Electrical igniter for gas engines A keystone to understanding by Magluvin

[Magluvin]

Was thinking a one point if the 2 flywheels, a physical one and the inductor, what would be the differences. Could we get this same advantage in the physical world?

First, lets look at what is involved with getting them spinning.

The physical fly, tend to take a lot of energy to get them started, like an inductor has impedance. But I see them different. The real flywheel to get it spinning up consumes a lot of energy right from the start, and the excess input it took to get it going, will no longer be needed once up to speed.
But the inductor consumes less in the time it is switched on, and the energy consumed or let through the circuit as the inductors wheel gets spinning, allows more energy to be pulled from the source till the wheel is up to speed( field peak= current peak).    So they have a bit of different attributes.   

I broke off from these experiments early on when I figured that, at the time, how can I get 6.32 v in a cap to deliver all of its energy back into the battery. I then knew that just getting over cop was not enough for looping.

But lately, I have bunches of ideas that stem from old projects. And I think I found the answer for the 6.32v conversion.

Charge 10 caps to that voltage from the battery with the circuit.
Put them in series and discharge into the batt.  63.2v  Now when this series cap discharges into the batt, once the cap is discharged down to 5v, we have put back more than we took to fill all ten caps individually.  The amount we figured we got for free was
1.32v x 10 =13.2v     13.2v-5v = 8.7 of the free got back to the battery + 100% of what was used for the process..   The more caps we charge and put in series for discharge, the more % of free we get to keep per discharge.  That last 5 can be joul thieved in. ;]   lol

Mags

[nul-points]

...
But lately, I have bunches of ideas that stem from old projects. And I think I found the answer for the 6.32v conversion.
...
Charge 10 caps to that voltage from the battery with the circuit.
Put them in series and discharge into the batt.  63.2v  Now when this series cap discharges into the batt, once the cap is discharged down to 5v, we have put back more than we took to fill all ten caps individually. 
...
Mags

hi Mags

yes, looks feasible - when i tried the old variation on this (charge 2 caps in parallel - reconnect 2 in series to feed back to battery) the battery just discharged! 

   http://www.overunity.com/index.php?topic=10231.0


perhaps 10 caps in series will behave differently?  looking forward to hearing how you get on

all the best
np


http://docsfreelunch.blogspot.com

[forest]

The problem is to KNOW exactly where is energy.If at 1/4 of resonant frequency period it is still in magnetic field of coil then we need to disconnect coil right then and got energy in capacitors in parallel.Many.
The dump excess into battery but not allowing to kill the dipole, meaning voltage in capacitors HAS TO be left at higher level then power source.
Our habits are against us,not circuits
http://www.freestacja.pl/wp-content/uploads/2011/02/radio_0.jpg

[Magluvin]

Hey Forest

Maybe you are missing my point..

lets say that 5v in the cap can be considered an amount of energy, the same amount at 5v, every time.

When we charge the cap to 5v, then disconnect the battery, the cap continues to load up to 6.32v.  The extra 1.32 is, well extra, free.

But having a way to force all that 6.32v worth of energy back into the batt may be done with a joule thief. If the JT can use the charge in the cap till the cap reaches .5v, then we are charging the battery with more energy than was taken from the battery.

Or like in my previous way, charge 10 caps with the circuit. Then all caps in series is 63.2v of which 50v is what the battery produced, and the 13.2 was free.  In this way we can just dump the series caps into the batt till the cap is down to 5v, equal to the batt. That would leave .5v in each cap.  6.32-5= 1.32    1.32-.5= .82v   
So each cap will have delivered 5.82v from each cap back to the batt, while the battery only provided 5v to each to begin with.
If it were 20 caps, then we will have gotten more of our free 1.32v back to the batt.   50 caps, even closer to 100% of the 6.32v back into the batt.

Took me a while to get the grip on it, but I find it solid.

Mags

[nul-points]

Hey Forest

Maybe you are missing my point..
...
When we charge the cap to 5v, then disconnect the battery, the cap continues to load up to 6.32v.  The extra 1.32 is, well extra, free.
...
If the JT can use the charge in the cap till the cap reaches .5v, then we are charging the battery with more energy than was taken from the battery.
...
Mags

hi Mags

i think that both you & Forest are essentially making the same point about collecting the stored energy: ie., when you transfer energy back to the battery, leave some in the cap(s) so the source dipole isn't killed


but the extra 1.32 is not showing free energy, i think

it's some energy from the battery that got stored in the field of the coil (as Forest mentioned) - caused by the series current flowing round the circuit. whilst the cap was getting charged

when the current flow was stopped, the coil 'flywheel' effect (which you've mentioned earlier) tried to maintain the current by using the energy stored in its field

this 'discharge' of the coil field energy got added to the energy in the cap - and it shows up as an extra 1.32V on the cap

so all the energy now in the cap has come from the battery

this must be why that old 'charge 2 caps in parallel, discharge as 2 caps in series' version (mentioned a couple posts above) just depleted the battery...

...no extra energy was getting returned to the battery - because it all came from the battery in the first place!


it looked such a neat idea too 
np


http://docsfreelunch.blogspot.com