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Author Topic: Checking out Joel Lagace - Hang on I'M SERIOUS!!  (Read 14570 times)

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They make 10000 Farad capacitors now ?  :o
It all started with ths video
https://www.youtube.com/watch?v=0c7llvTjlwQ
That is where the 10000F caps claim originated.


---------------------------
Electrostatic induction: Put a 1KV charge on 1 plate of a capacitor. What does the environment do to the 2nd  plate?
   

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hey aking

yeah, im not 'saying' f6flt is wrong. but, as we should all know, electrons have a neg charge vs pos, crt knowledge should solidify that. and a pos charge is brought about by atoms missing electrons. doesnt sound so far fetched. now comes the tough part.  why does a magnetic field come about when electrons flow through a conductor?

see, i have a 'theory' that ive only presented a couple times in the last 10years or so. the theory is based on why, depending on pole polarity of magnetic flux and the direction that flux crosses or cuts a conductor, does that determine the influence of currents in that wire, one direction or the other? how does that process work exactly. so this is what i have come up with....

im thinking that electrons are where that mag field derives from. also im thinking that electrons have a pos and neg side to them, with the mag field circling around the pos and neg axis of the electron.  sounds crazy, i know. but hang with me here...

when we apply current to a wire, we are told that electrons enter the wire from the neg of the source, through the wire towards the pos of the source. so lets just say as the current is low, fewer electrons flow and higher current, more electrons flow, in 1 direction. why do we have less mag field with low current, and stronger field with high current? ah, we say well possibly my theory that electrons, each one, and say just ones that are loose from their atoms, has its own field and when there is more current thus more electrons.  but how do the electrons know to align their individual fields with the fields of other flowing electrons?  this is where the pos and neg sides of the electrons theory becomes a possibilty.  we apply the source to the wire and electrons are flowing because of charge difference from one side of the wire to the other. if the electrons have a pos and neg side to them and they align themselves, while flowing in the wire, with the pos and neg input, then the mag fields of each will also be in alignment with each other. thus more current flow, more loose electrons, more magnetice field strength.  all this being said, the electrons pos and neg alignment properties with the polarity of the input must be greater than the mag fields of the electrons, as the individual mag fields between aligned electrons would oppose oneanother, as in repell.  like 2 magnets being brought together, side by side, they would repell.

not saying i am correct. just a theory that has stuck with me for a while. its my way of thinking why does current flow this direction or that direction depending on the direction the flux cuts the wire 'prependicularly'  thats the puzzle my brain is trying to solve. so if the idea of electrons having a pos and neg side electrical field and the mag field, b field, circling the pos and neg axis, then inducing current direction with flux cutting works similarly where the flux aligns the electrons in a polarity pos side this direction, neg side that direction. this would imply that the electron is also the source of electric field. drag the magnet across the wire, electrons align pos and neg in the wire, only to move if there is a load on the induced wire.

if i am totally wrong, then i dont get where the b field around the wire comes from exactly. it remains a big mystery the functions of this polarity, that polarity. this direction and that direction. all whether we are inducing the wire or running current through it. also if im wrong, then i have to think of the electron as just a neg particle that is just mysteriously manipulated as we seem to maybe know, as if we know some of how it works together in very basic form, but we dont know why it all works as it does. yet.





i have a small neon drive circuit from an older copy scanner machine. high freq, about 1kv.  so in the case of a cap with an av plug(2 diodes) and we connect the plug to just 1 wire of the neon trafo output, we are able to charge that cap.  was not in the belief that with the neon driver operating, no load, that the charges on one end of the trafo ouput winding, are only electrical fields on their own.  i was in the belief the electrons in that output winding, mind you, high volt, very fine wire and many turns, are being compressed from one end of the wire to the other, back and forth.  so when there is an excess of electrons at the av plug, they can release a bit of pressure into 1 plate of the cap through 1 diode, and also suck electrons out of the other plate of the cap, through the other diode when the pumping of electrons in the winding goes the other direction. thus allowing the cap to charge via 1 wire.  but if my theory is correct, then it could be just pos e fields at one end of the hv winding and neg e field at the other end, could be that electrons are just aligned in the unloaded winding and the source of the e field is all those electrons with pos and neg sides are aligned + - + -.  but again, if i am wrong, then i revert back to electrons being pumped from one end of the winding to thebother and back again, in order to legitemise the electrical charges at each end of the winding.

as for the device you have, i suppose you would need a lab grade current meter to see if the currents you have going to the battery through the tiny cap are actually coming from the wall.  tiny caps like that only allow small amounts of ac currents through.  like audio analog crossover networks where the caps reduce low freq currents getting to tweeters for example, or used to absorb high freq to gnd after a series inductor to keep high freq away from the woofer.

you might get better results by making an lc that is tuned to your house line freq and just tickling it with 1 wire from the hot lead in the wall. cant say that the house line meter would register that if you got high enough currents from the lc.

mags


   

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I've just had a thought about Lagace's 10000F comment.

