Now I have a 1/2 hp , 247 in. lb torque, ac gearmotor that bogs down under the loads I need it for , and I need a bigger motor. I found another
1/2 hp motor that has 487 in lbs , but am not certain that the higher torque rating alone is enough. So the question is : what is the relationship between hp, torque, and actual ability to perform high-load operations ?. This is (to be) set up powering a rolling mill. It was suggested that I get a .75 or 1 hp motor , but...thanks, Dar
If you can accept about half the speed of your current motor, it is OK. You would achieve the same if you could double the gear ratio on your existing motor.
It is a very simple and direct relationship: Power is torque multiplied by rotational speed. Depending in which measurement units you choose, you may have to multiply by some constant.
In other words: If you keep the power constant and gear the motor to run half as fast, it will produce twice the torque. And vice versa.
Dar
You need to know how much torque is necessary to drive your load, and how fast you want it to rotate. Add some torque margin (like 50%) and then pick a gear motor with matching specs.
Measure how much torque it takes to crank your mill with the heaviest load you expect to use.
Ignoring gear efficiency losses hp = (RPM * torque_in*lb) / (5252*12)
I calculated the efficiency on one gear motor listing as 69%, others may vary.
CarlBoyd
There is a table of torque, horsepower and RPM in the "POCKET REF", page 126. It begins at 1 HP but you could divide the 5 and 7.5 entries by 10. Torque in inch-Lbs =3D (HP * 63025) / RPM.
Turn the mill with a torque wrench to see how much you need.
jw
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