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

Jon Elson wrote:

>[snip]
>You can be sure this is what we are talking about by doing a power
>calculation. If
>the real resistance was 7 Ohms, the power loss would be I^2 * R = 18.5 *
>18.5 * 7 =
>2396 W or 3.2 HP. Since this equals the total power input to the motor
>(130 V * 18.5 A = 2405W) there is no power left to come out the motor shaft.
>Obviously a motor that absorbs 3.2 electrical HP and produces zero
>mechanical
>output is worthless.
>
>Jon
>

And using those formulas to calculate the resistance, voltage constant,
etc.:
Input power = 2405W
Output power = 1865W (2.5 HP @ 746W per HP)
Power wasted = 540W (ouch!)
current = 18.5A, I^2 = 342.25
R=P/I^2 = 540 / 342.25 = 1.58 ohms
voltage drop across the internal resistance = 1.58 * 18.5 = 29.23 volts
voltage from back EMF = 130-29.23 = 100.77
voltage / kRPM = 100.77/6.7 = 15.04

So, going back to the calculations for what you'll get driving this
motor with a Gecko:

Max voltage: 130V
Max speed: 6700RPM
Torque constant: 20.16 oz-in/A (assuming 373 oz-in continuous torque)
Continuous current: 18.5A
Voltage constant (calculated): 15V / kRPM (close enough)
Winding resistance (calculated): 1.58 ohms.

We already know the maximum "torque voltage" - it's 29.23V. So, you have another 50.77V that will give you speed.
50.77V / 15V/kRPM = 3384 RPM at full torque.

There are probably some losses that I haven't taken into account, but you could probably count on 3000 RPM, still with near full torque. If you want 200 IPM rapids, then you need 1000 RPM at the screw, so a 3:1 reduction.

- Steve