Peak or RMS

[Jonathan]

Just to confuse matters further

Which is the 4.2A on said motor? Peak or Phase? Why?

Peak is 1.414 * RMS (or put list the inverse RMS = Peak * .707)

...
RMS/Phase is the current you stepper motors run at under normal full current load (no micro-stepping no stand by mode).

It's rather odd that in the stepper driver datasheets they state the RMS current and the current multiplied by 1.414 for peak current since the current waveform varies so much from sinusoidal. It depends on the driver and the speed the motor is running at - on some drivers the current is just chopped, so without microstepping it's closest to a square wave. With microstepping the waveform approximates a sinusoid (but quite poorly at the low microstepping settings we use) and for better drivers at medium-high speed they do transform to a sinusoidal waveform, but still an approximation. The only waveform for which the peak value is 1.414*RMS is a sinusoid, so this formula generally doesn't apply...yet they still confuse the ratings by including it.

Anyway..

The current rating on the stepper motor is the maximum constant current you can apply to the coil/phase due to the thermal limit (i.e. in steady state). Since in the 3Nm motors this rating is 2.1A, they're 4.2A in parallel so you know the current cannot be greater than 4.2A so it's safe to use 4.2A peak (4.09A) is nearest on PM752.

4.2 * .707 = 2.9694A is your operating Amperage (RMS/Phase) IF that was peak; this section we are assuming RMS so you have reduced you usable power by 30%.

Power is proportional to the square of the voltage, (P=I^2*R etc) so reducing the current by 1/2^0.5 reduces the power by the square of that, so (1/2^0.5)^2=0.5. Therefore you have reduced the available power by 50%. However the motor torque is (roughly) proportional to phase current, so you would be correct in saying the torque is reduced by 30% in this hypothetical situation.