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#1
10-12-2010, 06:06 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Charge conserving Capacitive Spring.

Charge conserving Capacitive Spring

Or

The "Time Machine"

Concept:

http://i210.photobucket.com/albums/b...3/DSCF0071.jpg

Here can be seen the basic circuit.

C1 = Capacitor having movable platens.
C2 = Static capacitance condenser

Basic Embodiment:

Initial conditions:
Free Space Permittivity 8.8541E-12
Area Of Plates 1129420269
Initial Voltage (both caps) 10
Charge (C1) 100
Charge (C2) 100
Total Charge 200
C2 Capacitance 10

1) Capacitor 1 (C1) will be a parallel plate variable condenser with movable platens.

2) Connected in series to (C1) will be a capacitor of initially equal value (C2) and resistive load (R).

3) Initially both capacitors will be charged to an equipotential state, so that both have the same voltage across the terminals, and by virtue of having the same capacitance, they will also have an equal charge between them. (See graph for charge distribution)

4) Entering into the time domain, the platens of capacitor (C1) will beg separated, causing a decrease in capacitance via the equation:

As the distance between the plates increases, the capacitance of (C1) moves asymptotically towards zero.

5)Because of the decrease in capacitance (C1), with charge kept constant across the combined capacitor of C1 and C2(law of conservation of charge: Charge conservation - Wikipedia, the free encyclopedia ) The capacitance across combined capacitors (C1 & C2) will decrease and the voltage across combined capacitors (C1 & C2) will increase.

6) The rise in potential across C1, will cause a potential difference between C1 and C2, meaning they are out of equilibrium. Because of their tendency towards equilibrium, charge will move from C1 to C2 (the sum of both will always equal the total charge initially in the system). Because there is a conduction path through the load, the capacitors C1 and C2 constantly equalize, and as the plate are separated, a new higher voltage is attained across both capacitor simultaneously. This is shown in the graph below.

Here the values for the capacitors are 10 farads initially, then C1 decreases as discussed. Here we have capacitance of combined capacitor C1 & 2. It starts at 20 farads (parallel reference frame) and as C1 tends towards zero, the total capacitance moves asymptotically towards 10f.

Here is the total voltage across both capacitors. Starting out at 10 volts initial, as C1 tends towards 0, the voltage asymptotically moves towards 20V.

The charge distribution across caps (C1) and (C2) changes with respect to time, meaning charge has moved through the load in the equalization process. With initial conditions of C1= 10 Farads and C2= 10 farads, both have an initial charge of 100 coulombs on each condenser. As the capacitance of C1 (blue line) decreases, so does its charge, while the charge across C2 (pink line) increases. The sum of both at any one moment equals 200, or the initial condition of charge once again satisfying the laws of conservation of charge.

Finally we have the force between the plates of condenser (C1) as a function of distance or time:

This graph follows the force between the plates of condenser (C1) as they are moved apart and the voltage across both capacitors increases simultaneously. As you can see, while close, there is great force and while they separate, the force decreases. The behavior is that of an exponential decay, much like an inverse square field.

So what does this equate to?

By separating the plates of the capacitor, you have raised the energy of the entire system. In doing so you have caused a finite amount of charge to move across the load for a given time, giving you your power. The force exerted will be given by the integral of the equation for force between the plates, so that we can see the total force required to pull apart distance X.

What is interesting is that while it requires X force to pull the plates apart, they still have a tendency towards moving back together due to the electric forces. -X energy will be exerted by the system returning the plates to their initial position of 10 farads.

Energy to pull plates apart = X

Total energy = X+ (-x) = 0

For each investment of mechanical energy in the direction of plate separation, the system returns the energy in the next stroke returning to initial conditions.

For each stroke (moving the plates out, or moving the plates in) you move (Y) charge in time (T) where power equals (Y)*(T) = Watt

How quickly the plates can move appart or come together is a function of the TIME CONSTANT of the circuit.

This can be figured out through calculus solving the following KCL equations.

These will solve to give an equation similar to the form of

The heavier the load...the lower the resistance of the load...The faster the time constant.....The more charge moved in less time, equating to more power. Heavier loading = increase in power output.

If the load is an inductor instead of a resistor (perhaps a transformer)....heavy loading on the secondary equates to decrease in reactance of primary = very small time constants = charge moved very quickly = large power output.

If the load is an inductor like a motor. Because charge always stays constant, the motors BEMF is not allowed to cancel any charge via its counter generative properties, hence a standard motors BEMF will no longer affect its operation.

All energy used to cause initial separation of plates, and rise in potential of overall system, will be returned in the subsequent cycle as plates return to initial position and energy state of the entire system decreases. Net input theoretically = Zero

If all input is returned, and each cycle constitutes a movement of charge, in a definite time dictated by the impedance of the load, then the load dictates the time required to complete a cycle, and how much power is developed. Meaning While loading is constant, Time is the only variable consumed

The "heaviness" of the load dictates the time constants, the less the time constant, the more power developed

Put another way, Decrease in time constant = decrease in time domain = increase in possible frequency used. Increase in frequency = increase in power.
Plates can be moved acoustically, through Super Radiance and Phase transition effects, mechanically, etc.
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Last edited by Armagdn03; 10-12-2010 at 06:35 PM.
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#2
10-12-2010, 06:13 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
This is one method of accomplishing what Dr. Lindemann mentions here in the upcoming Renaissance workshop:

Quote:
 I am going to explain EVERYTHING I know about how and why Back EMF functions in electric motors, how it masks the real efficiency of these machines, and how to overcome it, even in CONVENTIONAL MOTORS.
Because charge is conserved through the motor, BEMF does not affect the process like it does within a traditional circuit.

