feet. Hence the ac- 



12 v 



celeration of the motor is given by 



?y C 



a=0'155Jlf ''- r.p.s. per second ...... (85). 



d 2 W 



Combining this with Equation 80 we get the value of /,-, 

 namely 



