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Author Topic: Electrical igniter for gas engines A keystone to understanding by Magluvin  (Read 250925 times)

Magluvin

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Hey Nul

I understand your reasoning, and I dont want to argue, and i know you are not trying to argue. =]

I am basing my thoughts on the idea that we have been fooled into thinking that things are the way they described them to us.

Like the Gabriel transformer. This was a Tesla idea in 1890.
And what we had learned about transformers is actually true. Its just we were never told we could separate the flux produced by the secondary from the primary flux core, of which causes the primary to draw more from the source as the sec current rises.  ;]

So we HAVE to question everything. Everything!

And if we have stored energy in the inductor, energy put there while charging the cap to 5v, energy beyond what it took to charge the cap, then you may be right. But Im not thinking so. =]

I am working on proof on this very subject as we speak. ;]

I look at it like this. In a battery, when we take energy from it, we are just per say taking electrons that are packed into one side, and allowing them to fill holes in the more empty side. We could say this in a few ways, but this is just 1 way.

If we fire up the circuit, slowing things down so we can watch what is going on, and we stop or disconnect the battery somewhere below 5v so that when the inductor stops spinning(per say), the cap ends up with 5v. Did we expend a caps worth at 5v of energy from the battery at the time we disconnected the batt at the point below 5v that finally reaches the cap?  How can we say for sure? Are you sure and ready to move on?   

When we slow things down, being that the inductor will not accept gobs of current and charge the cap instantly, we have to watch the current as it increases.  It increases more as the field builds in the inductor, never really getting to a current level that would be seen if we were just directly charging the cap from the battery. The only reason current is increasing is the fact that the inductor is determining how much gets through over time. But what is going in, IS what is coming out in reference to the inductors leads.
I dont believe that more current had gone through the inductor to get to the cap during the time the cap was charging WHILE the battery was connected. I dont believe it. Where did it go? Not in the cap. Are they packed into the inductor and slowly delivered to the cap after the batt was disconnected?

Take this example. Battery, cap, inductor and an led in series. the led will light till the cap is full and the led will go out due to no more current will flow against the charged cap. Well guess what? That cap is now charged, and will light the led all by itself when we reverse the polarity. We used energy from the battery to charge the cap THROUGH the led. A measurable amount of energy used in the led. But we still have that same amount of energy still in the cap after wards. We can use that energy 2 times, just about. We have a loss due to the led V drop, but that is just a function of the led to not conduct below that drop. Not a loss, just a barrier. That barrier did cause waste. But that is all in circuit design.

So get rid of the led and just replace with a diode and the inductor with a primary of a transformer. Now we can charge the cap through the transformers primary and still have energy in the cap for the other phase, and after that the cap is still charged, just in the opposite polarity. Like a controlled oscillation, stopping at the peaks and switching to give the other cycle, all from 1 half cycle charge from the battery that charged the cap in the first place.
If the transformer sec is loaded, then we will see the cap wont have too much left in charge, only because the draw on the sec caused a damping in the primaries ability to freewheel, or oscillate. No real loss, just the energy running in the primary was taken up from the sec on the other side, and the primary was stalled from freewheeling.

That is where the Gabriel transformer is a key element in all of this. The primary can be put into oscillation and not be killed by the sec current produced flux, because that flux is not instilled into the primary windings. So now we can have an extended oscillation of the primary while taking from the secondary. Oscillation is preserved and each cycle can produce power out without being killed off quickly.

If that doesnt make sense, I will come up with a way to describe it better. ;]  Or better yet Ill have to prove it.  ;]


mags
I have presented this before but its is hard to get most to listen.


Magluvin

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Hey Loner  ;]

Just as you say, I put some "maybe's"  and "possibly s" just in case i am wrong, even though at times I think Im right.  ;]  I have to because I dont think I would like the taste of my foot. ;]

When you spoke of the flywheel, I thought of my separate descriptions between the inductor and the physical wheel.

The physical wheel is different. A lot of energy is expended to get it started and that energy that is further put into the spinning, need not be as much once it is already spinning. And once it is at full speed, very little is needed to keep it there. ;]

The inductor on the other hand, requires voltage(pressure) to get it going, but the current is low in the beginning. The power is low when starting the inductor flywheel. As it speeds up, more energy is required to accelerate, more current flows, more power required.

So in my frame of mind, with a cap in series with the inductor, we are charging the cap, but the inductor is in the way of getting it done NOW. It is just a form of decreasing resistance in the path of getting the cap to become equal or leveling out with the source.

