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

[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.

[chrisofnottingham]

Even after looking at the pdf I'm still not quite sure what the point of all this is. Are you try to find an efficient way to transfer charge or do you think you have broken physics?

.

I've never heard of a 'conservation of charge' before. The only conservation here seem to be energy and how to prevent losses.

.

Don't forget that since Energy stored in a cap is proportional to V squared, the charge is unlikely to remain the same quantity throughout.

[introvertebrate]

@chrisofnottingham You've never heard of conservation of charge???

"the total electric charge of an isolated system remains constant regardless of changes within the system"

[chrisofnottingham]

@introvertebrate I'd never heard of conservation of C as a law. It might have physics applications but it doesn't seem to help as tool for circuit analysis.

.

It is telling me that if I charge up a cap with C coulombs, the net charge is really zero and is equal amounts of +ve and -ve charge (which we knew anyway). Then if I short it out, it will just be zero charge made up of zero + zero. Is that correct or did I miss something? In which case the net charge of all these circuits is always zero.

[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]

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.

[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]

This is far too rambling and you also don't provide a circuit diagram of the oscillator so it is completely useless as a video.

.

Re pdf Expt 1: Energy lost is heat due to 'resistance'. You say there is no resistor but what is the discharge current? With no resistance it is theoretically infinite. This indicates that the idealised circuit symbols don't reflect real life.

.

For instance we can connect a 12V and 6V symbol in parallel but it won't explain what happens with real batteries.

[introvertebrate]

@chrisofnottingham it isn't a super conductor... didn't have one handy at the time, so this circuit has parasitic resistance.

[chrisofnottingham]

@introvertebrate Yes, so that is where your i^2r losses will occur. The point being that the circuit symbols represent idealised components. Things like internal resistance of a battery or inductance of a wire are shown in an equivalent circuit when appropriate. Connecting two caps directly together in parallel is such a situation because otherwise it leads to infinities, which never happen irl.

[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.

[chrisofnottingham]

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

[introvertebrate]

@chrisofnottingham This video is two years old ^_^

[sziamiau1]

Pc= Uc*Uc * C(Farad) / 2

Pc Stat = Pc1 + Pc2

[scubasteve2469]

You need to use an oscilloscope, not a DVM.. This will help you understand what is really happening.

[kukulcangod1]

Introvertebrate , I regard your experiments as remarkable as well as your efforts I experimented shorting out capacitors to charge them but they lost their charge after a while this were the ones for disposable cameras 330 volts thing is the energy doesn't seem to be usable for other than flashes? Best regards

[panoarturo1]

New law of energy = ? The energy can be created, transform and destroyed.

[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.

[prsaledja]

please read at least one physics book before you start to use terms like free energy. if you read it everything will be  clear to you

[thejax]

THATS SO COOL I LIKE THAT..

[openmindmbj]

Great set-up!!

I agree, No OU...this time.

25% efficiency in set-up 1,

75% efficiency in set-up 2.

Considering more losses from the circuit in set-up 2 ,where did the difference (50%) come from?

If you used bigger caps, higher voltage, same circuit, would the efficiency difference be the same?

Thanks for sharing. Show us more.

[openmindmbj]

Could it have been the difference in the I*I*R loss in the wires?

[afdhalatifftan92]

I love your cat... LOL!

[TheIronStark]

not in this experiment

[afdhalatifftan92]

IMHO, he done nothing wrong...

He use two identical capacitor, thus, proving efficiencies is being raised...

[afdhalatifftan92]

Thanks for your patient guiding me...

I'm still learning as I grow... Lol!!

[klas2k]

Hmm - what i see i very interesting - but i do not, without having read the book, see that we get som form om mysterious energy - it is more about efficiency - and yes i read the pdf. I do not find it likely that 75% of the energy goes to heat unless you have very thin wiring/coils...

One thing to calculate with in these types of setups though is time...

/K

[afdhalatifftan92]

Time, the 4th dimension...

