My God… I want one.

Posted at 12:28 pm on Sep 20th, 2008 by Theo V

I wonder if this guy will make it into the Darwin awards (if he’s not careful!)
http://www.darwinawards.com/

Posted at 12:29 pm on Sep 20th, 2008 by Chris

damn… I was going to do this after reading up an a few of the layden jar configurations… doh!

Posted at 12:44 pm on Sep 20th, 2008 by Louis II

While blowing pickles and potatoes up is fun, the 24kj bank’s main intent is induction launchers. Previously, I used a 3.5kj system to launch
induction “rockets” with a 3.5″ diameter coil, as seen here:

http://www.youtube.com/watch?v=bebtDaZP0pQ

I’ve just built an 11″ diameter coil and should be able to send massive rockets up with this bank.

The noise during discharge of that much energy sounds like a military sniper rifle. With this many lytics in an array, they are more than capable of large pulse operations as seen already.

Posted at 12:46 pm on Sep 20th, 2008 by FastMHz

ok, newb question. I have around 2000 4.7uf 50v electrolytic capacitors. Could i (for ease of thought) say charge them all and link them in series to get 100,000V at 4.7uf? i’d love for somebody to point me to something i can do with them all. Thanks.

.dok

Posted at 1:01 pm on Sep 20th, 2008 by dok

In reply to above, yes, in theory you can series all of them for 100kv…but how will you charge them?

Posted at 1:23 pm on Sep 20th, 2008 by FastMHz

I was thinking charge them with some high resistance resistors between them then have spark gaps for discharged (they would act as high voltage switches) that should work right?

Posted at 1:26 pm on Sep 20th, 2008 by dok

charge them with a tesla coil *evil mad scientist laugh*

Posted at 1:37 pm on Sep 20th, 2008 by sly

to dok.
No, you will get less capacitance.
4.7uF/number of capacitors to be correct.

Posted at 1:53 pm on Sep 20th, 2008 by fhunter

I’m impressed that he was able to build that without killing himself. I am off to look for a HOWTO for making those rockets.

Posted at 3:16 pm on Sep 20th, 2008 by Edward Nardella

Is he high or something?

Posted at 3:21 pm on Sep 20th, 2008 by henk

He seems mad with power, not high.

Posted at 3:34 pm on Sep 20th, 2008 by BigD145

i must have on! it will go nicely with my potato cannons (only one of wich i use… in fact. my 4 inch internal diameter cannon i have only shot 1 time.)

Posted at 3:41 pm on Sep 20th, 2008 by mrbob1000

The induction rockets look pretty cool I wonder if the concept could be used in conjunction with normal solid or hybrid rocket motors.
Of course the timing and ignition system will have to non electrical.

Posted at 7:32 pm on Sep 20th, 2008 by nitori

also tok, at the maximum voltage of 50V per cap, those caps will collectively have 23.5J of electrical energy across them.

Posted at 7:42 pm on Sep 20th, 2008 by lambda

Sorry, that’s 11.75J

Posted at 7:43 pm on Sep 20th, 2008 by lambda

I presume fastmhz is planning on mounting this within the trunk / passenger compartment of a car / van and mounting the spark gap assembly inside a sturdy semi-elliptical or parabolic waveguide. At that point he would have a directional EMP cannon that should be easily able to knock out a police vehicle’s *, leaving the officer behind the wheel only able to shake his fist and yell out the window as fastmhz cruises on away.

Just make sure that waveguide is really of high quality; you wouldn’t want any back-scatter or nasty reflections coming back to you…

Posted at 10:00 pm on Sep 20th, 2008 by Dan Fruzzetti

dok, you should link them in parallel. Capacitors need to be in parallel to add like you are thinking. They add the opposite way of resistors. The equivalent capacitance of capacitors in series would be Ctotal = ((C1)(c2))/(C1+c2) start at one end and replace 2 with this equation then use that value and the next capacitor all the way till you only have one number.

Posted at 5:44 am on Sep 21st, 2008 by Johnny

Maybe someone else can tell me why people do this or if anyone else is annoyed by this too.

Why do you say 2400uf instead of 2.4mf? Same thing with wall adapters, its 1000ma why not just say 1 amp? We are using these suffixes for a reason right?

I’m really just curious if there is a reason and if it annoys anyone else? Perhaps I’m just strange.

