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current or voltage for rail gun?

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?
will this capacitor work? (http://nl.rs-online.com/web/p/products/7637858/?cm_mmc=NL-PPC-_-google-_-3_NL_NL_M_IP_and_E_Exact-_-cooper_bussmann%7Celectric_double_layer_capacitors&mkwid=sagNlIMw4_dc%7Cpcrid%7C90954405920%7Cpkw%7Cxv3560+2r7407+r%7Cpmt%7Ce%7Cprd%7C&gclid=CjwKEAiAoOvEBRDD25uyu9Lg9ycSJAD0cnByHZD584cPkCSCgOnFENsjk4lnghmHy3XmwBKaYeWquhoCukvw_wcB)

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

thanks.

sorry for my bad english.

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Jack A Lopez5 months ago

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

Sorry couldn't follow your link.

Joules is important so high volts and capacitance.

Try this one:

http://nl.rs-online.com/web/p/products/7637858/

It is the same link, but minus all the args after the trailing-slash-question-mark, and minus the question-mark too.

well if he could discharge that in one second he would have about 2 horse.

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.

;-)

It will ether do something neat or boring; but I doubt it will last long, it isn't a flash capacitor.

They would be best.

Downunder35m5 months ago

You might want to read up on rail and coil guns first before trying to make one.
For example why the military versions are all mounted on big vessels and vehicles ;)

iceng5 months ago

I managed to follow it.

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...