Quote from: poynt99 on April 17, 2011, 05:02:52 PM
No, the caps have an internal resistance, so we are back at square 1.

Well if we have a superconducting wire, why not a superconducting cap?  ;]  Square 0

If we look at your example, where the wires are superconducting, and the caps are not, then we may not end up with .707v from 1v. Correct ?

Listen. I know there are a plethora of equations that describe everything under the sun, well almost. But we should be able to describe all things that are calculable, in plain words. Like if 10uf 100v is equal to 20uf 70.7v instead of just 20uf 50v is 50% of 10uf 100v. It took a few pages to get that out.



Im still trying to get it right in my head as to these facts.

For example. If we have 2 caps 10uf at 70.7v, we can use one of those caps to charge an inductor enough to pound almost all of that 1 cap into the already charged cap at 70.7, all the way to near 100v.

Seems like there was enough to do the job, even against the charge, 70.7v, that was in the receiver cap already. One might think that it would take more that 10uf 70.7 to accomplish this feat.

But Im getting close I believe. These are all good things to have knowledge of here.

Tito said there is a gain in the Igniter circuit. But all considering, I dont see it yet. It is the same circuit as the ozone pat. And many have referenced it also as capable of OU.

Tesla stated in the igniter pat. that it produced and increased eff.  Once I had gotten that idea from reading the pat. I fell out of the loop here for a bit.  But that is not to say that the circuit with some changes just may do the deed.  We can only push forward and try things.

We continue on....

Mags

Quote from: Magluvin on April 17, 2011, 06:22:05 PM
Well if we have a superconducting wire, why not a superconducting cap?  ;]  Square 0

You were referring to a non-hypothetical scenario, and so was I in response.

Quote
If we look at your example, where the wires are superconducting, and the caps are not, then we may not end up with .707v from 1v. Correct ?

You will end up with 0.500V, no maybe.

Quote
Im still trying to get it right in my head as to these facts.

For example. If we have 2 caps 10uf at 70.7v, we can use one of those caps to charge an inductor enough to pound almost all of that 1 cap into the already charged cap at 70.7, all the way to near 100v.

Incorrect. If you start with two capacitors at the same voltage, how is the inductor going to get energized when there is no potential difference placed across it?

.99

Quote from: poynt99 on April 17, 2011, 06:51:39 PM
You were referring to a non-hypothetical scenario, and so was I in response.
You will end up with 0.500V, no maybe.
Incorrect. If you start with two capacitors at the same voltage, how is the inductor going to get energized when there is no potential difference placed across it?

.99

I actually have to go back to respond to the first 2 sentences, but for now on the 3rd sentence...

We could just discharge the cap directly into the inductor till the cap is empty, and then the inductor discharges in to the already charged cap. I see where you are coming from as if it were through the BC. But I have shown a BC circuit already that has a switch going diagonal across the same circuit, and a diode that crosses the other way, in order to be able to do either way.

In the BC we get left with a bit in the source cap. When we directly charge the inductor, we can get all of the source cap into the inductor.

Anyway sorry for the confusion

Mags

Yes,

I forgot about that alternative method, as shown in the document GL uploaded.

You are correct, I was clearly thinking only about the cap to cap with an inductor directly in between.

.99

Quote from: woopy on April 17, 2011, 06:11:02 PM
Wow

always more interesting here

I go,  on the air tank theory  and assume and accept that, the 50% energy lost is a real fact and that's it.

But the lost of stored energy is not due to the tubing transfer  (or wire resistance heat lost in the case of cap transfer ), but simply to the DOUBLE VOLUME for the air to expand.  So the air pressue is simply divided by 2 in each tank. Seems very logical.

And always because of the SQUARED calculation of pressure (or voltage ) the result is mathematically and always giving a 50 % in energy lost    always because 1/2 * C * V^2  formula   and that's it.

And now the question is , how and why ,   in the Believe Circuit (BC) of Magvulin    it seems  possible to improve this  result ? I mean we really    in real life    do not lose 50 % of energy

I mean and it is a fact  by using the BC we can reach in real life i insist , a 82  % (with very basic circuitery ) and  more transfer energy ( see all the above post ),

For instance if i put an air turbine in place of the valve in the air tank transfering tubing, and this turbine should activate a flywheel during the transfewr , do you think that the stored energy in the flywheel could be strong enough to go on emptying the source tank to almost 1 BAR and to recompress this air in the receiver tank to almost (let's say 90 BAR )  ?

yep

sliping is needed

good luck at all

Laurent

I with ya here Woopy

I see the direct exchange or connection as just a said pressure in a said volume that is allowed to expand into a volume that is double.  No need to figure in resistance heat losses.
We are just losing from allowing the pressure to be released in expansion. It is a waste to do so. ;]

I would like to see an experiment where superconducting components would show 70.7v in each cap from a cap of 100v.

If this were the case, would not the inductor and diode provide an improvement on this?  ;]  ou?


Gotta go do laundry   =[

Mags