Longer battery wires

[Bennett548]

Originally Posted by Al Skierkiewicz

OK,
running worse case calaculations, a four motor CIM drive will draw ~480 amps at stall. 480 amps will drop .48 volts for every 2 feet of #6, please include the red and black wire in your calculations. The 6 foot run would be 12 feet of wire added to the two feet on the battery and about another equivalent 2 feet for properly crimped connectors and terminals. That adds up to almost 4 volts dropped in just the wiring feeding the PDP whenever you start driving or reverse direction. Add that to the internal resistance of the battery and you potentially have only 3 volts available even on a fully charged battery.
Jumping back to my easy calculator the WIRE FOOT, every WF will drop 0.1 volt at 100 amps.
Battery Internal Resistance=11WF
1 ft. of #6=0.5WF
1 ft. of #10=1WF
1 ft. of #12=2WF
When you view your voltage logs, those short dips to the 4 volt level you see are real. Make your #6 run as short as possible. Often you can significantly shorten the run by simply turning the PDP 180 degrees.

But if there were only 3 volts at the motors, can they still pull 480 amps? I know it is a problem, but I don't know if it is that bad.