Measuring Energy From the Vacuum (Capacitor as source)

[eurogoldexchange]

no chewing gum

[NoctumDeVir]

I tried it with high voltage cap of 400V. Eff about 94-95% as it should be =(

[aether222]

2 ideas, first put a good diode or 2 between the charging cap and the inductor, this way you may be able to keep the inductor and charging cap connected, will simplify automation. (but maybe you won't get the free energy?)

Second, take say 10 capacitors or more, use 5 in parallel as the power cap, and 5 in series as the charging cap, if they are good capacitors and don't lose energy too fast once charged you can reconfigure your charged cap to be your power cap and visa versa.

[legalizeitnow09]

in order for me to see the validity of this being viable is showing all of your wires

[Subspace4d]

I exactly calculated the numbers. C1=7500µF; C2=30µF; U1(1)=25,88V; U1(2)=21,93V ; U2=188V

Useing the formula: Energy = C U² /2 you can calculate that.

dE1=7500µF ((25,88V)² - (21,93V)²)=0.7Joule

dE2=30µF 188V² =0.53 Joule

0.53Joule - 0.7Joule = -0.17 Joule

So we made a loss of 0.17 Joule in fact!

[Subspace4d]

Ups forgot to devide by two. But still tells that you have not overunity.

[NRGFromTheVacuum]

Correct,

This was an experiment to show measurements and the basic concept of EFTV systems. It was never intended to show "Overunity".

I also don't believe any calculations done here were correct. You can't possibly know how much energy is stored in the collector capacitor unless you make that energy do work.

Don't be fooled, basic capacitor equations wont tell you anything.

[Subspace4d]

One thing I have lerned from this is that if you ever want to transport most of the energy from one capacitor to another capacitor, then you should use an inductor inbetween that is switched via a switch. If your source cap is high capacity and your target cap is low capacity then the voltage will be also higher in the target one accordingly.

Those experiments where done by Nikolav Tesla. I read one of his patent and he did the same what you did.

[NRGFromTheVacuum]

Yes Tesla was the one who mastered this effect among other amazing effects he discovered..

One common misconception people keep having is that the energy is transported from one source to the next, this is untrue.

ALL of the energy used from the source capacitor is lost. I have measured this personally. The energy collected in the secondary capacitor does not come from the source. The energy from the source is only used as a tool to gather the energy in the secondary cap.

[Subspace4d]

Is this the Ed Grey effect? But where is the catcher electrode wrapping this?

[baloghcsongor]

Subspace4d . Yep, you're right about this one, but do this all again with data from 1:09! There's the magic moment.

[Subspace4d]

At 1:09 this happen. C1=7500µF; C2=30µF; U1(1)=25,88V; U1(2)=25,87V ; U2=25.70V

Useing the formula: Energy = C U² /2 you can calculate that.

dE1=7500µF ((25,88V)² - (25,87V)²)/2=1.9milliJoule

dE2=30µF 660.49V² /2=9.9milliJoule

9.9milliJoule - 1.9milliJoule = 8milliJoule

So we made a win of 8milliJoule !!! We made more than we used. Correct baloghcsongor! Magic is really happening! Thank you

[Paxmax]

@ 1:09 he just takes the wrong wire and hooks CAP2 to the source instead of cap.

Folks, there is no magic here.

[Subspace4d]

Just do this experiment yourself. It is easy to get some high voltage capacitors and then charge them like this. Then you can see for yourself if it works or not. Thanks!

[Paxmax]

Ofcoz you can pump up some exciting voltage with a boost forward converter. But there is no need to demonstrate it, since it's use and technology is widely known.

Think critically about the cap size shown here, that blue cap is not 40V 7500µF.

More like 40V and 75000µF.

[janne808]

I'm not making any assumptions or selling anything to you. You're jumping into conclusions on things you can't possibly know, the usual 'strawman' argumentation.