Yup, I agreed...

[angelcruzpr]

The schematic is included in the

Book : "Free Energy Generation Circuits & Schematics"

by Thomas Bedini & John Bearden

ISBN: 0-9725146-8-6

Publisher: Cheniere Press

[TommeyReed]

This does not show over unity, you have a lost in charge.

Volts x amps = watts

You're experiment is charging another cap with a lost of charge on the first capacitor.

[xaris106]

nice problem. its not easy actually. you can't use simple circuit theory. the 1st model you describe would require infinite current and also infinite rate of change of current. thus you have to go to maxwells equations and boundary conditions.I think the lost energy is actually EM radiation that can be modeled by a resistor.not sure though.

When you put the L now its not that hard to notice that the energy gets stored in the L and we have no radiation due to non huge rates of change of current.

[crippled82]

Just apply some formulas:

12.50 V on 10mF = .7813 J

out:

8.47V on 10mF = .3587 J

7.24V on 10mF = .2620 J

No OU :-(

or did I miss something?

[introvertebrate]

I think so ^_^

did you read the pdf in the description?

[crippled82]

apparently not :)

[Pirate88179]

Great video! This reminds me of the "Young Effect" as demonstrated on overunitydotcom a while back. All of your videos are interesting. Keep up the good work. Thanks.

Bill

[currentmachine]

not fake at all. good work. perhaps we should try to down transformes the flybackpuls to get more current.

[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.

[introvertebrate]

you are right... when the caps are in series there is less charge than the start...

[impaler999]

cute cat!

[dodoshlodo]

introvertebrate, i believe its possible to extract radiant energy from the vacuum and create a so called "free" energy machine. I am playing around with a radiant oscillator and have built a Bedini.

But this video does not show Overunty. It is well known that u can raise voltage in a cap by pulsing current through a coil.

Just showing a rise in Voltage is only half the equation, u need to measure Amperage. U have to show Power in Vs Power out, otherwise this is just misleading

[introvertebrate]

Did you read the PDF?

[gadzometer]

Hey Introvertebrate, doing a good job by the way. I am not one of those people trying to discredit or deter you in your research. A lot of people just like spiting out Electrical and Physics terms without having any actual experience. They think Google and Desertphile makes them an expert.

I read your PDF and it sounds like you are doing your homework. Keep experimenting and posting your questions and results on certified helpful forums. Then post your results on youtube for us all to see.

[baroutologos]

You can ignore my previous post since i carefully read your pdf included.

Too many so called OU devices utilize the high voltage - current spikes of the collapsing magnetic field to do the trick. Perhaps is a starting point. Who knows, until OU is demonstarted then is assumed not existing. (for conservative reasons)

keep the good work though

[baroutologos]

Nice video dude!

but not quite convinced... if you could only at the end of the experiment you connected those capacitors together... and showed that the "after experiment" voltage of each one was higher from the previously measure one by charging one capacitor and then connecting with the other... then perhapas i would believe :P

waiting ur answer

[fu2ah]

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

Umm, I have been staring at this quote for thirty minutes. Didnt you just prove it to me on video? One of us has a serious problem with reality.

[introvertebrate]

did I prove it to you on video?

[fu2ah]

volts, watts, amps, joules, power/time, charge/time... go to school, at least you are only dangerous to yourself and the other believing dimwits on youtube.

increase in volts is an increase in volts, its an increase in potential. its not an increase in power, its like getting an erection and having nowhere to put it.

suggested reading: desertphiles principia mathematica, and desertphiles natural philosophy.

[introvertebrate]

did you read the pdf I linked to in the description?

[Bonznumber1]

course he didn't that would show him he is dumb and why would he want to do that =) keep up the good work and when i get approved to post on the forums i have some questions and pics of my build of the fan version... it's not powering up =(

[AshMan32]

Farads0.01

VoltageJoulesSum

Va00.030.00

Vb012.180.740.74

Va17.240.26

Vb18.470.360.62

Efficiency0.84

Not OVER UNITY! False advertising.

