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

On Sun, 4 Jan 2009 01:51:53 -0800 (PST), Dar
wrote:

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.

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.

So the question is : what is the relationship between hp,
torque, and actual
ability to perform high-load operations

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.
--
RoRo

Dar wrote:
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

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

On Jan 4, 4:51*am, Dar wrote:
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

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 = (HP * 63025) / RPM.

Turn the mill with a torque wrench to see how much you need.

jw

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