CH. VIII THE FOECE FACTOR 173 



at 200 feet per minute, the tension of the line being 

 125 volts, the velocity ratio 75, the friction 1,400 pounds 

 at the rim of the rope drum, the diameter of which 

 is 36 inches. The resistance of the motor is 0'05 ohm. 

 Inserting these values in Equation 20, we find the in- 

 duction factor to be 4'45 and the final current 130 am- 

 peres. The force factor required simply to drive the 

 lift is thus 578 dynes. Since the total pull at the rim of 

 the rope drum is 3,400 pounds, this motor would develop 

 20 horse-power when running at full speed. 



If the lift had to act always at full speed, it would be 

 sufficient to describe the motor as a 20 horse-power 

 motor ; but this would give no information as to the 

 ability of the motor to accelerate, since at the moment 

 of starting the horse-power is nothing. We have then 

 to find the force factor required to start up the car, say, 

 in two seconds. To get up a speed of 200 f.p.m. in two 

 seconds we require an acceleration of 1'67 f.p.s. per second. 

 If we suppose that the total mass to be moved is five tons, 

 the accelerating current as given by Equation 83 is 22 

 amperes, making a total current at the start of 152 

 amperes. The force factor is therefore 676 dynes, or 17 

 per cent, greater than that required to raise the car when 

 once started. 



Suppose now that we had to reduce the time of starting 

 by one half, and to start up in one second. One way of 

 stating the case would be to say that we needed a ' more 

 powerful ' motor, but so long as the maximum load, the 

 friction, and the final speed remain the same, the maximum 

 horse-power required remains unaltered, so that we do 

 not need more power, but more force ; we require a motor 

 with a higher force factor. 



