Charge conserving Capacitive Spring.

[Armagdn03]

Yeah, I know these things need to be engineered correctly, but I have very limited time these days. I usually don't build "all out" until I see that there is something to be had. I'm sure I have missed some real gems that way, but I'm not funded nor have the man power to do everything right. Wouldn't it be nice to quit your day job and build/experiment on somebody else's dime...

I really appreciate all of the insight that you have given. You are a clear thinker amongst all of the chaos.

Dave

I hear ya.

My "lab" has once again been temporarily reduced to a card table, so is life.

I tend to put my all into ideas which I have really really thought out. 90 percent of what I do is proof of concept experiment. When I learn what I wanted to learn I move on.

For me its not so much having a free energy machine, as it is the fun of understanding the "how" of things.... Once this is accomplished the rest is cake as they say.

I think its fine that you guys are looking at the math of the instantaneous vs average, it is really good to defend a position on either side as it makes ya really work things out. Actually this prompted me to pick back up an old calculus text book and start brushing up! good fun!

[Web000x]

Sure you can work it out with the basic technique you've described (more than one way to skin a cat when it comes to math), I am not saying you can't. But you're missing some important points as to why it's not as simple as you've calculated.

Average Power means you need to know the average current and that means you need to know the time constant also since you are only working from Q.

Let me ask you some fundamental questions since you are using what you believe to be the average power.

So what is your average current (dQ/dt) that you are going to use?

What is your average current when the cap increases?
What is your average current when the cap decreases?

You are aware that these are not the same values? The voltage changes are different. The average I is different also. Even though the Q is the same!

You also can't get the average current unless you know the voltage it's changing from and to where it's going to. So I ask you:

What voltage is it changing from when the cap increases? Also what voltage is it changing TO when the cap increases?
What voltage is it changing from when the cap decreases? Also what voltage is it changing TO when the cap decreases?

You have to know the above answer to both caps!

Do you now see why your technique as you've used it is not appropriate? You've simplified it too much that it's basically wrong.

Also, R is not just the R of the resistor, the R is the entire resistances in your loop. This includes the resistance through the tube. Without knowing exactly what R is, we don't even know the real time constant.

Another point in regards to time constant (ie RC).... what exactly is your C?? Is it the static C? Is it the variable C? If so, which one Cmax or Cmin? Or is it some combination there of? Easy to make a mistake here. This alone should show you why the system is not as simple as you've made it to be. The time constant is different when the cap is charging versus when it's discharging.

A small change in time constant and you get different energy results. You're off by 20% in the RC and you're off 20% in your results.

So unless you can really answer these question to yourself and really understand what is happening when these charges are shuttled back and forth, you can't really use the simple method you've suggested.

It could be done with more work as I've pointed out above but there is a simpler method that does not depend on average I or the time constant as long as it's a lot larger than 5RC and that's to just look at the energy before and after the system changes.

And regardless of which technique anyone use, there is one answer and whatever technique used should yield the same answer. Agreed?

I think looking at the energy and going through each step of the process will yield the easiest and most reliable answer. Which is what I've done.

This is more in line of how I was understanding the system. I understand that one RC time constant is going to be lower that the other. So long as the frequency of change is two multiplied by the largest 5*RC, you would still have a consistent average amperage because of the nature of the Cmax/Cmin dwell time being equal.

I am okay with everything in this last post.

However, this was the post that got me confused to begin with:

If your systems parameters were really:

Vi = 7500 V
Cmax = 300 pF
Cmin = 10 pf
Co = 500 pF
f = 57 kHz

Your ideal output would be 5652 Watts.

I made some mistakes in my original calculations on my RC time constants, but they were not in the favor of more power. If you still believe the previous statement is correct, would you mind sharing your methods of calculation?

I'm glad most of this is being cleared up. I knew from the way that you post that you were intelligent.

Thanks

Dave

[SilverToGold]

Thanks Dave, I'm not that intelligent. I just put a lot of thought into this problem. I had the same thoughts as you originally and didn't understand why I was not understanding some things.

I applaud you for doing work and taking an interest in this also.