[Drevtoobe]

You are making a huge assumption, that this clip is "uncharted territory" that I "can't possibly know" about. I will take a guess that you have almost no knowledge of electronics, just what you have seen on YouTube, which for the most part will really pollute your mind.

[janne808]

You seem to have a need to discredit all this stuff as nonsense, something that will pollute our minds from THE TRUTH. The truth of simple lumped element model. I find this amusing.

Where am I making this huge assumption you're talking about? I'm not making any. You couldn't be more wrong with your guess.

[janne808]

@drevtoobe, yes i know the analogy. but it's just a simple analogy, just like ohm's law and rest of those simplified highschool level physics or university level electrical engineering tools.

[Drevtoobe]

Again, it's pointless for you to make the "simple" pitch, I'm not buying. What is simple is this little experiment. You have a belief from somewhere that something else is going on here that conventional science can't describe. You are absolutely incorrect in that assumption.

Before you can start to "push the envelope" you have to have a mastery of the current state of the art. It ls like that for anything.

[baloghcsongor]

Oh and consider this: if the oil cap had approx. 12.2uF instead of 30uF, the energy transfer's efficiency would be still slightly above 100%. And if I calculated correctly this tolerance of capacity is -60%.. not +-20%. If it would be +60%, the efficiency would be around 400% calculated from values at 1:11. At the other side, eff. would be just over 100% if the blue cap had as much as 18500uF. Hard to belive huh? Calculate it yourself!

[Drevtoobe]

Right, the big moment at 1:11 when the cap goes from 25.88 volts and the small cap goes to 256.2 volts. Stark got very excited about that. Sometime later I will try to crunch the numbers.

[Drevtoobe]

Here is the answer and it might surprise you: All calculations for the event at 1:11 are invalid.

The multimeter display goes from 25:88 volts to 25:87 volts. All this tells you is that there was a voltage drop between something like 0.002 volts and 0.0198 volts.

In other words, the voltage can be between two millivolts and 19.8 millivolts and the meter display would look the same. There is no point in even making the calculation.

[Drevtoobe]

Therefore you and Stark did not understand the limitations of your measuring equipment or you consciously or unconsciously chose to ignore that basic fact.

If you want to follow-up, you have to figure out a way to get around this measurement problem.

[Drevtoobe]

I explained 1:11 for you, the measurement taken is meaningless because both you and Stark could not understand the limitations of the digital multimeter. There is no COP > 1, it's all just a misunderstanding on your part. Care to comment?

[janne808]

@drevtoobe, you're giving the basic textbook model. there's a small error though, coils don't store energy as current but as magnetic field potential energy inside the coil.

like you're saying with your own words, there is high voltage at play here. do you really expect a simplified highschool level physics model to explain the complex dynamics of current flowing thru ionized air as is the case with these devices?

[Drevtoobe]

>>> @drevtoobe, you're giving the basic textbook model. there's a small error though, coils don't store energy as current but as magnetic field potential energy inside the coil.

They are one in the same. Current flowing through a wire produces a magnetic field around the wire. Coil up the wire and you get an inductor. I prefer the current-force voltage-velocity analogy. Do you know that analogy?

[Drevtoobe]

"I prefer the current-force voltage-velocity analogy. Do you know that analogy? "

I should also say that my analogy is indeed equivalent to what you said Janne808, the magnetic field represents potential energy, and the voltage represents kinetic energy. So the current through the inductor is equivalent to the magnetic field potential energy of the inductor.

[armakuni2000]

This experiment has a COP between 10 and 16 or moore please look at :youtubeDOTcom/watch?v=F3fxOqj­Ssyc

[baloghcsongor]

OVERUNITY! Thanks armakuni2000, I didn't pay enough attention on that one... 1:08 is the magic moment! If the multimeters are not lying and the capacitors have really those values (there are no internal problems that could cause one to have less capacitance) then NRGFromTheVacuum have achieved OVER UNITY (and lived to tell the tail :) )! So lets analyze the whole thing with the accepted and approved laws of physics so no one can question it... (continued)

[baloghcsongor]

(continuing) Well going to calculate the energy wasted from the blue cap. (input) and the energy taped in the oil cap.(output) and compare the two. If output exceeds input then we have over unity.

