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Author Topic: Cap to Cap dump  (Read 11246 times)

Offline webby1

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Re: Cap to Cap dump
« Reply #15 on: June 04, 2016, 02:52:56 PM »
I think that even in an ideal state the energy stored after the dumped between 2 identical caps will be 50% of what you started with,, that being without any heat loss.

Empirical data shows the end result,, and it also shows that for non-identical caps what the energy left will be and that matches what the formulas predict.

What the resistor is showing is the portion of energy that is being used to do something while the dump takes place.

Replace the resistor with a motor and you can extract that energy transfer as mechanical work,, the resistor extracts it as heat.

The assumption and conclusion seem to be what is wrong,, not the formulas.

If you can not separate the events,, what ever the "loss" transfer is and the final condition of transfer in total,, then what would you call it?

If these were not able to be separated then in the end it would be easiest to state that the energy is converted into the resistance of interaction between the caps,, then that could be later shortened into resistance between the caps and then that would end up being taught that the energy is burned up in the resistance.

Free Energy | searching for free energy and discussing free energy

Re: Cap to Cap dump
« Reply #15 on: June 04, 2016, 02:52:56 PM »

Offline webby1

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Re: Cap to Cap dump
« Reply #16 on: June 04, 2016, 06:20:41 PM »
Just a simple test,,

I have 4 identical caps,,

3 are connected in parallel through 2 motors,,

The 4th is charged up to 10V,,

I connect that 4th cap to the other 3 caps through another motor,,

all 3 motors spin and the final voltage on all 4 caps is 2.5V each.

So no matter what I do with the energy that is "burnt up in the resistance" I end up with the same voltage, aka the same stored energy.

Free Energy | searching for free energy and discussing free energy


Offline dieter

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Re: Cap to Cap dump
« Reply #17 on: June 07, 2016, 05:24:21 AM »
I tried to discuss this a year ago, but was told "heat dissipation, chapter closed!", by the usual sceptics.

Cap to cap dump is a weird thing because at the same time they are connected in parallel and in serial. However, current flows from high to low voltage, so we consider the dump a serial connection.

That means, the inner two plates of this connection practicly cease to exist and the two caps become one cap:

-||-  -||-  becomes: -| |-

While the voltage remains the same (the outer plates are + and -), the DISTANCE between the plates is doubled, meaning the capacitance is only a quarter of the original sum of both caps. hence, (capacitance/4) * ((voltage^2)/2) is left in total. If the two cap are separated, they will, ehrm, somehow, ehrm regain a charge separation on the formerly ceased inner plates.

So let's test this:

Let cap 1 be 1000uF at 10V and cap 2 1000uF at 0V

Total energy is: (1000*10*10/2) + (1000*0*0/2)=50'000

Now in series dumped, it became one cap:
500uF  *10*10/2 = 25'000

There you've got it. Half of the energy has vanished.

Of course, two caps with 5V held in series give 10V, just like batteries. So, no surprise there after separation

What is really fascinating about it is: what happens, if we INCREASE the capacitance? Is this "loss" phenomen reversible?

According to some sources it is possible to get an energy gain from socalled parametric capacitance.

That being said, IMHO one would have to dissipate the energy by a load after increasing the capacitance, and recharge the plates after decreasing the capacitance.

Although, where the energy comes from or goes to remains a mystery to me and is certainly a physical paradox that weakens the currently accepted laws further.

Offline webby1

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Re: Cap to Cap dump
« Reply #18 on: June 07, 2016, 06:04:37 AM »
The only time you loose 50% is when you are using identical caps.

The spreadsheet I use has a little more in it than the second one I posted but I think it still can show things correctly.

I do not think the "charge carriers" have much to do with the dump.  Kind of like does the magnetic field care about the voltage?? not really but the voltage does set the current that can flow,, so maybe the cap does not care about the current in the same fashion, the charge carriers set how much voltage there is.

The problem with that is that you need more electrons shifted to bring the caps up to the half energy value.  ???

The caps charge negative to negative and positive to positive,, the electrons run out of the negative plate and into the other negative plate which pushes electrons out of the positive plate and fills in some of the holes on the charged cap positive plate.

It is an interesting thing to ponder,,


Offline webby1

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Re: Cap to Cap dump
« Reply #19 on: June 07, 2016, 06:12:38 AM »
:)
If you start with a cap that has 100 times more capacitance than the non-charged cap you can have a 99% efficient interaction :)
:)

Just did that in the spreadsheet.