Happy B-Day Sucahyo!

Sorry i cannot make it to the demonstrations, sounds like fun.

This is also the same circuit I was referring to when I posted in the parametrics thread after Eric Dollard Spoke:

Quote:
 Imagine a circuit where action and reaction are separated by TIME. The charge in the system is always conserved and stays the same, The work itself oscillates, causing a rise in potential and current simultaneously in the load. You raise the energy state of the entire system and cause an internal imbalance. The system then seeks balance, and in so doing accomplishes work, over a set time. After equilibrium is reached, the system is at a higher energy state than in began at. The amount of time dictates the power at a given moment. Then the system is allowed to return to its original lower energy state (the reaction separated by time), causing another internal imbalance, which takes TIME to equalize, again causing the same amount of work to be done as the initial impetus. It would seem from this circuit that it is possible to create circuits in which time is the dominant factor in output to the load. Dare I say, it is possible to harness the action, reaction, and time is our "source".
Enjoy!

Andrew M.
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Last edited by Armagdn03; 10-12-2010 at 06:25 PM.
#3
10-13-2010, 03:38 PM
 Jetijs Gold Member Join Date: Aug 2007 Posts: 2,134
Makes sense and makes you think. What I don't get is why the separated capacitor plates would pull themselves back again? I guess if HV would be used, then the electrostatic forces could do that, but at low voltage I don't know.
Thanks,
Jetijs
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It's better to wear off by working than to rust by doing nothing.
#4
10-13-2010, 04:13 PM
 rave154 Silver Member Join Date: Jan 2008 Posts: 926
very thought provoking ARMA
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#5
10-13-2010, 05:14 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by Jetijs Great thread Armagedon Makes sense and makes you think. What I don't get is why the separated capacitor plates would pull themselves back again? I guess if HV would be used, then the electrostatic forces could do that, but at low voltage I don't know. Thanks, Jetijs
The voltage itself causes a mechanical action towards closure of the two plates.

In a simpler version of the demonstration an only one capacitor is used with moving platens. Charged, the plates are moved appart, and due to the conservation of charge the voltage must rise to compensate for the decrease in charge Q.

The energy stored in a condenser is given by U = .5C*V^2. Thus, as the capacitance decreases, the voltage increases, which is the squared unit within the energy equation giving rise to energy. The separation of the plates raises energy state, and the force required move the plates represents the energy expended to create this additional voltage / energy. Thus traditionally you have no energy gain. The resistance to moving the plates due to coulomb electric forces is in response to the gain in energy.

You put mechanical energy in, and get electric rise in potential energy out.

Were it not for this force, the energy gained would already be "free".

As you can see by the graph of the force between the plates, when the area, or the voltage is high enough, force can be rather great.

Here I have a large area, specifically 1129420269m^2 which I chose to give the capacitor a value of 10 farads. Unrealistically large. But one can easily decrease area and increase voltage I used only 10volts. To those "skilled in the art" the proper proportions will be taken into consideration.

What is interesting about this setup, is that all charged moved...and all mechanical force used to pull the plates apart...and cause charge re balancing...will be returned when you let the plates go, and the coulomb forces cause them to close once again.

In reality perhaps you are using several KV, and moving the plates only 1mm or so, Or not really moving the plates at all, but using a property such as optical coherence to achieve a reduction in capacity.
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Last edited by Armagdn03; 10-13-2010 at 05:24 PM.
#6
10-14-2010, 02:09 PM
 Michael John Nunnerley Gold Member Join Date: May 2008 Posts: 1,193
Hi, look at Dr. Stifflers last two videos, simplified S- Gate,

Mike
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#7
10-14-2010, 02:59 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Yes I was looking at that yesterday, thanks. While you may perceive similarities, i do not intend any similarity between this embodiment of this device and Dr. Stifflers work, as it does not use SEC theory. However, other embodiments may have similarities.
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#8
10-14-2010, 04:17 PM
 rave154 Silver Member Join Date: Jan 2008 Posts: 926
ARMA,

i just watched the MIT capacitor video you posted.

interesting, the current flow seemed somewhat proportional to the "speed" that the plates were being seperated at.

also, at the end, where it shows a plexiglass sheet being used as a "block"....thinking of those patents & inventors ideas where they attempt to alternately block & then allow magnetic flux by using a spinning disk with various materials glud to them ( im sure you know what i mean )... with this idea in mind could we not replace the plexiglass sheet with a plexiglass roter... with segments cut out that would give su the same effect as lowering and raising the sheet....in order to generate current each time.......the motor powering the rotor would not need to be powerful at all as almost zero torque would be needed ( perhaps a pulse motor since they usually only have a very small torque and can also capture the kickback spike too ).

thoughts?

