Re: [CAD_CAM_EDM_DRO] Re: Declining motor torque with lower voltage.

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