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Author Topic: The secret to Overunity  (Read 45796 times)

Offline sm0ky2

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Re: The secret to Overunity
« Reply #90 on: December 22, 2017, 02:18:47 PM »
@Chet


It’s much easier to obtain/produce graphite
Cheaper too. They do pretty much the same thing.
The advantages that come from graphene are tiny.
The real advantage is economic profitability.
That’s why research gets funding in this area.


Graphite batteries,capacitors, inductors, semiconductors,peltier’s etc.
perform almost as well in any situation we would use them for.

There are self-organizing carbon structures that out perform
graphene, but are increasingly more difficult to create.


We ‘can’ make graphene in the kitchen, as I and others have shown
But my experience tells me it’s not worth the investment.




Free Energy | searching for free energy and discussing free energy

Re: The secret to Overunity
« Reply #90 on: December 22, 2017, 02:18:47 PM »

Offline sm0ky2

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Re: The secret to Overunity
« Reply #91 on: December 22, 2017, 02:36:16 PM »
To Void:


Your difference in opinion sounds like it is derived from the
difference between the technician point of view and the engineer
viewpoint.


I’ll try to explain:


A technician will tell you that to charge a capacitor, you must do it
through a load. Such as a resistor or a lightbulb, etc.
As this applies to the application (in our modern circuitry)


An Engineer will tell you that a capacitor charges inductively.
A charge on one plate induces an equal and opposite charge on
the other plate.
As this pertains to the science.


Both statements are true, and in fact both occur in modern circuitry.
Simultaneously


However, the prior is not inherent of capacitors, only of our circuits.


Capacitor science is derived from condenser study.


Capacitor technical data is derived from modern application.
(done almost 2 centuries later)


What you are claiming to be “false” only applies to the modern application.
We can also charge a capacitor inductively by applying a potential to one
plate.
The other will induce an equal and opposite charge
as a function of the distance between them and change in voltage.


Charging a capacitor without a load has no measurable current.
it is so small it can only be calculated,
You can scope it with an adequately high-frequency oscilloscope.
and we see a sharp spike at the front, and a decrease in current
inversely proportionate to the voltage.
Current at 0 charge is ‘infinite’, current at max charge is 0.
Internal resistance to a single capacitor plate increases with charge.


The large spike (across a the charging time) of current is overwhelmed
by the increasingly low current as it approaches max charge.
This will continue until the potential between the charged plate and the
induced plate equals the potential between the source and our reference.
Resulting in almost no current at all.
Current is a function of time.
The time is so short it destroys the equation.
It is approximately 0 current.


Hope this helps you understand.


Online ramset

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Re: The secret to Overunity
« Reply #92 on: December 22, 2017, 02:53:44 PM »
Smokey
as usual I did not explain myself well
its in my head but doesn't always make it to the paper
 ::)

no I was referring to Roberts path, sourcing already made products from vendors

However
I know others are making it and experimenting with solutions and materials [Mwah]
and its nice to show the really young members of the planet [Children]
what they can do with their own hands[under supervision of course]

to make a change in the world , IMO too much button pressing and not enuff dirty hands
with the Children.

and for Clarity the energy density claimed by Robert is on a par or better than Li

and for additional clarity
we hope to qualify that claim independently here.

ps
gotta log out [storm heading this way




Free Energy | searching for free energy and discussing free energy

Re: The secret to Overunity
« Reply #92 on: December 22, 2017, 02:53:44 PM »
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Offline Void

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Re: The secret to Overunity
« Reply #93 on: December 22, 2017, 06:09:18 PM »
For those who want to claim overunity on something, I suggest they first prove by experiment that their system can perpetually run or self-run.
Proving that the output is greater than the input is not enough. Measurement errors always exist.
We have already seen many discussion threads on overunity spark gaps, overunity capacitors, overunity inductors, OU resistors, 100kW systems, 5MW systems, etc, etc.
I am not ruling out the possibility that self-running systems exist.

Hi blueplanet. Yes, being able to demonstrate a self-sustaining setup is pretty much the
benchmark for OU claims now, given all the many false claims and setups showing
very obvious improper measurements over the years. It is easy to make claims, but it is
a whole different kettle of fish to be able to back up those claims with a self-sustaining setup. 
This quickly separates the real from the fantasy. :)

P.S. If someone can show that they get a very much larger output power than input power
then that is not anything to sneeze at, but the next logical step would be to try to loop it and
make it self-sustaining if possible, as a truly self-sustaining setup is something that is pretty
hard to argue against, outside of the possibility of plain fraud.