If we forget that the inductor actually stores any energy that had gone through it,  and we see that it just has a decreasing resistance as the cap charges, we can then say that the cap, when it reaches the same level as the source, holds an equal amount of energy that was taken from the source.  Agreed?  ;]

Now we look at the energy stored, per say the flywheel is spinning, freewheeling weighted wheel. While the resistance is decreasing and the flywheel is accelerating, where can we say that energy was used to get it spinning?
 We could look at the action that is happening in the inductor is just as I described, a decreasing resistance value, and we could calculate that and be accurate as to the amount of energy being fed through the inductor and ignore the flywheel other than it is just an artifact of the current that had gone though the inductor.

I would say that in the series circuit presented, when current flows, that current is the same throughout the circuit at any given time.
I dont think anyone could argue this.

So I dont see the spinning up of the inductor as an energy loss between the source and the cap getting charged, just a delay of sorts.
I dont see it as a hole in the water pipe where we lost something before the water gets to the bucket.

I dont see it as a place in the circuit that took anything from the source, that didnt make it to the cap.

If it did, and this is key, then our current in the circuit would be different on the input lead to the inductor as compared to the output lead of the inductor.  But we know that current in the circuit is a constant per time of flow, especially being that the circuit has no branches and is just a closed loop.

As for the differences in voltage vs current in the inductor, that is easy.  Imagine the inductor being 1 ohm in resistance. That would be when the inductor is not in circuit.
So when we present a voltage across the inductor, the resistance (impedance) of the coil will be very high in the beginning, thus high voltage and low current. As the wheels get spinning that resistance decreases, more current flows and the resistance becomes even lower.
Once the resistance finally becomes 1ohm, high currents will flow and the voltage measured across that resistance will be low, when measured across the inductor, especially if the source cant hold its own against the very low resistance, and other resistances in the circuit will affect this also.

So, did we use any energy from the source to get the wheel spinning, as in actual energy taken from the source and not provided to the charge cap?  Did all of the energy that came from the source, from no charge in cap to full charge in cap, reach the cap? 

So in my circuit description in the above post, if we got rid of the diode that is across the bat/switch, and we disconnected the batt when the cap reached 5v(source =5v), did we consume more from the batt than it took to get the cap charged to 5v?  I say no. I say we are left with a spinning flywheel that has nowhere to put the energy instilled in the inductor. And I think its free if we capture it properly.

Lets get rid of the cap and the diode across the batt/switch so we have batt, switch inductor and diode in loop.  Now when we close the switch, the inductor starts spinning and currents are accelerating.
But now we are just discharging the battery, and getting the wheel spinning. In the circuit we still just have a decreasing resistance seen in the inductor, and electrons are just filling holes in the + side of the battery. Our loss is the battery just having a load on it that varies as the circuit continues to flow current.
We can say that we stored the energy in the inductor, but only to the point that the inductors field cannot increase any longer because the source cannot provide more pressure than 5v. Beyond that, the battery just sees a resistance load, and the action of being drained because the batteries - and + want to equal out.

If we have a water fall and we can figure how much water per min is falling, then we add a paddle wheel under the falls, can we still say that the amount of energy from the source had a loss or conversion of energy between the amount of water that begins to fall and the amount that made it past the paddle wheel to the river below? No, it is the same amount. Plus we got the wheel spinning. =]
This is not a perfect example, but it gets the message across.

=]

Mags

Magluvin

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And,  what really might blow your mind is my late thoughts on heat developed in circuits. I dont have it all worked out in my head yet, but it just may also be an artifact of current flow through a resistance.

What does the heat offer to the circuit other than a change(higher) in the resistance that it came from? It is more like a physical flywheel as in the power is high before the heat, lower resistance, and as the heat increases, the resistance gets higher and currents lowered. But I still think that we didnt loose anything creating the heat, especially when the heat is a desired effect.

But in a circuit that doesnt have the intentions of creating heat, the heat that is created in the circuit is just more of an obstacle to current flow rather than a loss, due to higher resistance and lower current flow.  ;]

One thing though, the heat stored or created would need another device to return its energy back to the system, but the inductor can do it all on its own.  =]

Mags gone crazy   : :-*

gcramer789

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HI  every one

forest

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Mags

Does it matter if we use circuit that  use a bit of battery power which has to be recharged afer a month of continous usage ? I don't think so.
Anyway it's a basic circuit.It will work IF we left voltage on cap above battery voltage and discharge the rest BACK into battery GROUNDED with a really good ground with IRON.
The faster we do it the longer it will work on the same battery.
It has to do with battery, it won't work without it. In fact if we SHORT battery or CAP very quickly when it is GROUNDED while applying higher voltage (even HV) at positive terminal it will recharge ! Just a small idea  ::)
nothing important ...

woopy

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Hi Mag and all

Nice theory i will reread to get it totally.