I am still quite surprised by the energy I calculated but if I went step by step and showed you how I got those values, I think you will be in agreement with me that is in the ball park.

The calculations will only work for sudden capacitive change systems like we have here. It will not work for a mechanical system like Chris Carson.

The problem with this is that I will have to do some drawing and somehow figure out how to get them on this post to make sure things are crystal clear.

Told you I'm not that smart

[SilverToGold]

This is the basic diagram of the circuit. There is a resistance in the tube as well as the load. This is important to consider since when the capacitor is changing, the amount of power through this circuit loop will transfer the power in proportion to the resistance's values.



The voltage source is assumed to be a high voltage sinusoidal AC signal. The diodes are there so that the tube lights up on the positive part of the cycle and OFF on the other part of the cycle. This is needed to allow time for the variable capacitance to completely change value.

Should look something like the clipped signal below but the signal should be something more like a 3kV max and running around 30 kHz.

[7imix]

Silvertogold, great post, thanks! I plan on building this very soon, once I get some parts.

[SilverToGold]

I actually wouldn't recommend anyone building this circuit until they've really gone through and gained a good understanding of how this thing should work. The reason being is that if you do, you probably won't get it "working", get frustrated and just think it's another "free energy" device dead end.

This circuit works on some pretty basic traditional electrical principles that are fundamental. There are no "radiant spikes", "scalar waves", "voodoo meteors from Mars" or some other non quantifiable mumbo-jumbo in this circuit.

But to get it working you've got to understand how it should operate from a traditional electrical component view. Because the design and operation of a working device takes some thought and time to do it right.

I have not built or tested this circuit in full because I don't have the parts to do it properly and from what little I have done with an actual build, I know that it needs lots of attention to extreme high voltages.

So understand the circuit fully before you try building it.

Silvertogold, great post, thanks! I plan on building this very soon, once I get some parts.

[tao]

Some words of advice...

Gentlemen,

I just thought I'd give you some advice, so that your plasma tube capacitor experiments and not done in vain.

You need to ensure that your chosen gas-to-plasma conversion has 'enough free electrons available' for a given voltage. What I mean is, if, when you ignite your gas and produce the plasmas in your tubes, you need to make sure that that plasma has enough free electrons such that the plasma can PROPERLY act like a 'metal plate' FOR a desired VOLTAGE. For, if you don't have enough free electrons in your generated plasma, then you will have a limit to what VOLTAGE you can charge your setup to. I hope that makes sense.

The number of free electrons in an plasma tube is based on a number of factors, but to generalize, and keep things 'less boring' I shall provide the general equation to determine the correct Torr value that your tube need to be such that it has 'enough' free electrons to reach your desired operating voltages.

P=Pressure in Torr of desired gas(neon, argon, etc)
Er=dielectric constant
V=Voltage
a=thickness of dielectric
d=diameter of plasma tube

P=(1.73e-7*Er*V)/(a*d)

Knowing this, if your going for anything above even 500V across your 'plasma capacitor', your going to require at least a Torr of 25, at least. For more voltage, you'll need a much higher Torr value.

I decided to write this post because many are using fluorescents as their tubes, but the problem with this is the Torr values of fluorescents is VERY LOW and therefore, the voltages you'll be able to reach ultimately will not be what would be desired, and now you know why.

I can not take credit for all of the above, for it was JLN that gave me the design parameters from which to calculate the above.

I'd also like to mention, some are talking about running these solid state plasma capacitor models at 20khz and above, but I have heard that at 20khz and above that there is not enough time there for the plasma to even fully extinguish and as such, this would eliminate the possibility of the device to work. As such, and as even JLN mentioned, a desired operating frequency much lower than 20khz, ie. more near to 1khz is desireable.

Good job so far guys.

Tao

[Armagdn03]

I actually wouldn't recommend anyone building this circuit until they've really gone through and gained a good understanding of how this thing should work. The reason being is that if you do, you probably won't get it "working", get frustrated and just think it's another "free energy" device dead end.

This circuit works on some pretty basic traditional electrical principles that are fundamental. There are no "radiant spikes", "scalar waves", "voodoo meteors from Mars" or some other non quantifiable mumbo-jumbo in this circuit.