It could be that he has 4x2,500F caps in series.
Therefore Cap C3 would be only 625F
« Last Edit: 2025-02-26, 05:26:29 by Aking.21 »


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Electrostatic induction: Put a 1KV charge on 1 plate of a capacitor. What does the environment do to the 2nd  plate?
   

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I've just had a thought about Lagace's 10000F comment.

It could be that he has 4x2,500F caps in series.
Therefore Cap C3 would be only 625F

looks to be in series.  also looks to be 10,000F also from the closups. i wonder if there are caps out there being labeled wrong intentionally to sell them.  i havnt seen any for sale.  titles of products for sale that say 10,000F but when you look, they only have maybe 3000F or less. anyway, i still want one of those 10000F caps. ;)

why he says they are 10000F but doesnt verbally specify they are in series makes me wonder whats the gist, or maybe he just forgot to say it at the time. dunno

the use of a smaller capacitance for cap C is the only way he could get these numbers. especially when the circuit drawing shows all caps same value. why change that in the demo. just too thick on boloni so far.

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I updated my post #114 with some additional data (33H and 48H) and copied it here, including the graph.


 
                C1                             C2                            C3
start          2.7V                         0V                             0V
6 hours      1.704V                     1.156V                       2.08V
11 hours    1.618V                     1.194V                       2.13V
24 hours    1.455V (105.85 J)     1.255V  (78.75 J)        2.18V  (59.4 J)
33 hours    1.349V (90.99 J)       1.29V    (83.2 J)         2.2V   (60.05 J)
48 hours    1.249V (78 J)            1.311V  (85.94 J)       2.22V  (61.6 J)
58Hours     1.201V   (72.12 J)     1.305V  (85.15 J)       2.23V  (62.16 J)

Its new to see that C1 now has less energy in it than C2.

But overall we still see no gain as we started out with 364.5 Joule in C1, and now we got a total of 225.54 Joule in the 3 caps.
This however is more than what we could expect when paralleling 2x 100F caps which should end with half of the starting 364.5 J being 182.25 Joule!




I agree that 10000F caps are very hard to find / get now (i could not find any, the skinny ones like Joel shows), so i wonder where they came from already 2 years ago.
With other words, its almost certain they are not 4x 10000F caps in series (meaning 2500F) IMO.

Itsu
« Last Edit: 2025-02-27, 09:48:40 by Itsu »
   

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Itsu, that's an interesting result..

It shows that something is going on which possibly defies Kirchoff's law.

I would like to bet that a repeat of the experiment will show a greater gain.
BTW Lagace said this only works for cop>1 with supercaps.

I don't know if mine qualify, as they are from an electric bus at 2.5v 2600F each, but I'll give it a go.
They are Maxwell Boostcaps Type BCAP0010 A08

I think your results are encouraging and some proof for my hypothesis that repeated pulsing can
 extract energy from the electrons surrounding the device.
I  have finally built my F G device, so can move on to serious stuff.

Thanks, once again for your input, Itsu.




---------------------------
Electrostatic induction: Put a 1KV charge on 1 plate of a capacitor. What does the environment do to the 2nd  plate?
   

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I agree that those results are interesting and that something is going on, but i am not sure what.

The fact that both 100F capacitors C1 and C2 suppose to settle out to half of the C1 initial energy level (365.4 Joule) thus 182.25 total, but that we ended up together with C3 to a total of 225.54 could be considered IMO as a gain.


Your caps will qualify i think, why not, the question is only if you need a higher capacity for C3 or a lower on, so try both cases if possible.

I think i will return to my 220V / 8V transformer and repeat the experiment to see if the lower C3 capacity (4x 100F in series, thus 25F) provide similar results as with the 1:1 CMC transformer.

Itsu

   
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I agree that those results are interesting and that something is going on, but i am not sure what.

The fact that both 100F capacitors C1 and C2 suppose to settle out to half of the C1 initial energy level (365.4 Joule) thus 182.25 total, but that we ended up together with C3 to a total of 225.54 could be considered IMO as a gain.


Your caps will qualify i think, why not, the question is only if you need a higher capacity for C3 or a lower on, so try both cases if possible.

