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Author Topic: Yet another Free energy Awnser...But a really good one ;)  (Read 11206 times)

Offline armagdn03

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Yet another Free energy Awnser...But a really good one ;)
« on: August 07, 2007, 06:27:39 PM »
A while back, I got interested in the C-Stack http://jnaudin.free.fr/cstack/index.htm . This device represents a 100 percent separation between voltage and current from the source to the load. While this may not mean too much to you at first, it literally translates to free energy. If all you are transferring from the source is voltage with no current you have zero power consumption from the source?..Yet still have a voltage across and a small amperage on the load. Think about that for a second, immediately off the bat, we have free power with this device, yet nobody seems to recognize this.  The importance of separation of voltage (since it is ?Free?) is outlined in detail in Tom Beardens ?The Final Secret Of Free Energy? 1993, which is a must read for anybody shuffling through these forums. The C-Stack represents a solid state replacement for fancy spark gaps in Teslian technology, in Ed Grey technology and in countless other devices that rely on oscillation for the separation or addition of current and voltage.

I would like to expand upon this device and hopefully give people some direction so that useful devices may be built.

First let me re-introduce you to Tank circuits so that we are all on the same page. There are essentially two kinds of tank, or RLC circuits. First we have a series circuit, where an inductor and a capacitor of equal Impedances are hooked in series with a power source. Upon hitting the predefined resonance, an interesting thing happens, we essentially have a short circuit. We achieve resonance, but we are consuming the most power possible. We do have another option. We can use a Parallel setup where we have an inductor, a capacitor, and a power source hooked together in parallel. At reaching the predefined resonant frequency an even more interesting thing happens?. We have an infinite impedance!!! If you haven?t drawn the lines together yet, that means that in an ideal circuit with no resistance, when all components reach oscillation at their resonant frequency, zero amperage is drawn. Zero amperage means zero power is needed to keep the circuit in oscillation!!!!

So what does this mean to us? It means that if we take this setup, we now have a circuit that consumes no power, but yet creates a sinusoidal pressure (voltage) change. Okay?.so what? Well now is where the C-Stack comes into play. Since the C-stack consumes now amperage, only voltage, its ideal power source would be a parallel resonant RLC circuit. In fact, we could replace the capacitor in the circuit with the C-Stack, find its capacitance, and away we go. We now have a voltage source with no power driving a C-Stack that outputs power from its inner plates!

Now since there is no such thing as an ideal circuit because there will always be some resistance, the RLC circuit will dissipate some power, not zero. Well as long as the C-Stack can output more power than is needed to maintain oscillation we have???..OVERUNITY!!!! The power needed to maintain oscillation is very small. The power put out by the C-Stack is directly proportional to the voltage across its plates. This means that if we up the voltage at the source, it will still consume no power, and yet the power at the load (what ever is hooked up to the inner plates of the C-Stack) will increase.

This is a very simple device to test and build. Unfortunately I am in the market for a new signal generator and oscilloscope, so I am limited in my experimentation.

Lets run over what we have again:

1)   RLC circuit, at resonance = infinite impedance = no power consumption = separation of voltage from amperage

2)   The C in the RLC will stand for C-stack (although traditionally it stands for capacitor) The C-Stack takes a voltage (with no amperage) and creates power.

3)   Some of the power can be fed back into the RLC circuit to maintain amplitude lost due to natural resistance.  Any extra is ?Free energy? The amount of free energy is limited only by the amount of voltage we use.

TO do this we need only a signal generator, a high voltage C-stack (build your own! They do not exist on the market, and a little engineering now how and research can help you optimize design, or you can cheat and message me and I will tell you ;) ), an inductor with impedance to match the C-Stack, and some sort of load (I would suggest some sort of step up voltage multiplier circuit, so that we can feed some power back into the RLC to keep the amplitude of the sine wave steady)

Think about this!!!! Don?t just glance over this and say?.hmmmm interesting, I wonder what will ever become of that. This is about as simple as it gets, An AC signal, a couple metal plates, and a trip to radio shack should do the trick!

Now the real trick, is how do we get an AC wave for ?free? We must expend some energy in its creation?.I think I have an answer, or as close as one can get, but I would like to see what other people think first

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Offline tinu

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #1 on: August 07, 2007, 09:16:19 PM »
Very good topic, armagdn0!

My humble suggestion to the members not familiar with C-stack would be to take it one spoon at a time. The C-stack is an excellent learning tool! That?s its main use, imho.