But to get it working you've got to understand how it should operate from a traditional electrical component view. Because the design and operation of a working device takes some thought and time to do it right.

I have not built or tested this circuit in full because I don't have the parts to do it properly and from what little I have done with an actual build, I know that it needs lots of attention to extreme high voltages.

So understand the circuit fully before you try building it.

Absolutely! I could not agree more. The fundamental concept behind this system is what is interesting. To say it bluntly, 'I know its there, I just need to understand it well enough'.

[Armagdn03]

Gentlemen,

I just thought I'd give you some advice, so that your plasma tube capacitor experiments and not done in vain.

You need to ensure that your chosen gas-to-plasma conversion has 'enough free electrons available' for a given voltage. What I mean is, if, when you ignite your gas and produce the plasmas in your tubes, you need to make sure that that plasma has enough free electrons such that the plasma can PROPERLY act like a 'metal plate' FOR a desired VOLTAGE. For, if you don't have enough free electrons in your generated plasma, then you will have a limit to what VOLTAGE you can charge your setup to. I hope that makes sense.

The number of free electrons in an plasma tube is based on a number of factors, but to generalize, and keep things 'less boring' I shall provide the general equation to determine the correct Torr value that your tube need to be such that it has 'enough' free electrons to reach your desired operating voltages.

P=Pressure in Torr of desired gas(neon, argon, etc)
Er=dielectric constant
V=Voltage
a=thickness of dielectric
d=diameter of plasma tube

P=(1.73e-7*Er*V)/(a*d)

Knowing this, if your going for anything above even 500V across your 'plasma capacitor', your going to require at least a Torr of 25, at least. For more voltage, you'll need a much higher Torr value.

I decided to write this post because many are using fluorescents as their tubes, but the problem with this is the Torr values of fluorescents is VERY LOW and therefore, the voltages you'll be able to reach ultimately will not be what would be desired, and now you know why.

I can not take credit for all of the above, for it was JLN that gave me the design parameters from which to calculate the above.

I'd also like to mention, some are talking about running these solid state plasma capacitor models at 20khz and above, but I have heard that at 20khz and above that there is not enough time there for the plasma to even fully extinguish and as such, this would eliminate the possibility of the device to work. As such, and as even JLN mentioned, a desired operating frequency much lower than 20khz, ie. more near to 1khz is desireable.

Good job so far guys.

Tao

Great information! Thank you very much for the leg work. These design parameters are absolutely fundamental. For example, I have been designing an experiment to measure the quench time, but was looking for a ballpark, you might have saved me alot of time!

[SilverToGold]

Thanks for the post Tao.

I don't like just taking people's words for anything, so I looked a little into it and found this PDF. Looks like the 1000 Hz was taken from this line:

http://www.ece.vt.edu/ece3354/labs/ballast.pdf

"Magnetic ballasts are operated in 50/60Hz line frequency. Every half line cycle, they re-ignite the lamp and limit the lamp current. Although magnetic ballasts have the advantages of low cost and high reliability, there exist at least three fundamental performance limitations due to the low-frequency operation. First of all, they are usually large and heavy. Second, the time constant of the discharge lamps is around one millisecond, which is shorter than the half line period (8.3ms for 60Hz line cycle), so the arc extinguishes at line voltage zero crossing, and then is re-ignited. Figure 4 shows the measured voltage and current waveforms of an F40T12 lamp operating at 60 Hz."

A time constant of 1 ms means 1000 Hz.

I find Figure 5 on page 7 very interesting. The current goes to zero but the lamp is supposedly still ionized and on?

I also found this contradictory statement from someone that sounds like he knows a thing or two about these lamps.

Florescent lighting? - Page 2

"The time scale for changes in the electron density of a fluorescent lamp is far less than 100 usec, less than 10 usec under some conditions, so for all practical purposes a 50 Hz or 60 Hz power source has a fixed output voltage over a 100 usec time period."

This means frequencies as high as 100,000 Hz are possible.

