[Al Skierkiewicz]

Keith,
I have talked about wire losses and voltage drop for years here on CD. One of the easy to remember items is the "wire foot". Simply this is a simple calculation you can make in your head. As previously stated, #10 wire is .001 ohms per foot. Well, at 100 amps, that one foot chunk of wire will drop exactly 0.1 volts. In terms of 'wire feet' here are some other equivalents.
#6= 0.5 WF/ft.
#12=2 WF/ft.
#14=3 WF/ft.
Victor=6 WF
Talon and Jag=4 WF
Battery=11WF
0.1 volt drop/WF/100 amps
So all you have to do is just add up the losses in WF and that will get you an approximation of the loss (keeping in mind both wires need to be counted).
As an example, #6 wiring from battery to PDP is typically 1 foot from battery to APP connector and 2 feet from there through the main breaker to the PDP. So that is 6 feet of #6 at 0.5 WF=3WF or 0.3 volts at 100 amps. Add 11 WF for the battery and you get 1.4 volts. Typical FRC robots at worse case will draw 500 amps so 5 x 1.4 volts=7 volts drop at the input of the PDP. And yes, that is real, the PDP will only see 5 volts at the input with a fully charged battery. That is why the PD of old had the boost/buck power supplies that were capable of running down to 4.5 volts.