David. D
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#9
10-14-2010, 05:40 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
@Rave154

Please note patent 3,013,201 "Self Excited Variable Capacitance Electrostatic Generator"

Or 3,094,653 "Electrostatic Generator"
Or 3,175,104 "High Voltage Electric Generator"

Or note the Evert Motor, which is a similar electrostatic version of the "inverse transformer" I posted a while back.

evert rotor tech

These would all be viable choices for a method to vary the capacitance. Personally I would like to tend towards a solid state method of variable capacitance because it would reduce build time, effort, and money. Two strong candidates for doing so exist that I know of.

In the solid state embodiment, the "force between the plates" analogy falls apart, because there is no real change in distance between the plates. The mechanical concept of manual plate separation is one embodiment of the concept of shuttling charge between two capacitors one of variable capacitance. I included it here because it is easiest to imagine the forces on the plates in this state. How when you separate, the charge will move to redistribute, how the voltage of the combined cap will rise towards a 2x potential, how the time constants dictate the forces between plates at any point in time etc. Conceptually it is easy.
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#10
10-14-2010, 08:40 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by rave154 ARMA, i just watched the MIT capacitor video you posted. interesting, the current flow seemed somewhat proportional to the "speed" that the plates were being seperated at. thoughts? David. D
Actually, the plates in the beginning of the video are being kept at constant voltage. Thus as the plates are separated, charge must leave the capacitor to satisfy the constant voltage criteria. As the man turns the knob, the distance between the plates dictates the capacitance, and thus the charge on them, not the time in which the separation took place. However, the quicker the man turns the knob, the more coulombs per second are moved which equates to higher power at any given moment.
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#11
10-15-2010, 05:28 AM
 Mario Senior Member Join Date: Apr 2007 Posts: 423
Hi Andrew,

great thread, thanks! I would like to share a thought. Say we make our variable cap in a very crude way by putting two caps in parallel and vary the capacitance by connecting them in series (a bit like the scalar charger). If the series caps total voltage is 20V and they get discharged into a cap sitting at 10V half of the energy is lost. Are you suggesting that this doesn't happen if the capacitance is changed gradually (stepless)?

Also, how would you vary the capacitance solid state? I know about varicaps but am not sure how you would use them in this application. I suppose the easiest way would be to have a cap with a capacitance that could be varied with a control voltage that wouldn't consume almost anything...

regards,
Mario
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#12
10-15-2010, 02:32 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by Mario Hi Andrew, great thread, thanks! I would like to share a thought. Say we make our variable cap in a very crude way by putting two caps in parallel and vary the capacitance by connecting them in series (a bit like the scalar charger). If the series caps total voltage is 20V and they get discharged into a cap sitting at 10V half of the energy is lost. Are you suggesting that this doesn't happen if the capacitance is changed gradually (stepless)? Also, how would you vary the capacitance solid state? I know about varicaps but am not sure how you would use them in this application. I suppose the easiest way would be to have a cap with a capacitance that could be varied with a control voltage that wouldn't consume almost anything... regards, Mario
When the capacitor in my scheme reduces its capacitance, if there were no path for it to discharge into (Capacitor no 2) it would rise its voltage to a very large height. For example. if Q=VC and your initial voltage is 10, and your initial capacitance is 10, you would have a Q of 100. Now you reduce your capacitance to 1. Because your charge is constant, your voltage must now also increase 10 fold to 1000v so that 100Q = 1C * 100V.

But you DO actually have a discharge path for it, into C2, which is also 10 farads. So...we must re calculate.

C1 has now 100V at 1farad,

C2 has 10V at 10farads, for a total of 200Q between the two capacitors.

The combined capacitor (C1+C2) has a capacitance of 11Farads, with a Q of 200...

solving for V=Q/C we get its steady state voltage to be 18.18v. This shows us that as C1 tends towards 0, the combined capacitor (C1+C2) tends towards 10 farads, and the equation V=Q/C = 200q/10f = 20v, and as you can see the system tends towards 20volts.

In one case we have a rise to 100v, and in the other because of the shared C2 (which works as a discharge path) it only rises to 20v. It would be the same situation if we were to let C1 become a 1 farad capacitor, then throw a switch and let it discharge into C2, it would discharge from 100 to 20 volts, for a loss of energy....so yes you are correct...but...

Lets consider the energy state of the whole system, Combined capacitor C1+C2 went from 20farads at 10volts, to 11farads at 18.18 volts. The energies are then (according to E = .5CV^2) 1000 joules for the initial state, and 1817.81 for the “reduced capacitance in C1 state”. So you have risen the potential of the entire system (by a factor of 1.817), and…..

The Q on C1 was 100Q, after it "discharges" into C2, it now has a voltage of 18.18, with a C of 1farad, giving a Q of 18.18. So 81.82coulombs of charge were moved between the capacitors (100q-18.18q).

So you have risen the potential of the entire system AND moved 81.82Coulombs of charge.

