The secret to Overunity

[sm0ky2]

rolled condensers are the same as a flat plate.
(when using them for their capacitance factor)



Anything other than parallel, includes angular equations.


Within the pi function:
Starting from the center, induction decreases with increasing radius.
They are truncated at some outer radius where the changes in induction
respective to each next outer wrapping become negligible, and it is sealed off.


We also have the convenience of a large surface area packed into a small space.


For the purposes of discussion, there is no real difference between the forms.
Just trivial changes to the formula to maintain perspective accuracy.


The difference comes into play when we use the inductive relationship to charge
one of the plates for free, by charging only one plate. (Or layer)


Remember, that each layer is still the same plate.
If we ignore the angular inductive aspect of it being rolled
It is the same as if we placed two parallel plates of equal area
the same distance apart.


We can engineer capacitors and condensers of infinite shape and design
That perform exactly the same. (under most conditions)


Where their differences come into play is in the changes to the induced charge.


There's only one important question to ask yourselves.


What is the current of charge induction?
i.e.: when we charge a plate (or later) how much "current" moves to induce
an equal and opposite charge on the other plate (or layer)?
[ignore the modern technique of grounding the opposite plate]


And how does this numerical value apply to the "current" drawn from discharging
the capacitor/condenser?

[Tajerek]

There's only one important question to ask yourselves.


What is the current of charge induction?
i.e.: when we charge a plate (or later) how much "current" moves to induce
an equal and opposite charge on the other plate (or layer)?
[ignore the modern technique of grounding the opposite plate]


And how does this numerical value apply to the "current" drawn from discharging
the capacitor/condenser?

charging current across a capacitor is  I=V/R * Exponential (-t/RC)
It is ONLY the combined resistance of the circuit that consumes current, which is equivalent to R+ESR . The cap itself needs only voltage to charge.
In practice you cannot really remove resistance but you can lower it to a very small value where virtually just high voltage with very small current is able to charge up the cap.
The charge on the cap is Q=C*V

some people argue that cap takes long time to charge , that's probably because some caps have bleeding resistor (such as microwave oven caps) or high ESR. That resistive part is the one that slows the charging and consumes power P=I*V. not the charging of cap.

Those who argue should really know the science and understand the formulas otherwise y'all be shooting darts in the dark with only 'practice' wishing for overunity because there are infinite ways to experiment.  And if by any luck you get it you won't be able to reproduce it, prove it, or explain it.

[Tajerek]

There is a huge difference between debunkers who dismiss everything related to OU out of hand,
and healthy skepticism where one neither dismisses nor believes unless something can be proven
by proper bench testing. IMO, the OU research area needs much more healthy skepticism and a lot less people
just making claims without being able to demonstrate anything to support what they are saying.

I don't need to prove textbook formulas. Unfortunately many people want to debunk known textbook formulas as wrong and their lab tests as more correct. it is astonishing to me how people see the formula that says charge on a capacitor Q=CV and still want to convince me that it depends on current.
with that being said, I have tested what I am saying and confirmed it. But I thought sharing the basic formulas is more valuable than showing a working OU that no one knows how it work and why. Understanding the theory of OU can make 1000 experimenters come up with their own applications of it with their own OU devices.

EV gray, Don Smith, Kapanadze...etc showed OU working devices but people can't replicate them. Why? because no one know HOW they work and what's the theory behind . I just revealed the theory and it is up to anyone who wants to apply it or not.

[peper10]

charging current across a capacitor is  I=V/R * Exponential (-t/RC)
It is ONLY the combined resistance of the circuit that consumes current, which is equivalent to R+ESR . The cap itself needs only voltage to charge.
In practice you cannot really remove resistance but you can lower it to a very small value where virtually just high voltage with very small current is able to charge up the cap.
The charge on the cap is Q=C*V

some people argue that cap takes long time to charge , that's probably because some caps have bleeding resistor (such as microwave oven caps) or high ESR. That resistive part is the one that slows the charging and consumes power P=I*V. not the charging of cap.

Those who argue should really know the science and understand the formulas otherwise y'all be shooting darts in the dark with only 'practice' wishing for overunity because there are infinite ways to experiment.  And if by any luck you get it you won't be able to reproduce it, prove it, or explain it.

That can be observed with a simple joule thief loading a cap and discharging it manually...  As the voltage raise in the cap, the joule thief raise in frequency due to the LESS RESISTIVE CHARGE of the cap..  Good point Tajarek

[Void]

I don't need to prove textbook formulas. Unfortunately many people want to debunk known textbook formulas as wrong and their lab tests as more correct.
it is astonishing to me how people see the formula that says charge on a capacitor Q=CV and still want to convince me that it depends on current.

Hi Tajerek. It has already been pointed out to you that you are making major mistakes in
your assumptions and interpretations. The total charge on a capacitor at a given point in time
is equal to C x V, yes, but that in no way means or implies that to build up that charge on
the capacitor that no current has to flow. You are showing that you have no understanding
of what the formulas represent.

with that being said, I have tested what I am saying and confirmed it.

There is no possible way that you have tested this, because if you did do some actual testing
you would quickly realize that what you have been saying here is false. To charge a capacitor
requires a flow of current (flow of charges). The formula which you yourself have posted for the capacitor
charge current clearly shows that the higher the applied voltage, the higher the initial capacitor charging current will be.

For you to keep saying that when you apply a high voltage to the capacitor there will be little or no capacitor
charge current is not only obviously completely at odds with the formula which you posted, but it shows that
you have not even the most basic understanding of simple electric circuits and how capacitors work. If you did,
you would not be making such obvious errors.

I won't waste further time on this, as anyone with even just a very basic understanding of capacitors will be
able to see that what you are saying makes no sense.

All the best...