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# current or voltage for rail gun? Answered

Hello i've ben wanting to make a powerful rail gun.
I have ben checking out capacitors but don't know which one to buy.

what my main question is,
should i get a high capacitor with high voltage discharges?
or should i get a high voltage capacitor?

and is current more important than volts in a rail gun?

thanks.

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## Discussions

I think this question of, "current more important than volt[age]", is not a useful question.

If you want a guiding principle for railgun design, I suggest, "Fast is better than slow."

To say that in other words, you want your stored energy to convert to kinetic energy in a small amount of time.

I think that design consideration is going to make you want a smaller capacitance C.

The time constant for a RC circuit is:

R*C, (R times C)

The period for a LC circuit is:

2*pi*(L*C)^0.5, (two times pi times square root of L times C)

For both of those, large C is making the time response slower.

Since energy stored in a capacitor is U=0.5*C*V^2, the only way to store large energy in a small capacitor, is to make V^2 large; i.e. large initial voltage on the capacitor.

That's the only way to do it using a single capacitor.

If you have the freedom to rearrange the capacitors, like from wired in parallel, N*C, to wired in series, C/N, prior to discharging, that is another way to get large V and small C. A Marx generator,

https://en.wikipedia.org/wiki/Marx_generator

does exactly this.

I probably should link to the Wiki articles for RC, LC, RLC circuits too.

I know that's maybe kind of leap of faith to say that a railgun is like a RLC circuit, but I am guessing this is a good approximation.

https://en.wikipedia.org/wiki/RLC_circuit

https://en.wikipedia.org/wiki/RC_circuit

https://en.wikipedia.org/wiki/LC_circuit

The Wiki article on "Pulsed Power" might have some relevant info too, since a railgun is a pulsed power machine,

https://en.wikipedia.org/wiki/Pulsed_power

Also

https://en.wikipedia.org/wiki/Railgun

Well, you know, timing is part of the problem.

That little table of specs, on that page, says this thing has internal resistance of 3.2 milliohm. I mean that seems really small, but what number do I get for Rs and C multiplied together?

V=2.7;C=400;Rs=3.2e-3;

Rs*C = 1.28 s

I mean that's the magic time constant, for an ideal R and C, shorted together; i.e. the amount of time it takes for the voltage on the capacitor to fall to 1/e its previous value.

This might a lower limit on the amount of time it takes to discharge this thing? But I am not sure about that. I mean, I am guessing the load that will discharge a capacitor fastest, is an ideal short, something with zero resistance and zero inductance. Although that's just a guess.

Moreover, I am kind of wondering if it is even like, safe or sane, to intentionally short a capacitor of this kind. I mean that stored energy,

0.5*C*V^2 = 1458 J

has to go somewhere.

I probably should read the data sheet first.

;-)

Energy in a capacitor is C*V*V/2, so Joules of energy increase as the Square of the Voltage..

Your Bussmann 400F 2.75V capacitor only deliverers "several amps for milliseconds" .

BTW the sliding resistance can easily use up over half of your tiny 2.75v...

Yes you will be paralleling several capacitors BUT a rail gun will need almost a Hundred Amps.

https://en.wikipedia.org/wiki/Railgun

This guy used three 120uf 300v capacitors.

https://www.instructables.com/id/Make-poweful-a-gr...