Yet another Free energy Awnser...But a really good one ;)

[armagdn03]

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.

[tinu]

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

[armagdn03]

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.

[armagdn03]

 

[Unicron]

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!