In the mean time i have made a small experiment after my last video and the correction i have received about it..

I decided to charge a cap with a battery at 10 volts and than to discharge it in parallel in a same value (63 volt and 47 micro F) cap and the efficiency is about 50 %.  See test 1 in the pix 1.

Than i redo the same but this time the cap is discharge through the diode and inductor (always MOT primary) and this time the efficiency of the transfer is about 74 %.  See test 2 in the pix 1

So without the freewheeling the direct transfer is 50 % efficient and with the freewheeling (that is to say with one diode and one 220 mH inductor and all the added resistance  of them)  the efficiency climb to 74 % that is to say  1.48 time better.

What do you think of this result :o

good luck at all :)

Laurent

Magluvin

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hey Woopy and Loner

Woops, well I see that in the first circuit, when you discharge from the first cap, 10v, into the second cap, you end up with 5v in each. This shows that the 2 caps still have the total energy taken from the batt. No loss, just a transfer of energy.  ;)

But in your second circuit, if Im seeing it correctly, 8.4 in 1 cap and 7.9 in the other?  That IS more than what was taken from the battery. Absolutely, no doubt.  Do me a favor and check the polarity one each cap in both circuits to see if there are differences, just polarity once the circuit performs its function.

I have had a cap that I was discharging to another cap, through and inductor and diode, that the cap that was discharging, went full reversal of polarity than it started with, all due to the inductors inertia. But the charge is still good, just reversed polarity.


The first circuit had 100% transfer from the batt. But your second circuit is well beyond 100%   ;D ;D ;D ;D ;D

Im at lunch, will be back later.  Great show woops.

Hey loner

I will have to get back at ya this evening on that.  In basic terms, I think that in our case, not in every circuit with inductors, the circuit objective is to charge the cap when all is said and done. Once the cap is full, lets say equal to the batt voltage or more, once it happens and the circuit stops conducting, the function of the circuit is complete for that cycle. Lets say that small losses in the circuit are negligible. but where are the losses really?

 Just because the cap did not get charged instantaneously, Does that mean we lost anything along the way? The resistance(impedance) is only slowing the process down, the process of charging the cap, as compared to direct charging. Where do we lose any energy from the source in that time, just because the impedance only slowed down the process of getting the cap charged? Just because the current was slowed down during that time? Thats not a loss. ;]

Im not a poet, nor the best writer or speaker. But once I do get it all straight and in my mind, my words will be like a good song.  ;]

Be back later.  thanks for the hard work woops, and you are doing awesome. ;]

And thanks loner for really trying to get what Im putting forth. Its hard to get a grip, and I fall from such great heights at times myself, but then I remember the details that got me here, and it all makes sense again.  ;]

Mags

woopy

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Hi Mag

Please read better my post . In the second circuit the result is 8.4 in a cap and 1.9 in the other (not 7.9). Which should be slightly over the 10 original voltage. but it is probably due to the scope shot imprecision.

But there is something anyway and i have tried a crude explanation see the pix. I think that if you try to replicate my drawings in the real life  (for instance the blue color should be water ) it will be impossible to recreate a potential difference with the flywheel (because what you explain in your previous post ).

But in electrical reality it is possible and this is fantastic  ;D. It seems that the transfer with freewheeling effect "IN ITSELF" is 130 % efficient.

Will reread your post to better understand.

Good luck at all

Lasurent

woopy

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Hi Loner

thank's for answering

As i also like the inertial real force, and  i would be very interested in the Amish method of pumping by inertia.
Have you some thread or link to visit on this subject ?.
Perhaps you have already seen but i am also involved in a one pulse per revolution bouncer motor ä la Milkovic or Mattew Johns and looking for a way to better use of the huge vibrating power of those kind of machine .Just for info, i include here a video but don't want to distract too much.
http://www.youtube.com/watch?v=Awci9aWLPhI

But i think that the real physical flywheel has not the same working model as the electrical freewheeling. I think that in physical model if there is any OU to find, it will be in the centrifugal force but until now i did not manage to get anything extraordinary from my experiment. I hope it will perhaps come (Amish pump).

