[bearwen]

I have a newb question I have figured out the torque of my motors with my lead screws and well converted it to lbs/in and I have come up with 514lbs/in I am using 262 oz/in motors and a 1/2-10 lead screw(please correct if I am wrong). The question I have is well what does this tell me? I know that as motor spins torque drops so lets say I drop 75% I will have 128lbs/in left does this mean I can move 128lbs per inch? What about load on the spindle from cutting say 1/4in deep how do I know ( or calculate) what I can do and at what speeds? I am only wanting to figure rough estimate so I dont go trying to cut something with to high of feed rate at a given depth and brake something ( basiclly looking to figure the rough capabilities are of the machine).

Thanks
Berry

[NC Cams]

Torque is force time distance.

Thus a 200 oz-in motor will devlop the same resistive force at the end of a 12" long lever arm as a 100 oz-in motor will at 24" long lever.

If you hooked the 100 oz-in motor to a 2:1 gear box, the force at the end of a 12" lever would be the same as a 200 oz-in motor w/o a gearbox.

Or if you put a 50 oz force at the end of a 4" lever, you'd stall the 200 oz-in motor.

Figuring out the torque/friction/resistance deal with a lead screw is a bit more involved. I'd suggest a trip to the library for a decent machine book.

Try: Design of Machine Elements by M.F. Spotts

Mine is relatively old BUT the laws of physics and machine elements haven't changed....

WORK is NOT the same as POWER.

(Edit: see this link
http://www.merkle-korff.com/formulas.asp#con
for more info)

Lift 3300 lbs 10 feet and you have done 33000 ft-lbs (or is it lbs-ft) of work.

but

Lift 33000 lbs 1 foot in 1 minute and you've consumed 1 hp worth of power

You have to go thru ALL the math for all the linkages and take into consideration ALL the loads/forces (including friction) to answer your question and I don't believe you have all the input data available so as to be able to do so...

[ger21]

I think you figured out something a little wrong. The easiest way to get a rough idea of the force you'll have, is to look in Nook's acme catalog at www.nookind.com

That will tell you that , if your using a plastic nut, than you need .039 oz-in of torque to move 1 pound. So, with your 262 oz-in motor, you have 262 oz-in / .039 /16oz/lb which = about 420 lbs of force. 25% of that is 105lbs.

You didn't say what kind of machine this was, or what your driving the motors with, and at what voltage. If it's a router, most people want to make rapid moves as fast as possible. Go too fast and you may end up with far less torque than you think.

How much torque you need to cut 1/4" deep can vary widely. It depends on a lot of factors. Material being cut, diameter of tool, spindle speed, tool geometry, tool sharpness.....

Tool manufacturers will usually publish recommended chip loads for tools, which will get you in the ballpark as far as spindle speeds and feed rates are concerned. Exceed these numbers (for chip load), and you may break tools. If your tools are dull, they may break if pushed too hard. If the spindle is underpowered, you can break a tool if the spindle bogs down.

[jeffs555]

Berry,
I think you have a few things wrong. The "514lbs/in" and "128lbs per inch" should be just plain pounds, they are forces, not torques. Also, you appear to have neglected friction in the leadscrew nut. The figures you have would be correct for ballscrews, but for acme leadscrews, you need to take friction of the nut into account. The 514 lbs would be about 420 lbs with a plastic nut, or 350 lbs with a bronze nut.

I don't have the equations for calculating cutting force, but do know that when using a hand router, I am never pushing with anywhere near 100 lbs of force.

Jeff