[cts_casemod]

Thats just what the literature tells you. But from that you cant calculate the torque.

let me clarify:

Theory says that torque falls by the square of the voltage, so technically at twice the nameplate speed one would have 25% torque as the V/Hz ratio has now fallen to 50%.

Some sources disagrees and state that you'll have half of the torque, since the horsepower is constant. Clearly something is not right here. How can you have the same horsepower if the torque has fallen to 25 and the speed has doubled!? Perhaps one could double the slip, which makes a lot of sense, but this is never mentioned in technical papers.

The idea is to keep the rotor power constant. For that, if the motor is operating at half the V/Hz ratio, one doubles the rotor frequency, by increasing the slip.
The net result would in fact be constant torque, since the primary V/Hz ratio and hence the flux has fallen to 1/2, but the rotor flux is kept constant, but I haven't found any sources mentioning anything about the slip, so for the time being, this is just me putting 2+2 together.

Adding to the problem is how reluctance will play here. Not only the V/Hz ratio is increasing, but the impedance of the motor (XL) is also increasing. More impedance = less current.