Wiring Rules

[Al Skierkiewicz]

Originally Posted by mechanicalbrain

I know it's a pain to calculate,due to a bunch of dependent variables, but I've always wondered how the resistance gained from lower gauge compares to the loss of power due to weight. I guess the lesson hear is to plan your robot to use as little wire as possible (including enough slack that nothing gets pulled).

A rule of thumb I have been using is the "wire foot". At stall several of the kit motors draw about 100 amps. For #10 wire (resistance =.001 ohm) solving for Ohm's Law, V=RxI every foot of wire is equivalent to a 0.1 volt drop. Remember that you have two wires, so if the wire run to a motor is 4 feet, you have eight feet of wire, 0.8 volts at stall will be dropped in the wire. So for our wire sizes (rough approximations) 1 foot of #6 is 1/2 wire foot and #12 is 1.5 wire feet, and #14 is 2.5 wire feet.
As to weight, there are several factors like strand count, type of insulation and temperature rating of insulation that add to the values. Suffice it to say that #6 wire is 50% larger in diameter than #10 and #12 is only 20% smaller in diameter. I have found that there is negligible weight savings of #12 over #10 but a huge savings in voltage drop. #6 will have half to voltage drop but twice the weight. For that reason, I would tend to keep #6 runs short (battery to fuse panel(s) and main breaker) and keep #10 for all motors except the globe. The overall series resistance of this choice actually goes down due to the splitting of the current loads. Remember all robot current flows through the #6 primary wiring but only single loads flow through the individual branch circuits.