If you move that in one second, you have moved 81.82coulombs per second, or you have 81.82 watts of power average. If you had higher resistance load and it took two seconds to move the charge you would have an average power of 21.82/2 = 40.91 watts of power average….but over 2 seconds. Same work done, only TIME has changed depending on the impedance of the load…..

According to conservation of energy, the work you get out electrically, is equal and opposite to what you put in mechanically. So you put in 817joules of mechanical energy to separate plates, and you get 817 joules of energy in rising the potential of the system, and move your charge per unit time. When you let the plates go, the process reverses, and you get a 817 joule drop, and move another charge per unit time.

What you put in you get back out, and work is done.
To be continued on next post (this one is getting long)
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Last edited by Armagdn03; 10-15-2010 at 04:46 PM.
#13
10-15-2010, 03:34 PM
 rave154 Silver Member Join Date: Jan 2008 Posts: 926
looking forwards to the continuation
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#14
10-15-2010, 04:17 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Now continuing this and considering force, lets say we start out at our conditions of

10 volts across a 20 farad (C1+C2) capacitor with a charge of 200Q. The Separation of the plates is .001meter and our enormous area of 1129420269m^2. The force between the plates is now 500,000newtons.

Now we have to pull hard to separate our plate to .002meters.

This places our capacitance from 10 farads to 5 farads on C1. This makes our combined capacitance (C1+C2) 15 farads. Its voltage goes from 10V to 13.33.

The new force on our plates is now 222,222.22Newtons. N

Now our distance goes from .002 to .003, Capacitance goes to 13.333 For (C1+C2), Voltage goes to 15v, and the force on the plates will be 125,000, and this process will continue…

…..all the way to say .01m, where it has moved 10x further apart than initially. Here we have our voltage of 18.18, and a force between the plates of 16,528Newtons.

We have moved 81.82coulombs of charge, our force has gone from 500,000 to 16,528 or .03% of the initial force.

Now the force of the 16,528 although a small fraction of what it was, is still in the direction of closure. If let free the plates would of their own forces move together. From .01 to .009…. At this point, the voltage drops from 18.18 to 18.00v to 20,000newtons of force. Then to .008meters, where the voltage drops to 17.77V and the force grows to 24,691 Newtons.

As you can see, if let do its own thing, the device will want to return to its initial voltage of 10V across 20 Farads. As the plates move closer and closer, both physically, and to these values, the force grows quickly,
16,528.9256 at .01m
20,000.0000
24,691.3580
31,250.0000
40,816.3265
55,555.5556
80,000.0000
125,000.0000
222,222.2222
500,000.0000 at .001m

As you can see, as it moves towards its natural tendency, which initially was working against us, when let go, it works for us, growing the force between the plates, closing them faster and faster, moving more and more charge. In fact every step we went through in their separation is now re-lived as they close and we move ANOTHER 81.82 coulombs of charge. How quickly this charge is moved is given by the RC time constants. The heavier the load, the quicker the time, the quicker the time, the more coulombs per second, which equates to wattage.
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#15
10-15-2010, 05:53 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
An interesting story from:

http://www.i-am-a-i.org/free-energy/index.html

Quote:
 One of my electronics college instructors worked for a high power AM radio station when he was younger. He told us the story of what happened when a million volt capacitor connected to a low voltage variable tuning capacitor. (A capacitor stores electrical charge. The variable tuning capacitor is used in this instance to keep the radio station on frequency. This type of variable capacitor was two sets of parallel mounted metal plates. They were arranged such that one set could move freely between the other set without touching.) The story goes, that the charged million volt capacitor got connected to the normally low voltage tuning capacitor accidentally. There was a loud "thunk". The result was the instantaneous forces generated from Coulomb's Law was so great, that afterwards, the formerly moveable plates were pushed together by such a pressure that "It was impossible to take the tuning capacitor apart."
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#16
10-16-2010, 10:16 AM
 shubhamforme Member Join Date: Mar 2009 Posts: 40
new solid state very easy overunity generator "fleet"

go here and check out the simplest overunity generator the "fleet"
thx
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#17
10-16-2010, 02:24 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by shubhamforme joule ef radiant charger/oscillator go here and check out the simplest overunity generator the "fleet" thx

I was excited to see someone post on my thread! but....
How does that have any relevance here?
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#18
10-16-2010, 02:58 PM
 DrStiffler Silver Member Join Date: Mar 2009 Posts: 948
Quote:
 Originally Posted by Armagdn03 An interesting story from: Electric Field Transformer
@Armagdn03
I would not waste time on this story and what it presents. I worked in the broadcast industry as a chief engineer for AM,FM and TV for transmitters from 5kW to 50kW and never saw a 1MegVlt capacitor in any of them, not to mention that long before the high voltage mica capacitors used in transmitters reached say 10kW, all commercial broadcast transmitters were under crystal control.

This does not have much to do with what you present, but I wonder about all these stories that take people into directions of wasted time and thought.

Now for you presentation, well why has it never been shown to really work? Wonder if there could be some truth to the work required to move the plates under heavy charge? because I think these small pulse motors everyone seems to be building could work nicely for the application, if, there was not the inclusion of a drag inducement.