[introvertebrate]

Hi Ash... I take it you read the description?

[AshMan32]

Hey Intorvertabrate I sent you a YouTube email. Read it.

I just did a calculation of the COP of this experiment. Its 1.5! U = 1/2*C*V^2

[minteko]

very good

[tonytvtech]

no no no look up capacitor charge and energy, simply explained it takes much less power to charge a cap the first 1/2 than it does the second 1/2. study capacitor charging. Just like it takes less power to fill a water tower the first 1/2 than it is to fill the second 1/2. use commonsense.

[introvertebrate]

That's a bit over simplified, but yes I understand. So what is your point and how does it effect the experiment?

[pauldude000]

Then you did , with all BS aside, demonstrate something unique here. Good job!

[pauldude000]

The difference in voltage between the caps does demonstrate one thing which you may have missed. Simply that you are working with an assumed 10,000 uf per cap, which is almost guaranteed to be untrue.

Electrolytic caps (especially the cheap ones) vary in capacitance to the stated capacitance by a factor of +-10% in value. One could actually be a 9000uf and the other 11000uf, with a stated rating of 10000uf on each, and still be in parameter.

The only way to check is with a capacitance meter.

[introvertebrate]

I do have a capacitance meter and I made sure I choose two fairly matched caps. I also did the first experiment as a control test ^_^

[marthale7]

I am wondering... if you took the two in series at the end and put another capacitor in could you grow the potential again? ( Hook the 15 V at primary, and another empty cap as output from your circuit )

Awesome test.

[Drevtoobe]

"Does anyone know a good method for calculating a battery's internal resistance?"

You get 17 Brownie Points for asking a smart question. The link was useless.

You take a medium-sized 12-volt sealed battery. You measure the open-circuit voltage. Suppose you know that about one amp is a good 'medium' load for that sized battery. So you use a standard 10-ohm 10-watt resistor. It has to be 10 watts because the power dissipation is going to be about 14 watts.

[Drevtoobe]

You connect the 10-ohm resistor to the battery for a few seconds and take a voltage reading.

Let's keep it simple for this example and say you measured 10 volts. Therefore the current is one amp. Also, to keep it simple assume that the open-circuit voltage was 12 volts.

So, the battery is modeled as an ideal voltage source in series with the internal resistance. When you did the test, the internal resistance caused a 2 volt voltage drip and the current was one amp.

R = V/I = 2/1 = 2 ohms.

[Drevtoobe]

When a battery is freshly charged, the internal resistance is low. As it gets progressively discharged the internal resistance starts to go up. That is why you see the voltage dropping over time under load.

When you start and end various tests you should consider reporting and comparing internal resistance measurements, it's very useful data about the condition of the batteries. "Fluffy" should be eliminated as a term.

As you might imagine smaller batteries have higher internal resistance.

[Drevtoobe]

When you measure the internal resistance of a battery, then you know what it's short-circuit current would be without having to short it out. You can also see how a nearly dead battery with a high internal resistance will produce little current when you short it out. You can also get a handle on how much a given load is likely to affect the battery if your know its internal resistance beforehand.

Hope that you liked this book!

[Drevtoobe]

Bonus round: If you put a 2-ohm load on the battery then you will get the maximum power out of the battery. When the resistance goes up or down from 2 ohms the power output of the battery goes down. The 2-ohm load is the 'impedance match' for the battery. Whenever you are loading a battery or pick-up coils or whatever, the power you measure into the load is not necessarily the maximum power. Your load may be way off relative to the internal impedance of your energy source.

[rich0319726]

Well, if it's impossible to prove in a video, there is a 1 million dollar prize available in the USA. Why don't you go claim that, LOL!

