Re: [CAD_CAM_EDM_DRO] Re: Thinking of retrofitting

Fantastic definition of how lb-ft oz-ft oz-in etc works.
It's all starting to come back now.

Thanks Mike.

Regards Jess

----- Original Message -----
From: "Michael Johnston" <mike_johnston@...>
To: <CAD_CAM_EDM_DRO@yahoogroups.com>
Sent: Thursday, April 29, 2004 4:47 AM
Subject: RE: [CAD_CAM_EDM_DRO] Re: Thinking of retrofitting


> Torque is a force (oz, lbs, Newtons, etc.) applied over a distance (in,
ft,
> meter). To put it another way, suppose you have a motor with a 12 inch
arm
> attached to the shaft and the arm is currently horizontal. If you hang a
1
> oz weight on the end of the stick, the torque produced at the shaft would
be
> 1oz x 12in = 12 oz-in. If you double the length of the stick to 24 inches
> and hang the same 1 oz weight on it, you would have 1oz x 24in = 24 oz-in.
> Likewise, if you doubled the weight on the 12 inch arm to 2 oz you would
get
> 2oz x 12in = 24 oz-in of torque at the motor's shaft.
>
> So, if you start with 1076 oz-in and divide by 12 (12" per ft) you will
get
> a measurement in oz-ft (or ft-oz means same thing). 1076 oz-in / 12 =
87.7
> ft-oz. If you then divide this by 16 (16 oz per lb) you will get your
> ft-lbs measurement. 87.7 ft-oz / 16 = 5.6 ft lbs.
>
> Mike
>
....snip



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