People have tried 10k RPM motors driving disks etc., and I have never heard of one going farther than the initial work and observation. Theory look nice, yet what is the missing part?
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#19
10-16-2010, 05:20 PM
 Mario Senior Member Join Date: Apr 2007 Posts: 423
Hi Andrew,

I understand the theory you show, I did a test with four 47'000uF caps. Two are permanently connected in parallel and can be considered one 94'000uF cap (C2), the others are first in parallel, then in series, I call them C1 having a capacitance of either 94'000uF or 23'500uF. First step:

C1 being in parallel gets charged to 10V and so does C2. Total capacitance of C1+C2 is 188'000uF which at 10V gives a Q of 1'880'000 Joules.

Step two:

C1 one gets connected in series and becomes a cap of 23'500uF at 20V. If C1 discharges into C2 we have a total capacitance of 23'500uF+94'000uF=117'500uF. Q can not change so 1'880'000J/117'500uF=16V which we should be reading by measuring the voltage across the system. In theory.... but in the real world there are only 12.06V. The rest is lost, which is 25%.
If we then reconnect the C1 in parallel energy flows back from C2 which is at 12.06V to C1 which sits at 6.03V(parallel). The system levels out at 9.18V. And so on, in every passage energy is lost till all the caps are empty.

This is what I have also seen in many Tesla switch style or Scalar switch style circuits I have been working on over the last years. To make a simple example, if you have a source of energy which charges a cap which then gets discharged into a battery, the higher the voltage of the cap (battery is 12V) the more inefficient the energy transfer becomes. If I let the cap charge up to say 24V or even higher before dumping it to the battery a lot of the energy is lost. If I dump the cap only one or a couple volts over the battery the trandfer is way more efficient.

What remains to be tested is what happens if the capacitance of C1 is changed gradually and the voltage of C1 never goes much higher than C2's voltage. This would require a very low impedance load causing only a small voltage drop, meaning a VERY high amp low voltage load.

As you correctly say time is important, the faster the energy gets shuttled back and forth (low resistance/impedance load) the more power can be taken from the system. But I must say that in all my attempts I have never been able to get more energy out than I had to put in to make up for the lost energy.

regards,
Mario

P.S. about the "fleet" circuit which btw doesn't belong in this thread... This simply looks like a blocking oscillator circuit. Depending on how you connect the load it is a buck-boost converter or a boost converter if you connect the load between the output diode and the negative rail of the circuit. The second wire of the bifilar is simply the trigger coil which allows for self-oscillating operation, nothing magic about this bifilar configuration. Never found this to be overunity...
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#20
10-17-2010, 02:27 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
I see what you are saying, however your described system is much different from mine, Let me explain with energies.

Say you have your 2 47'000uF's is hooked in parallel to 10 volts. Your energy is then .5(.094)*10^2 = 4.7 joules,

Now you switch them to series to have .0235 farads at 20 volts.
.5(.0235)*20^2 = 4.7 joules,

In your cap switching scheme you have a set of capacitors witch start at energy level X, then undergo a change, and end up back at the very same energy level.

In my set up we start with a capacitor at 20Farads, reduce to 11 through mechanical action.

the math here works to: 0.5*(20)*10^2 = 1000 joules initially
because Q=VC= 200Q. So when we reduce to 11 farads, we get
200=11V so new V = 18.18v. So when we are done we are at a total of

.5*(11)*18.18^2 = 1817.81 Joules

After the mechanical input of energy, I am at an energy state that is 1.87 fold larger than the initiall state!

What you put in mechanically in this case you get out electrically

Both conservation of charge and energy are adhered to.

But now consider the state of the system, when let alone, free to do what it likes without outside influence and force, how does it react?

As I have shown, its tendency is to move backwards towards initial conditions, returning the energy imparted to it.

As for losses, if the conduction paths are insulated, and the switching ONLY affects capacitance, and never comes into contact with the shuttled charge which is moved indirectly by capacitance, and is therefore indirectly switched, losses can be kept VERY low.
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#21
10-17-2010, 02:49 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by DrStiffler @Armagdn03 I would not waste time on this story and what it presents. I worked in the broadcast industry as a chief engineer for AM,FM and TV for transmitters from 5kW to 50kW and never saw a 1MegVlt capacitor in any of them, not to mention that long before the high voltage mica capacitors used in transmitters reached say 10kW, all commercial broadcast transmitters were under crystal control. This does not have much to do with what you present, but I wonder about all these stories that take people into directions of wasted time and thought. Now for you presentation, well why has it never been shown to really work? Wonder if there could be some truth to the work required to move the plates under heavy charge? because I think these small pulse motors everyone seems to be building could work nicely for the application, if, there was not the inclusion of a drag inducement. People have tried 10k RPM motors driving disks etc., and I have never heard of one going farther than the initial work and observation. Theory look nice, yet what is the missing part?
Are you perhaps goading me into revealing my method?

