Charge Anomaly ... 12v + 0v = 15v???

Changed the title since the original was a bit controversial.

Simple demonstration of Overunity in regards to the LAW of Conservation of Charge.

As always, I do not claim this is proof of overunity, as I believe that it is impossible to prove in a video.

First I show a simple experiment using two capacitors; one charged and one uncharged. Then I discharge the charged capacitor into the uncharged and read the voltage.

I then perform the same experiment, except this time I discharge the charged capacitor through my toroidial solid state oscillator based on one of John Bedini's designs. Then I read the voltage again.

This clearly demonstrates that there is more charge available after the experiment.

Here is the write up for the experiment that goes into more detail
http://www.box.net/shared/z82ur8o9b5

*** AND JUST TO MAKE IT PERFECTLY CLEAR... I AM NOT SAYING THIS CONTRADICTS THE LAW OF CONSERVATION OF ENERGY. CONSERVATION OF ENERGY AND CONSERVATION OF CHARGE ARE TWO SEPERATE LAWS. ***

Category:

Science & Technology

Tags:

• overunity
• bedini
• solid
• state
• toroid
• toroidial
• demonstration
• proof
• law
• conservation
• charge
• coil

[plopfizzgargle]

You actually have less of a charge at the end of the experiment than at the begining. Don't confuse "Voltage" with "Charge". Capacitance in Farads (C) is the charge in Coulombs (Q) Divided by the voltage (V). In your first experiment the .01 Farad capactors are placed in parallel which gives you .02 F. C=Q/v so Q=C*V, or .02 F*12.5V=.25 C. After the bit with the "ocillator" you place the caps in SERIES. That gives you .005 F. So .005 F*15.6 V=.078 C. Much less than you started with.

[KBendix]

Who said that? Energy can be taken. "Ether", solar wind, heat... Free energy doesn't mean creation of energy. That means you can get it for free.

[Steve042176]

What would it do through a rodin coil?

[TheBblindman]

Sir i have to apologize for the post before! I did the energy conservation law on your first test when the capacitors get to same voltage , which is normal. The initial energy is 2x compared to final Energy, THAT'S REALLY IMPOSSIBLE, the losses in that circuit can't come to 1/2 of initial Energy NO WAY. No "scientific" explanation for that besides IT'S RADIATION LOSSES, that's BULLSHIT, the law of conservation is a flaw. I'm SORRY AGAIN FOR BEING SO NARROW MINDED! i need your email address pls.

[TheBblindman]

This is NOT overunity: C1 initial at 12v, Einitial =C1/2x12^2+E2i(=0) = 144 x constant

at end: Eend=E1(8.5^2xC1/2) + E2(7^2xC2/2) = about 125 x same constant

SO U GOT A COP of about 86%

Guys like you are either stupid or intentionally misleading.

Only way to get costfree energy is by TAPPING RADIATION, like Tesla did or by using very sharp transient pulses at discharge, it's said that a condenser stepcharged is using less energy for charging as per discharging sharp. I have to check that too.

[introvertebrate]

@chrisofnottingham This video is two years old ^_^

[chrisofnottingham]

@introvertebrate I guess because I don't think of circuits in terms of charge conservation it just isn't a problem for me.

.

I get that in parallel there is no path for two caps to build up charge and that this law is probably related to Kirchoff's current law at nodes. But resonant circuits are famous for having large circulating currents in isolated branches. I can't remember all the details these days but thinking about charge conservation looks like a misleading approach to me.

[chrisofnottingham]

@introvertebrate I'm confused. If you know all this why why were you doubting it in the first place?

[introvertebrate]

but in the second experiment, a larger part of the energy is conserved, but when you calculate the charge in coloumbs, it is greater than when we began.

[introvertebrate]

it is a physical law, and yes, the law needs clarification which was the point of the demonstration. You are seeing it out of context.

Charge can be measured in coloumbs, and if you were to take the law literally then it is saying that you can not connect two capacitors together and end up with a greater number of coloumbs than when you started. When you connect two capacitors together like at the start of the video, charge (in coloumbs) remains the same though half the energy is lost.

[introvertebrate]

@chrisofnottingham It would lead to infinites if the componants were ideal and the wires wire completely free of resistance.

but as you said this can not happen. But due to the parasistic resistance, half the energy will always be lost when you connect two identical capacitors directly together. R does not equal zero even when it is effectively a dead short.