If I have an industrial sized turntable and I know the weight of the thing and the diameter and the weight of the load it’s carrying, is there a quick and dirty way to estimate the torque required to turn it at a given RPM or is this major engineering any way you slice it?
I get calls all the time with questions like, “I have a 2ft turntable that weighs a total of 10 lbs and I want to turn it at 5 rpm.”
I understand that there will be bearing friction and perhaps other considerations for a precise answer, I’m just looking to find a ballpark answer to help the backyard engineers that call me because I sell gear-motors.
Let’s assume the motor will be directly connected to the turntable shaft and we’re not going to get any torque multiplication via a chain or belt drive.
I’m thinking I should just assume that I’m trying to*** lift ***a 10 lb weight on a 2ft arm and go for 20 ft lbs of torque. I’m sure it’s overkill, but is it way out?
I will be watching this thread as well. I have a similar type thing I have to work out from time to time. The way I have been doing it is to find the center of mass on the object I am turnng and calculate how many foot pounds of KE it would take to accelerate the center of mass to my desired speed. I use feet per second. This seems to work for me but I don’t know if it is correct.
The bearing friction is not only a necessary consideration; it’s the only one. On a frictionless turntable, you could get up to any rotational speed you want, using any amount of torque you want. If the turntable is very massive or large, or the torque is very low, then it will take a long time to get up to speed, but it’ll still get there eventually.
This is true theoretically, but it’s no help in the real world. If a motor can’t get to running speed in a pretty short time, it overheats and shuts down. “Getting there eventually” isn’t an option unless you have some sort of clutch arrangement… and I’m not looking to go down that path.
For the sort of numbers you describe, I agree with Chronos. It’s all friction. Assuming a 10 lb plate with a diameter of 2 feet, and assuming your motor can deal with the heat load from a 10 second spin up time, then frictional torque will dominate as long as its above 0.1 oz-ft or so.
I am an engineer, not a physicist - so this answer is more practical than theoretical. Please use consistent units - because this where a lot of folks go wrong.
First - calculate (Wk2) i.e. the Moment of Inertia of the disk. =(MR^2)/2 in lb-ft2 . For your example 10 x 1^2/2 = 5 lb-ft2
Assume a start up time of 5 seconds - a typical start up time for a motor. Using this approximation, you can now find the torque that will be needed to accelerate the disk from 0 to 5 rpm in 5 secs.
So, ta (Acceleration time) = 5 seconds
**Delta N ** (Speed change in 5 seconds) = 5 rpm / 5 sec = 60 rpm/min
Then the accelerating torque Ta is given by
This is the torque for the turn table - for the motor shaft torque - you need to divide by the gear ratio ( if your motor speed is say 1800 then the gear ratio is 5/1800)
Add a factor of safety of 2 to 4 to account for bearing/other losses (also motors sometimes are incorrectly labeled for torque) and you should be good to go.
Another simple way will be to calculate the rotational (kinetic) energy in the disk spinning at 5 rpm. Assume the motor will have to deliver this energy in 5 second - and now you have the motor power requirement. Use a factor of 2 to 4 to account for startup current / losses - and you have the motor power.