[Al Skierkiewicz]

Marshall,
If your battery is 11.5 volts the motor will run slower than if the battery was 12 volts. In a normal robot, if one motor is turned on it will run at a specific speed that varies with battery voltage. However, as more motors are turned on, more current flows through the primary wiring (#6) and the PD. Even if the battery remained at 12 volts (which it can't in the real world), the current flowing through the #6 and PD will cause a voltage drop across the resistance contained in those devices. If four CIM motors were at full throttle with the robot pressed against a field border and the wheels are not turning, then the motor are in stall at 129 amps per motor. All things being equal, 500 amps through #6 wire drops 0.25 volts/ft. In a typical robot, with 5 ft of #6 (remember both the red and black wires need to be included) plus the series resistance of the battery and the series resistance of the PD you would be lucky to get 7 volts at the output of the PD per motor. You can fudge your way to a prediction if you show the motor as a resistor but the motor is not a true resistive load. Using Ohm's Law, 12 volts at 129 amps would be about 0.010 ohms for the motor.
I tend to use worse case analysis since in my work I need to find what might have caused a failure to prevent a repeat in the future. Although, under normal operation you would not run at 129 amps per CIM, during parts of the match most people will. This occasional high current is what is tripping the over current sensor in the Jaguars. The FETs can handle temporary over current and the Victors have been doing it for years. It is the sustained over current that causes problems. Keep this in mind, a motor that is in stall is one in which current is applied but the motor is not turning. Every motor is in stall when you start it. Move the robot with a four CIM drive and you will draw 516 amps for a short period of time when it starts to move.