All the best...

Offline Void

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Re: The secret to Overunity
« Reply #94 on: December 22, 2017, 08:31:53 PM »
Hi sm0ky2. I have already explained to you that my comments were to Tajerek,
and are in response to the specific circuit diagram he posted and the specific comments
he has made here.  I understand well that there is something called electrostatic induction, but
my comments are in direct relation to specific things Tajerek has been saying, not on other
approaches or other possibilities.

I have spent many years doing my own experimenting in this area, and my approach has been to start
with very basic concepts and do various tests so I understand well how things really work, and then
to move step by step into different variations and observe how things really work in those different
variations. After a person has done many hours of experimentation along these lines over many years,
they can often quickly see where people are going wrong who do not actually try things out and who
just make statements based on assumptions and incorrect understanding and that sort of thing.
Again, my comments have been in direct response to specific things that Tajerek was saying here.

The unfortunate reality is there are a lot of very 'mistaken' people (to be kind about it) in the OU research area.
A common pattern you see from these people is they typically just can't back up what they are saying with an actual
sound test setup. They often make excuses why they can't demonstrate anything to back up what they are saying,
or may show some shoddy setup along with improper measurements to try to back up what they are saying, or insist the
problem is with other people for not blindly accepting what they are saying. :)

In reality, the only realistic way to be sure about something related to OU being authentic or not is
to test the heck out of it and to always be on the watch for things that you might have overlooked or
places where you may be making an error or incorrect assumption. If a person doesn't have that mindset when
they are doing OU experimenting then I think they are lost before they have even started. You must be willing to
closely examine and question everything, and be willing to check things over many times to try to find places where you
could have gone wrong, etc. IMO, this is the difference between someone who is truly open to trying to get a real
understanding of things, and someone who is just looking to promote some pet ideas or beliefs they may have
without being willing or capable to do the work to see if there really may be anything to it or not or to accept
feedback on what they are saying or doing as well.

Anyway, enough on this, what I am saying should really be old hat to anyone who has been at this
for any length of time anyway. Newcomers to the topic should consider what I am saying here though
if you want to try to avoid being lead down the garden path and end up wasting a lot of time. :)

All the best...


Free Energy | searching for free energy and discussing free energy

Re: The secret to Overunity
« Reply #94 on: December 22, 2017, 08:31:53 PM »
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Offline Tajerek

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Re: The secret to Overunity
« Reply #95 on: December 22, 2017, 11:26:40 PM »

We can also charge a capacitor inductively by applying a potential to one
plate.
The other will induce an equal and opposite charge
as a function of the distance between them and change in voltage.


Charging a capacitor without a load has no measurable current.
it is so small it can only be calculated,
You can scope it with an adequately high-frequency oscilloscope.
and we see a sharp spike at the front, and a decrease in current
inversely proportionate to the voltage.
Current at 0 charge is ‘infinite’, current at max charge is 0.
Internal resistance to a single capacitor plate increases with charge.


The large spike (across a the charging time) of current is overwhelmed
by the increasingly low current as it approaches max charge.
This will continue until the potential between the charged plate and the
induced plate equals the potential between the source and our reference.
Resulting in almost no current at all.
Current is a function of time.
The time is so short it destroys the equation.
It is approximately 0 current.


Hope this helps you understand.

That's what I was trying to say. And explained in my previous message with formulas that you need current from the source only in proportion to the circuit resistance. If you remove the resistance component you literally don't need current to charge a cap. It is the voltage pressure across the plates that causes the charges that are already on the plates to be displaced, when the charges are displaced there appears current as a consequence. Current is not the cause of the charging cap but the consequence. The current doesn't come from the source but is created by the cap when it is charging, and the cause is the voltage.
Obviously we cannot have zero resistance that's why small current is still needed, but in theory a perfect cap doesn't need current to charge but only voltage.

As example Take a high voltage cap of 1uF and ESR=0.2 ohm  it will charge up in 5*R*C as known by time constant formulas (this is science not me claiming) . It means you fully charge it up within 1 micro second ! And you charge it up to 63% in 1/5th of a microsecond. The charges are displaced from one plate to the other so fast that you see large spike of current but that's not really drawn from the source but displacement of charges from 1 plate to the other. Again it is the resistive component of your circuit and cap that's consuming current/power.