I think i will return to my 220V / 8V transformer and repeat the experiment to see if the lower C3 capacity (4x 100F in series, thus 25F) provide similar results as with the 1:1 CMC transformer.

Itsu

Instead of C1/C2 losing half of their energy via resistance in the wire, some of the energy is transferred via the inductive impedance of the transformer, effectively storing it in C3 instead of dissipating it as heat.
   

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Instead of C1/C2 losing half of their energy via resistance in the wire, some of the energy is transferred via the inductive impedance of the transformer, effectively storing it in C3 instead of dissipating it as heat.

the 50% loss is not due to resistance.


mags
   

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one way to think about the possiblity of resistance as being a loss in the cap to cap scenario, is to look at other ways resistance causes losses in other circuits, even simple circuits.

look at a 10v source to a lightbulb.  add more resistance in series with the bulb, what happens?

in the cap to cap deal, try changing that resistance.  super cool super conducting wires and the caps for the least amount of resistance,  then try no cooling and the lowest resistance possible.  then add 1000ohms.   then 1megohm.

the outcome of say starting with10v will end in 5v in each cap every time.  not a microvolt lost. this example above, simply put, should encourage one to seek elswhere for the apparent losses we are discussing.

now.  why did we lose 50%???


mags
   

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still trying things with the cap to cap and orbo in place of the transformer.

below from left to right is caps A, B and C. 

the best transfer of cap A to B through the toroid winding was so far with 2uf plastic dip caps.  the cap to cap potential and currents are quite different than pulsing the toroid from a capable source. the immediate dip in voltage across the winding, cap to cap, forces me to try higher voltages in order to get as much of the magnets flux out of the core to cut the output windings as possible. pulsing the winding with even short duty cycles works better. like i said earlier, will need to rewind the toroid for more inductance to try to improve things. the caps in the pic are 53uf 450v.  i get good output from the output winding but the cap to cap transfer suffers due to too low inductance.  the amount of energy stored in the inductance reaches peak to quickly, and by the time the caps get closer to equilibrium, the energy in the inductor is diminished.


im going to wind a new one and see if it improves anything.


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one more thing.. earlier i said the 2uf caps self drain too quickly.  the meter i was measuring that voltage with must have a low input impedance.  when i tried my fluke meter, 10mohm impedance, the drop was still there but not near as bad.  the other meter i one of the new tweezer meters.  works great for many things.  but this was an issue measuring stored voltage on small caps.

mags
   

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Instead of C1/C2 losing half of their energy via resistance in the wire, some of the energy is transferred via the inductive impedance of the transformer, effectively storing it in C3 instead of dissipating it as heat.


the 50% loss is not due to resistance.


mags


Thanks web000, thanks Mags,

not sure what it is here, but looking at this wiki page: https://en.wikipedia.org/wiki/Two_capacitor_paradox  science tend to believe its due to resistance and radiation (RF).

Itsu

   

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one more thing.. earlier i said the 2uf caps self drain too quickly.  the meter i was measuring that voltage with must have a low input impedance.  when i tried my fluke meter, 10mohm impedance, the drop was still there but not near as bad.  the other meter i one of the new tweezer meters.  works great for many things.  but this was an issue measuring stored voltage on small caps.

mags

Good call Mags,  so we have to check what the influence of the used DMM's are on the measurements.  O0
   

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I changed in my last setup the CMC 1;1 transformer for the 8V / 220V transformer, the rest is the same, see diagram:



I recorded the first hour the voltages every 5 minutes, then let it run overnight and toke again a measurement after some 12 hours, hence the somewhat skewed graphs.

The result can be seen here:



No similar increase on C3 as with the CMC transformer now.

So the impedance / inductance of the transformer plays a major role in the transfer to C3.

Itsu

   

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Thanks web000, thanks Mags,

not sure what it is here, but looking at this wiki page: https://en.wikipedia.org/wiki/Two_capacitor_paradox  science tend to believe its due to resistance and radiation (RF).