From my own experience I used an LC meter connected to my variable cap while it was being lighted by a 30k Hz high voltage source. The meter samples at a rate of 10k Hz and was not able to get any sort of stable reading. It jumped wildly. To me that indicates that the capacitance was changing faster than the 10k Hz sampling.

So I question the limit of 1000 Hz being some real limit. One also has to take into account the effects of the variable capacitor plates on the electron density time constant. It may make it worse or better.

Gentlemen,

I just thought I'd give you some advice, so that your plasma tube capacitor experiments and not done in vain.

You need to ensure that your chosen gas-to-plasma conversion has 'enough free electrons available' for a given voltage. What I mean is, if, when you ignite your gas and produce the plasmas in your tubes, you need to make sure that that plasma has enough free electrons such that the plasma can PROPERLY act like a 'metal plate' FOR a desired VOLTAGE. For, if you don't have enough free electrons in your generated plasma, then you will have a limit to what VOLTAGE you can charge your setup to. I hope that makes sense.

The number of free electrons in an plasma tube is based on a number of factors, but to generalize, and keep things 'less boring' I shall provide the general equation to determine the correct Torr value that your tube need to be such that it has 'enough' free electrons to reach your desired operating voltages.

P=Pressure in Torr of desired gas(neon, argon, etc)
Er=dielectric constant
V=Voltage
a=thickness of dielectric
d=diameter of plasma tube

P=(1.73e-7*Er*V)/(a*d)

Knowing this, if your going for anything above even 500V across your 'plasma capacitor', your going to require at least a Torr of 25, at least. For more voltage, you'll need a much higher Torr value.

I decided to write this post because many are using fluorescents as their tubes, but the problem with this is the Torr values of fluorescents is VERY LOW and therefore, the voltages you'll be able to reach ultimately will not be what would be desired, and now you know why.

I can not take credit for all of the above, for it was JLN that gave me the design parameters from which to calculate the above.

I'd also like to mention, some are talking about running these solid state plasma capacitor models at 20khz and above, but I have heard that at 20khz and above that there is not enough time there for the plasma to even fully extinguish and as such, this would eliminate the possibility of the device to work. As such, and as even JLN mentioned, a desired operating frequency much lower than 20khz, ie. more near to 1khz is desireable.

Good job so far guys.

Tao

[SilverToGold]

Edison famously said, "Genius is one percent inspiration and ninety-nine percent perspiration."

Tesla's response, recalling the time he spent working for Edison, was, "If Edison had a needle to find in a haystack, he would proceed at once with the diligence of the bee to examine straw after straw until he found the object of his search. I was a sorry witness of such doings, knowing that a little theory and calculation would have saved him ninety percent of his labor."

I truly recommend people go through some thought and look at the simple parallel capacitor connected through a load and do some simple calculations. Prove to yourself that this system works. Don't spend money and effort replicating this without KNOWING what you are doing before hand and this cost you about ZERO dollars to do.

All you need to understand this circuit is this:

1) Charge Q is conserved.
2) Energy is not conserved while the capacitance changes value.
3) Energy is conserves when the charges move from capacitor to capacitor.
4) Energy of a Capacitor = 0.5*C*V^2
5) Q=VC, V=Q/C

This is all you need to know to study this system and get a basic idea of how it works and why it works. This knowledge will allow you to get a system working and a rough estimate of how much power you could expect to generate.

Although we are talking about high voltage systems here, remember that Eric Dollard used a vibrator, some capacitors and some car batteries to power all the electrical needs of his car! Once you study this system and do the basic work, you will understand exactly what he was talking about and how he did it. He DID NOT directly use any high voltage sources. You don't need super high voltages to get this system "synthesizing" energy but you do need to do your homework!

Absolutely! I could not agree more. The fundamental concept behind this system is what is interesting. To say it bluntly, 'I know its there, I just need to understand it well enough'.

[Armagdn03]

Edison famously said, "Genius is one percent inspiration and ninety-nine percent perspiration."

Tesla's response, recalling the time he spent working for Edison, was, "If Edison had a needle to find in a haystack, he would proceed at once with the diligence of the bee to examine straw after straw until he found the object of his search. I was a sorry witness of such doings, knowing that a little theory and calculation would have saved him ninety percent of his labor."

