Electronic – DC motors – Torque/Speed Relationship

I have a section here in my lecture notes that's implying the startup torque is equal to the loading torque but that makes no sense to me.

And you are right. The airgap torque has to be higher than the load torque in order for the machine to accelerate. Alternatively, the airgap torque has to be lower than the load torque for the machine to accelerate in the backward direction.

Just the fact that the motor is at standstill does not imply zero torque. It implies that the load torque is equal to the airgap torque, zero or non-zero.

The static load could be X [Nm] and the airgap torque could cancel it out if it were also X [Nm]. Many loads have the load torque proportional to the square of the shaft speed. Hence, the load torque is zero at standstill. This means that your DC motor can develop any torque equal to \$K_t I_{winding}\$ up to rated winding current \$I_{rated}\$. The startup torque of the motor would be equal to \$T_{start} = V_{dc}/R_{winding} * K_t\$ [Nm]. Note that the current \$I_{winding}\$ would likely be higher than the rated current if not actively controlled.

If the motor windings are initially disconnected or shorted (provided non-zero winding resistance) and the shaft speed is zero, then such state implies zero load torque.

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.