Now anyone who tells me that cap doesn't charge fully in 5*R*C then they don't know science...

The current across the cap is i=C*dv/dt  That means it is resulting from the rate of voltage change across the capacitor it is not the source current.
I hope this clears the confusion for Void who has wrong assumptions thinking that the current C*dv/dt is required from the source, while as formula clearly says it comes as consequence of the voltage increase across the cap as I just explained :) that is in fact the secret to OU :)

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.

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.


Offline Void

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Re: The secret to Overunity
« Reply #96 on: December 23, 2017, 12:09:42 AM »
That's what I was trying to say. And explained in my previous message with formulas that you need current from the source only in proportion to the circuit resistance. If you remove the resistance component you literally don't need current to charge a cap. It is the voltage pressure across the plates that causes the charges that are already on the plates to be displaced, when the charges are displaced there appears current as a consequence. Current is not the cause of the charging cap but the consequence. The current doesn't come from the source but is created by the cap when it is charging, and the cause is the voltage.
Obviously we cannot have zero resistance that's why small current is still needed, but in theory a perfect cap doesn't need current to charge but only voltage.

Hi Tajerek my friend. You keep saying the same thing, but unfortunately you seem to be unwilling
to give due consideration to what people are telling you in response. :) You ignoring where people
are pointing out how you are wrong is not helping your situation at all. :)

Based on your circuit diagram, you appear to be talking about charging a two plate type capacitor
from an rectified DC input voltage source. The applied voltage is the 'pressure' (AKA electromotive force) which
causes the charges to start to move from one cap plate to the other.  This movement of charge from one plate to the
other is a current. Packing electrons into one plate from the other plate takes an expenditure of energy. This power
consumption is represented as the applied voltage times the capacitor current at any given point in time. The formula for capacitor
current which you posted clearly shows that the higher the applied voltage, the higher the initial capacitor charge current
will be for a given circuit. There is simply no denying that. Just look at the capacitor charge current formula.

This capacitor charge current will fall off in an exponential curve which is accounted for in the cap current formula
by the exponential component in the formula. Nothing you have shown or described in your circuit diagram
would seem to do anything to bypass this normal type of capacitor charge behavior. Anyway, you have made it clear
that you are not interested in listening to reason about this, as this has been pointed out to you several times already,
so I'll leave it at that.

In order to have a chance of seeing OU, it would seem a setup would have to be doing something very out of the ordinary.
I see nothing out of the ordinary in your circuit setup, and your basic premise is clearly at odds with the formula
for capacitor charge current. In other words, you are not making sense. :)

All the best...



Free Energy | searching for free energy and discussing free energy

Re: The secret to Overunity
« Reply #96 on: December 23, 2017, 12:09:42 AM »
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Offline Tajerek

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Re: The secret to Overunity
« Reply #97 on: December 23, 2017, 02:56:29 AM »
Hi Void , you missed the point you are thinking in the traditional way of conservation of energy which leads to "overunity doesn't exist".
All your assumptions are based on a false premise that "if there's work it means there's equivalent expenditure which it has to come from the source" that negates any chance of believing in OU every existing. I am wondering what you are doing in this OU forum then  ???

I will explain where you make the mistake
The applied voltage is the 'pressure' (AKA electromotive force) which
causes the charges to start to move from one cap plate to the other.  This movement of charge from one plate to the other is a current.
This is correct , I agree with you here. that there is current and I said it serveral times and in my last message. I used the term 'displacement' current. I say it doesn't have to come from the source.

Packing electrons into one plate from the other plate takes an expenditure of energy. This power
consumption is represented as the applied voltage times the capacitor current at any given point in time.
This is a misunderstanding. It applies to resistive loads not capacitors. Capacitors do not perform any work and don't consume power. Do your research. https://www.crazyengineers.com/threads/how-to-calculate-power-in-capacitor.12638/

The formula for capacitor
current which you posted clearly shows that the higher the applied voltage, the higher the initial capacitor charge current
will be for a given circuit. There is simply no denying that. Just look at the capacitor charge current formula.
that formula says i=c dv/dt and it doesn't say the higher the voltage the higher the current  :D it links the current in the cap directly to the rate of change of the voltage of the source. This current is NOT the current of the source like you think it is.
If it was like you think it means the source voltage and the source current will have to be linked by this formula which is absurd. to explain the absurdity i will take example of a linear rising voltage at a rate of Const*t. since we are using very high voltage to Const is very high as voltage will jump from 0 to high value across the cap when it is charged. such as 3000v in 1 sec. Then Const = 3000
take a cap of 5 uF as example.

i=C*dv/dt  = 0.005* d(Const*T)/dt = 15A

This means when you charge a 5uF cap with 3kv it draws 15A. But what if you are using a voltage source of 3kv and 100ma ?? does it make sense that the cap will draw 15A from 100ma source ?? that's absurd.

