Re: Declining motor torque with lower voltage.

Hey,
Mr. Elson knows his stuff, listen to him. However, I would like
to add, peak torque values mean almost squat. Your 600 oz-in peak
motor probably puts out around 150 oz-in "continuous" torque. This
is the magic number. Furthermore, don't ask your motors run up
to "max RPM" either. Use around 75-80% of max RPM in your rapid
calculations. Good luck.
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--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Polaraligned"
<polaraligned@o...> wrote:

>
>
> > 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
>
> So Jon, what your are saying is that I could direct couple my 600
> oz-in (peak) servo's and they would produce 1000+ lbs of linear

force

> on my Bridgeport. And 1000 lbs is a good value to be at for the
> X and Y axis. I would also then get 240 ipm rapids.
> Am I correct? Everyone seems to be using a lot more power to the
> table than this.
>
> Scott