I talk from experience and I found C-stack to be tricky, at a first glance, at least to my way of thinking, which is pretty much common-sense (I hope ;)). But then, after experimenting and doing a little bit of paper work, well? I will not spoil for now your pleasure of testing. Just ask me if you have any question.


A word of caution: not all ?given truths? on the page are real truths. Take just two of them:

1. ?In a conventional capacitor, if the capacitance is known, and the voltage is known, and the capacitor is allowed to charge to that voltage, then the energy stored must be .5*CV2 . This was not the case.? Really?  ??? Why not?! This is precisely the case, but just not fully understood by the author at the time he wrote that part. Check it out.

2. ?A C-stack can be configured to step-up or step-down (the voltage) simply by which set of leads you select, inner versus outer respectively.? Sorry. It can be used to step the voltage down but not to step it up. Actually the author corrects itself later on: ?This (step-up) appears only to work at its resonant frequency, however. (You can expect that the resonant frequencies of devices on the scale that I built will be in the 3 - 4 MHz range, larger devices will be less).? Now of course the last case is possible at resonance but ask yourself where this resonance really occurs. Is it in the device itself? Well, urrr?  :-\

Tinu

Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #2 on: August 07, 2007, 10:19:17 PM »
 ;D
« Last Edit: February 25, 2008, 04:09:26 AM by armagdn03 »

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #2 on: August 07, 2007, 10:19:17 PM »
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Offline tinu

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #3 on: August 07, 2007, 11:49:04 PM »
So, you are a physicist! That makes two of us.
I?m glad for having you here and hopefully so will other members be.

I said I will not spoil everyone?s pleasure for testing and I?ll try keeping that promise but still giving you some hints. Maybe not all of them tonight (it?s already over midnight here and I type slowly in English?) but at least some.

Firstly, consider various equipotential areas in a flat capacitor and define their relative potential difference to the main (outer) plates. Then take two of this imaginary areas at your choice (actually they are planes, parallel to the outer plates, right?) and put metallic inner plates in their place. Now you have a C-stack. But nothing really changed, right? (You can check by various measurements, on the outside plates in cc but the voltmeter needs to be a really good one, with very high impedance ? the cheapometers will be misleading). Do not touch the inner plates as of yet and do not measure the voltage on them because this is going to change your measurement. Just check that the outer plates still keep the same voltage.

So, by inserting parallel inner plates into the stack or by removing them, nothing changes. The outer capacitor is the same, it has the same capacity and it will behave exactly the same as before inserting the plates, in all respects.
The ?effects? appear when you short-circuit two inner plates. I will leave you to do the equations; they are really high-grade, no much painful work here. So, given the outer capacitor initially charged at a specified voltage and then disconnected from the power supply, the main questions you have to deal with are: How much charge is moving between the inner plates when you connect them, using a wire, a load/resistor, a voltmeter (see?) etc.? Why exactly so much and not more or less charge? After the equilibrium is established, what is the voltage to be measured between the outer plates? Why it has decreased?  :o Ooops, I?ve already said too much.

As results from the above, most of my experiments were conceived and conducted in pure cc and using a bare (but good) voltmeter. (I might have used two of them but that?s all). Why would you need a signal generator and a scope to check the above? It?s fancy, I know, but not really necessary. I?ve tried the C-stack response in ac also but it confirmed what I?ve already found from cc tests. And cc measurements confirmed the known equations (C = Epsi x Surface / Distance). Given the Epsi not well known for various materials one may use, some measurements are still needed to simplify the whole work& calculus but again, the device is 100% obeying the simple equations of capacitance (flat capacitor) and of charge conservation.

For the plates I?ve used 2 regular double-sided circuit boards and one stripped of Cu for the inner insulator. All in all, the whole setup would take minutes before going to take measurements. Yet, the measurements may look very misleading if the above rationale is not properly assimilated. That?s why I said that the main use of C-stack is mainly for learning purposes.

That would be all for now.
If not enough, please ask for more.
Or comment on where I was too laconic.

Tinu

Offline Unicron

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #4 on: August 08, 2007, 12:16:57 AM »
from what i read in the explanation of the C-stack, the plates not connect to power supply should first (draw on their own) electrons from somewere (maybe to get the EMF in balance) before it can discharge. isn't this correct?


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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #4 on: August 08, 2007, 12:16:57 AM »
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Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #5 on: August 08, 2007, 12:35:15 AM »
Ah the discussion gets interesting!

