To answer your question, yes you can run two DC motors of A4988 Module because the output stages that are driving a bipolar stepper winding are H-bridges and you need two of them (Check here for more on bipolar motors) .
Controlling two motors with different speeds is out of mind. Because there's no different pins.
Now to answer that how would you control the speed. You need to pulse step pin.
You might need to set the micro-stepping modes. Like 8/16 micro-step per step.
Check the current graphs. Because you are approximating sinusoidal waveform.
Motor might start at some value of current(which can be controlled by pot as you mentioned)
EDIT : Quick Google search gives me this link
Good Resource. Let me bookmark.
The DC motor parameter you are interested in maximizing is variously called "starting torque" or "stall torque" or "locked rotor torque", or similar. It is the torque (measured in foot-lbs or inch-lbs) created by the motor when the rotor of the motor is prevented from moving. (This is done in a test bench by jamming the motor shaft mechanically and making the torque measurement - e.g. with a torque wrench.)
This "locked rotor" condition is what the motor will experience momentarily when it starts from a completely stopped condition. E.g. when your electric bike is at a standstill and you press the accelerator.
Some DC motor data sheets will state Locked Rotor Torque directly, others leave it to a simple calculation you must perform. In this case you must know the "torque constant" of the motor, which is usually present in the data sheet.
The torque constant is a fixed value for a motor which tells you how many ft-lbs of torque the motor produces for each ampere of current passing thru its armature winding. In other words, ft-lbs per amp.
So, if you know the torque constant all you have to do is measure the current flowing thru the motor at stall (e.g. at the instant you apply power to it) and you will know the resultant torque by simple multiplication ( torque constant x amps).
Typically, you will use the same voltage supply for starting the motor that you do for running the motor. In your case this seems to be 24 Volts. So, in most applications the stopped motor will be started by applying the supply voltage directly across the motor's terminals. What is the starting torque in this common condition? Very simple, use Ohm's Law to figure this out. The motor's data sheet will usually list the motor's "internal resistance" or "winding resistance". In a motor like yours it will likely be less than one Ohm. According to Ohm's Law divide the terminal voltage (24 V) by the Winding Resistance and you will get the starting current in Amps. Now, multiply this calculated starting current by the Torque Constant and you get Starting Torque - the torque the motor will produce while the rotor is not moving - exactly what happens for the first instant you apply voltage to the motor before the rotor actually starts to turn. See? It's a pretty straightforward calculation. If you can make a non-intrusive amperage reading in the motor leads (e.g. with a DC current clamp) you can verify the electrical part of the calculation pretty simply.
As a practical matter, typical DC motors will perform closely to the theoretical calculation I described above. The problem comes in actually delivering the full voltage and amperage to the stalled motor. The amperage will be high for a motor like yours. You state "15-20 Amps", but it may be even greater when you do the calculations from the data sheet values. This means you have to use very heavy gauge wire to allow enough voltage to actually reach the motor without significant voltage drop in the supply wires. Plus, you've got to have a voltage supply that can actually deliver that much instantaneous current.
So that, if you are trying to select a DC motor to get the maximum starting torque at the starting condition, you will look for one with the highest Torque Constant and the lowest Winding Resistance. It's that simple.
Best Answer
You'd have to use a harmonics analyzer or high-resolution scope as suggested in this article to measure the apparent power with PWM. Also from that (2001) article I gather there's no industry standard for measuring efficiency of DC motors driven by PWM...
In general, you can measure losses with the calorimetric method, but that's very cumbersome for motors. I only found specifics for (AC) induction motors, which being larger and more expensive are probably worth the hassle. For those they are worried whether inverter [PWM-]generated sinewaves (which contain a lot of ripple noise) cause significant extra heating, which is similar but not quite the same problem with what you're asking.
Image from this paper, which also details the issue of having to couple a mechanical load while decoupling it thermally:
Regarding:
Given the dearth of publications on this (for PWM-driven DC motors)... I would venture a guess the answer is usually "no"... however, I found a few publications dealing with the latter, the most recent one I found being combo paper that [despite its title] also studied efficiency of DC motors in a section, which I'm quoting here, and it's indeed (like @JonRB said) mostly and issue of eddy-current losses:
So yes, theory predicts that the losses in the motor are greater with PWM than if you supplied it with DC (of the average value of the PWM). (In those calculations they assume that PWM source voltage drops to 0 during the off period, but since the windings are indeed inductors they provide their own averaging affect).
And the paper also has some experimental data to back that up (however this part is somewhat confusing to me, see my comments after the following quote):
It's not really clear to me if this "full bridge dc–dc converter" they used has its own output filter or not. Anyway those extra 15W losses on a 2kW motor is however under 1% (ok at 1/4 power so half duty it would be 3%)... so I suspect that's why you probably don't hear much about it. But it's not clear to me if that is with a pure PWM drive or if their DC-DC converter has its own filter, in which case the ripple sent to the motor would be a lot less... so the losses would perhaps be more significant without it. (Normally a full bridge dc–dc converter would have an LC output filter, but perhaps for the purpose of their experiment it doesn't have that so that it's the same signal shape [i.e. dropping to 0 when off] as in their theory section? I couldn't figure from the rest of the paper what this dc-dc converter was exactly.)
Actually one reason I couldn't find more materials is terminology. A lot of the motor people refer to PWM drive as "chopper controlled" instead. With that in mind, more can be found:
As high frequency as possible it seems. That was among their conclusions & recommendations. However, back in the day they could only test up to 400Hz... so presumably more can be said; alas the much more recent paper (quoted previously) that used a dc-dc converter didn't mention its frequency.
This info on frequency choice is repeated in a 2004 book that explains it by "reduced harmonic content of the current at higher frequencies"", and which cites precisely that 1982 DOE/NASA report in support. Apparently no more recent research had been done. I'm hoping that with the trendy electric vehicles research nowadays, more would exist on this, but insofar I wasn't able to locate much more.