It is a tradeoff, hence actual wound-rotor motors expose the rotor windings so that a resistance can be added during start, and tapered off or just removed when at speed. The resistance added is actually not tremendous, usually from 1 to 5 times the DC resistance of the rotor.
The initial problem, is that at at start, a typical induction motor can draw up to 1000% rated current. For large (> 200 HP motors) this can severely stress the supply. Pullout torque (torque required to either start the rotor, or drop it out from near synchronous speed) is directly related to slip. Increasing resistance in the rotor, effectively increases the slip, which increases the torque.
As more resistance is added to the rotor, the peak torque curve is moved closer and closer to zero speed. That is the sweet spot for starting a wound rotor motor; lowest inrush current, highest slip, highest torque. Adding more resistance will reduce the available torque, as the slip begins again to decrease.
You can run Wound Rotor motors in a variable speed mode, if you can control the resistance, but that is generally very inefficient.
Consider that basically, an induction motor at standstill is a transformer, with an essentially shorted secondary. When power is applied to the stator, a voltage is induced into the rotor, which, being shorted develops the current which creates the magnetic field to pull the rotor along with the stator's rotating field.
Okay, since the induction motor is a transformer when starting (at zero speed), there are reactances to deal with. The reactance actually causes the induced voltage (and current, and generated magnetic field) to be out of phase with the stator field, generally lagging by about 90 degrees, so the magnetic interaction between the rotor and stator is fairly weak.
If pure resistance is added to the rotor circuit, the phase lag starts to grow smaller. (Note that different constructions of the rotor bars, with different resistances are used to permanently affect the torque curves of many motors). Add enough resistance, and the phase lag reduces to the point of what the motors design slip is, which corresponds to its maximum design torque.
The problem with leaving the resistance in circuit (aside from power dissipation) is that as the motor speeds up, and approaches its synchronous speed - slip, now you have advanced the rotor/stator magnetic phase, resulting in reduced torque at the shaft.
Calculate separately the powers you need for
- acceleration (simple Newtonian physics)
- hill climbing (ditto, once you decide what speed you want up what gradient),
- rolling resistance (you'll need to research that, for the ground surfaces you're crossing. It depends on speed, ground surface, type of tyre, wheel size etc, you'll find plenty of info online to get you started)
- Air resistance (again, some research needed, but you can estimate a Cd and you know the frontal area from the bodywork design. Again there's plenty of info online)
- Losses in the gear mechanism
Add these together and you have a minimum figure for the required power, you probably want to add 10% to 50% as a safety margin. (As you probably don't need full acceleration uphill, exercise judgment whetehr to add both of these numbers, or the larger of the two).
Torque (Nm) is simply power(W) / rotational speed (radians/second).
Torque at the motor for a given velocity obviously depends on wheel size and gearing.
Best Answer
One of the several important advantages of the three-phase, a.c. induction motor is the grace with which it delivers a variable torque. You did not mention the motor's number of poles or whether your power was of the 50-Hz, Old World kind or of the 60-Hz kind used in the New World. In the Old World, the maximum (not minimum) speed of a typically configured induction motor is 3000 rpm, and this only if it is a two-pole motor. Four-pole motors are more common: their maximum speed in the Old World is 1500 rpm.
For the sake of answering, let us suppose that you are in the Old World and that your motor has two poles: maximum speed, 3000 rpm. (If in the New World, you can scale this answer's speeds by [60 Hz]/[50 Hz] = 1.20: maximum speed, 3600 rpm.)
For moderate values of torque—usually up to about 115 percent of the motor's full rated load—the motor's speed will decrease slightly and approximately linearly as the mechanical load to which the motor delivers power demands torque. For example, if the motor turns at 2800 rpm at full rated load, then it will turn at about 2850 rpm at 75 percent of full rated load, 2900 rpm and 50 percent, 2950 rpm at 25 percent, and (almost) 3000 rpm when unloaded. At loads greater than 115 percent, the motor's speed will decrease more than the linear model predicts, until the motor's breakdown torque is reached (the exact value of breakdown torque depends on the motor, the ambient temperature and other factors, but 200 to 230 percent of full rated load might be a typical figure). At breakdown torque, the motor will spin down to a stop, thenceforth acting as an inductive winding and delivering the motor's rated locked rotor or starting torque to the load.
I give all this detail for three reasons. First, because it may contain the answer you seek. Second, to explain why there is no single torque, but are many torques involved. Third, to convey the crucial concept that, within the motor's operational domain, it is the mechanical load, not the motor, that determines the torque.
If you really need to measure the actual torque accurately under specific conditions, then use a dynamometer. However, if an approximate indication of torque is all you need, and if you are operating in the motor's normal, spun-up state, at no more than 115 percent of full load, then you can approximate the torque pretty accurately from the motor's measured speed and its nameplate data, using the linear relationship described. The approximation is so much easier than the dynamometer that I would recommend the approximation in most cases.
Note: A few, unusual three-phase a.c. induction motors do not have any breakdown torque. In other words, for a few induction motors, the locked-rotor torque is the maximum torque the motor can deliver. In the United States, such motors usually read "Design: D" on their nameplates. This is probably not the case for you, but I thought that I should at least mention the matter.