Given a 50% drop in measured voltage across the motor during stall, does that mean the setup is still not good enough to measure the Back EMF and resistance correctly?

Your supply setup has always been less than marginal. You're getting closer to something that can actually test the motor but you need to limit the run (or stall) time. In a perfect setup, there would be NO voltage drop from no load to stalled. That isn't practical unless you want to use a car battery and booster cables.

Your voltage source has an internal resistance (Thevinin resistance), your wire has a resistance and the motor has a resistance. You would want the sum of the internal resistance plus the wire resistance to be much less than the internal resistance (factor of 10?) but that's just not going to happen.

You have the motor internal resistance and that's all you need to compute the back emf versus RPM where RPM is dropping as a result of increasing load (some kind of Prony brake or whatever). We've already discussed using applied voltage and running current to calculate back emf. You have everything you need.

Motors, in general, aren't intended to be stalled and they self-destruct pretty quick. The outlier being torque motors but that's not what we're talking about.

So, take your internal resistance as 0.09 Ohms and go to work on the rest of the simulation. Just realize that if you didn't have that Thevinin resistance and wire resistance, at 12V applied, the current would be 133 Amps. Even at 4V, the current would be nearly 45 Amps.

What the manufacturer was telling you was how to compute the Thevinin equivalent resistance which is a whole lot like this back emf thing.

Consider a 10V battery with 0 Ohms internal resistance and put a 100 Ohm resistor in series with it. Now put the whole thing in a black box (yes, it has to be black, the instructions say so). You measure the open circuit voltage and, voila' you get 10V because your meter impedance is MUCH higher than the 100 Ohm series resistor.

Now, put a 10 Ohm resistor across the terminals and measure the voltage again. You will get 0.9091V (almost every bit of the battery voltage is dropped internally). Since you know the voltage across the 10 Ohm resistor, you know the current through the 3 devices in series. 0.09091A. You also know that the internal resistor, at 0.09091A dropped 9.0909V and dividing 9.0909/0.09091 gives 100 Ohms for the internal (Thevinin equivalent) resistance.

So, what they told you to do is set a voltage and measure it unloaded. Then apply a 30A load (however you want) and measure the voltage again. There will be a lower voltage and using the ideas above, you can calculate the Thevinin resistance.

http://www.facstaff.bucknell.edu/mastascu/elessonshtml/source/source2.htmlMy numbers above were deliberately skewed. It would be more reasonable for the black box internal resistance to be 10 Ohms and the load to be 100 Ohms. But, in your motor case, your equivalent resistance (power supply plus lead resistance) IS 10 times higher than the load resistance.