[introvertebrate]

lots of reasons ^_^ have you read the terms and conditions of the overunity prize? It is a bit more complicated than just making a simple one-off desktop model. Any, what's a million dollars worth these days? £3.50? lol

[tonytvtech]

but you had me for a minute there nice magic trick tho

[tonytvtech]

ok, This is the explaination. your 15 volt measurement was a measurement of 2 capacitors in series...  as we all know how to calculate capacitance in series. you have lost energy in that because your capacitance has been lowered. another words you no longer have 12 volts at 10,000 ufd you have 15 volts at 5000 ufd.. that is a big loss not overunity.

[Drevtoobe]

You said:

start with

1,000 microfarad - 100v ( 5 joules )

1000 farad - 12v (72,000 joules )

which would discharge to

1,000 microfarad - 12.000088v (0.0720010560039 Joules)

1000 farad - 12.000088v (72001.0560039 Joules)

so 4.927998943961 joules left the charged capacitor but only 1.0550039 joules made it's way into the charging cap. Which means 21.4% of the energy required to charge the original cap to 100v made its way to the charging cap.

or in other words 79.6% of the energy was lost.

[Drevtoobe]

So on the mechanical side, how do you envision a perfectly inelastic collision? That's easy: When the two masses hit, there is a layer of putty that deforms and they end up stuck together. The deforming putty absorbs the energy and gets hot as a result.

So look at your example above: A 1 kilogram mass traveling at 100 meters per second has a perfectly inelastic collision with a 1000 kilogram mass traveling in the same direction at 12 meters per second.

[Drevtoobe]

That is almost the same as a one kilogram mass smashing into a brick wall (for all practical intents and purposes) at (100 - 12) = 88 meters per second. ALL of the energy in that smash-up is lost, the one kilogram mass is embedded in the soft putty on the wall.

So yes, your and the boys on the Energetic Forum arrived at the interesting conclusion that cap discharges into batteries where the cap voltage is only a few volts above the bat voltage are inherently more efficient.

[Drevtoobe]

My 'smashing masses' analogy was done in an attempt to have you *SEE* it in your mind as opposed to just accepting it as one more 'rule of the road' to be aware of.

You short two caps together and you get a quick impulse of current and the voltages are the same. Two masses have an inelastic collision and you get a quick impulse of force through the putty and the velocities are the same.

Can you see that? Being able to see that is the difference between being an engineer and a technician.

[Drevtoobe]

Just for fun, watch me do a quick calculation of the percentage of lost energy: Since the 1000 Kg brick wall is so heavy, I can ignore the fact that it is going to increase in velocity just a bit when it is hit by the one Kg mass at 88 meters per second.

Therefore the quick lost energy percentage quick calc is: (88^2/100^2)*100 (proportional to the square of the velocity or voltage)

That equals 77.4% of the energy will be lost in the collision. Your true calculation is 79.6%. Not too bad...

[Drevtoobe]

Wow, I have a thread on the Energetic Forum! lol Yes indeed I had a fight with Aaron about how inductors work, and guess what? I was right and he was wrong. Let that sink in.

For what's it worth, I got an A+ in my Pulse Circuits course, no shit! I did all the labs where we played with inductors and measured energy, looked at waveforms, calculated and measured time constants, etc.

BTW, how come Mr. Bedini charges $400 for a "CD Clarifier" when you can's clean bits?? lol

[Drevtoobe]

I can assure you that I am not a "disinformation agent". I don't get any MIB emails with assignments to go after somebody that self destruct after 10 seconds!

Yes, I thought a Bedini motor was a "magical magnet" a.k.a "Perendev"-type motor when I fist spoke to you, now I know better.

Anyway, in the Energetic Forum you guys have your "own language" when you talk about this stuff, I speak a different language.

For the analogy below, I really hope it helped, cause it's the real thing!

[Drevtoobe]

Seph, Your friend Aaron's account was suspended, any idea why?

[introvertebrate]

Yeah, I noticed that. Youtube haven't given a reason for suspending his account, nor are they willing to communicate. All his videos were his own and none of the content was of an adult nature so there is no ligit reason for closing his account. The only guess I could make is that it was because he was promoting the energetic forum.

