That’s a great question. The crux of this comes down to the motor controller. Given a controller with no current limits, commanding 100% output with a 12V bus will apply 12V across the phases of the motor. You’ll end up on the “theoretical” motor curve. For something like a NEO 1.1, at 1.5N*m of torque, the motor will draw 77A (1.5/0.01958) and spin at 3,740 RPM. The motor would provide 589W of output power and lose around 341W in the stator as a result of free speed losses and resistive losses.
That’s the traditional math that motors.vex.com will help you with. However, current limits are going to throw this off completely - which is what the paper is about!
First of all, your stator current times the kT of the motor is the maximum torque of the motor (Region 1 in the Limited curve). So if your stator current limit is lower than 77A, you’re simply not going to be able to provide that 1.5N*m in the first place. Stator resistive losses are equal to I^2*R, where I is your stator current and R is your phase resistance (stall current/12V, a constant for each motor). We can conclude that your required torque will determine your required current and how much power the motor will dissipate outputting that torque.
Assuming you select a stator current limit of 80A so that you can actually even sustain that kind of torque, your supply current limit will kick in next. 589W of output power and 341W of losses is 77A at 12V! That means that if you have a supply limit of 40A, the motor will need to throttle the speed significantly and reduce its output power to stay within its 480W limit (40A*12V).
Plugging in an 80A stator limit and 40A supply limit to my calculator, we get the following motor curve for a NEO 1.1:
The motor would thus only spin at 930 RPM when supplying an output torque of 1.5N*m, with an efficiency of ~30%. If we gear down the motor 2:1 with the same current limits in place, we actually get a lot more speed (and thus power output):
2280 RPM! Given that power = speed * torque, this means that a gear ratio of 2:1 in your example would give us 145% more output power, and efficiency jumps to ~80%! And because most losses are I^2*R losses, we also actually lose less power to heat in the process, as the gear ratio lets us get more torque for less stator current.
TL;DR: plug in the current limits and gear ratio into the calculator, select your motor, and follow the torque curve down until you hit your desired torque. The speed at that point is the speed you’ll get with the parameters. I have added this information to the google doc.