But Today, i am absolutely sure that the ELECTRICAL freewheeling transfer (or pumping)  is far more efficient than the BARE transfer.

And the work here is to understand HOW and  WHY as usual

good luck at all



laurent


Magluvin

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hey all

While Im at work, Im thinking all the time, and not about work. The work just automatic and my brain is on the bench. =]

Sorry for the mistake on the 7.9 value. Imagine the smile on my face at the time. But when I went back to work, I remembered that this config can only produce so much, as is. So when you replied that I read it incorrectly, I felt assured that we are still in the ball park. ;]

Lets say that if through testing that we are certain that we are at least getting more out than in, we can then at least have something that will help our beliefs in what we are searching for. We now know for certain that it is possible. We can hold it in our hands, finally.

This is a very basic circuit. Not any complication to it. This is a circuit that Rosemary should put on the table. It would not be as hard to defend the theory against the wolves.  Simple and to the point.

In Woopys second circuit above, was thinking that of the voltages he had shown, the diodes drop could be figured in and his numbers show accordingly.  So I suggest using a higher voltage, maybe 20v as the source. I believe our output numbers will increase for the good and help lower the percentage of the diodes drop on the whole.
If your source were 2v, that diode drop will affect the output in a huge way. Our numbers will be far short of 100%, as the diodes drop could be .7v to upwards of 1v. So 20v will show an improved output ratio. ;]

As for losses, it is weird I admit that I see things in what could be in a different light.

But my theory that the inertial forces that is built in the inductor come at only the cost of the changing impedance effect on the circuit, and not a loss of power, can very well explain why we are able to get more in the cap, after we disconnect the battery at the point that the cap has reached battery voltage level of 5v.

Run this through your noggin. ;] 

If we hit the switch and wait for the cap to reach 5v, and then we very quickly removed the whole inductor from the circuit. Would you say that if we measured the cap and it has 5v measured and the source was 5v, did we lose anything in the transfer? Do you believe that some of the current that was flowing in the inductor(conductor), some how kept something that never got to the cap?

Well I do believe that electrons and or charge is compressible. Could it be that some electrons, per say, were compressed in the inductor, and when we took the inductor away, we lost those extra electrons that it held and the battery is missing those plus the ones put into the cap? Is that the loss you are speaking of.  ;]

Well if we consider the source as having potential for motive force, and the cap has potential of accepting that potential from the source, I could say that the source is pushing and the cap is pulling. Just those functions should be like a push and pull situation on the inductor, there by where was there really any compression?

The source sees the cap as a destination, but the impedance of the inductor is like breathing through a coffee straw to begin with. No mad rush of current, just an increase as the current flows through the inductor. Why would we call impeding or resisting a current from flowing a loss? i would say its conserving energy by not draining the battery so fast. And thats all the source sees in the inductor is a changing resistance.

If we have a 12v battery and a 12v motor that does some job.
We connect the motor to the battery and the motor does work, just as it should.
Now we add a 0.5ohm resistor in series with the motor. Now we connect the circuit and we notice that we are not getting as much work done. Oh, we have a loss in the resistor, and thats why the motor isnt putting out as much work now.  loss?
No.  There is just not as much current flowing now due to total resistance of the motor and resistor. AND less is being pulled from the battery. We are not pulling the same amount of power as before with just the motor as a load.

Loss?  Ok yes there is loss. Loss to the output of the motor because for the amount of current flowing through the circuit, the resistor is helping to determining the current in the total circuit. It is lowering the total current. We are drawing less from the battery because the total resistance is higher. 

It is not as if we are pulling the same amount of current as with just the motor and battery alone, with the resistor added.
And we have just losses in the resistor? so thats why the motor doesnt put out as much work because of losses?
"The resistor is just a limitation"  not a loss.

If we add the resistor, less will be pulled from the battery. "LESS". ;]
Thats why the motors output is not as high, because we have added a limiter, a valve that is half closed. We didnt lose anything other than what the eff of the motor allows as physical output. Thats just the fault of the motor design. We only lost a desired effect from the motor, BECAUSE WE ARE NOT PULLING ENOUGH CURRENT FROM THE BATTERY TO RUN THE MOTOR PROPERLY.  ;)   This is not a loss, but a limitation in the application.

We only lost the desired amount current to run the motor properly for its job title. It doesnt mean that electrons and or charge is being spilled on the ground and wasted.