The real answer is that your source remains 3kv 100ma and STILL you will see the current shooting up to 15A but it is NOT related to that 100ma of the source. That's the secret of excess power i've been trying to explain :)

This capacitor charge current will fall off in an exponential curve which is accounted for in the cap current formula
by the exponential component in the formula. Nothing you have shown or described in your circuit diagram
would seem to do anything to bypass this normal type of capacitor charge behavior.
current will go down exponentially so what? that's not what my circuit is trying to show. my circuit is an illustration that you can draw small power to store high energy into a cap, then covert that electric energy into useful form.

Offline blueplanet

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Re: The secret to Overunity
« Reply #98 on: December 23, 2017, 04:42:23 AM »
I agree.

There have been a lot of real overunity devices around. There have been a lot of great ideas posted on the internet. Unfortunately, these real devices attract no attention. Whenever you disclose a self-running device that has been running for several decades, the ordinary people will look for hidden wires underneath the carpet. The credibility of free energy science has gone down to almost zero.

Instead, the devices which never fail to attract public attention are those overunity spark gaps, 5MW devices, overunity resistor, abnormal supercap, etc, etc. Some individuals are even trying to release disinformation to the public in a way to discredit the science of free energy.   The self-running benchmark will certainly clear a lot of doubts.


Hi blueplanet. Yes, being able to demonstrate a self-sustaining setup is pretty much the
benchmark for OU claims now, given all the many false claims and setups showing
very obvious improper measurements over the years. It is easy to make claims, but it is
a whole different kettle of fish to be able to back up those claims with a self-sustaining setup. 
This quickly separates the real from the fantasy. :)

P.S. If someone can show that they get a very much larger output power than input power
then that is not anything to sneeze at, but the next logical step would be to try to loop it and
make it self-sustaining if possible, as a truly self-sustaining setup is something that is pretty
hard to argue against, outside of the possibility of plain fraud.

All the best...

Free Energy | searching for free energy and discussing free energy

Re: The secret to Overunity
« Reply #98 on: December 23, 2017, 04:42:23 AM »
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Offline Void

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Re: The secret to Overunity
« Reply #99 on: December 23, 2017, 04:55:12 AM »
Hi Void , you missed the point you are thinking in the traditional way of conservation of energy which leads to "overunity doesn't exist".
All your assumptions are based on a false premise that "if there's work it means there's equivalent expenditure which it has to come from the source" that negates any chance of believing in OU every existing. I am wondering what you are doing in this OU forum then  ???

Hi Tajerek. Your statement that I am thinking "overunity doesn't exist" is a complete fabrication on your part.
I have never said anything even remotely like that, and I have pointed out that I have done experimentation
in this area for years. Please do not put words into my mouth that I have clearly never said.


This is a misunderstanding. It applies to resistive loads not capacitors. Capacitors do not perform any work and don't consume power. Do your research.

Good grief man. You are mixing apples and oranges. You posted a link to a capacitor in an AC circuit (sinewave) 
which does not consume power, that is correct, but I was clearly talking about the power   
consumed from the high voltage DC power source when charging the first capacitor in your circuit
diagram to a DC voltage.  Very obviously I wasn't talking about a capacitor in an AC circuit.
Sorry, but for you to mix up the difference between charging up a capacitor with a DC voltage (which is what
we have been talking about all along based on your circuit diagram) and how a capacitor behaves in an AC circuit
with a sinewave applied to it shows that you have no idea what you are talking about.


that formula says i=c dv/dt and it doesn't say the higher the voltage the higher the current  :D

The formula I am referring to is the formula for the charge current of a capacitor when a DC voltage source
is applied to it to charge up the capacitor, which you yourself have posted previously:
I = (V/R)  x e^-t/RC
This is the formula for determining the charge current in a capacitor that is being charged with a DC voltage V.
As you can see this formula contains Tau (RC). This is the formula you must use for determining the capacitor charge current
during the 5 Tau charge period when charging from a DC power source V. That you did not realize something this basic again
shows that you do not know what you are talking about.

