Re: Declining motor torque with lower voltage.

--- In CAD_CAM_EDM_DRO@yahoogroups.com, Jon Elson <elson@p...> wrote:

> cnc_4_me wrote:
> >
> >1) You have a treadmill motor rated 130VDC, 2.5HP 6700RPM, and
> > 373 oz-in continuous.
> >
> >2) The motor is run from a 80VDC power supply.
> >
> >
> I believe the treadmill motor in question is a PM motor. The

torque is

> proportional
> to current, the speed is proportional to voltage. The speed would

fall

> to 61.5% or
> 4123 RPM. The torque available would still be 373 Oz-In.
>
> Jon

I looked this over again and have a new conclusion…As you say torque
is proportional to current and I agree with this. But I came to a
different conclusion on the end torque with lower voltage.

Since you are running the motor aprox 38% below rated voltage you
cannot push the amps thru the motor to get full torque.

I left out the current ratting before but here it is 18.5 amp.
Armature resistance would be 130vdc / 18.5A = 7 ohm. At 80vdc and 7
ohm you can only make 11.4 amps. The torque per amp is 373 0z-in /
18.5 = 20 oz-in. At 11.4 amps max current you get 11.4 x 20 oz-in =
230 oz-in.

If we use your speed reduction number above 61.5% and multiply it
times the torque of 373 oz-in we get 229.4 oz-in. Seems to agree
with ohms law calcs.

Wally