So if we look at the inductor in the same view, it may come clearer to you that the varying resistance of the inductor in the circuit is not a losing deal. It is just in control(valve) of the currents in the circuit and thats it. Where is the loss? Especially if we have a gain.  ;)

I think it would be a hard argument for someone to come in here and tell me where the losses are, and how we lost them.  I dare ya!  ;D
I just may be able to fend for myself in debate with only what I know now, and find some good friends on my side along the way.


I can see losing rf through holes in our capture device and not being able to recapture and bring it all back to where it came from. But in a basic circuit, Im seeing things differently, and seem to have explanation for so called anomalies we are encountering, and the explanations seem to fit. So far. ;]

So until we have a better explanation or reason to not believe, then I say we go with it for now and see what comes of it.

I am going to name this circuit, The Believe Circuit.  Because thats what it does.  =]

Back in a bit.  Showwa and eats.   

Mags
« Last Edit: March 30, 2011, 05:59:37 AM by Magluvin »

forest

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Maybe it's not different flow around circuits, Loner, just we use parts and arrange them to use original flow to kill the source  ;D

What is the reason of dropping voltage across resistor ? heat generated ?
What I believe is crazy and I cannot prove it.
I believe that we are using the original energy flow to set an opposite dipoles in circuits which affect power source (either by chemical reaction or by lenz law).
Electrons are just small magnets which flips by this energy flow and generate whatever we wish and has nothing to do with killing the source dipole.They are just here and always ready to pass energy in chunks while flipping either magnetic side or electric.Unfortunately the only tip which points to that is Mr. Dole work and Don Smith schematic.While in area of "not killing the dipole" Don schematic shows a resistor divider without loosing power inside resistors.That's the only explanation I found for Don Smith method of lowering voltage down in output side of his device. I think Dole could confirm that using HV resistors divider.

And finally when we are using that original power without killing it we would be happily able to pass it in circle.Like my neverending thought about : "what is going on with electricity returned back to power station?".

Magluvin

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hey Loner

Lol  dont take my post as me arguing with you. ;]  It may have seemed that way. What I am saying is for everyone that reads it.  ;]

I will stick to just the inductors flywheel idea for now and Ill get back to losses and things later.  My statements on losses and where the energy went if not to the cap, is all based on if any of the circuits energy were actually converted to another energy(inductors production of mag field) then the gain must be just magical. But if the circuit doesnt consume(lose) energy in the production of that field(flywheel), then that can explain why we got more out, because we produced that field for free, and the circuits reaction to that field build was just a lengthened time period to charge the cap due to impedance(changing resistance and or "negative resistance").  =]   Just a theory is all Im presenting.

I read in a thread here the other night, dont remember which, but someone stated that when it comes to inductors and transformers, information is a little on the light side. I agree.  I yahoo searched inductance. I found that answers.com had a few definitions from different sources.
http://www.answers.com/topic/inductance

They are not all coinciding informations. What is the truth? What is all the confusion?  Well im determined to find out. ;]

Think about it. If tesla had invented a transformer in 1890 that was capable of more out than in, but all we ever seen is transformers that ALWAYS pull more in than out, even with no load  ;), I say there is a coverup.   And if that is so, what else do we need to question?  Im asking some of those questions, whether they are absolutely valid yet, time and testing will tell.

Im like everyone else. If we have a gain, where did it come from? And maybe the answers are not necessarily from the vacuum or aether. maybe it is just a function that common physics does not describe. And probably most or all people into physics, just go by the book and never notice anything otherwise.

Woopys latest test where he charges a cap from the source and uses that cap in the Believe Circuit as a source, leaves us with something to be considered.
We now have a finite amount of energy as the source in that cap, and we still obtained a gain. Did we lose anything?
Im not saying losses do not exist.
What I am saying is, some things or functions just might not be or have a loss as we might think from what we are told. ;] 

Im going to get into some bench work this evening and try not to post long non understandable information for now. I may be premature as to anything that I have stated. But consider it all just theories at this time.

Stick around Loner. I think it will sink in as we go.

I think your catching on forest. ;]

Mags

forest

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inductance = flywheel or spring valve (a valve having a rotor inside or a spring with a piston inside) - can oscillate naturally (inertially) or forced
capacitance = elastic bag - can oscillate when pressure ratio in and out is specific to the elasticity and size
resistance =  valve which can also resonate if flow speed match the resistance of valve (like water in fast opening valve)
voltage= pressure of ether flow
current = amount of ether chunks flowing in time or just amount
electrons = small standing waves having both magnetic and electric side at 90 degrees


correct me and comment  :)

IMHO There is no need of COP>1 , COP=1 is enough if we learn how to compress flow before load and decompress when flowing throught the load back to the source. Load effects are not connected to source dipole energy but are alone produced by electrons.Crazy,yeah I know....

