On Sunday 16 March 2008, Patrick Ferrick wrote:
>Hello again,
>
>[Machine-related, but off topic, so I'm really looking for someone to steer
> me toward a good reference on the subject, or to communicate off-list.]
>
>I'm thinking about replacing my Z-axis leadscrew with a larger diameter one
> that will have less backlash.  I'm trying to figure out how to calculate
> how much (linear) force a certain stepper will provide with the new screw.
>
>The screw's pitch is such that one turn gives 0.2" of travel, or 5 turns per
> inch. That would seem to give it a mechanical advantage of 5, but what has
> me a little bit confused is the translation of a torque (from the motor) to
> a linear force (by the nut).
>
>I believe it is correct to say that a 200 oz.-in. motor, for example, can
> exert a force of 200 oz. (12.5 lb) at a distance of one inch from the
> center of its shaft.
>
>Since the leadscrew has a mechanical advantage of 5 and is 0.5" in diameter,
> does that mean that I should expect (200 oz. / 0.5") * 5 = 2000 ounces (or
> 125 pounds) of push?
>
>thanks again,
>Pat

You might be able to get into that region with ball screws, but with acme, nuh 
huh.  I have a 425 on my z axis, with a 17 tooth sprocket on the motor, and a 
30 tooth on the ball bearing supported nut carrier, with a 1/2" acme of 10 
tpi doing the honors.  With that, I can just hit 155 pounds of z push on the 
bathroom scales.  By your frictionless reasoning that might SWAG at 7500.  
There's a lot of friction loss in screws when under load.

>....................
>Patrick Ferrick
>Town of Webb School
>Main Street
>Old Forge, NY  13420
>
>(315) 369-3222
>(315) 369-6216
>....................
>
>
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