Re: formula ?

Hi Jon,

I thought about this as I was trying to conceptualize it.

If my 50 oz/in stepper takes 2,000 steps per inch, then that SAME
2,000 steps per inch generate so much force, regardless if it is on a
rack or a leadscrew or timing belt.

Friction and alignment and other losses will effect the delivered
force, but theoretically, it is the same.

If you want more, get one of them big A$$ steppers 802 oz stepper
from Dan.


Dave

--- In CAD_CAM_EDM_DRO@y..., Jon Elson <elson@p...> wrote:
> batwings@i... wrote:
>
> > I don't think this is correct.
> >
> > At 06:14 AM 4/15/02 +0000, you wrote:
> > >docholliday01201 wrote:
> > >
> > >> once again, who knows the formula for converting rotational
torque at
> > >> the motor-to linear force at the table ??
> > >
> > >Its really simple. Consider the leadscrew to be a drum pulling
a string,
> > >where the circumference of the drum is equal to the lead of the
screw.
> > >If you have a 5 TPI screw, that is the equivalent of a drum with
> > circumference
> > >of .200 in, or .0637" diameter or .0318" radius. If the motor
produces 10
> > In-Lb,
> > >that is the same as 10 Lbs pull at a radius of one inch. This
will translate
> > >to 10 / .0318 = 314.16 Lbs linear force on the table! Pretty
amazing, these
> > >ball screws!
>
> How can it not be correct? Two eqivalent mechanisms (neglecting
friction)
> should produce the same results. A drum pulling string or a screw
moving
> a linear slide by the same distance per revolution, have to be
equivalent.
> (This is only true for ballscrews, because other screws have
significant
> friction which increases under load.)
>
> Unless someone can point out where I went wrong, I stand by this
> calculation.
>
> Jon