HP is the rate at which work is done. Torque is a measure of rotational force required to do work. As long as you don't try to take a deeper cut as you slow down, you do not need more horsepower.
I did not say you need more HP. I said when you slow down, you need more torque to have the same HP. There is no mystery to to this. HP is the product of torque and rpm. Do you disagree with that formula?
Slower speeds mean less chip removal per unit time therefore less hp is required.
That would be true if diameter was the same. In discussing a lathe's available power over its speed range, I would have thought that implied different diameters. Usually you would usually keep the cutting speed, FPM, roughly the same and vary speed with diameter. As diameter goes up (i.e. speed goes down to maintain the cutting speed), torque must go up to have the same removal rate in cubic inches per minute (i.e. HP or work).
Maybe an example would clarify things in case I'm not presenting this clearly - the following assumes with a 50 thou DOC and .005 feed rate
Scenario 1)
1" dia, speed 400 rpm. = 104 feet per minute. Removal rate (or work) = about .3 cubic inches per minute
Scenario 2)
6" dia, speed 66 rpm. = 104 feet per minute. Removal rate (or work) = about .3 cubic inches per minute
Scenario 1 & 2 both require the same power (HP) and do the same work (approx .3 cubic inches per minute). Scenario 2 requires 6x the torque as Scenario 1 (as both scenarios MUST have the same power and speed has gone down so torque must go up)
What is there to disagree about? The formulas are well established. I don't want to take the OP's topic further astray, but suffice to say the 10ee drive gives much better performance than a constant torque VFD