Torque Requirements

[Soulbear]

HI Everyone,
Got a little query regarding Torque Requirements for my DIY Cutting Lathe Turntable. After several dead ends with respect of powerful motors (I think I may have gone a little overboard specification wise, some that I looked at would probably have raised the Titanic!!) At the risk of looking stupid I have settled on a 2600RPM synchronous geared motor rated at 32W. With its 41.66:1 reduction it gives an output speed 0f 62.4RPM further reduced via belt pulley to the Turntable Drive Spindle. I've made the following assumptions calculating that 32WX41.66 gives a theoretical usable output of 1.33KW Assuming at a worst case a 40% efficiency through gearbox, bearings and drive pulley I should have around .533KW delivered at the spindle, which to my limited understanding seems plenty. Have I missed anything ? Thanks, Soulbear

[markrob]

Hi Yes,

The gear reduction does not in any way change the power output of the system. In fact total shaft power power available will be reduced by the gear box efficiency. Torque is NOT power. Torque times RPM is proportional to horsepower which = about 750 Watts. So, using a motor with a 100% efficient gear box, if you reduce the shaft speed, you increase torque in proportion to the reduction ratio, but at the same time, you reduce shaft speed by the same amount and power remains the same. In the real world, you will lose some torque due to losses. For reference, the 6N used a 1/25 HP motor (about 30 watts). This is in the range you specified. However, note that the 6N platter also needs to drive the overhead. If you don't need to do this, you may be able to get away with less power. If the cost and size is not too big of a deal, you are better off oversizing.

Mark

[Soulbear]

Thanks for the input Mark, the 10 T.P.I. CutterHead Feed Leadscrew will be powered with a variable speed 24VDC Motor driving it, so the Turntable Motor only has the single duty. Soulbear