Itsu

if you can think about it like 2 5gal buckets instead of caps.  even air tanks.  pressure is pressure and currents are currents

5 gal water in bucket A.  bucket B is empty. at the bottom the buckets are connected with a hose with an on/off valve.
open the valve.  wait.  in the end, howany gal do we end up with in each bucket?

add resistance.  open the valve just a little bit. same result, just longer time to complete.  kinda with the super caps. get 2 bu kets with a million gal capacity and repeat the experiment. how many gal would we have in both containers after the long time of waiting for the result?  same results, shorter time.

we lost 50% because we didnt do anything with the energy 'transfer'.  but here, in your experiments, you are using the transfer to. harge cap C. and still only have 50% left in caps A and B. but the addition of cap C brings us above 50%.

like if you do the cap to cap with an inductor of proper value and a diode, you can transfer all of cap A to cap B, minus only the diode drop voltage what ever it is.  and the masters of the universe will say, oh, the use of the inductor negates the resistance losses.

colombs law should be understood.  if you could count the electrons taken from one plate to another, then with the capacitance  value we can calculate the exact voltage in that cap. 

does resistance remove electrons from the current in the circuit??? did the resistance of the water valve create any heat losses and change the amount of water we ended up with in the buckets?? 

if we dont use the energy transfer, then we just reduced the pressure, in each case, and wasted that energy. used energy to fill up those caps and buckets, but chose to just kill off half of the energy to fill the first one in the first place. and no water was spilled on the ground. no electrons left the system, nor gobbled up by the resistors. we could put a water motor in place of the valve and drive a flywheel and take all of bucket A to bucket B.   this could be a useful thing.  say we had 2 60 gal drums. each drum is 1000ft away from each other.run a long hose( already primed with water) with a valve and turn it on.  how many gal would we end up with in each drum?  30 each?   ok. so we got 30gal of water transfered from here to there. but if we put the water motor with flywheel, we could possibly transfer all of the 60 gal 1000ft.  same as the inductor builds the mag field(flywheel), expanding, then the collapsing field pulls all the rest of capA charge and pushes it into capB. no resistance loses.   infact if you can time the on off of the switch, transistor, perfectly, you can get rid of the diode and have no losses at all....


i dont know why they say it is resistance that causes the 50% loss. if Colomb had it figured out back then, then why the deception, or they just really dont know.  what might we be missing if we dont know the truth here? is it a coverup, or a widespread mistake?

mags
 


   

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Mags,

the wiki page is not for nothing called the "capacitor paradox"  ;)

There are countless discussions to be found about this subject, so i do not want to go there and just follow the general consensus.

Itsu
   
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When the circuit has resistance, half of the energy is lost to that resistance.  Take a large value cap and discharge it through a resistive load into another cap.  Feel the heat generated from the resistor.  Use a scope probe across the resistor and set up math functions to integrate the power wave and see the energy dissipated in the resistor.  It will be half of the starting energy of the charged capacitor.  The only time there is a charge transfer paradox is when there is no resistance in the circuit and all of the components are ideal.

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thats just it. even ideal, the loss would be the same.  still 5v in each cap still 5gal in each bucket. still 5lb of air in each tank.

if it were ideal, would we end up with 7.07v in each cap from 10v? could we get 7.07 gal in each bucket from 10gal??

just for example, say we have 2 known value caps. and with that known value, we can say that if there were 1000 electrons pulled from 1 plate to the other and that would calculate out to 10v with the value of the cap.

                                           
now we do cap to cap. now we have only 500 electron count in each cap.  we now have 5v in each cap. we did not lose electrons. resistance did not gobble them up. resistance, heat, no heat, only slows down the flow of electrons, not take them away from the system.

if we start 10v, we would need to end up with 7.07v in each cap to calc to 100% total, no loss.  well where did you get those extra electrons from to acomplish that??  the ideal world gives them out for free i suppose.  maybe in that world we could start with 10gal of water and end up with 14.14 gal total in 2 buckets.

one issue is, people seem to trust wiki above all knowledge, even though some stuff has been known to be wrong there. need to get back to basics and read from the actual texts of these guys that put it all together for us 100yrs back and more. Tesla. Faraday. Colomb. etc.

again, if i used an inductor in series with the 2 caps, if i can get all of the 10v from A to B, in the real world mind you, then there must be some overunity happening there according to your knowledge, beeing all those resistances in the wires, inductors and caps oh my.
and if your correct, imagine the cap to cap through the inductor in an ideal situation. must be overwhelming the amount of overunity happening now.  ehhh. wrong. still 10v this cap to that cap.

mags





 
   
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As Itsu has said, radiation resistance is another place for the energy to go.  When you have zero resistance, your RC time constant is zero and therefore the time it takes to transfer that energy into the other cap is instantaneous.  Powerful capacitive discharges create a rather large wave in the ether.

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Hi Itsu,

In your Reply #71 https://www.overunityresearch.com/index.php?topic=4757.msg114752#msg114752  you used the 8V / 220V transformer already and gave the transformer specs.

The 220V coil on it has 509 Ohm DC resistance versus the CMC's 1 Ohm, this surely causes a relatively big loss in charging C3. Also the wall plug-in transformer core can have core losses at 26 kHz. 

