# I want to build a portable railgun and I'm a complete rookie [closed]

I'd like to enter the world of electronic hobbies by making a railgun, and so far I know that I understand the basics of the device and that I need high amounts of direct current.

However, I am puzzled, as to where to begin calculating. I'd like to set myself X amount of joules/kjoules of electric energy and work from that, but I know stuff like rail length and how the rails push eachother apart would affect performance, so I'd like to know;

• How Farads relate to Joules (I know the J/V^2 formula but that wasn't clear enough)?
• Should I opt for a single powerful capacitor or a bunch of smaller ones, especially since I want to unload the charge as fast as possible?
• What kind of conductive metal should I use that would be both conductive enough for acceleration and durable enough to resist the recoil and other forces involved?
• How should I cool it or deal with the heat?
• Would there be much heat to begin with in a portable railgun?
• Would ABS coating be a proper body for the gun, protecting the user from the lethal amperages that will be involved?
• How exactly does rail length help the case, as I would use a bipod for a long rifle if that will increase my efficiency?
• Would it be a better idea to have conductive bullets, insulated bullets with discarding and conductive sabots, or a stable aperture that will simply stop at the end of the barrel by whatever means I can come up with while letting the unbound insulated bullet out?
• Should I keep my circuit simple as to just use switches, a battery, capacitor, and some LEDs, or are there more efficient circuit components?
• I'd like to have a control mechanism that lights the Red LED when it is charging, green when it is charged, how can I do that? In my simple sketchup of the circuit I have V-switches (don't know the technical name) aligned in a way that the lights should light correctly during the course of my planned three stage trigger, but better ideas are obviously welcome.
• Will a capacitor lose its current when neither of the ends are connected to a complete circuit but both of them have wires attached to them? I am guessing not but better safe than sorry.
• Is there a way to make sure the capacitor has depleted its entire charge?
• What would be a good idea to improve safety and efficiency?

• That's too many questions. Can you separate them out? Some of them are likely to be answered already here or on Wikipedia etc. – RedGrittyBrick May 15 '13 at 10:45
• Final question - don't build a gun if you want to improve safety but assuming you still will, what sort of current are you expecting to deliver into your rails and what sort of mass do you intend to accelerate and to what speed? – Andy aka May 15 '13 at 11:25
• As I have said, I don't know much to begin with, but for the mass of the projectile I am assuming not more than 9-10 grams as I will try to replicate the 7.62x51mm round without the cartridge, propellant and such. The speed is something I decided not to try calculating, because most of it will be lost thanks to the friction in the rails. I have no idea what the exact amount of current would be, but I imagine I'll use around 16.000 microfarads in total. But really, I could make up my mind if I knew the formulas. – lychnus May 15 '13 at 11:40
• @lychnus, as a guess based on coil guns I've seen to approach the energy of 7.62x51mm you'll probably be looking at something you need to cart around in the back of a trailer. The portable ones I've seen are more like the energy of throwing a dart. – PeterJ May 15 '13 at 11:51
• Oh yes, I just meant that for size, not actual power. I simply want to see the railgun's advantage over coilgun as the rail length has a bigger impact on acceleration than on coilguns. – lychnus May 15 '13 at 11:57

A 10g projectile moving at 100m/s has a kinetic energy given by 1/2 m (v^2), ie 50J.

Assume 10% efficiency, so you need to supply 500J from the capacitor bank. Assume we're using your 16milifard suggested number. Rearrange F = J/V^2 to V = sqrt(J/F). That yields a comparatively modest 176V. Of course, that's the energy in fully charging them, and you can't use all of it due to the voltage drop.

Assume a 50cm rail and a stationary projectile to start (in practice they're launched in mechanically). By s = (v1 + v2) t/2 that gives us 1/100s time of acceleration. That yields an instantaneous power of 50kW and current of ~280A. Delivering that kind of current requires extremely low resistance wiring and busbars and will be quite hard to achieve in a portable design. I suspect that a practical design will use a higher voltage to offset this problem and offset the voltage drop as the capacitors discharge. Perhaps only the top 10-20% of charge can be used effectively.

Putting switches in the power path is to be avoided as the current will tend to weld them shut.

Heat issues: one shot will dump a few hundred J of heat into the rails and air around them. Air cooling will do for low rates of fire.

Safety: is a relative term. You're never going to get this thing CE approved. Shrouding the whole thing in ABS (or PVC piping) to keep fingers away from high voltage, temperature, and moving parts is a good start. I'd apply an insulating varnish or conformal coating to all exposed metal on the internals (except rail active surfaces) as well, as an attempt at double-insulation and to keep out moisture.