Trying to decide what guage wire to use for 40A motor circuits. Even though <R87> allows 12 AWG wire to be used, I’m thinking of using 10 AWG. I like to make things better than the minium. Am I just wasting money on bigger wire, or do you think its worth it? I also believe it will crimp getter. What do you use?
Brian
Team 1225 Team Mentor
From the breaker switch to the distribution block, and from the distribution block to the maxi block, we use the required 6 AWG wire. From that maxi to the victors, we use 12 AWG. In both cases, the gauge of the wire is overkill for safety reasons. However, if you would rather use the heavier gauge wire because it’s easier for you, then use that.
Wire with larger cross section has less electrical resistance per unit of length, and therefore drops less voltage and dissipates less power than wire with smaller cross section. In Europe and most other parts of the world, wire cross section is expressed in square millimeters. In the US it is more often expressed in AWG (American Wire Gage) units, a logarithmic scale defined by this formula:
AWG = 10 + (10 * log10®), where R is resistance per unit length, measured in Ohms per thousand feet of wire.
So 1000 feet of 10 AWG wire is equivalent to a one Ohm resistance. Scaling to dimensions more appropriate to FRC, ten feet of 10 AWG wire is equivalent to a 0.01 Ohm resistance. A current of 40 Amperes flowing in that wire will create a 0.4 Volt drop, and dissipate 16 Watts. The same 40 Amperes flowing in ten feet of 12 AWG wire will create a 0.63 Volt drop, and dissipate 25 Watts.
So using 10 AWG instead of 12 AWG keeps your wiring cooler and makes more power available to your motors; the trade off is higher cost and weight of the thicker wire.
My team uses 10 AWG for our 40 Ampere circuits.
Expanding on Richard’s post above, I like to the use the analogy of the “Wire Foot”. This is an easy way to calculate out losses on the robot. The Chalupa and FP motors have stall currents near 100 amps. #10 wire has about .001 ohm per foot, which at 100 amps drops .1 volts. Add up the length of wire (both sides, black and red, please) and you can easily see how much drop you will encounter for your design. A few fudge numbers for you as well…one foot of #6=.5 wire feet, #12=2WF, Victor=2WF, battery internal resistance=11WF. Good connections should have very little resistance but bad crimps and loose hardware can account for as much as 1 or 2 WF per connection. Add in 1WF for breakers (total) just to be on the safe side. Once you have made the simple calculation then compare it to the motor graph for your motor. If you cannot supply a full 12 volts then the preformance of your motor slides down the graph to a lower performance area of the graph. If you had designed for peak torque, addig 6 ft. of #12 will not allow you to achieve your design goals.
Thanks for posting about your Wire Foot method of analyzing electrical system losses, Al. I think you have mentioned it a few times previously on these boards.
Maybe you have already posted this elsewhere, but if not it would be very helpful if you provide a list of WF values associated with typical FRC electrical system components, and some more examples of how your method can be used by FRC teams planning their electrical layouts. Is this already a CD White Paper?
I have only made this part of my presentations at Championship. If you think it would be helpful, I will assemble a simple table and an example or two. I need to add something for this year’s presentation anyway. Thanks!
Thanks again, Al. I’m sure that many of us will look forward to seeing your table and examples when you have them ready. If I get to go the Championship this year it will be fun to attend your presentation.