As you can see from the formula, the lower 'R' is (in your circuit there is no actual R, but there is a diode which
has some losses plus the ESR of the capacitor), the higher the initial charge current will be, and this charge current
will fall exponentially from the maximum as the capacitor charges. So, definitely, the higher the source DC voltage
is, the higher the initial capacitor charge current will be, and it will fall off exponentially from there over the period of 5 x Tau.

By the way, when you are charging a capacitor from a high voltage DC power supply, which usually have
fairly low current capacity, typically what happens is the output voltage of the high voltage DC power supply
drops when it is charging the capacitor due to the high internal resistance of the high voltage power source, so
the actual cap charge current in this situation will be limited by the internal resistance of the power supply. However, 
clearly power is consumed from the high voltage DC power supply when charging the capacitor. It doesn't make any difference
what the actual DC supply voltage is, power will still be consumed when charging a capacitor to a DC voltage.
This is very, very basic stuff.

Tajerek, sorry man, but you clearly have no idea what you are talking about. For absolute certain power is consumed from
the high voltage DC power source when the capacitor is being charged up with this DC voltage applied, regardless of the
applied voltage.  Anyone with even the most basic understanding of capacitors should be aware of this. 
Enough with the nonsense already.  :o





Offline sm0ky2

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Re: The secret to Overunity
« Reply #100 on: December 24, 2017, 03:18:17 AM »
Void, try charging your capacitor from a high voltage source that is capable
of greatly exceeding the charge current of the equation (max value on the data sheet).


What is the charge time?

Free Energy | searching for free energy and discussing free energy

Re: The secret to Overunity
« Reply #100 on: December 24, 2017, 03:18:17 AM »
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Offline Void

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Re: The secret to Overunity
« Reply #101 on: December 24, 2017, 05:05:39 AM »
Void, try charging your capacitor from a high voltage source that is capable
of greatly exceeding the charge current of the equation (max value on the data sheet).
What is the charge time?

Hi Sm0ky2. If you have a high voltage DC power source with very high current capability,
and you have very little resistance and component losses, then the capacitor initial charge
current would be very high. To have a high voltage DC power source that can supply that
kind of current would be quite dangerous to work around. I am not interested in trying it. :)

People have used microwave oven transformers as a fairly high voltage power source with
fairly high current capacity, and if you use something like that (careful, it's dangerous) with a suitable
high current rated HV diode to rectify the AC to DC, even a relatively large capacitance HV cap can be charged
to a high voltage very quickly, and the initial capacitor charge current would be relatively pretty high.
Doing this still consumes power from the DC power supply however in proportion to the voltage you are
charging up the capacitor to, regardless of the power supply voltage you are actually using. 

All the best...

Offline SolarLab

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Re: The secret to Overunity
« Reply #102 on: December 24, 2017, 05:26:47 AM »
F.Y.I.

Vyacheslav Gorchilin shows [proves] mathematically that excess (free) energy is achievable.
He also presents several methods/techniques and suggests some practical implementation
approaches. Animated Mathcad on-line graphs are made available to aid in the process as well.

His approach to the solution [proof] is rather brilliant IMHO!

From:  http://gorchilin.com/articles/
(use translation (flag in upper right hand corner) if required - unfortunately Google Translate will not properly
display the formulas, however Yandex - the flags - will work OK)

    Energy parametric RLC-circuits

Free energy in a parametric RLC-circuits of the first kind of the second order
http://gorchilin.com/articles/energy/RLC_5?lang=en

"In this work we consider the electric circuit containing the nonlinear reactive elements: the
capacitor and inductance. Their nonlinearity is determined by the parametric dependence:
the capacitance — voltage on it, and inductance from the current flowing through it. The
resistance is constant, but even if its value changed, for example, from the time the proof is
not affected . According to the classification here will be considered generators of the first kind
and second order. In real devices the parametric dependence can be only one element
capacitance or inductance, but here we show the General case and prove that these elements
are independent from each other to influence change in the efficiency of the second kind. "

Insights:  " Insights on circuits of the second order similar to those done on the circuits of the
first order. However, repeat them.
In this work we proved that it is impossible to gain energy in parametric circuits of the second
order in the full cycle (FCC) because the energy dissipated in the resistance is always equal to
the energy expended by the power formula (5.8).

But if the cycle is incomplete, the receiving gain becomes achievable task
.
 