The energy stored in a capacitor is given by the formula: W=C*U*U/2 (google it or get a book). With words: Energy(Joules)=(capacitance(Fa­rads, not microF!)*(voltage on the cap(Volts))squared)/2. And we should analyze the worst case(when the voltage starts to drop) just to make sure. (continued)

[baloghcsongor]

(continuing) Starting energy in blue cap: W1=0.0075F*sqr(25.88V)/2=2.511­654 J. Remaining energy in blue cap: W2=0.0075F*sqr(25.86V)/2=2.507­7735 J. The energy wasted: W1-W2=0.0038805 J=INPUT. Energy in the gray cap: W=0.00003F*sqr(25,37V)/2=0.009­6545535 J=OUTPUT. Woops! Thats almost 3 times MORE! 2.48796637 times to be precise. So the efficiency=248.796637% and this is the worst case remember? Calculating right after the switching the eff. is around 500%. Try it out Id say

[Drevtoobe]

The way this experiment works is very simple. The battery transfers some of it's energy into the coil, current x voltage x seconds. That energy is stored as current running through the coil. When the coil is disconnected from the battery it instantaneously generates a high voltage because that's how all coils work. It's identical to what's going on in a car ignition system. Some of the high voltage in the coil then gets dumped into the very small cap.

[Drevtoobe]

The high voltage will ionize the air for a fraction of a second and that will conduct current into the cap. Since the cap is so small it shoots up to a high voltage. If the target cap was the same size as the source cap, they voltage would only go up a small amount. The wire disconnects from the small cap and it retains the high voltage.

If you could make your measurements properly and do the calculations, you would prove that the conservation of energy was taking place here.

[Drevtoobe]

The first big problem is that the tolerance for capacitor values is probably +/-20%. None of the calculations will be accurate if you don't know the precise values of the capacitors. Using 1% precision resistors you could measure the time constant for each capacitor and get a pretty accurate measurement.

You make your measurements factoring in the cumulative error effects associated with all of your tolerances. Do it right and you will prove conservation of energy.

[Drevtoobe]

V = L di/dt, the voltage across an inductor is proportional to the size of the inductor times the rate of change of current through the inductor with respect to time.

When the moving wire scraping across the metal plate disconnects the coil, it causes a very high rate of change in the current, and the formula above kicks in. The voltage shoots up in reaction. i.e.; reactive power. Stick your fingers across a 12-v relay coil and disconnect the battery and feel it for yourself.

[baloghcsongor]

Drevtoobe, at first I was thinking the same thing: the inductor builds up a magnetic field around it and when the circuit brakes, the coil shouts back a high voltage spike that (by ionizing the air and creating a spark) gets in contact with the capacitor and charges it. BUT how come the energy built up in that capacitor is greater then the energy lost from the initial cap (the energy built up in the coil's magnetic field when connected)? Explain that if you can by conventional physics.

[Drevtoobe]

"BUT how come the energy built up in that capacitor is greater then the energy lost from the initial cap (...)? Explain that if you can by conventional physics."

That's your assumption but in fact you are making the wrong assumption based on the data you are working with and you're a newbie to electronics. I could crunch the numbers but perhaps later, not tonight. I urge you to thing critically about this issue. For example, are you sure the big cap really drops by exactly 0.01 volt?

[Drevtoobe]

>>> do you really expect a simplified highschool level physics model to explain the complex dynamics of current flowing...

Better to say a simple university-level model. Inductors and caps are described by differential and integral equations. They describe and match the dynamics perfectly. When you say "the complex dynamics" you are trying to imply that this is "new territory". That's your big error, this is not new territory.

