Power Draw

[ThunderPorpoise]

Our team has been plowing ahead but we've run into a bit of an problem.
The motors that we are using for our catapult are behaving a bit inconsistently because, we believe, the draw from the battery is momentarily too great. We've thought about capacitors, but we're not sure they're legal. Has anyone else run into a similar problem or does anybody have an idea for how to solve this issue?

[cgmv123]

Capacitors are not inherently illegal, but they would be considered custom circuits, and as such, can not "alter power pathways" including those feeding motors and/or motor controllers.

The best thing you can do is monitor the battery level in code using DriverStation.getBatteryVoltage();, and account for a lower voltage by increasing power. Older batteries can also lose voltage faster, so only use healthy batteries in competition.

[Ether]

The motors that we are using for our catapult are behaving a bit inconsistently because, we believe, the draw from the battery is momentarily too great.

Without further detail, it sounds like your motors may not be geared properly.

You can reduce momentary current surges by using a slew rate limiter in your code.

[Gregor]

Have you calibrated your speed controllers?

[Al Skierkiewicz]

Capacitors are legal for noise suppression only. You could not find a big enough capacitor to hang on a motor to help if brownout is the issue. And then it would only work if your motor was just spinning in one direction. Large caps usually are not bipolar. It is possible that you are tripping the breakers feeding the motors or that the controllers are faulting. A common issue for long runs is to use too small a wire gauge. There can be considerable voltage dropped across smaller guage wire.

[Ether]

We've thought about capacitors, but we're not sure they're legal.

No, they're not FRC legal in that application.

Yes, in theory you could put a capacitor across the motor controller's power inputs.

But let's do a very rough back-of-the-envelope calculation.

Let's say you wanted the capacitor to provide 40 amps for 1/4 second to mitigate the current surge. That's 10 Coulombs of charge.

Let's say the capacitor started at 12 volts (fully charged) and you want the capacitor's voltage to be no lower than 11 volts after discharging that 10 Coulombs.

What's capacitance would be required?

C = dQ/dV = 10/1 = 10 Farads.

Let's put 10 Farads into perspective.

See that big blue capacitor at the top of this image (http://upload.wikimedia.org/wikipedia/commons/9/91/Electrolytic_capacitors.jpg)? It's 45,000 microFarads. It's 5.7 inches long and 1.7 inches diameter.

It would require 220 of those wired together in parallel to give you 10 Farads.

Or, for only $250, you could wire 10 of these (http://www.chiefdelphi.com/forums/attachment.php?attachmentid=16274&stc=1&d=1392659295) in parallel to get a 10Farad 20V capacitor.

[Al Skierkiewicz]

I wonder what the ESR would be?

[Ether]

I wonder what the ESR would be?

Good point. I was looking at very rough back-of-the-envelope best-case calculation, to show how impractical it would be.

I updated the supercap to emphasize that.