Has anyone done testing to see the voltage drop per foot of 12awg vs 10awg wire? We used 10awg this year on all motors but wasn’t sure if someone has data that shows how much drop there is per foot. What about 14awg vs 10awg?

You might take a look at this simulation:

(sorry, I don’t know how to link to the previous post)

Using this tool, you can examine the impacts of changing from 12 to 10 AWG or from 12 to 14 AWG. You can also look at impacts of changing from 6 to 4 to 2 AWG, as well.

Hope that this helps!

Already had this tab in my browser from some shunt calculations this afternoon.

Each 3 AWG increase in wire diameter [decrease in AWG number] doubles the cross section, and cuts the resistance and voltage drop for a given current in half. Each 2AWG step up increases cross section by ~59%, and decreases resistance and voltage drop by ~37%.

Please search for my other entries over the years. However, a simple rule of thumb is a wire foot, WF. Simply stated, a wire foot is a one foot piece of #10 wire. When passing 100 amps, it will drop 0.1 volt.

#6 wire is 0.5 WF/ft.

#10 is 1 WF/ft.

#12 is ~ 2WF/ft.

The battery has 11 WF of internal resistance.

The main breaker when new is about 0.1 WF and each crimped terminal is 0.1 WF worse case.

The PDP varies for each terminal. The amount of copper loos increases as you move across the PDP from the battery terminals.

Speed controllers also have some loss due to internal resistance. Check the spec sheet for the device you are analyzing.

One simply needs to add up the WF to determine the loss. Remember both the red and black wire. Please remember that the battery can produce up to 600 amps when fully charged but the terminal voltage will be reduced by the voltage drop across the internal resistance.

Al’s WF is essentially one mΩ. Converting some of the even entries in the table I linked and rounding to two significant figures:

Admiral Hopper had a famous demonstration in which she held up a foot of wire and called it a “nanosecond”.

So, Big Al’s wire-foot (aka milliOhm) is in excellent company.

I’m a bit confused by this. Mainly the wire foot and this part

6 wire is 0.5 WF/ft.

10 is 1 WF/ft.

12 is ~ 2WF/ft.

You mean AWG of wire when you say number right? and when you say 10 is 1WF/ft I’m not following to much.

1WF = 1mΩ

Thus:

That is to say: ten gauge wire adds one miliohm of resistance for every foot of length that a current must pass through.

People (including customers) have asked me why the latency to command receipt and verification to our GEO spacecraft can take so long (specifically around 250 ms). “Can’t you push this up to 5Hz??” “Not unless you can change the speed of light…”

ns,

I started using this analogy many years ago in an attempt to get entry level electrical students to understand the loss in wire is a function of the wire size and it is real loss. One foot of #10AWG has a series resistance of around 0.001 ohm or 1 mohm. When passing 100 amps through that wire, it will drop 0.1 volts for every foot of wire, black and red. Ohm’s Law is V=I*R. So when looking at robots that are browning out is essential to examine the losses in the system. The WireFoot is a simple way to do this. The battery has 11WF internal so at 100 amps, it will lower the terminal voltage by 1.1 volts. Yes that means that at full charge the battery will read 10.9 volts when delivering 100 amps. So as you start adding up the losses, it is easy to see why some robots are prone to brownouts. Some typical items that catch teams are losses in the range of 200-500 WF in things like loose connections on the battery, poor crimps on #6 wire, loose terminals on the main breaker or the PDP. Even if all of these connections are great, high current can also lower the available voltage. 500 amps across the internal resistance of the battery can drop the terminal voltage by 500 * 0.011 ohms = 5.5 volts. The fully charged battery under this condition will be 12-5.5=6.5 volts.

Now you have heard of fires and smoke on FRC robots right? Well another version of Ohm’s Law relates to power. P=V*I. So for the 500 amp example above, the 5.5 volts dropped internal to the battery results in 2750 watts. Yes that is the equivalent to 27 100 watt light bulbs giving off the same heat inside the battery. Or 500 amps at a loose battery connector could develop a similar power which would result in some really high temperature at the connection. That temperature would be high enough to set flammable materials on fire. That is why teams should never use paper based tape to insulate high current connections like the battery terminals.

Now in real world FRC robots, it is rare to develop that kind of power, but even 100 amps can raise temperature to excessive levels, resulting in smoke or flame. I have seen SB50 connectors melt under conditions where the terminals were not properly crimped or the connector not fully mated.

I think the most telling thing about this simulation isn’t the difference between 12awg and 10awg wire (which is relatively minor), but the difference between an internal battery resistance of 0.015 Ohms, and 0.025 Ohms. If ever you needed evidence that the quality of your batteries matter, here it is. :eek:

Bingo. We retire anything that tests over .021 or so.

Yeah, we made the mistake of ordering some “cheap” batteries from Amazon this year that all shipped with an internal resistance between 0.021 and 0.027. Needless to say we did not use them on our competition bot. We later ordered some from AndyMark that were much better (0.015).

Chris,

What device are you using to measure internal resistance?

The battery beak makes a calculation based on measurements using two resistors as I remember. The difference between 0.011 and 0.015 is not big enough to make a serious decision on battery condition. Many teams use a West Mountain Radio CBAIV. We use this device and it can be set to conditions that manufacturers use to determine the condition of their batteries. This device produces a constant load based on current and then graphs the terminal voltage over time. I use 7.5 amps and 10.5 volts as the cutoff for the test. This will take the better part of two hours to run each test but you can then use the data to make a determination of the internal resistance by calculating resistance using the terminal voltage and current. The CBA does an actual amp hour rating and the 7.5 amps will closely match the standard set of curves for this battery type. Which is even more helpful is the ability to save tests for the same battery and overlay them with the current test.

What do you measure it with and whats the process behind it?

Battery Beak. We have observed short run times and ‘slow’ performance to batteries that show this sort of internal resistance. For our usage it matches with the ‘two year max’ rule of thumb on competition batteries.

It’s not super formal or scientific, but it’s better than nothing.