Rushing.
More later maybe ...
Note that "stall torque" is often used to mean locked rotor 0 RPM torque BUT you use it in the sense "dropout torque at a given speed". That's entirely fine as long as you note that some references will mean the former and not the latter.
Criticism (kind / constructive) welcome.
Written at a rush and unchecked. Better is possible.
See writer "Toper925" comment here
He notes:
Where:
p is the number of pole pairs
λ is the amplitude of the flux induced by the PMs in the stator phase
Lq and Ld are the q and d axis inductances
R is the resistance in the stator winding
iq and id are the q and d axis currents
I'd need to read more on what he said to make total sense.
Stall torque is when torque is not sufficient to "pull in" the next rotor pole piece using the available magnetic field.
SO, I'd expect
More pole pairs better. I'd expect better than linear gain as distance halves with doubled pairs BUT magnetic force at worst falls as distance cubed (only a considerable number of magnetic pole diameters away so not in most sensible motors), comes closer to falling with distance squared as gap falls to near pole width and at best can only approach linear at close proximity. SO more ples should give less interpole distance so ... (but pole sizes are down so ...).
Torque = power per rev. If the power falls faster than RPM your margin is dropping until you reach the point of no pull in. At a quick glance I think that this is what this man here is alluding to about half way down below the graph. Leading to ...
If you have a power curve you also have a torque curve as the two are related by motor rpm. (Torque = k x Power / RPM). If you have a speed-power plot for you load you should be able to overlay this on torque curve and see where load torque is > generated torque. This will be better than real world (probably).
Lowest R should help as it allows greatest I but this is really a secondary effect for two motors with the same power at the same RPM.
Induced flux should play an immense part. I'd expect non saturating magnetic (eg steel) cores to provide superior results EXCEPT if you can get all gaps so small that field is well maintained by magnet. Rule of thumb is you can get about 0.5 Tesla at an airgap of 1/2 a magnet diameter using a top class NdFeB magnet. Say N52? N45 won't be too bad.
Note that the US process NdFeB magnets are cast but ground and sintered subsequently and are inferior in max possible flux to the Japanese versions. This should all be covered in the flux spec.
You seem to be confused about what you want. If you want to decrease the motor speed, but you still want maximum torque, then you must apply full rated electrical power to the motor, and put a mechanical brake on the motor until it slows to the speed you desire. Or, you must somehow make your motor less efficient. I don't think that's what you want.
Think of it this way: electrical power is the product of current \$I\$ and voltage \$E\$:
$$ P = I E $$
Mechanical power is the product of torque (\$\tau\$, in newton-meters) in and angular velocity (\$\omega\$, in radians per second):
$$ P = \tau \omega $$
A motor is an electrical to mechanical power converter. The mechanical power always equals the electrical power after losses.
Furthermore, current is proportional to torque, because the more current you apply, the stronger the magnetic field inside the motor, and the attraction between the motor's poles becomes greater.
If the mechanical and electrical powers are correlated, as are the current and torque, then voltage and speed must be, also. And they are, because the faster the rotor spins through the stator field, the greater back-emf it will generate. This is Faraday's law of induction.
So, if you want to decrease speed, decrease voltage. If you want to decrease torque, decrease current. If you increase torque (say by putting a brake on the motor), you are increasing motor torque. But if you don't change the supply of electrical power, then the mechanical power also won't change. If torque increased, the only way to keep mechanical power constant is to decrease speed, so the motor slows down.
There is one kink here: as torque goes up, current goes up. The resistive losses in the motor also go up, because the windings have some resistance, and those resistive losses are proportional to the square of the current:
$$ P = R I^2 $$
So, as current goes up, the resistive losses increase, making the motor a less efficient converter of electrical energy to mechanical energy, because some of that electrical energy is now creating heat. If you stall the motor, then the motor reaches 0% efficiency: speed is zero, so mechanical power must be zero, but the motor is drawing a ton of current, and there is a voltage drop over the winding resistance, so electrical power is very high.
Interesting fact: if you can make a motor with no winding resistance (or other losses), and you connect it to a perfect voltage source, then the speed regulation (how much speed changes with torque) is perfect. That is, the motor won't slow down if you try to stop it: it will just draw exactly enough more current from your battery to keep spinning at the same speed, no matter what.
PWM is all irrelevant to this. PWM motor control is just a way to efficiently apply less than the full battery voltage to the motor. It works because a PWM driven motor is equivalent to a buck converter. Changing your PWM duty cycle is equivalent to changing your supply voltage:
The maximum torque you could have (which you will get when the motor is stalled) is limited by the current your power supply can supply and the losses in the motor, just as it is without PWM. Your PWM driver might add a bit of resistance to the circuit, reducing the current and torque a bit, but usually this isn't significant compared to the resistance of the motor windings.
Best Answer
Looking at the datasheet, while it doesn't lie, it is definitely on the edge of misleading.
Note that the "max power" stated occurs at almost exactly half the unloaded RPM and half the stall current. This is indeed the "max power" point for such a motor, but the datasheet fails to mention that it is also the nominal 50% efficiency point, thus dissipating 12V*68A-337W = 479W in that tiny motor - destroying it, probably in minutes.
(Ideally, exactly half the power would be delivered, about 400W shaft and 400W heat, but the motor isn't ideal).
The motor is probably suitable for 100-150W continuous output and 200-250W short term.
So practically you must operate the motor at the upper end of the speed range, and if the speed falls below (say) 70% of the unloaded speed (or the current rises to 30% of the stall current) then - unless this is strictly temporary, like starting a heavy load or hitting a chilled spot while machining a cast iron surface, you need to cut the current and protect the motor.
Then the question of which side of the torque speed curve doesn't apply - unless the protection has tripped, you should be on the high speed side.
You can get circuit breakers that will allow short-term overcurrent. These are "motor rated" or Class C breakers for the AC motors used in most machine tools. I don't know of anything suitable for 12V DC though. I'd be looking for a 12V DC supply that can be set to trip if its output exceeds 40A for more than a couple of seconds. And as Olin says, if you want to monitor it yourself, measuring the current is definitely the way to go.