As to your inquiry as to whether there may be work required to move heavy charge, yes, there is, this is the force need to overcome the coulomb forces, and I provided a graph for that. What is not included which is very important is MASS of the moving plate. The surface area to mass ratio is always terrible. You have much more mass than you can ever have capacitance, and so the dominant force in moving the plates is using the inertia of the plates themselves. (from my understading) And if there were additional drag, this would be an important discovery in and of itself. also, looking at the derivatives or slopes of the movemnt of a rotor vs RC time constant, if the Time constant is at any point slower physical movement of the plates, then additional drag will be felt. so you must look at the slopes through a range and make sure parameters match well. But I would never take this route. But say:

-----ll-----ll---------

you have two caps, 4 sets of plates, inner and outer. If you remove the inner plates, you have much much less capacitance, because the remaining capacitor is dictated by the two outer plates. But how to physically remove the inner plates????

With a modified version of:

Capacity Changer

Why not add and remove a virtual plate. This of course throws out all of the "force on plates" mumbo jumbo, but that was still important to realize the mechanism in its simplest embodiment.
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Last edited by Armagdn03; 10-17-2010 at 02:57 PM.
#22
10-18-2010, 09:45 AM
 Mario Senior Member Join Date: Apr 2007 Posts: 423
Quote:
 Originally Posted by Armagdn03 I see what you are saying, however your described system is much different from mine, Let me explain with energies. Say you have your 2 47'000uF's is hooked in parallel to 10 volts. Your energy is then .5(.094)*10^2 = 4.7 joules, Now you switch them to series to have .0235 farads at 20 volts. .5(.0235)*20^2 = 4.7 joules, In your cap switching scheme you have a set of capacitors witch start at energy level X, then undergo a change, and end up back at the very same energy level. In my set up we start with a capacitor at 20Farads, reduce to 11 through mechanical action. the math here works to: 0.5*(20)*10^2 = 1000 joules initially because Q=VC= 200Q. So when we reduce to 11 farads, we get 200=11V so new V = 18.18v. So when we are done we are at a total of .5*(11)*18.18^2 = 1817.81 Joules After the mechanical input of energy, I am at an energy state that is 1.87 fold larger than the initiall state! What you put in mechanically in this case you get out electrically Both conservation of charge and energy are adhered to. But now consider the state of the system, when let alone, free to do what it likes without outside influence and force, how does it react? As I have shown, its tendency is to move backwards towards initial conditions, returning the energy imparted to it. As for losses, if the conduction paths are insulated, and the switching ONLY affects capacitance, and never comes into contact with the shuttled charge which is moved indirectly by capacitance, and is therefore indirectly switched, losses can be kept VERY low.
Hi Andrew,

I think your comparison between our two setups is not correct. With your setup you start calculating with 20 farads which is the total of C1+C2, and in the second step with 11. When you calculated my setup you started with C1 only, not the total of C1 and C2. Let me recalculate my setup:

First step:

C2=same as above

Energy of initial state is .5(.188*10^2) = 9.4 joules

Total charge is .188F*10V=1.88 Coulombs

So by applying the calculation of your setup in the second step to my setup:

The two .047F caps comprising C1 are switched to series giving a capacitance of 0.0235F. Capacitance of C2 is still 0.094 Farads so the total system capacitance now is 0.0235F+0.094F= 0.1175 Farads

Q remains 1.88 Coulombs so 1.88/0.1175=16V

The energy when we are done: 0.5*(0.1175*16^2)= 15.04 Joules

So from the calculation I should have risen the energy of my system by a factor of 1.6. But the truth is you don't get 16V across the system, only 12.06V in real world.
Again, it could be different if C1's capacitance is changed gradully, but I don't have a variable cap I could experiment with... Have you actually verified this on an existing setup?

regards,
Mario
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#23
10-18-2010, 03:57 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
@mario

I see your calculation now, yes that looks correct and I only did half the calculation.

I guess I do not understand how you have a missing 4 volts. Perhaps it would have to do with actual switching of the charges, whereas my unit, The switch only changes capacity, never touches charges being shuttled.

Also, how many diodes do you have in your system, 6-7 diodes makes a 4+v drop across a system. When you are dealing with low voltage, the .6v diode drop is a significant percentage of your total system.

The reality Vs. Practice, has not been a problem for me thus far in my experimentation, and the relationships from the books seem to match reality quite well. In fact there are many patents based on this very principle:

3,094,653 Electrostatic Generator
3,013,201 Self Excited Variable capacitance
3,175,104 High Voltage Electric Generator
3,412,118 Variable capacitor Electric Power Generator
6,936,994 Electrostatic Energy Generators
4,095,162 Capacity Changer

and here is one from our friend JLN Jean Naudin

Patent application: 10472714 Electrical power source.

Quote:
 To understand the mode of operation of this invention, a few well known principles of classical physics need to be recalled. If a metal plate capacitor is charged by means of a voltage source, and if the metal plates are moved away from each other after the capacitor has been disconnected from its sources through a switch, there is an increase in voltage at the terminals of the capacitor resulting from the law of conservation of charge Q= CV…. The internal plates are movable. When they are removed, after the switch S1 has been opened, there is an increase in electrostatic energy that is localized in the capacitor formed by the remaining electrodes. The system is therefore an energy multiplier.

