casselli Posts: 1, Reputation: 1 New Member #1 Sep 13, 2010, 10:38 AM
How do I figure out how much torque I need to move a turntable measuring 15 inches
The turntable measures 15 inches and will have about 20 lbs of distributed load. I want to drive it with a dc gearhear motor via a belt to the central shaft which measures 1 inch in diameter.
 ebaines Posts: 12,132, Reputation: 1307 Expert #2 Sep 13, 2010, 11:43 AM

If you mean that the turntable is 15 inch diameter, and is oriented vertically to lift 20 pounds of load carried on the rim (like a hoist), then the torque is 15/2 inches x 20 pound = 150 pound-feet. You would actually need a bit more to acount for start-up loads (how quickly mist it spin up to speed with this load) and friction. However, if you mean that the turntable is laying horizontally, then all you need worry about is fricton and how fast you want it to be able to spin up to speed (rotational acceleration). So please clarify how this turn table is being used.

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