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Having a cylindrical iron container with wall thickness of 20 cm, and a total weight of 50 tons when filled with its cargo, how would you be able to calculate its required voltage and current input in a vertical railgun system, with a needed exit velocity of 12km/s? For example, the rail diameter would be 80cm. There would be 4 in total, 2 positive terminals and 2 negative terminals, connected to each other independently, through an 'insulated channel', so it doesn't make contact with the container itself.

Also, is there any formula that can be used for calculating current and voltage required for a certain exit velocity, or exit speed having a certain current and voltage input? Would it depend on many other factors, if so, which?

If any more specifications would be required for such a calculation, please make some as an example, it would be very helpful.

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    \$\begingroup\$ If the rails aren't making contact with the container, then how does it work? The whole idea of a railgun is that current passes through the projectile. You haven't mentioned if you have a pusher or something that sits behind your cylindrical iron container. \$\endgroup\$ – Los Frijoles Mar 3 '18 at 23:59
  • \$\begingroup\$ @Los Frijoles I am sorry to not having included it in the post, but the iron container would have on its sides concave contacts, which would allow current flow from each rail to its corresponding rail \$\endgroup\$ – PlanetGazer8360 Mar 4 '18 at 0:02
  • \$\begingroup\$ Just for giggles, the Navy uses 10 mega-joules to 100 mega-joules to launch 12 inch and 16 inch projectiles at a range of 20 miles. They do not discuss barrel velocity or materials used. \$\endgroup\$ – Sparky256 Mar 4 '18 at 0:18
  • \$\begingroup\$ Unless we can find a ref article, this question is too broad for a simple answer. Maybe @jonk can tackle this. \$\endgroup\$ – Sparky256 Mar 4 '18 at 0:21
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    \$\begingroup\$ Don't you know how to compute kinetic energy? \$\endgroup\$ – Sunnyskyguy EE75 Mar 4 '18 at 0:24
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This page should help you, it's pretty much the exact calculation you're looking for.

https://www.wired.com/2014/08/the-physics-of-the-railgun/

Note that the projectile itself MUST be conductive, as this is the way that railguns work.

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