Charge conserving Capacitive Spring.

[shubhamforme]

new solid state very easy overunity generator "fleet"

http://www.energeticforum.com/renewa...tml#post113228


go here and check out the simplest overunity generator the "fleet"
thx

[Armagdn03]

http://www.energeticforum.com/renewa...tml#post113228


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?

[DrStiffler]

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?

[Mario]

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...

[Armagdn03]

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...

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.

[Armagdn03]

@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.

[Mario]

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:

C1=.047F+.047F=0.094Farads)
C2=same as above

meaning we start with .094 +.094=0.188 Farads

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

[Armagdn03]

@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.

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.

[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

[Armagdn03]

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.

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.

[Armagdn03]

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.

[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

[Armagdn03]

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

[Armagdn03]

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.

[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

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