If what I argue were not the case, and your experimental results were valid, there would be a rather large discrepancy. If you start with 16 volts, and you end up around 12, there is a 25% discrepancy between theory and practice. Were this the case, all of these patents would be invalid, based on their primary working principle. And basic relationships would need to be re-established, which would have profound effects in and of itself, however all experimentation I have done, and research I have seen points to the relationships being correct.
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#24
10-18-2010, 04:18 PM
 Mario Senior Member Join Date: Apr 2007 Posts: 423
Andrew, my setup was extremely basic and can be replicated in 5 minutes (maybe less ). I made C1 and C2 comprised both of two 47'000uF caps. C2 is permanently in parallel thus a 94'000uF cap. C1 is the same but its two 47'000uF caps can be either in parallel or in series.
When all is in parallel they all get charged to 10V. The switching of C1 from parallel to series I did manually with a couple of wires! No diodes or voltage drops. I did it a few times, the result being the same. Btw, you could use wires or resistors or a lamp to make the connection, the result stays the same, the discharge just takes longer depending on resistance, but what in the end is left in the caps is the same.

Like I explained earlier, the higher the voltage difference when discharging from one cap to another, the more energy is lost. Something like an impedance mismatch type loss.

regards,
Mario
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#25
10-18-2010, 05:55 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by Mario Andrew, my setup was extremely basic and can be replicated in 5 minutes (maybe less ). I made C1 and C2 comprised both of two 47'000uF caps. C2 is permanently in parallel thus a 94'000uF cap. C1 is the same but its two 47'000uF caps can be either in parallel or in series. When all is in parallel they all get charged to 10V. The switching of C1 from parallel to series I did manually with a couple of wires! No diodes or voltage drops. I did it a few times, the result being the same. Btw, you could use wires or resistors or a lamp to make the connection, the result stays the same, the discharge just takes longer depending on resistance, but what in the end is left in the caps is the same. Like I explained earlier, the higher the voltage difference when discharging from one cap to another, the more energy is lost. Something like an impedance mismatch type loss. regards, Mario
@ Mario,

Hello again,

I was very perplexed by your suposed loss of energy, so I pulled out my calculator and did the calculations based on what you have given me....here are the results....

There are 4 caps, C1, C2, C3, C4,
These are connected so that:

C1-2 (series) = .0235 Farads @ 20volts
C3-4 (parallel) = .094 Farads @ 10 Volts

Charges:

C1-2 (series) = 0.0235F*20C =0.47Q
C3-4 (parallel) = 0.094 * 10F = 0.94Q

Joules:

C1-2 (series) = 4.7Joules
C3-4 (parallel) = 4.7Joules

These are combined to give a resultant parallel capacitor (C1-2)+(C3-4)

The resultant capacitance is 0.0235 + 0.094 = .1175C
The charge across the resultant capacitor is (C1-2) +( C2-3) = 0.47+0.94 =1.41Q

Knowing Q=CV we plug in 1.41Q=0.1175C * V
Solving, V=Q/C we get 1.41/0.1175 = 12V

Thus your result of 12V is in accordance with theory

I suspect you had an error in calculation at some point.

Quote:
 Like I explained earlier, the higher the voltage difference when discharging from one cap to another, the more energy is lost. Something like an impedance mismatch type loss.
Also this is not correct. Think of capacitance as area, and charge as a finite compressible unit. The higher the voltage, the more charge compressed into a given volume, the more pressure released when allowed to occupy more area (added capacitance) You can take an energy ratio, of energy in the initial caps, vs energy of the resultant cap and know how the energies compare.

For example, two equal caps one charged one not, discharing into each other will result in a 0.5 Energy ratio, meaning 1/2 of the energy was lost.
In the example you gave me, the ratio is 0.9 meaning 10% of the energy was lost.
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#26
10-18-2010, 05:56 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by Mario Btw, you could use wires or resistors or a lamp to make the connection, the result stays the same, the discharge just takes longer depending on resistance, but what in the end is left in the caps is the same. regards, Mario
This is at the very heart of the device. The heavier the load, the lower the resistance, the lower the resistance, the more easily charge moves per unit time, more charge per unit time = more watts.
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#27
10-18-2010, 07:17 PM
 Mario Senior Member Join Date: Apr 2007 Posts: 423
Hi again Andrew,

I'm not sure where the error is, I'll re-check later as right now I don't have the time to do so, but I think we agree that there is a loss as you correctly pointed out with the example of discharging one charged cap into an equally sized one that's empty, where 50% of the energy is lost.
I'll make another example. Say you have an energy source that charges a cap to 50V and then discharges it periodically to a 12V battery. Let's say the input power is 10W. The battery will charge at a certain rate. If we use another supply that charges the cap to 15V and periodically discharges to the battery, with the same 10W input you will see the battery charge way faster. I have quite some experience with this type of circuit so, the closer you get to the battery voltage the more efficient it gets. The same is true if you have a cap instead of the battery.