[chrisleckie]

the question is not of voltage, it is the over all energy it takes to charge the caps. you may gain voltage but will lose amperage thus.. 0 overall energy gains.

[Drevtoobe]

Okay... the timeout buzzer has sounded... lol

The answer is that when you connect the two caps together like Introvertibrate does in this clip it like the charged capacitor has smashed into the uncharged capacitor (electrically) and they end up stuck together (wires connected). Some of you may have noticed how the calculations below are eerily similar to the measurements made in the first part of this clip.

My example was intended to help you visualize one cap "smashing" into the other cap.

[Drevtoobe]

In addition, this is not an interesting coincidence that they are so similar, they are in fact identical, and you are simply "changing dimensions" when you go from one to the other. It's a two-way street, the masses model the caps and the caps model the masses.

Here are the "dimensional" connections:

velocity = voltage

mass = capacitance

energy = energy

the law of conservation of momentum = the law of conservation of charge

force = current

Put on your tin thinking caps! lol

[Drevtoobe]

Something for everyone to ponder:

Perhaps some of you did physics labs where you worked with the very low friction rail that you can slide masses on and work with springs, etc. It uses compressed air to reduce the friction to a min just like an air-hockey game.

There is what's called a perfectly elastic collision: Two equal masses, one stationary, one moving. The moving mass A hits the stationary mass B. A stops, B continues on at the same speed as A, like a billiard ball, no lost energy.

[Drevtoobe]

Then there is a perfectly INelastic collision: Equal masses A and B. B is stationary and A hits B at velocity V. This time they stick together and now move forward as a pair.

Most of us know about the law of conservation of momentum. Let's say A and B have a mass of M.  The momentum before the hit is MV. After they hit and are stuck together, the moving mass is now 2M. Since the mass has doubled, the velocity must reduce by 1/2 to conserve the momentum. Therefore the new velocity is V/2.

[Drevtoobe]

Now look at the summary for momentum and energy for the two cases:

Perfectly elastic:

Momentum = M * V, before and after hit.

Energy = 1/2 * M * V-squared.

Perfectly inelastic:

Momentum = M * V before hit.

Momentum = 2M * V/2 after hit. (before = after)

Energy = 1/2 * M * V-squared before hit.

Energy = 1/2 * 2M * (V/2)-squared after hit.

The same as: Energy = 1/4 * M * V-squared after hit. (NOTE: THE ENERGY HAS DROPPED BY 1/2 AFTER THE HIT)

[Drevtoobe]

Just to summarize in plain English: Two equal masses have a perfectly inelastic collision as described above.

The moving MASS loses ONE-HALF of it's VELOCITY when it sticks to the second MASS, and the ENERGY in the system drops by ONE-HALF. Also, the MOMENTUM stays the SAME.

Does anybody have any thoughts about this???

[dingusmungus]

Any chance you could isolate the caps from one another with your osc and coil rig? In other words can you make one cap drain until the circuit no longer works and then measure just the output cap. If you could do this in a isolated system you would have something unfortunately you loose a lot of capacitance when you put caps in series. So while the caps have more voltage they have less amps. Please keep up the great research though!

[introvertebrate]

Yes, that is how the circuit is conventionally, but that doesn't demonstrate the charge anomaly so I modified it for this experiment. I agree there is less energy at the end though very little energy is lost compared to simply shorting the caps together.

[Drevtoobe]

"but that doesn't demonstrate the charge anomaly"

Hey, Seph. Like I explained there is no charge anomaly. You are welcome to rebut my points if you want to.

[Drevtoobe]

Seph! I think that I solved the enigma associated with losing half your energy:

Instead of shorting the caps together, suppose that you put a 100 ohm resistor between the two caps and waited for the voltage to be the same on both caps. You should end up at the same place as compared to shorting the caps together. If you change the 100 ohm resistor for a 1K resistor the same thing should happen but you will have to wait a while longer.

