Stepper current in partial-step positions

This is partly a question to Mariss and Dan Mauch... It concerns how much
current stepper drivers apply to the motor windings during partial steps.
This started with a conversation in the yeagerautomation group, where a person
mentioned running 3 Camtronics 5A drivers in half-step mode off an 8A power
supply, and the question arose as to what was the max current draw on the power
supply, 15A, 21.2A, or 30A.

When the motor is at a full-step position, of course, only one winding is
energized, and it should be getting the full rated current of the motor (Imax).
This current causes a power dissipation within the motor of Imax*Rwinding, and
it is the motor's ongoing ability to dissipate this power (heat) which
determines the motor's max current rating in the first place.

Suppose instead the motor is at a half-step position. In this case, both coils
are getting the same current input, producing a magnetic field at the half-step
orientation.

The question is: How much current should each coil get?

If the magnetic fields of the two windings just add up as vectors, then to get
the same net field (and thus the same torque) as at the full-step positions,
each coil should be getting 0.707*Imax (0.707 being approx. the square root of
0.5).
However, if the driver does this, each winding is dissipating
0.707*Imax*Rwinding, for a total power (heat) production of 1.414*Imax*Rwinding,
which exceeds the motor's rated ability to shed heat.

To avoid overheating the motor, the total current in both windings should never
exceed Imax, so in the half-step positions, each coil would get 0.5*Imax, and
the half-step positions would only have 70% of the holding torque that the
full-step positions do.

A similar analysis can be made for the 1/10th-step positions in a 10-step
microstepper driver.

So what fraction of the limit current do drivers supply at the partial-step
positions?
-Kevin Martin