Coil Gun - High V, few MFR but very high power (in theory); need opinions
(2200pf / 1.000.000.000) * 0.5 * (15000 * 15000) = 247,5 joules!


Sorry to disappoint but pF means divide by 1,000,000,000,000 which makes 2nd stage energy: -

E = \$\dfrac{C V^2}{2}\$ = \$\dfrac{2200 \times 10^{-12} \times 15,000^2}{2}\$ = 0.2475 joules

For the first stage the energy is \$\dfrac{2200 \times 10^{-6}\times 450^2}{2}\$ = 222.75 joules

I guess if you use two 2200uF caps this comes to about 445 joules.

The other thing to say is that you should never run a capacitor at or close to its rated voltage - it will fail much more easily - run about two-thirds for reliability but who ever believed a gun was reliable was stupid so what the heck!!

If the bullet weighs 10 gramme (?) with 500 joules imparted into it, it should achieve a velocity of: -

speed = \$\sqrt{\dfrac{2\times 500}{0.01}}\$ = 316 m/sec. Pretty fast but the real problem is getting the induction coil to impart all that energy into the bullet and more likely it'll be about 25% so the velocity might be more like 150 m/sec. Good luck.