Free Energy | searching for free energy and discussing free energy

Re: Cap to Cap dump
« Reply #19 on: June 07, 2016, 06:12:38 AM »
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Offline dieter

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Re: Cap to Cap dump
« Reply #20 on: June 07, 2016, 06:59:05 AM »
edit: oops, I was wrong, you are right in that the charge is distributed in parallel, although the energy "loss" follows the logic of serial connection.

Offline webby1

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Re: Cap to Cap dump
« Reply #21 on: June 07, 2016, 01:06:59 PM »
Would it cost more or less to "sweep" the electrons over to one side while the two were connected?

Free Energy | searching for free energy and discussing free energy

Re: Cap to Cap dump
« Reply #21 on: June 07, 2016, 01:06:59 PM »
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Offline dieter

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Re: Cap to Cap dump
« Reply #22 on: June 07, 2016, 05:32:37 PM »
How would you do that anyway? Most likely not by turning it 90 deg...

You could put a diode between them. If you then increase the capacitance of one, the energy would partially flow to that one and could't flow back once the capacitance is back at the initial level.

Whether this would cause loss or gain, I don't know.

However, a simple experiment would be a cap that is made of 2 big plates. when you squize them together the capacitance is increased.

I once made a test with two caps that were connected alternatingly in serial and parallel, using a 555, driving 3 relays, just to see if there is a voltage building up (as described in one paper), but other than like 50 millivolts or so nothing happened.

I didn't tried a diode tho, maybe that was my mistake.

The idea was to connect an arial to one end and ground to the other end of this oscillating capacitance, in order to pull the energy right out of the environment. Romantic idea, didn't work so far.

Offline webby1

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Re: Cap to Cap dump
« Reply #23 on: June 07, 2016, 05:47:34 PM »
The picture that cam to mind was of a large wheel with just under 1 turn on it for the plates,, and then combs to "brush" the electrons over,,

I have been playing with 2 caps hooked together by 1 plate and charging up one of the caps to see if I can move something,, I had a small change show across the connection but that was most likely a connection issue.

Another one I had was the same large wheel,, charge the plates and then roll them up individually on a smaller wheel,, I guess that the mechanical cost would be about whatever change in energy would be.

Offline tinman

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Re: Cap to Cap dump
« Reply #24 on: June 12, 2016, 08:36:30 AM »
Even Hyperrphysics seems to imply that an ideal cap and connection is not possible, and half the energy is always dissipated.

Another good page.

And this. He seems to indicate that even an ideal cap transfer will result in 50% loss of energy, but he doesn't say where the lost energy goes.

This page may be of interest.

I would think this is enough to determine that the loss is not due to resistance.
I feel the loss is due more to the slowing down of the electrons,and resulting in those electrons having less impact energy upon the receiving capacitor plates,even though the amount of electrons remains the same. This is why inductive transfer is more efficient,as the electrons are accelerated to a higher speed(higher voltage),and those higher speed electrons hit the plates of the receiving capacitor with a higher energy state.

I have carried out some tests today,where the resistor is replaced with a small DC motor,and as expected,the transfer from one cap to another is more efficient,and we end up with more than half of the starting stored energy. What is also interesting,is that i can place a load on that motor by way of either an air screw,or small generator,and the end resulting energy in the two caps remains the same as that without having a load on the motor.

We know by placing a load on the motor,will make that motor work harder,and there for dissipate more heat-and so with the primary capacitor. But this extra heat being dissipated by the motor and primary cap,has no effect what so ever in the end result--as in we still have the same amount of energy left in the two cap's that we would if the motor is not loaded.

If the missing energy was lost due to resistive heat losses,then loading the motor would have increased those loses--but it did not.
Along with your statements above Poynt,i think it is safe to say that the energy is not lost due to resistive heat--although a very small amount may be.


Brad

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Re: Cap to Cap dump
« Reply #24 on: June 12, 2016, 08:36:30 AM »
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Offline Dog-One

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Re: Cap to Cap dump
« Reply #25 on: June 12, 2016, 11:24:21 AM »
@tinman,

I'm very happy to see you participate in this discussion and conduct tests yourself.  Given your time at these sort of things, I'm a little surprised you have not posted a video of this particular test years ago.

The burning question I have to ask, having seen these results for yourself, if by moving charge from a full capacitor to an empty one conserves charge, but loses energy, is it safe to say this experiment demonstrates the loss of energy thereby breaking the conservation of energy law in a closed loop scenario?

