Coil Gun - High V, few MFR but very high power (in theory); need opinions

I'm doing an electronic related course in school. I really want to make something amazing using theory I'm getting everyday. So when I learned about coil guns I knew I'd want to make very powerful, yet portable one.

I read that it's better to make two stage coils rather than one. I have doubts about my calculations for the second stage because they seem too good to be true.

The first one has high current and medium voltage so I searched and saw some nice 450V 2200uF, so I guess two of those would do the trick according to this formula:

$$0.5 \cdot C \cdot V^2$$

With that I'll get around 445 joules.

The second coil should be high voltage, low current (I read something to do with saturation or something) so I found those caps at 15KV (15000V) 2200pF. And my mind exploded after doing the math (the same formula):

$$\dfrac{2200pf}{1.000.000.000} \cdot 0.5 \cdot 15000^2 = 247.5 \; joules!$$

So from what little I know and from the price of those tiny caps I wanted to buy like 20. I tested in an Android simulator (Everycircuit) and with 20 of those I get a current surge in the coil at 20 amps. They are also very quick to reload.

But when things are too good to be true I need to hear some opinions of experienced people. So do you think this will work? Because if it does, that means I'll get almost 5000 Joules just from the second stage.

(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.

• ho thanks Andy aka. I knew something wasn't right ^^ i totaly forgot the nano.. and idivided like it was mF uF pF And thanks for the voltage tip I guess I'll try to see caps at 5000v
– Aero
Jun 25, 2014 at 20:15
• Thanks again for the velocity tip, I'll keep it in mind.
– Aero
Jun 25, 2014 at 20:21
• Anyway for free shipping I'll have to buy 4 of those. so I guess I will achieve a reasonable speed, I still have to find good caps for the second stage with high voltage
– Aero
Jun 25, 2014 at 20:23
• Making the 2nd stage high voltage - is there a sound reason behind this dude. I can't see it to be honest!! Jun 25, 2014 at 21:19
• @Andyaka: The pulse in the second-stage coil needs to be much shorter than the first stage. Assuming the coils are basically identical, this requires higher voltages to cause the coil current to rise and fall faster. Jun 25, 2014 at 21:50