Machining forces

WARNING: this is off the top....

I wish to shear some plastic area at a known rate. To compute this area, I need the speed at which the chuck turns.

Let me suppose the tube spins at 1200 rpm. This is 20 rev/sec, shearing a width of 1 mm and depth 0.5 mm and length pi X 50 mm X 20 per second

Let me take a wild guess that the shear strength of the material is

A crucial question is the effective area over which the instantaneous shear takes place. Supposing (out of the blue) that the instantaneous area is (1 + 0.5 mm)^2, the force would be 2.25 X 4.2 newton = 9.5N

Force times distance gives work: 9.5N X 3meters = 28.5 joules per second which is a power of 28.5 watts.

I wouldn't trust this power estimate as far as I could throw it to the round file. So I would next consult cutting speeds in Machinery or some such....

Brian Whatcott Altus OK

Thanks, I'll go over this to see if I get it. I'm essentially after a 'rule of thumb' to quickly prove whether this operation can be done in the field with a battery drill. I haven't the means at the moment to do a practical trial.