I mention this because I think we would need to consider any FG output energy leakage into the circuit via the switching transistor IRF540.
This leakage may add energy to the system and during the 48 hour run time it resulted in the data you included in Reply #129 yesterday.

Data sheet includes typical capacitance graphs versus drain to source voltage, say at VDS=2V the gate to drain capacitance (Crss=Cgd) is around 600 pF, and the other
capacitances are also included in the snapshot on Figure 5 attached from data sheet https://www.vishay.com/docs/91021/irf540.pdf

Although you did not run the test for 48 hours with the 8V / 220V transformer like with the CMC one, it is possible that the energy leakage from the FG in the same circuit
was fully consumed by the losses in the 8v / 220V transformer so the "extra" energy (or the strange behaviour) you experienced with the CMC transformer simply could not manifest (especially not in C3)
(here I assume the
"extra" energy enters the system from the FG via the IRF540's capacitances).

Comments are welcome from everybody of course.

Gyula
« Last Edit: 2025-02-27, 19:56:15 by gyula »
   

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As Itsu has said, radiation resistance is another place for the energy to go.  When you have zero resistance, your RC time constant is zero and therefore the time it takes to transfer that energy into the other cap is instantaneous.  Powerful capacitive discharges create a rather large wave in the ether.

Dave

im not saying heat and radiation are not being produced.  but i am saying that the 50% loss is not due to that.  we are only talking cap to cap here.  its not a circuit typically used for anything in the industry.  what we are not looking at is the fact that the electron count is always going to be the same in order to calc the voltage very precisely in a known cap value. if we cant get a grip on the analogy of voltage pressure vs pressure of the weight of water and even air pressure as being the same, then explain thise differences to me. all 3 of those pressures can do similar work. more work the higher the pressure and less work the lower the pressure. and those works can all be measured in the same watts. those stored energies can be calculated in the same joules.

 so what if we are all possibly missing out on something in all we do here just because of this one paradox that may not be mainstream correct?  whyvwould my claim be simply dismissed with a wiki post, but we all spend so much time on this questionable claim in a vid?

mags
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I get what you're saying but in conventional circuits we are indeed losing energy to resistance or RF.  Where the energy comes from in these OU circuits is the million dollar question.  You said that 50% of the energy is not lost due to resistance but what I see on my bench is that it is indeed lost to just that.

Something energetic is coming from a low impedance (zero resistance and/or stout copper busses) when a capacitor is discharged through them.  See Gerry Vassilatos Secrets of Cold War Technology where Tesla talks of the stinging rays from large cap discharges into low impedances.  Energy is leaving the circuit boundaries.

Here is a quick test I did earlier with two 5600uF caps and a 30 ohm resistor.  I charged the C1 up to 10V and C2 at 0V.  I connected the resistor between the positive terminals of the caps and put a current probe on the wire and a voltage probe across the resistor.  F3 (Purple) is a math function of CH2*CH4.  I took the area of F3 as indicated under the measurements tab and it aligns almost perfectly with 1/2 of the energy from the initial charge on C1.


CH4 - 1mV = 1mA

Initial Energy in C1 - (10^2)*(5600uF)/2 = 280mJ

Area under F3 = 137mJ

Dave
   

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i agree. in conventional circuits, everything that is in line between the source and the load, can be a loss, but only because of voltage divisions preventing the load from seeing max voltage from the source. 

but in this case, lets say we charge cap A to 10v, then just short it out till 0v.  did we lose 'all' the energy because of resistance? or did we lose what we once had by just letting it go?  we simply discharged the cap. if there was radiation because we discharged the cap, that radiation wasnt the reason the exess electrons in the neg plate were transferred to the electron depleted positive plate for an end result of 0v, zero energ left. and also, IF there were no radiation nor any resistance, would that 10v cap still just discharge to 0v?  or would you expect a better result than a 100% loss?
if so, explain. ;)

mags
   

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We haven't even considered what effect earthing one of the caps would have, nor have we considered what effect pulsing the earth ground would have.
Don Smith and Kapanadze both use grounding. It's an interesting topic that I will be looking at eventually.
It is also interesting to see what effect the MOSFET has.
Kapanadze told me that the negative terminal of a car battery acts as an earth. (thus avoiding any mains leakage).
I'm nearly ready for testing, but the big caps I  use require a lot of messing about to keep them in place.


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Electrostatic induction: Put a 1KV charge on 1 plate of a capacitor. What does the environment do to the 2nd  plate?
   
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