If you agree the energy escapes in some fashion where the energy loss cannot be measured, can we speculate from this example it may be possible to reverse this action and recover energy in excess of what exists initially?

I ask this question because I have always thought if you can lose energy from a system where you have taken every possible precaution to account for it, then you have also isolated some form of gateway where a process can be formulated to reverse the action.  The conservation of energy law runs both ways--if you cannot gain energy, then you also cannot lose it.  If it is proven this example loses energy, then the whole conservation of energy law becomes invalid unless specifically restricted to certain situations.

Just my feelings.  I'm not taking sides one way or another on this, because I do not have the necessary tools to prove beyond any doubt we in fact have a condition where energy is actually lost.

M@

Offline tinman

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Re: Cap to Cap dump
« Reply #26 on: June 12, 2016, 12:27:56 PM »
@tinman,





If you agree the energy escapes in some fashion where the energy loss cannot be measured, can we speculate from this example it may be possible to reverse this action and recover energy in excess of what exists initially?



Just my feelings.  I'm not taking sides one way or another on this, because I do not have the necessary tools to prove beyond any doubt we in fact have a condition where energy is actually lost.

M@

Quote
I'm very happy to see you participate in this discussion and conduct tests yourself.  Given your time at these sort of things, I'm a little surprised you have not posted a video of this particular test years ago.

Oh but i did--long ago when i knew very little about it,and my youtube channel was a different name. But being very new to the game,i screwed up all the accepted terms we use here today,so i wont bother posting it here. But just watching it again,i did notice a big effect that took place.that i will try again now i know more than i did back then. Time to throw together another pulse motor lol.

Quote
The burning question I have to ask, having seen these results for yourself, if by moving charge from a full capacitor to an empty one conserves charge, but loses energy, is it safe to say this experiment demonstrates the loss of energy thereby breaking the conservation of energy law in a closed loop scenario?

Well if what the people say in the links Poynt provided,and where half the energy would be lost even if the resistance value was 0,then one dose have to wonder as to where that energy went?,as if the resistance value was 0,then no energy can be dissipated as resistive heat loss--as has been assumed until now.


Anyway--i have to hit the workshop after watching my old video,as i seen something in that that i missed back then--or should i say,did not know what i was looking at at the time lol.

Quote
I ask this question because I have always thought if you can lose energy from a system where you have taken every possible precaution to account for it, then you have also isolated some form of gateway where a process can be formulated to reverse the action.  The conservation of energy law runs both ways--if you cannot gain energy, then you also cannot lose it.  If it is proven this example loses energy, then the whole conservation of energy law becomes invalid unless specifically restricted to certain situations.

Well,lets wait and see how i go with this pulse motor test,as we can isolate a third cap from the system,and do comparison runs between two configurations.


Brad


Offline pomodoro

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Re: Cap to Cap dump
« Reply #27 on: June 12, 2016, 01:50:22 PM »
Anyone who wants to know what really occurs during a cap to cap discharge you need a good quality oscilloscope , two low loss NON polarized  caps and an air wound inductance wound with thick copper wire. The oscilloscope probe is lightly coupled to circuit with a small few pF cap. One cap is charged to 100V or so and discharged  into the other cap through the inductor.  Once you observe what happens, you decrease the inductance further, and so forth, until you have just a straight wire. Aim initially for a resonant frequency well within the bandwidth of the oscilloscope, a few MHz is usually OK, using the usual formula 1/(6.28xsqrt(LC)) keeping the ratio of L/C >200. You may then see where the lost energy actually goes.

Online verpies

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Re: Cap to Cap dump
« Reply #28 on: June 12, 2016, 05:11:31 PM »
I would think this is enough to determine that the loss is not due to resistance.
Not only due to resistance in realistic cases.
In an ideal cap2cap transfer (without intermediating resistance and inductance) infinite current would flow initially and the entire energy loss would be due to EM radiation.

Free Energy | searching for free energy and discussing free energy


Offline webby1

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Re: Cap to Cap dump
« Reply #29 on: June 12, 2016, 07:30:42 PM »
Indeed,,

What you do not use you throw away,, but as I have asked many times before,, WHY? why throw it away if you do not need to.

Does "capacitance" care about the charge carrier or only the charge,, meaning does the movement of the charge carrier matter.

What if the charge carrier never leaves the plate?

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Re: Cap to Cap dump
« Reply #29 on: June 12, 2016, 07:30:42 PM »

 

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