Re: formula ?

Hello,

And don't forget to de-rate that figure by the friction of your 12/13
allthread drive...

Maybe 30% efficient (if that), so now you're at 153 lbs. Hmmmmm...

You know, I'm a bit in favor of the "Mariss" approach myself; that is
to use an experiment in addition to the calcs.

I wonder what the "real world" result will be if you put a "fish
scale" on the driven part of your system... Run it slow, and let us
know, please. (I'm guessing less than 100 lbs actual)

Okay, that might be a big fish scale<G> but even some workout weights
and a rope and pulley will give us some idea of where "reality" lies

Hope this helps,

Ballendo

--- In CAD_CAM_EDM_DRO@y..., Jon Elson <elson@p...> wrote:
> dave_ace_me wrote:
>
> > Hi Jon,
> >
> > great explanation. I couldn't get the concept that the lead of
the
> > screw was part of the circumference part of the equation.
> >
> > OF course that makes sense, a 1/2-5TPI 1/2 screw or a 2"-5tpi will
> > still move the table one inch for 5 revolutions. so the actual
> > diameter or circumference of the screw is not important.
> >
> > I noticed that we rate steppers in oz/in but power in inch/lb.
> > Can I assume that 16oz/inch is the same as 1in/lb?
>
> I believe this is correct. There are so many torque measurement
units, its
> a mess. N-M, N-cm, gm-cm, oz-in, lb-ft, .........
>
> > That would make a 50oz/in stepper 3.125 inch/lb. Fits close to
your
> > 10 inch /lb value. Yours seems to be 150oz/in.
> >
> > so my 50in/oz at 2:1 is 6.25 in/lb on the leadscrew. The 13tpi
> > (0.0769"circumference/0.02448dia/0.012243radius) then 6.25 /
0.012243
> > = output force
> >
> > 6.25 / 0.012243 = 510 pounds of force. That kind of force will
bend
> > the noodle of a leadscrew I am using.
>
> Yup, sounds good. but, remember, stepper ratings are good at zero
speed,
> only. They drop off very rapidly as speed increases.
>
> Jon