The energy is lost in the wire resistance.

[introvertebrate]

Hi Drev,

Yes, still trying to account for the lost energy. One theory goes that the voltage sloshes between the two caps, backwards and forwards at incredible frequency until the voltages equalise and half the energy is dissipitated in heat through resistance regardless of the value of resistor.

Though I have a problem with that theory. You could prevent the "sloshing" by discharging the cap through a diode. However I tried that earlier this week and ended with 2 caps with 6 volts.

[Drevtoobe]

The "sloshing" theory is applicable to another setup; when you connect a charged cap to a coil that happens. Cap to cap you just get a big spike of current and it's all over.

There is a mechanical analogy for this that I can state if you want, but whenever I do this I don't think most people "see" it.

[Drevtoobe]

Hey Seph, just a few more thoughts.

"in the first experiment I simply discharge one cap into the other with no load and negligable resistance, and ended up losing half my energy :("

That is somewhat of an enigma, how come you loose half of your energy? I am not sure what the answer is but with some digging you could probably find it. I think it has to do with the fact that some of the charge moves from point A to point B, and that takes energy.

[Drevtoobe]

For example when the caps are connected, there is a brief spike of current of hundreds of amps or more, and we know that the wire resistance will dissipate energy proportional to the square of the current, (P = I-squared * R) so that may or may not be the reason.

Without showing the calculations, if you assume that you can go from one cap to the other with zero loss of energy here are the numbers:

[Drevtoobe]

Start with 10,000 uF @ 12.45 Volts.

Start energy = 0.775 Joules, charge = 0.1245 Coulombs

End with two caps of 10,000 uF @ 8.8 Volts

End energy per cap = 0.387 Joules

Charge per cap = 0.088 Coulombs

Total charge both caps = 0.176 Coulombs

So you can see that you have increased the number of coulombs, with a corresponding decrease in voltage for the same total energy.

The way to go from high voltage - low current (charge) to low voltage - high current (charge) is via some sort of transformer.

[introvertebrate]

I'm still not sure whether a transformer is capable of this. Doing more research but I think experiments may be in order.

[Drevtoobe]

Just remember the Golden rule for any transformer: High voltage low current on one side, low voltage high current on the other side, for the same energy for both sides.

I am going to slip in a mechanical analogy: A car's transmission. No sheit, a car's transmission behaves exactly like an electrical transformer. Gear ratio is the same as turn ratio.

[harmoniser]

Would it be possible that in the second experiment, when you add the oscillator, the second cap is somehow damaged by the oscillating/AC current and thus gets less capacity, thereby "fills up faster" to a higher voltage?

[introvertebrate]

Highly unlikely as I have performed this experiment many times with the same and different caps and they all appear to be in perfect condition.  But you are right. The only way to end up with a higher voltage in this experiment (according to convential physics) is if the caps are different capacities.

[Careica]

Nice proof! I should fix my SSG motor back and try tests again!

[Drevtoobe]

Great little experiment Introvertebrate because you have to think about it for a while to figure it out.

Just a minor correction that doesn't really affect the discussion: Q = V*C (I don't know where OmegaThe1 got the "2".

For the energy stuff that everybody is pretty much in agreement on, the key fact is that the energy in a capacitor is proportional to the square of the voltage. That's why there is no energy gain in part 2 where you measure 7.24 and 8.47 volts.

[Drevtoobe]

Here is the tricky part:

"In this deomstration I am apparently breaking the law of conservation of charge because it states "the total electric charge of an isolated system remains constant regardless of changes within the system.""

I was also thrown off until it hit me. The key word is "total." The total charge on the capacitor when you start is zero. Remember both plates, positive and negative charge. Which of course means before and after the total charge is zero. The law is satisfied.