If reactive elements contain potential energy in the beginning of the
cycle (PCCIE), the gain can be found using the formula (5.12). If the energy in the reactive elements
is at the end of the cycle, then the conditions for receiving allowances, we can find from formula (5.14),
and the increment of efficiency by (5.16).

You need to understand that there is a mathematically strictly proved potentially achievable values of
the increment of energy, part of which, in the real reactance, can be spent inefficiently, for example, on
heating. However, on the basis of evidence about the energy increment in the fractional cycles, one
can obtain special cases for engineering calculations, which, in turn, will allow you to build a real device
with high efficiency.
Additional materials and some special cases with examples from real wireless components you can
see here. "

Application to the proof of parametric RLC-circuits of the first kind
http://gorchilin.com/articles/energy/RLC_51?lang=en

Parametric inductance in PCCFE

In this case, without knowledge of the schema and according to   of the power source,
the increase in efficiency we can not calculate, but we can search conditions for     from
the formula (4.15). Assume that our coil is described by the parametric dependence of
inductance from its current next next:
         {see text for formulas - unfortunately they can not be reproduced here}
For rows with a large value of the degree calculations are either very bulky, or, in principle,
not derived analytically. But for example, it will be enough and second degree.

Reference:
{further background study}

SAINT-PETERSBURG   STATE  INSTITUTE   ACCURATE   MECHANICS   AND   OPTICS
(TECHNICAL   UNIVERSITY)
Department of Electrical Engineering and Precision Electromechanical Systems
Yu.M. Osipov
FREQUENCY   AND   TEMPORAL   ANALYSIS  STATIONARY   AND   TRANSITIONAL 
CHARACTERISTICS   LINEAR   ELECTRICAL   CIRCUITS
A manual on electrical engineering and TOE courses
Part two
St. Petersburg  2002

3.3 Transient processes in second-order circuits
http://ets.ifmo.ru/osipov/os1/3_3.htm
Translation to English (Google)
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Offline sm0ky2

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Re: The secret to Overunity
« Reply #103 on: December 24, 2017, 06:38:07 AM »
Hi Sm0ky2. If you have a high voltage DC power source with very high current capability,
and you have very little resistance and component losses, then the capacitor initial charge
current would be very high. To have a high voltage DC power source that can supply that
kind of current would be quite dangerous to work around. I am not interested in trying it. :)

People have used microwave oven transformers as a fairly high voltage power source with
fairly high current capacity, and if you use something like that (careful, it's dangerous) with a suitable
high current rated HV diode to rectify the AC to DC, even a relatively large capacitance HV cap can be charged
to a high voltage very quickly, and the initial capacitor charge current would be relatively pretty high.
Doing this still consumes power from the DC power supply however in proportion to the voltage you are
charging up the capacitor to, regardless of the power supply voltage you are actually using. 

All the best...


Losses are negligible, and at high enough voltages there is no resistance in the circuit
External to the capicitor


The only impedance is then the internal resistance curve of the capacitor that defines charging current


Without adding extraneous components, and making assumptions outside of the situation.


What I am asking is, what is the time-frame over which the charging current exists?
And what portion of this time does the current hold a ‘high-value’?


My HV source provides infinite instantaneous current.
And exists as only potential except the moment charge is drawn from it.
Current, defined in Amperes, is a time-derived quantity.


The magnitude of the time defines the order of magnitude of the current.
Here’s a hint: 1 Amp is 1 coulomb per second.


If the total charge time is 1/82 of a second,
Current is only of appreciable value for 1/6000th of a second
after which internal resistance of the capacitor reaches such high value
that the rest of the charging time, current is calculably minuscule.


So what exactly is the charge current you keep arguing about?
And how does the power rating compare to the output of the capacitor discharge?




Offline Void

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Re: The secret to Overunity
« Reply #104 on: December 24, 2017, 07:23:43 AM »
Hi sm0ky2. The formula for the capacitor charge current makes it quite clear what will happen.
I am not going to keep repeating myself, and this is very basic and well established stuff anyway.  ;)
If you doubt it, you can always experiment and see for yourself how things work.
A very simple setup to see what happens is if you charge one capacitor from another capacitor, whether
at high voltage or not, the source capacitor will lose charge (a loss of energy) to charge the other capacitor.

Be very careful if you mess with high voltages, especially at higher current capacity. Not recommended. It can certainly be lethal.

All the best...


 

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