Electronic – ny way to determine power rating of a potentiometer

I am surprised that you cannot find a rotary encoder good to one degree, but I do not know a whole lot about them. A 10 to 1 gear train can improve the resolution by a factor of 10. Or a stepper motor can be used as a rotary encoder with 360/(steps per rotation) accuracy. A 1.8 degree stepper is almost there.

Each of the 3 power rails (3.3V, 5V, +12V) are connected to a fuse, then to a binding post, while the negative binding posts go to ground.

This sounds like the problem. Keep in mind, voltages are differences in potential between two points. There's nothing special about ground, it's just an arbitrary point which we pick. It's 0V because the difference of something with itself is 0.

You can call anything you want "ground". That's why this circuit works. If you call the -12V output as "ground", then everything else is 12V higher. That's including what was previously called "ground": now it's 12V, because it's 12V more than what you are now calling ground.

Now consider what you've done:

^{simulate this circuit – Schematic created using CircuitLab}

The "ground" of the power supply connects all of the voltages supplied together (connections labeled B and C). The output voltages are relative to this. Notice how the -12V (V4) makes a negative voltage because it's positive side is attached to "ground".

Then, you attached the negative binding posts of all the supplies together. Largely this is redundant: you are duplicating connections B and C. But you are also adding connection A.

See the problem? You've shorted out V4. A wire has ideally zero resistance. By Ohm's law, the current that will flow is:

$$ \frac{12V}{0\Omega} = $$

In reality, the wires used to make this connection actually have some very small resistance, and you get a whole ton of current. This far exceeded the current the voltage regulator can handle and the smoke got out.