[Drevtoobe]

When you think about it further, your oscillator is really a transformer in action. It can take a small "amount of charge" at high potential and turn it into a much larger "amount of charge" at low potential. I am intentionally using the quotations because we now know that we are talking about net zero charge.

Shucks, no energy over-unity, no extra charge being absorbed from the quantum vacuum flux, but none the less a nice thought-provoking experiment!

[introvertebrate]

Thanks drev... it's nice to get some good feedback from you...

though do you think it is possible to use a step up transformer to reproduce the effect? I'm not so sure.

Also there is no conventional transformer effect in the circuit. It is a trifilar toroid. One wire is the trigger with approx 12kohms resistance, the other two wires are primaries. No secondary. All I'm doing is pulsing one cap into the other cap through an inductor.

[introvertebrate]

in the first experiment I simply discharge one cap into the other with no load and negligable resistance, and ended up losing half my energy :(

but in the second experiment I'm doing EXACTLY the same thing except the oscillator is between them. So now the electricity has to go through several hundred feet of magnet wire, resistors, transistors and diodes. Logically, more energy should have been dissipitated out of the system in losses.

[introvertebrate]

but in the second experiment I only lost 11% of my joules. To me, and you may not agree, this indicates an open system.

[Drevtoobe]

I don't know if a step-up transformer would reproduce the effect or be much different from your oscillator system. Since the cap acts exactly like an air tank in this situation it's the amount of charge (or air) in the cap that determines the cap voltage; V=Q/C.

Your trifilar toroid is essentially a fancy transformer: Two or more coils whose magnetic fields are coupled together so one can influence the other and thus transfer energy from one to the other(s).

[Drevtoobe]

"All I'm doing is pulsing one cap into the other cap through an inductor."

Do you mean sort of like a Bedini circuit? The nice thing about that circuit is that the collapsing field in the coil can charge a cap independent of the voltage in the cap, the sky is the limit! It may help explain how you got your voltages in part 2.

BTW, there are single-coil "auto-transformers" that have a common terminal and two taps on different "rings" allowing you to boost/decrease voltage.

[introvertebrate]

I don't understand what you mean the total charge is zero... could you elaborate please?

[Drevtoobe]

Sure, the capacitor sort of has an electrical Yin-Yang. One plate has an excess of electrons for negative charge, and the other plate has a lack of electrons for positive charge. The -ve and the +ve cancel each other out so that if you are standing outside the system you see a net zero charge, or you could say the summation or total charge is zero.

A simple example: A person on a bicycle-generator charging a large capacitor. There is no input or output of charge, yet the cap gets "charged".

[OmegaThe1]

Q = SQRT(2*E*C)

E is energy stored in the cap.

[OmegaThe1]

Introvertebrate, this is really great and I've never seen such simple and clear presentation of OU before.

Can you do a test somewhere where there is no possibility of the coil picking up mains emf?

And can you please post your circuit diagram and component values for everyone to replicate ?

Thanks ! :)

[introvertebrate]

Hi Omega, glad you liked it! However, I ust point out there is a significant difference between conservation of charge and conservation of energy.

In this deomstration I am apparently breaking the law of conservation of charge because it states "the total electric charge of an isolated system remains constant regardless of changes within the system." So no matter what the oscillator was doing, it is impossible for me to of gained in voltage using two equal capacitors.

[introvertebrate]

However, Just because I have gained in charge, doesn't mean I gained in energy (joules). In the first experiment I actually LOST HALF of my energy but conserved my charge. In the second experiment I GAINED 25% charge and lost ONLY 11% of my energy (joules). This implies that this is not a closed system and that energy is entering the system from outside of the circuit.

[introvertebrate]

anyone can replicate it, though I'm afraid that I can not post the exact schematic I am using. Any solid state oscillator should produce the same effects and there are some good equivillant schematics posted on the Energetic Forum.

[OmegaThe1]

Ok after some calculations, it appears that its not OU afterall....