I think you are suggesting that if the load inserted into the discharge path is of very low impedance (like the primary of a transformer) and the frequency is high enough, the energy extracted by the transformer can be more than the energy lost by the process we discussed earlier. This is what I believed also (still do somewhere ) but in the many setups I've built I always got proved wrong. Would really like to see this work though!

regards, Mario
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#28
10-18-2010, 08:20 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 Originally Posted by Mario Hi again Andrew, I'm not sure where the error is, I'll re-check later as right now I don't have the time to do so, but I think we agree that there is a loss as you correctly pointed out with the example of discharging one charged cap into an equally sized one that's empty, where 50% of the energy is lost. I'll make another example. Say you have an energy source that charges a cap to 50V and then discharges it periodically to a 12V battery. Let's say the input power is 10W. The battery will charge at a certain rate. If we use another supply that charges the cap to 15V and periodically discharges to the battery, with the same 10W input you will see the battery charge way faster. I have quite some experience with this type of circuit so, the closer you get to the battery voltage the more efficient it gets. The same is true if you have a cap instead of the battery. I think you are suggesting that if the load inserted into the discharge path is of very low impedance (like the primary of a transformer) and the frequency is high enough, the energy extracted by the transformer can be more than the energy lost by the process we discussed earlier. This is what I believed also (still do somewhere ) but in the many setups I've built I always got proved wrong. Would really like to see this work though! regards, Mario
@Mario,

Here is where the flaw in your calculations lie.

Say you have your 2 .047F at 10 volts called C1 and C2
C1 = 0.047F, 10v, 0.47Q
C2 = 0.047F, 10v, 0.47Q

Added in parallel (C1+C2) you have Capacitor C1-2 which equals 0.094F @ 10V =0.94Q
Added in series (1/C1)+(1/C2) = 1/C12 ccapacitor C1-2 = .0235F @ 20V = 0.47Q

Thus the series added capacitors have a charge of 0.47 not 0.94,
which changes your capacitor Ctot (all capacitors together) to .0235C+.095C = .1185C, and a charge total of .47+.94 = 1.41Q, Thus Vtot = Qtot / C tot so…. 1.41Q / 0.1175C = 12V
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#29
10-18-2010, 10:21 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 918
Quote:
 but I think we agree that there is a loss as you correctly pointed out with the example of discharging one charged cap into an equally sized one that's empty, where 50% of the energy is lost.
@Mario,

There is a very big difference between discharging one cap into another cap and what I am proposing. It may sound subtle but there is a big difference.

In the setup you are describing, you take a 10f cap at 10v, reduce to 1F and get 100V. as you can see from this equation,
http://i210.photobucket.com/albums/b...45595943_1.jpg
the force between the plates is much much larger than in my setup. In mine, you have 10 reduce to 1 for a rise to 18.18v and a movement of 80 some odd coulombs. Force is 5x less, between plates, and work = force x distance.

if you do all the math, you will find that the mechanical work you put into moving the plates = work done by moving the plates. Work = force x distance, and above I gave both distance moved and forces. you put mechanical in, you get electrical out. There is no loss. if get 10 watt seconds out electrical, you put in 10 watt seconds mechanical.

But after you have done that, you let go of the plates, and you get 10 more watt seconds electrical as they want to move back into their original position.
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#30
10-19-2010, 08:30 AM
 Mario Senior Member Join Date: Apr 2007 Posts: 423
Quote:
 Originally Posted by Armagdn03 @Mario, Here is where the flaw in your calculations lie. Say you have your 2 .047F at 10 volts called C1 and C2 C1 = 0.047F, 10v, 0.47Q C2 = 0.047F, 10v, 0.47Q Added in parallel (C1+C2) you have Capacitor C1-2 which equals 0.094F @ 10V =0.94Q Added in series (1/C1)+(1/C2) = 1/C12 ccapacitor C1-2 = .0235F @ 20V = 0.47Q Thus the series added capacitors have a charge of 0.47 not 0.94, which changes your capacitor Ctot (all capacitors together) to .0235C+.095C = .1185C, and a charge total of .47+.94 = 1.41Q, Thus Vtot = Qtot / C tot so…. 1.41Q / 0.1175C = 12V

Andrew, I'm starting to get a headache, somewhere we don't meet...

What you calculated still isn't right. Let's get on the same page:
I have 4 caps. C1 is made of two 47'000uF caps, C2 also. C2 never changes its configuration so we can look at C2 as a permanent 94'000uF cap.
Only C1 changes its configuration from parallel to series and back, like if it were a variable cap but with only two positions. I used 4 caps so that C1 and C2 are equal in capacitance at start. Maybe you misunderstood that C1 and C2 never get connected in series, like they don't in your setup either. Only C1 changes. So at start we have:

C1 = 0.094F, 10V, 0.94Q
C2 = 0.094F, 10V, 0.94Q

When C1 is in parallel we have capacitor C1-2 which equals 0.094F + 0.094F = 0.188F @ 10V = 1.88Q
When C1 is in series (0.0235F) we have capacitor C1-2 which is 0.0235F + 0.094F = 0.1175F.
Now you tell me what resulting voltage I should obtain when C1 (0.1175F) discharges into C2 (0.094F) .

regards,
Mario
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