CIM motor + Electrical outlet = ?

[Vikesrock]

Originally Posted by Ryutso

Well what I planned to do was use 4 similar motors to drive a winch of sorts, that would lift my LCD TV up and down a stationary pole that I have in my room.

I asked around and found that the CIM's would lift the appropriate 25 pounds that my TV and 4 CIM's weigh. The only problem is, I don't want to have a car battery sitting at the foot of my bed.

CIM peak power = 321 W
321W * 4 = 1284 W
Assuming gearbox efficiency of something extremely conservative like .5
1284W*.5 = 642W

25 lbs * 10 ft (tallest most bedroom ceilings would be) = 339 joules
Power = Work /Time
Time = Work / Power
339 Joules / 642 W = .53 seconds

This means that at half of peak power 4 CIMS will lift a TV of 0 height from your floor to your ceiling in .53 seconds. I guess my point is, how did you decide you needed 4 CIMs for this job?