Energy stored in capacitor equals 0.5*C*V^2

and Q = 2*E*C

you started with: (the first figure is V, next is C and last is E)

12.18, 10k = 0.741762

0, 10k = 0

total in system = 0.741762 J

Q = 0.1218 Coulomb

[OmegaThe1]

Contd: after connecting through the oscillator 8.47,10k = 0.3587045 7.23, 10k = 0.2613645 total in system = 0.620069 J Q = 0.1113 Coulomb (if we put back 0.620069 J in just ONE cap somehow, so that its a fair comparison)just shorting the caps gives: 6.24,10k = 0.194688 6.36,10k = 0.202248 total in system = 0.396936 J Q = 0.0890 Coulomb So in all cases E & Q are reduced ! Sorry :(

[OmegaThe1]

Q = SQRT(2*E*C)

typo

[introvertebrate]

Forget joules for the time being... i'm not talking about joules ^_^

but i think you made an error somewhere about the total charge in the system...

8.47v, 10kuf = 0.0847coulombs

7.23v, 10kuf = 0.0723coulombs

which equals a total of 0.1570 coulombs

(started with 0.1218 coulombs)

[OmegaThe1]

I'm not so sure, if we can add charges like that

What I did was added the energy instead, of both caps and assumed that this energy is present in only one cap, which will give a charge of 0.1113 in that cap. Fair ?

(if we put back 0.620069 J in just ONE cap somehow, so that its a fair comparison)

[OmegaThe1]

When you just connect the caps together, following your method

before

12.46*10k = 0.124

after

0.0624+0.0636 = 0.126 coulombs

which is more. Do you see it now?

Its doesn't mean we can generate charge by just shorting two caps together. It confusing to me, but there is something wrong in adding charges like that on paper. Someone HELP :D

[pauldude000]

This is the right method, as the total energy of any system is the combinative of the present energies of each part of the system.

In essence, just like two equal storage batteries, the total energy of the two batteries is the addition of the energy in each battery. Caps are no different in this manner.

Just like the battery scenario, the total output energy would be limited to the average difference between the two caps, as they would first equalize charge upon connection.

My two bits.

[gmeatdaddy]

Great video Sep, Try adding a small lightbulb in series with the cap when it discharges to 2nd cap and let us know if the ending voltage is the same.I have tried a 12v bulb in series with the cap on chrge from the battery and it lights for a second and also lights again when dischargin cap to a second cap but I never measured the ending voltage.Ah more experiments to do.thx for sharing.

-gmeatdaddy

[selamatg]

Hi Seph,

Very good presentation, seems that real OU, but we only talking about voltage. How that can be OU ? that another question.

Nice work

[OmegaThe1]

Energy stored in capacitor equals 0.5*C*V^2

Here, C = 10000 uF, V(before) = 12.18 V and V(after) = 15.61 V, so

E(before) = 0.74 Joules

E(after) = 1.22 Joules

Gain = 0.48 Joules, which came out of nowhere !

Efficiency = 1.22*100/0.74 = 165%, which is OU.

[OmegaThe1]

Sorry that calculation is wrong. The capacitance of the two capacitors connected in series would become half (5000 uF), so the E(after) is actually 0.6 J and there is a loss.... not gain.

[marthale7]

As I watched your video I was thinking, that coils can store energy like a cap maybe the coil has stored voltage from a previous experiment?..... wonder what would happen if you short both caps then... hook them up to the circuit. I am hoping that you picked up radiant energy... One could take a relay system and milk this all day long..

[introvertebrate]

Hi Mart,

a coil can store energy though only while current is flowing through it. Once the current is removed the field around the coil collapses and then there is no energy stored. Unlike a capacitor, a coil can not indefinatly store energy. It is possible that there may be some capacitance in the circuit but this shouldn't have a significant effect with 10,000uf caps!

Good luck with your SS ;)

[FormalScience]

Nice, very interesting...

5000stars*

[demios99]

Sweet presentation Introvertebrate! Beautiful toroid, by the way! Keep on going.