Imagine if you will a c stack with plates 1 2 3 and 4. 1 and 4 are the outside plates, and 2 and 3 are the inside plates. Imagine what will happen if you charge the outside plates. one of the outside plates will be positive, well call that 1, and one of the plates will be negative well call that 4. So far so good.....

Now what is happening to the inside plates. Plate 2 (closest to plate 1) will be close to a positive charge, and plate 3 will be close to a negative charge. Now we have a C stack BUT THIS IS ONLY HALF OF THE GLORY OF THE C STACK!!!!

To really see what this sucker can do, lets keep using our imagination here. What happens if we put a diode inbetween plates 2 and 3. If it is pointed in the right direction, electrons will flow from one to the other. in this case, since plate 2 is closest to plate one it will want to accumulate a negative charge (oposites attract). Conversely, plate 3 will want a positive charge since it is closest to a negative one. So now lets recap. We have plate 1, positive followed by plate 2 negative, followed by diode, by plate 3 positive, followed by plate 4 negative.

What happens if we discharge plates 1 and 4? the electrostatic fields that held the charges in place on plates 2 and 3 are no longer there! the charges on 2 and 3 will want to neutralize. BUT, they cannot now because the diode is there preventing that. So lets put another diode in the picture to allow a discharge, and lets throw a load in the whole mix. Now lets throw an AC signal across plates 1 and 4. Charging of the plates on the outside causes a charge on the plates on the inside. Once the electrostatic field goes to zero, the inner plates will discharge in one direction via diode two and go through the load.

The problem before that people had with the c stack, is what to do with the power from the inside plates. We need an AC signal since it represents a charge and a discharge, that way the presence, the lack there of of the electrostatic field will cause a charging (through diode one) and discharging (through diode 2) of the inner plates powering a load.

If you have trouble with this explanation please tell me so, ill put up a picture

The effects can be greatly increased as well through different means which I can expand upon later, but ill let this be digested first

Offline tinu

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #6 on: August 08, 2007, 12:43:36 AM »
Already done! All the way around. ;D

But go for it! We keep in touch.
Bed time...

Tinu

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #6 on: August 08, 2007, 12:43:36 AM »
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Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #7 on: August 08, 2007, 12:44:33 AM »
 ;D
« Last Edit: February 25, 2008, 04:10:33 AM by armagdn03 »

Offline Unicron

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #8 on: August 08, 2007, 12:52:25 AM »
Ah interesting, but please armagdn03 put up a picture, Hard to keep it all in right now!

Free Energy | searching for free energy and discussing free energy

Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #8 on: August 08, 2007, 12:52:25 AM »
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Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #9 on: August 08, 2007, 03:57:46 AM »
Here is a photo so you can better visualize.

(http://i210.photobucket.com/albums/bb275/tortuga0303/cstack.jpg?t=1186537589)

in between each plate you can imagine a dielectric.

Now to maximise this set up I imagine the following.
The dielectric between plates 1 2 (3 and 4 as well) will have dielectric constants as close to 1 as possible. This allows for maximum transfer of static field between the inner and outer plates, I.e maximum pulling power on the electrons in plates 2 and 3.
The dielectric inbetween plates 2 and 3 will have a very high constant. This blocks alot of the electric field between the inner plates making its capacitance increase greatly. now we have good capacitance on the inner plates, which are strongly being affected by the outer plates, and we have an optimum combination for power generation.

Tinu, you said that much of what I have described has already been done...could you share a few results so that I may learn from your mistakes before making my own? I would love to know the shortcomings of this device.

Also different variations are possible.
For example take the following image

(http://i210.photobucket.com/albums/bb275/tortuga0303/cstac2.jpg?t=1186538079)

The plates and components are labled differently which makes it a bit confusing from the previous description, but the setup is the same, except that in this case, the inner plates rotate in and out of the outer plates via a shaft.

With this setup, you do not have to supply a frequency to the outer plates. Instead, you could have a constant voltage source, say a high voltage capacitor that keeps voltage on the outer plates. now the charge discharge cycles are dependant on rotation of the inner plates. the load could be an electric motor to turn the shaft.

Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #10 on: August 08, 2007, 04:09:17 AM »
Also, tinu, you say that the voltage on the outer plates decreased when you short ciruited the inner plates....how is this possible? is it due to the fact that now that the inner plates are electrically connected they circumvent the dielectric inbetween them, thus transfering the field to the outer plates thus having the oposing field vectors cancel? (making it appear as if there is less voltage, but there is still the same ammount of energy conserved.) could it be due to the impedence of your voltmeter? (I now you said that you used one with very high impedence, but just asking) could it be from ionization? I dont see how voltage can just dissapear, unless the total energy is concerved, yet the voltage simply doesnt seem to manifest itself.