Posted at 12:35 pm on Sep 21st, 2008 by SidusNare

I accidentally discharged 400uF @ 450V capacitor bank once. the EM pulse froze my Amiga 1200 at 2 meter distance and damaged hard drive, my PC was undamaged 4 meters away, my neighbours had their PC’s reset or frozen.
if fasttmhz does not want EM pulse then he should put the capacitor bank in faraday cage and twist the discharge wires together.
higher voltages should be better for EM pulse generators, lower for coil guns and can crushers.

fastmhz, be careful with that, it can kill.

Posted at 12:49 pm on Sep 21st, 2008 by space

If you set this thing up with a big electromagnet at the end you could make a gnarly fucking EMP…

Posted at 5:51 pm on Sep 21st, 2008 by Joey

“Why do you say 2400uf instead of 2.4mf? Same thing with wall adapters, its 1000ma why not just say 1 amp? We are using these suffixes for a reason right?”
You are forgetting the bit about significant digits. 2.4mF is not the same as 2400uF. 2400uF should be written as 2.400mF which means you’ve actually spent an extra character just to talk in units which you then have to convert back to uF to make any comparisons anyway (there aren’t a lot of capacitors int he world that are even 1mF big).

So, to answer your question, you talk about things in units of the significant digits. 99% of the time 134.567 amps will be written as 134,567 milliamps, and 134.567000 amps will be written as 134,567,000 microamps.

Posted at 8:51 pm on Sep 21st, 2008 by Insipid Melon

@Sidusnare: using mF as a unit is confusing, because it has historically been used for both “millifarad” and “microfarad”.

@Johnny: Ct = 1 / ( 1/C1 + 1/C2 + 1/Cn ) is mathematically the same, but calculating for a large number of identical capacitors is much easier; Ct = 1 / ( 1/C * N )

Your method is also correct, and can also be simplified for identical capacitors:
Ct = ( C ^ N ) / ( C * N )

Posted at 2:46 am on Sep 22nd, 2008 by Sparky

I, too, would giggle like a schoolgirl if I was discharging enough energy to kill me several times over. I mean… seriously, that’s some /energy/

Posted at 4:53 am on Sep 23rd, 2008 by srilyk

The discharge is too fast for a regular camera to do justice in slow-mo. He needs to contact the guys at the new TV show “Time warp” and see if he can get them to record it for their show at up to 20K fps.

You’d have to guarantee not to EM blast the equipment though

Posted at 12:59 pm on Sep 23rd, 2008 by phorce1

Now there’s a thought…I’d *love* to see this thing fire in slow-mo recorded at 20k fps!!!

Posted at 4:44 pm on Sep 23rd, 2008 by FastMHz

+1 creepy high-pitched maniacal laughter

Posted at 12:24 am on Sep 24th, 2008 by witchdoctor

I came

Posted at 5:38 am on Sep 25th, 2008 by fox64

THIS VIDEO IS SO FUNNY

Posted at 12:11 pm on Sep 25th, 2008 by isa de leener

Couldn’t you have used like 2 supercaps of 3000F 2.7V instead, wich is also like 22~24kj ?
Aren’t they fast enough to discharge maybe? Do you really need the 4.5kV?

Regards

Posted at 5:52 am on Sep 28th, 2008 by camel2

Yes, the kilovolts are needed because of Ohm’s law. 2.7v would simply heat the coil and very slowly discharge. The high voltage causes more current to be drawn, the discharge to happen much more quickly, and therefore the magnetic field to be much higher.

Posted at 2:06 pm on Sep 28th, 2008 by FastMHz

hate to be the one who pays his power bill

Posted at 10:38 am on Sep 29th, 2008 by aj-kun

where did u get those capacitors and how much did they cost?

Posted at 10:08 pm on Sep 29th, 2008 by jon

How can you calculate the power output (Joules) of any capacitor bank?

Posted at 1:34 pm on Nov 10th, 2008 by Sieg

if there is any short circuit in one of the capacitor connected in paralle. will it affect the other capacitor?. if your ansswer is yes . how can one rectified the fault?.

Posted at 6:31 am on Jan 12th, 2009 by NAPOLEON

what are the requirement need in construction of 33kv to 11kv capacitor bank.

Posted at 6:38 am on Jan 12th, 2009 by NAPOLEON

@sieg – To calculate power of a capacitor, you take voltage and multiply it by itself, multiply the product of that by the capacitance in farads, and divide by two, to get the energy in joules.

Simplified, this is V^2 * C / 2 = J

Where
V = Volts,
C = Capacitance,
2 = 2
J = Energy in Joules

Posted at 12:45 pm on Apr 21st, 2009 by Christopher Karr