what ya all think ;)
ist
sounds like we would have to do some impeadince matching for the amps to drive the cell does the ohms of the cell increase with the temp of the water??
ist
http://www.overunity.com/index.php/topic,4574.0.html
Ive tried it on a higher powered amp. I need to get a smaller cheapo one.
@ist just a thought,
since the current usually goes up in a cell, that might be declining resistance/ total impeadance...
i did test this out along with g.t and i tried it with a cheap ass amp and i think with some mods done to an amp it would work, but its even hard to find a prof. who will work on a car audio amp just to do a simple repair, let alone figure out a way to make it produce gas, i still like the whole idea though, its a shame it would be so damn hard
Howdy Y'all,
Maybe there is a professional EE right under your noses who has been suggesting using analog dissonant waveforms and audio frequency amplifiers for months and no one noticed. Hmmm, I wonder where he could be? He was talking about using one analog frequency at a frequency that would stress the water molecules and another dissonant frequency to crack the water molecular bonds. He also had suggested using a audio power booster to drive the electrolysis cell. I think this same guy came up with using a ultrasonic transducer to help shake the mineral deposits of the cell. Now who the hell was that? Oh well, I dunno...
Blessed Be Brothers...
Howdy Y'all,
Here's a thought, design your own frequency generators and audio amplifiers. Easy...
Mo Later...
I took the liberty to do some quick math.
I drive a volvo with a 2300cc engine. It runs on LPG.
When I drive down the highway at 100 km/hour it uses about 1 liter of liquid lpg for every 10 kilometers. Doing 1,67 km per minute, it takes 6 minutes to consume 1 liter liquid lpg. 1 liter liquid lpg turns to 270 liters of gasious lpg under the hood in the lpg-converter .
So every minute the engine needs 45 liters. With a fuel to air ratio of 1:15 the engine needs also 675 liters of air to burn it.
1 liter of lpg gas has about 50000 Joules in it. So 45 L has 2,25 Mjoules.
So the engine takes 2,25MJ / 60 = 37,5 kWatt of fuel.
You will need a BIIIIIIIIIIIIG car amp ;D ;D
Dont connect it to your subwoofer 8)
Greetings Spider
Quote from: Spider on June 30, 2008, 06:19:30 PM
You will need a BIIIIIIIIIIIIG car amp ;D ;D
Dont connect it to your subwoofer 8)
Greetings Spider
or a small amp that can miniupulate big power ;) ;) ;D
ever hook a car amp to a mot?? through a hv diode to a cap then to a cell?
me eather yet!!
ist
:)
It gets even worse....From this 37,5 kW the engine only uses 20% to maintain speed. IC efficiency sucks...
Considering the engine is capable of putting out 80 kW on the flywheel, you would need 400 kW of fuel input to get the same performance...:)
Thats a lot of overunity.....
Dont mess up your tweeters too ;)
Greetings Spider