Electrical igniter for gas engines A keystone to understanding by Magluvin

[woopy]

Thank's Gyula

very good explanation as usual.

So as we have no sinusoidal AC  but more a pulse DC in the believe circuit, how is the really important improvement in the transfer cap to cap possible.?
What is the "mechanic" of the flywheeling effect in the diode- inductor pair in the" Believe Circuit "of Magluvin?

thank's

Laurent

[poynt99]

So as we have no sinusoidal AC  but more a pulse DC in the believe circuit, how is the really important improvement in the transfer cap to cap possible.?
What is the "mechanic" of the flywheeling effect in the diode- inductor pair in the" Believe Circuit "of Magluvin?

thank's

Laurent

From this post:
http://www.overunity.com/index.php?topic=8841.msg281256#msg281256

Specifically:

Just remember; a capacitor or an inductor can store and release energy, whereas a resistor can do neither, it can only burn off energy. A piece of wire at low frequencies is purely resistive, so it burns off energy when connected between a source cap and a charge cap. When you replace that piece of wire with an inductor, no energy can be burned off so it stores it, then releases it.

.99

[Magluvin]

Seems funny that if there is resistances in any circuit, to be able to transfer 1000v from a 10uf cap to another 10uf cap, we can get up to 99.9% transferred. And usually the source cap holds the balance left over of which voltage wise, the totals add up to 1000v.

Seems like resistance is just non existent in these cases.  ;]

We do have current flow. Why no losses in inherent resistances in the circuit?  CURRENT FLOWED. With apparently no heat being generated at all in any resistances.  Think of the BC's function. The source cap is going down in voltage and the receiver is going up. Some where along the way, both of those caps reach 500v at the same time while one is going up and the other is going down!!  Are you making a connection?

Here in the BC we have a point that if the caps were removed from the circuit, each would hold 500v, just like a direct transfer, just like the 50% heat losses in a direct transfer. So can we say that the inductor stored the other 50% while still having the 50% heat loss?

Try the circuit on a bench. No recycle cutoff, just hold the switch till transfer is complete. 10v 1000uf source on a BC with a 0v 1000uf receiver. When the source cap reaches 5v, the receiver is also 5v, no doubt. 

So think about this. We got the flywheel going very well during the first half of the cycle, all while losing 50% in heat, and still able to transfer near 100% of source to receiver. And the source contains the balance when all is said and done.

SOOO,  during the first half cycle of the BC, we have expended 50% in heat, AND got the flywheel going enough to complete the transfer.

Now, IF we have expended 50% in heat during the first half of the cycle( proof of 500v in each cap at the same time during the cycle resembled the losses physics quotes, No?  =] )  And we got the flywheel going during that half cycle, where did the expenditure into getting the inductor going come from? Can you spot or show that it came from, umm  from,   well where?  Free?  ;]

Its a bit silly isnt it?  ;]



I get the phase thing, but over time, the current flowed. Resistance should have intervened somewhere along the way during a near 100% transfer. But we suffer super resistance losses in direct transfer. Monster losses compared to microscopic. 
Still smells fishy.  ;]

Mags

[poynt99]

Sorry,

You've lost me mags.

Could you formulate one or two questions directly pertaining to what you are wondering about, or what your confusion is about? I'm sure I can answer your questions if you could just nail down exactly what you are asking.

Thanks,
.99

[woopy]

Hi p.99

My question is very precise

I think that in the "Believe  Circuit " we are not on an AC system, but in a DC system. So the reactance of an inductor , Which in AC system can transfer energy without lost of energy  ( ok Guyla that is not totally true, because the VAR, but i try to get my response as simple as possible  ), because the sinusoidal gain and lost due to the sine wave (pos and neg),   does not apply here. Right ?

In a DC system, the inductor effectively stores energy and than can  release it but certainly not so much as it originally stored it. Because inherent resistance,  Right ?

The inductor in the "BC" is installed  between 2 cap and is connected  to those 2 cap with  resistiv wires , one wire on each side of the inductor . Right ?

As those 2 pieces of wire, has a small but real resistance , they should from themself destroy 50 % of the tranfered energy anyway,    as they do in a normal transfer ,  i mean , without the inductor . Right ?

So if i don't miss something , in the BC (Believe Circuit ) the inductor has not only to be almost free  working (i mean almost without any resistance ,that is no lost, which is surely not the case) , but it has also to recover the lost of the 2 adjacent wires,        which in normal transfer, should  destroy as i insist to remind ,50 % of the transfered energy. Right?


And finally i think that we have to consider not only the inductor , but the  "mariage " or wedding  pair of inductor and diode. because in my real life testing this effect  NEEDS the pair. Without inductor and only a diode= no improvement of the transfer.
without diode and only inductor  = no improvement of the transfer.

thank's to all for the good work

I am very happy to learn so much with you

and good luck at all

Laurent