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Overunity Machines Forum



Electrical igniter for gas engines A keystone to understanding by Magluvin

Started by Magluvin, March 01, 2010, 01:30:50 AM

Previous topic - Next topic

0 Members and 3 Guests are viewing this topic.

poynt99

In regards to offering any help with the switches etc., there is this:

Why do you need so may stages?

Obtaining 350V in a 80uF capacitor can be achieved with one stage, with the source capacitor being 10uF at 1000V.

Can you think of any adjustments to the original switch that would yield this result?

.99
question everything, double check the facts, THEN decide your path...

Simple Cheap Low Power Oscillators V2.0
http://www.overunity.com/index.php?action=downloads;sa=view;down=248
Towards Realizing the TPU V1.4: http://www.overunity.com/index.php?action=downloads;sa=view;down=217
Capacitor Energy Transfer Experiments V1.0: http://www.overunity.com/index.php?action=downloads;sa=view;down=209

nul-points

Quote from: poynt99 on April 06, 2011, 09:37:34 AM

Obtaining 350V in a 80uF capacitor can be achieved with one stage, with the source capacitor being 10uF at 1000V.

.99

say it was actually possible to transfer 4.9 Joules of energy starting from 1kV on a 10uF cap, to give 350 Volts on 4x 20uF in parallel, without any significant I*I*R loss in the wiring/switches (leaving up to 0.1 Joules [141 V] on the source 10uF)

you've ended up with less charge-separation at your output - and you haven't achieved any work

so what?
"To do is to be" ---  Descartes;
"To be is to do"  ---  Jean Paul Sarte;
"Do be do be do" ---  F. Sinatra

wayne49s

Hi,

I just read this thread and wanted to make a comment on the capacitor energy issue. From the energy formula, the energy goes up as the voltage sq. and only proportional to the capacitance. So you were wondering where the energy lost came from when charging a capacitor from a capacitor. The opposite was always a problem for me. Charge 2 capacitors in parallel, and then connect it in series. By doing so, the voltage is double, but the capacitance is halve, so the net energy gain is 200%. If you did the same with 3 capacitors (3 parallel, than 3 in series ), the net gain is 300% (32/3). So "playing around" the parallel & series connections will give you any gain you're looking for. So where did the extra energy come from by playing with this switching?

As far as the formula 1/2CV2, it is based on work to move a charge in an electrostatic field. The sq appears in other formulas like kinectic energy= 1/2 mv2.  E=mc2.

/Wayne


Magluvin

hey guys
I may be eating my feet

I have been working on this solution. I may be all wrong here.

My stages got to me.  I have to sit down with this tonight and see what I was thinking.

I have been at it for 40min here at lunch and Im just screwing up trying to go so fast.

if it was all wrong, Im sorry for any inconvenience. =[

I keep getting my caps divided and multiplied all screwed up.
Ill go over it tonight.

Im on it, right or wrong, I will tell. 

Gota get back to work.

mags

forest

Mags,wayne

The problem is with equations definitely. If you assume energy transfer without loose and without gain , just pure energy conservation , from one larger cap into two smaller or from one full into empty one then we could use other equations and the result is broken charge conservation. Charge is lost during operation.
Something is really fishy. Let's assume that really 50% is lost as a heat. Heat is EM radiation. With coil acting as flywheel we have no such problem , that means inductance can recover that energy like radio antenna back into electric current.
That 50% difference is very suspicious to me.I would expect the difference varied by resistance in path of transfer, not just exact 50% always.