A constant current means, for an ideal motor, a constant torque. This is approximately true for real motors. It doesn't matter what you attach to the motor, or how fast it's turning.
What you seem to be missing is Newton's second law of motion. It states that force is the product of mass and acceleration:
$$ F = ma $$
The constant current you supply to the motor is one force. The weight opposes that force. The difference is the net force, \$F\$ in this equation, and \$m\$ is the mass of the weight, plus the mass of the rotor and the string and everything else the motor must move.
You set current to be sent to the motor so that the torque applied is 10 in-lbs without any load.
Not possible. There is nothing for the motor to "torque against". This is the mechanical equivalent of trying to develop 10 volts across a dead short. The motor will rapidly spin at its maximum speed, and the back-EMF will rise to the driving voltage such that your driving electronics are unable to supply enough voltage above the back-EMF to make enough current to have that much torque.
Let's just say you determine how much current is required for 10 in-lbs of torque, and you drive your motor with a constant-current supply set to that.
What happens when the torque from the weight/load is 5 in-lbs?
Assuming that the rotor and the string are massless and frictionless, the weight will be accelerated upwards by the net 5 in-lbs of torque (motor's 10 in-lbs, less 5 in-lbs from the weight). The rate of the acceleration is determined by the mass of the weight and Newton's law above.
As the speed of the motor changes (the weight is accelerating), the back-EMF also changes. Your constant-current supply to the motor will have to apply an increasing voltage to maintain the same current. Electrical power thus goes up, as does mechanical power.
What happens when the torque from the weight/load is 10 in-lbs?
Motor torque balances weight torque. However fast the weight is moving (if at all), it keeps doing that. Newton's first law applies.
What happens when the torque from the weight/load is 15 in-lbs?
The weight will accelerate downward, overpowering the motor. However, it won't be a free-fall. The motor cancels some of the force of the weight, resulting in a slower acceleration downwards.
If the weight overpowers the motor, then eventually it can get the motor to run backwards, relative to the way it would run if there were no load. When this happens, the back-EMF now adds (instead of subtracts) from the voltage you apply to the motor. At some point, your controller, which is attempting to maintain a constant current, must apply a negative voltage to maintain that current. In other words, the back-EMF is sufficient to create the necessary torque on its own: your controller must oppose it.
This is perfectly symmetrical with the first case, where the motor was overpowering the weight. In that case, electrical and mechanical power went up (without bound, if you let them). In this case, electrical and mechanical power go down (negative, if you let them). Energy is conserved because you are changing the gravitational potential of the weight.
The need to resist the back-EMF usually means storing electrical energy in a capacitor or battery, or using it to heat a resistor. If you can't do this fast enough, then the motor will create more torque than your desired 10 in-lbs, and you have hit the limits of your "constant current" driver.
Further reading:
Best Answer
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:
There really is no single equation that fits all states of a PMSM but this one works in general:
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.