RE: [CAD_CAM_EDM_DRO] Re: Ounce-Inches, WAS Frozen stepper motor shaft

The torque required to provide a force of 10 lbs at a distance from the
center of rotation of 2 ft is equal to 10lbs * 2 ft = 20 ft-lb.

This same torque will provide a force of 20 lbs at a distance of 1 ft.

Or 40 lbs at a distance of 1/2 ft.

In each case, you multiply the force by the distance, and you get the
torque.

Note that by convention, torque is usually written as "ft-lb", while the
identically appearing units of work are usually wriotten as lb-ft. When
you are using other units, though, torque is usually seen written as
"oz-in", for example. Technically, either is correct, because a time b =
b times a.


Andy Wander
Verrex Corporation

PS the units you referred to, lb/ft, are used, as far as I know, only in
describing a spring constant.

-----Original Message-----
From: CAD_CAM_EDM_DRO@yahoogroups.com
[mailto:CAD_CAM_EDM_DRO@yahoogroups.com] On Behalf Of Bill Yates
Sent: Sunday, January 22, 2006 11:14 AM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: [CAD_CAM_EDM_DRO] Re: Ounce-Inches, WAS Frozen stepper motor
shaft

Hello Andy,

I'm sorry but I am getting a bit confused now...

For example, if a motor is just able to overcome a force of 10lb applied
at a distance of 2ft from its axis, then how would you calculate the
motor's torque?


Kind regards,

Bill

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