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #10 on: August 08, 2007, 04:09:17 AM »
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Offline tinu

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #11 on: August 08, 2007, 09:19:36 AM »
Also, tinu, you say that the voltage on the outer plates decreased when you short ciruited the inner plates....how is this possible? is it due to the fact that now that the inner plates are electrically connected they circumvent the dielectric inbetween them, thus transfering the field to the outer plates thus having the oposing field vectors cancel? (making it appear as if there is less voltage, but there is still the same ammount of energy conserved.) could it be due to the impedence of your voltmeter? (I now you said that you used one with very high impedence, but just asking) could it be from ionization? I dont see how voltage can just dissapear, unless the total energy is concerved, yet the voltage simply doesnt seem to manifest itself.

Yes, that?s the sad point, preventing any OU in the device. The voltage on the outer plates decreases when you short-circuit the inner plates because the E field between the inner plates becomes zero. Once E is zero, the equivalent distance (D) of the dielectric (C=Eps x S /D) decreases also, because the distance between the inner plates does not matter anymore (it?s like there is no distance there, from the poit of view of the outher plates). Then the capacity of the outer plates will increase like 1/D. But the total charge on the outer plates (Q) remains the same, as long as they are kept in an open circuit . Because Q = C x U, having an increased C invariably leads to a decrease in U.

Alternatively, in order to check the above, try the following setup. Charge the outer plates (a small 9V battery will do). Place a microammeter in between the battery and the outer plates and watch it carefully. Normally, it will measure (if sensitive enough) a very small leakage current through the dielectric. Now, short-circuit the inner plates. You?ll see a clear increase in the current through the ammeter, thus clear evidence that the voltage across the outer plates has dropped, according to the above explanation. When you short-circuit the inner plates, the outside power source is doing work, to compensate for it. So, what you get from the inside in terms of energy is either paid for from the battery (or, if the battery is disconnected, is paid for from the total energy stored on the outside plates) ? hence, no free lunch.

Sorry for the bad news.
The experiments still may be very exciting but imho an OU device has to burn something. That something may be anything (at least in principle) but no nothing. Unfortunately, C-stack device has nothing to burn, if regular dielectric is used for building it.

Tinu

Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #12 on: August 08, 2007, 03:19:23 PM »
I fully understand what you are saying. But think about what you have just told me. By short circuiting the inner plates you essentially trick the outter plates into thinking they are closer together than they really are. Now you have increased the capacitance of the outer plates. Because of this, you will see a drop off in voltage, and if connected to a power source, you will see a draw of electrons (amperage) due to the fact that for the given voltage (9 in your case) the outer plates can now hold more charge. Do I have it right so far?

Well I believe what you have done is acurate and correct, however what I believe that you have concluded is that this will continue to happen in each cycle. Take for instance the C-Stack motor I have described above. When the inner plates enter the outer plates, the voltage on the outer plates will drop, but there is no electron movement. energy is concerved in the outer plates (and capacitor connected to them). If more and more rotations ocur, there will not be a continual voltage drop. If there wasy, backing up a bit, we could take your setup with the nine volt battery, and through action of shorting the inside plates over and over, we could create a capacitor that could hold infinite charge!!!!!!

I believe what is happening is that before the inner plates are shorted, we have 1 capacitor setup at a particular voltage. When we short the inner plates, we have another setup where the particular voltage drops, and teh capacitance increases hence your amperage readings from your nine volt. but I do not believe that the voltage will continue to drop over and over again as you repeat the process or that would mean a transfer of electrons between the inner and outer plates, which I believe to be imposible.

This is why I would like to try these expiriments either with a signal generator, or the rotating c stack shown above. This way we are not "tricked" by taking a look at only one "cycle" of the process.

Offline armagdn03

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #13 on: August 08, 2007, 03:22:40 PM »
 ;D
« Last Edit: February 25, 2008, 04:09:44 AM by armagdn03 »

Offline Unicron

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Re: Yet another Free energy Awnser...But a really good one ;)
« Reply #14 on: August 08, 2007, 04:48:04 PM »
I think it wil not work, becouse of the lack of electrons.... the EMF of the outer rods keep the inner plate electrons in its place, therefore it will not flow.
and if you short the inner plates you'll probable bring in in balance.

If my thinking is wrong please correct me!

 

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