Re: Declining motor torque with lower voltage.

Since I made that post I researched how to calculate "torque to
linier force." And came up with the following…The 90% efficiency
number is realistic, I got it out of a ballscrew catalog. The
equation agrees with the numbers Rockford ballscrew gave me so I am
confident the equation is correct. The 50% efficiency number is just
something I found in a forum and have no way to know if it is correct…



T = Torque in in-lb

L = lead, 5 TPI screw = .2" lead. 10 TPI screw = .1" lead.

W = Weight in lbs

E = Efficiency, ballscrew = .9, acme screw aprox .5



Amount of torque needed to move a weight on a screw.

T = (W X Lead) / (2pi x Efficiency)

Example, 750Lb knee, 5TPI ballscrew

(750lb x .2) / (2pi x .90) = 26.5 lb-in = 3.0Nm




Amount of linier force in Lbs a screw will make, with a given torque
on it.

Force (lbs) = (in-lb x 6.28 x Efficiency) / Lead

Example, using numbers from above equation, to cross check.

(26.5 x 6.28 x .9) / .2 = 749 Lb


Wally

--- In CAD_CAM_EDM_DRO@yahoogroups.com, Jon Elson <elson@p...> wrote:
> cnc_4_me wrote:
>
> >I was talking a 9 x 42 bridgport knee...Big difference wow....
> >
> >I have not learned the calculations for force thru a ballscrew
yet.
> >That would tell us how much torque you really need...
> >
> >
> >
> I use a totally screwy way to do this, but it works for me. The
standard
> Bridgeport leadscrew and many ballscrews are 5 TPI, or 0.200" lead.
> A pulley that would advance a string .200" per turn would have a
pitch
> diameter of .200 / (2 * Pi) = 0.0318" Obviously not a really
practical
> pulley,
> but it does do the same rotary to linear conversion as the leadscrew
> (neglecting friction). So, if you have torque on the leadscrew in
Inch-Lbs.
> just divide by 0.0318 Inches to get linear force in Pounds. If you
want
> X linear force in Pounds, multiply by 0.0318" to get Inch_Lbs. of
torque.
>
> If you need 1000 Pounds of linear force (cutting force + friction +
> acceleration)
> then 1000 *0.0318 = 31.8 In-Lb, or 508 In-Oz. That would
presumably be
> the PEAK torque requirement for X and Y on a mill. But, it could
be
> close to this
> as the CONTINUOUS torque on the knee, unless you had a
counterbalance
> system.
>
> In the above examlple, if you had a 250 Oz-In peak torque motor,
then
> you'd want a
> 2:1 belt reduction ratio, at the minimum.
>
> Jon