Quote from: Loner on April 14, 2011, 02:31:01 PM
Forest, good point, except, the field is "Free" only when not changing.  To alter the field strength or size requires input or output, depending on direction.  Conventional theory states this is exactly in for out, minus circuit losses, but as is slowly being noticed, that is not always true.  (Remember, transformer...)  Defining that point is why I'm watching this very informational discussion.

(To EE's, I understand this concept is not acceptable.  Live with it.  sorry for that remark, but true reality is always stranger than fiction.)

I think Tesla understood it also and instead of fighting against changing which is useless he make it freely changing and the effect "escaped" from circuit as 50% gain, not loss 
You see if this 50% is free gift then the clever man will try to divide part  which is inside wire from the part which is around and collect both . Now you see what next ,right ??

wow moving fast here

thank's to all

@ Mag i really be with you in your post 688

now just before going to sleep    1  more question

if the connecting wire for the  DIRECT transfer (without inductor and diode,   that is cap to cap)  could be supraconductor  (i mean no loss in  at all ), would we also get 50 % energy   lost in the transfer ? I think yes and you ?

good night at all

Laurent

Quote from: poynt99 on April 14, 2011, 01:57:29 PM
It hasn't been my intention to spell everything out completely for you and Laurent.

I have been providing you with bits and pieces of key facts so that you may contemplete them and hopefully figure it out on your own.

You may still disagree, but I assure you that everything I have told you is correct.

.99

It may have sounded that I was charging you with neglect, but we are trying to look deep here. Just in case you have not noticed. ;]
The Point I was trying to make was, we are looking beyond just simple explanations. It may be that I just find out the hard way. ;]

So I was exclaiming, that your answer was not as definitive as what we were looking for. And as you said, your intention is not to be spelling it all out for us.   So we continue......

Mags

Quote from: Magluvin on April 14, 2011, 05:18:08 PM
It may have sounded that I was charging you with neglect, but we are trying to look deep here. Just in case you have not noticed. ;]
The Point I was trying to make was, we are looking beyond just simple explanations. It may be that I just find out the hard way. ;]

So I was exclaiming, that your answer was not as definitive as what we were looking for. And as you said, your intention is not to be spelling it all out for us.   So we continue......

Mags

From my perspective, I offer an insight and some key information, then I see if that is understood. If it appears so, then I move on and offer more information based on questions that arise from the previous info. As we proceed, we can go more in-depth if required. That is how I gauge the information I offer.

So far no one has acknowledged or been able to demonstrate that they understand the fundamentals I have presented so far....and you feel you are ready to jump into more detail already?

.99

Quote from: forest on April 14, 2011, 02:18:29 PM
Mags

Simply speaking : magnetic field around wire should be a free gift.Why ? Because how do you explain that magnetic field is stable if you run circuit a minute or  a day or a year. It does not add to itself, it's a response of ether to balance electric current density inside wire.

Hmm. How can I say it in another way.

My serious question is, if we compare the 2 circuits, the first circuit using an inductor/diode, and the second with just a resistor.

I cannot say that everyone understands, that during the period of discharge from the source cap, when the 2 caps, in either circuit, lost 50% at the "point" that the caps reached 500v each, During that time in both circuits is 50% is gone, hastalabyby 50%.

All well knowing that at this crossover point, circuit inductor/diode is not done yet, as the inductor is rolling and has enough energy to suck the source cap just about dry, and pump it into the receiver cap. The inductor is loaded with enough to cover those losses, ALREADY LOST IN PROCESS, by nearly 100%.   If I never looked in the middle of the process, and just knew what was in and what was out, we would not be filling these pages today with this stuff.  =]

But I did look in the middle. Things dont add up.

Answer this at this time. Would you agree that when all caps are at 500v at an instant in time, can you say that only so much energy from the source caps(both circuits) have delivered the same amount of energy? Has to be. Because if we removed those caps from the circuits when they are at the crossover point, All caps will measure 500v each. We have to agree on that. Other than that, we still have the inductor spinning a way in the inductor/diode circuit.

If we could swap out the inductor at the crossover point while the circuit is in the middle of transfer, when both caps are at 500v, from the inductor/diode circuit into the resistors place in circuit 2, I bet that inductor would take that 500v in the source cap and pump it to the receiver cap. And our caps left in the disassembled inductor/diode circuit will be standing with a 50% loss. And that 50% loss occurred during the charging of the inductor. So why do we not have any MORE energy lost from the disassembled circuit, the energy that got the inductor going in the first place?

So during this one way discharge, during the time that the source caps are declining in potential, where can we say that energy beyond the 50% lost already, got the inductor to its peak at this time? Just because the inductor was in the circuit, should we say that the 50% loss came from getting the inductor going? What a coincidence. ;]

I suppose that the magnetic field in the inductor just reaches out and just nullifies any currents running through all resistances in the circuit and reduces heat losses in those resistances to virtually nil.

I only see the 50% loss. And I see we have a flywheel all spun up while were losing the 50%, the same exact 50% that is lost while doing a direct transfer.


You know, if we do a direct transfer from cap to cap, we have a high freq oscillation. If we are able to disconnect the caps at a certain time, a time when that discharge induced oscillation only makes it to the peak of the other phase, 1 half cycle, I bet that we see a near 100% complete conversion. We still have inductance, just small, thus high freq osc.  And we simulated a diode by disconnecting at the reverse phase peak.  Same as inductor/diode BC.

We didnt lose 50%?  Only if we let it oscillate to its death.

If we add just an inductor and no diode, we get a longer, lower freq oscillation.

The only reason we lost 50% in a cap to cap transfer, is that we let it happen. We let the oscillations continue to decline till the caps are at 500v each.

If we have a larger value resistor involved in a cap to cap direct transfer, there could be no oscillations, at least not noticeable with the naked eye. We should just see dc declining in the source and inclining in the receiver. resistor-damper.




Mags   ;]