Quote from: poynt99 on April 13, 2011, 07:26:22 PM
Mags,


So, since the R of the wiring has not changed, but the current has been reduced by a lot (when we utilize an inductor), there will be less power dissipated in the R of the wiring.

.99

Hey Point

Well, if we added a 5kohm resistor to reduce the current, or even a lot, 5mohm, and we let a 10uf 1000v transfer to another 10uf 0v and we wait till they level out to 500v each. Should not our losses be different here with the reduction of current? As you described. ;]

Mags

Exactly Mags, exactly

This is the problem. Why loss is always 50% not matter what resistance is used ? Looks like a law NOT RELATED to resistance, something is cleverly hidden.
And ohm law is probably around it to cover.
We expect that  the amount of resistance would modify loss amount but it only change the time length of event.


Thus I conclude it has to be an oscillation which loss 50% between two caps and strangely enough this is the same circuit Tesla used to generate longitudinal radiant energy as presented by Dole.

If you consider equation Q= C*V and suppose energy is not lost during transfer you will end up with charge loss.

Now consider what I shown in my circuit mod. continuous oscillations when circuit is unpowered.

What I propose is radical : 50% of energy flow around circuit
and that part is lost and that loss is modulated by oscillations (imagine flow of water between two containers with open valve - it is not a single step transfer, it always roll back and forth before going into rest state) - the output is heat which is radiant energy !
Now the problem is that our "adored" Maxwell theory hide that and cover in many places , and it only stick out from various holes as a BEMF or reactive power or heat loss.

of course I must re-think this idea much more but the obvious connection with Tesla "harpin circuit" is striking.

Hi all

thank's for your very informative answers and patience.

I very slowly make up my mind on this subject. Not intuitive at all. But very interesting.

@Forest

i think you speak of this circuit and the high frequency (green ) on the left. I got also this type of frequency after the transfering (in the middle is the inductor trace on the BC). The amplitude of the frequency depends at what voltage you stop the source cap and leave the inductor work alone going on the transfer. Notice that i tried to stop  this frequency by shorting the inductor (upper small circuit)

hope this helps

Laurent

Quote from: Magluvin on April 14, 2011, 01:00:10 AM
Hey Point

Well, if we added a 5kohm resistor to reduce the current, or even a lot, 5mohm, and we let a 10uf 1000v transfer to another 10uf 0v and we wait till they level out to 500v each. Should not our losses be different here with the reduction of current? As you described. ;]

Mags

No, the loss will be 50% again.

The goal involves more than just reducing the current; you need to shift the balance as to where the energy is being absorbed. When you add the inductor, you are shifting the balance of absorbed energy away from the circuit resistance to the inductor and it's magnetic field, where it gets stored then transferred to the second cap.

.99

hey Point

I hear what you are saying. No ill will toward you, but Im not seeing the logic in that.
See, it is also important to all whom are reading this. If I had not made my previous post, the the only explanation others would have from you is that the inductor reduces the current in the circuit to the point that very little is lost in heat related losses.....


Mags,

Yes, current still flows through all the resistive wiring, but it will be a much lower current compared to that if the circuit consisted of the wiring and caps alone.

P = I2R correct?

So, since the R of the wiring has not changed, but the current has been reduced by a lot (when we utilize an inductor), there will be less power dissipated in the R of the wiring.

.99


But now there is more involved according to your last post. ;]
But I still find your reasons incomplete. 


Consider 2 caps, 1 with 1000v and one has no charge. If we use and inductor and diode, we have almost a complete transfer from source to receiver caps.
Now consider the same 2 caps, 1 with 1kv and one at 0v, but we have a resistor "only" between the 2 caps. This resistor will be of a value that in comparison to the previous consideration above, the resistor cap setup will make full transfer, losing 50% and both caps will end up with 500v, but in half the time that the circuit with inductor/diode makes its complete transfer.

The reason for half of the time will be clear in a moment. 
We set these 2 circuits side by side. We close the switch on both.

The source cap will start losing voltage potential in the inductor/diode circuit as fast as the resistor circuit. By the time the resistor circuit only circuit reaches 500v on each cap, the inductor/diode circuit will also have 500v on each cap.

There is a point in time in the inductor/diode circuit, that the caps, source and receiver are at 500v each, while the circuit is still completing, or in the middle of, its near 100% conversion.
That means that the caps, in total, at that time, are at the 50% loss point. Just like the resistor circuit. The same dang loss in time.

We can all agree that the inductor is building its field during the time that the source cap is of higher voltage than the receiver cap.
And once the source cap, in the middle of the near complete conversion, is equal or less than the receiver cap voltage, there is no longer a build up, but a decline(collapse)in the inductors field.

So during the time it takes for the caps in both circuits to reach 500v each, that is the time during the transfer that the inductor was storing its energy.

So now if we have just the inductor/diode circuit, and we close the switch to get things rolling, we have another switch across the inductor and diode also. So when we see the 2 caps reach that 500v each at the same time, call it a crossover point, we close the switch shorting out the inductor/diode. The caps will come to rest at 500v each, due to the elimination of the inductor/diode from the continuity path of the caps.

So here, our caps, in total have suffered a 50% loss, just as the resistor circuit did, at this point in time.

But now, since we closed the switch across the inductor/diode, the inductor is still spinning, in the same direction through the diode and switch, looped. It still wants to push current, till it dies down due to resistance which puts a damper, and limits the current flow.

If we added a 1kohm resistor in that loop, the inductors freewheeling action would die quickly due to the limitation of the resistor to pass the current that the inductor would like to continue. Damper

So if in the inductor/diode circuit, we can say that during the time that the caps reached the crossover point, that was when the energy was stored in the inductor.

If we got that inductor going during that 50% loss time, when the caps reached 500v each, then where did the energy come from to induce storage into the inductor? Was there a larger loss from the source than 50% during that time?  I say not. Or can we say that getting the inductor rolling doesnt cost anything here?  Fishy  ;]

We already see that in both circuits, when activated at the same time, they both reach 500v in all caps at the same time. So at that point, we have an equal loss in both circuits. But in the inductor/diode circuit, we have the inductor still going.

So how do we assess where the energy that was stored in the inductor came from?

If in both situations we disconnect the caps from the circuit when they are all at 500v each, we have a similar situation with the 50% loss but the flywheel keeps on rolling with the capability to recover most all of that 50% loss.

I dont think things are working they way you describe. Not to be a dick, just to be truthful. =]  or maybe there IS more than you have described. ;]

Respectfully
Mags