[janne808]

How about a setup of oil cap containing the initial charge and a bigger farad cap being initially empty? I'm not sure if I follow your rationale of non-equilibrium energy compression ratios.

[NRGFromTheVacuum]

Using the oil cap for the initial charge would not be wise, its rated at 4000+ volts about 250 Joules. Two capacitors the same size, one being of much higher voltage than the other may work though.

[NRGFromTheVacuum]

A capacitor has two major values, Voltage and Farads. The Voltage is energy, the Farads are the amount of time that energy can be released for. So if you have high Voltage with low Farads, your compression ratio is high and so is your gain in energy. A low Voltage with high Farads would be a very low compression ratio, and the result generates only amperes, which is the measurement of wasted energy. You get no gain and no free energy unless you use high Farad to low Farad / low voltage to high.

[janne808]

Try it with two same sized capacitors.

[baloghcsongor]

Lets analyze the test that resulted in a charge of 188V on the 30uF capacitor. Since the energy stored in a capacitor is given by the formula: W=C*U*U/2 the starting amount of energy in the blue cap is W1=0.0075(Farads)*sqr(25.85)/2­=2.505834375 Joules. You make the connection, then the blue cap. goes down to 22.9V. The remaining energy in the cap. is W2=0.0075*sqr(22.9)/2=1.966537­5 Joules. So the amount of energy that goes into the system is W1-W2=0.539296875 Joules.

[baloghcsongor]

continuing..

Now lets see if the energy on the 30uF cap. goes greater then this value W3=0.00003*sqr(188)/2=0.53016 Joules :( thats 0.009136875 Joules less energy out then in.

Overall efficiency on THIS test: Wout/Win*100=98.30578009561060­4085180356366797% NOT OVERUNITY

The idea though is interesting, keep testing, and if you go over 100% please post a video of your work Its too close man 98,3% :) Dont give up! (at 1:47 it looks like 37,3V in stead of 373, correct me if Im wrong)

[NRGFromTheVacuum]

You are correct 37.3 volts not 373.

There is also less output because of the test equipment I have connected.

[Zut50]

An unconventional form of energy? WTF?

[NRGFromTheVacuum]

Yes! "Unconventional" meaning your not use to it, or you have never been taught anything about it.

It follows different laws then electricity, and exhibits effects generally not talked about by anyone in the field of electrical engineering.

Yet its written about in Maxwell's original quaternion equations and Dirac's original theory.

[Termiic]

Great experiment!!!

You don't even need iron-doped Al wire, it can be done more conveniently with a regular Cu or Al wire, you just need a lots of it. REALLY BIG inductance, that is. Basically what we're after is taking so-called 'reactive power' from the source battery (wich doesn't kill the dipole) and then turning that into 'real power'...

[overunitydotcom]

Well, Just try to see, how much voltage your blue cap increases, if you put your charge from the 30uF cap into the blue cap!

If the blue cap charges up higher as the starting voltage you have shown, that it indeed works OU !

Regards, Stefan.

[overunitydotcom]

Now the blue cap lost at this moment just about 1 Volt, so 22 Volts were left in the cap.

When I did not make any calculation error the

blue cap wnet from 1.98 Wattseconds to 1.81 Wattseconds, so it just lost only 0.17 Wattseconds.

So you have a COP of:

2 Joules output / 0.17 Joules input = 11.7 !

Great results !

Regards, Stefan.

[overunitydotcom]

Great results, you are well overunity with these measurements.

If the chargeup cap in the background is 30 uF,

then in one test, where the blue cap had 23 Volts

you charged up the 30 uF cap to 373 at minute 1:47 and this is an energy of about 2 Wattseconds( Joules).

[overunitydotcom]

Was the backside capacitor the 30 uF type ?

[NRGFromTheVacuum]

Yes the blue capacitor is a 40v 7500uF and the back one (big silver one) is a 30uF 4000v oil filled capacitor.