I truly recommend people go through some thought and look at the simple parallel capacitor connected through a load and do some simple calculations. Prove to yourself that this system works. Don't spend money and effort replicating this without KNOWING what you are doing before hand and this cost you about ZERO dollars to do.

All you need to understand this circuit is this:

1) Charge Q is conserved.
2) Energy is not conserved while the capacitance changes value.
3) Energy is conserves when the charges move from capacitor to capacitor.
4) Energy of a Capacitor = 0.5*C*V^2
5) Q=VC, V=Q/C

This is all you need to know to study this system and get a basic idea of how it works and why it works. This knowledge will allow you to get a system working and a rough estimate of how much power you could expect to generate.

Although we are talking about high voltage systems here, remember that Eric Dollard used a vibrator, some capacitors and some car batteries to power all the electrical needs of his car! Once you study this system and do the basic work, you will understand exactly what he was talking about and how he did it. He DID NOT directly use any high voltage sources. You don't need super high voltages to get this system "synthesizing" energy but you do need to do your homework!

I do not assume that the vibrator was causing a variation in capacitance! Very well could be an inductive component.

I have been exploring this more in depth and will start a thread on it soon when I have a good amount to share which is useful.

[SilverToGold]

Interesting, I guess that's a possibility.

Here's Eric's post on this.

"I have a device, built for the Army Air Corps during World War 2, A/N number PP-18/AR Power Converter, which self-sustains the electrical system in my car. It uses the same theory of operation as Chris’s device but involves a different mechanical implementation utilizing a vibrator, several capacitors and 12V and 24V batteries that are connected in parallel through the device, rendering them as one."

Eric clearly states that the caps are connected in parallel to the batteries and that it is based on the same theory as Chris's variable cap device.

As I've stated and seen in the equations, the voltage on the static cap (if the cap is large enough) does not vary very much (less than a volt in correctly designed systems) so it could easily be replaced with a battery. Delivering a current on one cycle and being recharged on the next.

I guess that power converter could be an inductive device but I don't think it's the parametric variance part of the circuit. I believe it's the load component that converts the power into a more useable form - hence the term "power converter". Also he explains that the parallel circuit is connected through this power converter, this places it exactly where the load should be.

One thing I don't think many people get that Eric pointed out was that the unit for current (Ampere) and the unit for voltage (Volts) are not comparable in size. A fairer scale representation is that 1000V is more equivalent to 1A. So people aren't use to working with high voltages but ok with high amperages by scale.

I think we are seeing this with the capacitance variable machines vs inductive variance machines. People don't know exactly how to work with these high voltages.

I do not assume that the vibrator was causing a variation in capacitance! Very well could be an inductive component.

I have been exploring this more in depth and will start a thread on it soon when I have a good amount to share which is useful.

[SilverToGold]

I look forward to seeing you post on this topic as usual.

I'm studying a little on this inductive variance machines as you've posted to this thread. Thanks for the information. I guess I should also review that video Eric and Peter did on that inductive machine, been a while since I saw it and that was before I had any idea about parametric variation and energy synthesis.

I wonder if Bedini's SSG machine is not just an inductive parametric machine? The energy is transfered from one battery to the next but as Bedini always said, the extra energy is in the turning wheel. Could that extra energy be conversion from the variable induction?

I do not assume that the vibrator was causing a variation in capacitance! Very well could be an inductive component.

I have been exploring this more in depth and will start a thread on it soon when I have a good amount to share which is useful.

[boguslaw]

I think it doesn't have to be parametric in this sense even. Take a DC-DC converter for example. One coil when shorted convert large current low voltage into spike of high voltage low current.But what if we make a little twist and create some strange coil.Bifilar,trifilar, multimulti etc.
The only remaining aspect is to find a way to forward all spikes in one direction , store in capacitor and discharge again into the same coil and again and again as we have hit the barrier - everything is full of moving current in one direction ... The only our enemy is Lenz law and schematic thinking.

Think about Tesla and snow ball running down the hill.