CH. X TIME CURVES 225 



travelled is 1,309 feet, and the time is 51 '7 seconds. Full 

 speed is 37 '6 feet per second, and the rest of the distance, 

 717 feet, is covered at this speed in 19 seconds, making 

 in all 82 seconds. 



Example 52. A lift has to be designed to start 

 from rest and raise an unbalanced weight of 1,500 pounds 

 through 15 feet; the frictional torque on the drum 

 axle is known to be 8,000 inch-pounds ; the total mass to 

 be moved is 2-5 tons ; the tension of the line is 125 volts ; 

 the accelerating current is 25 amperes ; the drop at full 

 speed is not to exceed 10 volts ; the diameter of the rope 

 drum is 36 inches, and the velocity ratio is fixed at 70. 

 To find the induction factor that will cover the distance in 

 the shortest time. Using Equation 102, we find that the 

 induction factor has to be 2'72. The acceleration is 

 2 g 14 f.p.s. per second, full speed is 5*7 feet per second, 

 and the total time is 4 seconds. The torque on the 

 drum axle due to the unbalanced load of 1,500 pounds, 

 together with the frictional torque, make up 35,000 inch- 

 pounds, giving a final current of 131 amperes. As the 

 drop is limited to 10 volts, the resistance of the motor 

 must be 0'0765 ohm. The total current at the start 

 will be 156 amperes. 



If an induction factor of twice the above value were 

 taken, we could reduce the final current to 65 '5 am- 

 peres. The time would be increased to 5'6 seconds. 

 For M=8'16 the running current would be 43'7 amperes, 

 and the time 10 seconds. 