And a question at the end ; what will happen if I connect a capacitor charged to higher voltage to the battery in series (+ - + -) and load across both ? Will charge equalize ?

woopy

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Hi all

just a small video of my understanding and questions on the subject

good luck at all

laurent

http://www.youtube.com/watch?v=CjOG7OJslgA

Magluvin

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Hey all

Woopy. You are doing great things.  ;)   I have to check the calculations. I think there is something we are missing in figuring this out.

I had the same thoughts as you when I just ran it thorough my head. If we have 2 10uf caps and charge one up to 20v, and we short them as you did, each cap will contain 10v, as you have wonderfully shown.
Then I thought, well what if we put the 2 caps in series now to get our 20v back. BAHH  or like you groaned in the vid, MMMMMMMM!   ;D   
I now have 20v, but in a 5uf cap!!!!   What?????

BAGHHH!!  cant be!   

Then I climbed out of hell, where I thought I was for a moment and regained level ground.

This is a very important discovery, as you have experienced it and I had thought it. I like this. We work well together and communication is good.

This will be a VERY touchy subject, VERY.   Not right now. ;]

So back on earth,  we have 2 10uf caps with 5v in them.  At first we only had 10v in 47uf.  But now we have 5v in 94uf.  The difference is, we cant run a 10v device from the 5v in the caps, but we can run a 5v device on the 94uf at 5v, for just as long as a 10v device on a 47uf cap at 10v.  Get it? All being that the devices consume the same power levels. i can full describe it if you wish, but I think its good so far. So we are just not calculating nor thinking it properly the way we are.

But that loss that you calculated and I thought, well where did the power go in our thoughts and calculations?  I think its still there, we just shouldnt go about dealing with trying to get the 10v ability back after conversion to split the voltage.  You and I both know that if we use some energy into trying to put the 2 5v caps all back into one, there would be considerable LOSS =] in the work it took to do so.

But now for the super nasty thoughts. lol   

If we have 20v in a 10uf cap and another 10uf that is empty, then we connect them to get 10v in each 10uf cap.  If we put the caps in series and we only have 20v in a 5uf cap, we DID lose, a lot. We could never get the 20v 10uf amount of energy from a 20v 5uf cap. That is nuts woop. How dose that kind of loss happen?
What if it is a trick that is a real trick, the effect of energy actually disappeared in thin air and we were left with only 2.5 level in our water jugs instead of 5?
Here is the nasty good part, I hope. If we can make that energy just disappear, gone, nada, neva commin back, cant have it, well maybe just maybe we can double it by another way. It only seems logical that if the energy could just vanish into thin air, that there could be a way to increase it, out of thin air.  Lol, that is just mind boggling.  If we had never seen this issue, maybe we would never think of a possible way to accomplish the same trick but in reverse.

I may be very wrong here an we may have suffered a HUGE loss. But freekin how?  I know you are felling this right now woops. I feel ya.

We have to investigate this. At least its not boring.  ;D

Like in your vid, you said very low resistance to connect the caps, so little loss, try this woops, do the same thing but with a resistor, and check to see if we lost anything. If the resistor is high value, it will take some time for the caps to level out. Tell me how much we lost.  hehe =]  If we lost anything, then heat is a loss. If you end up with 5v in each cap, what can we say about heat? Was it free?  I havnt tried this as Im really being spontaneous with the thoughts here and you have the test setup with all the same parts as before.

If the caps have less than 5v after then I submit that the heat was energy taken and converted into heat . But what if those caps level out to 5v woops? Do we have something here or what?   ;)

I have said that I wasnt completely sure about heat not being a loss of energy from the source, but if you would, entertain me.
I will still maintain the flywheel as free till we discover otherwise.

Imagine a battery and we have a resistor that is getting hot, emitting heat, do less electrons make it to the positive side of the batt than what came out of the negative? Or are we just discharging the battery and heat is a free byproduct?  Remember, the hotter it gets, the less we pull from the battery. Would we not think that the hotter, the more current? =]
I dunno. Maybe Im just nuts and should be sent out for an afternoon of electroshock treatments. 

I just may be eating both of my feet tomorrow for lunch. I will clean them well in the morning, and bring some A1 sauce.  ;]

But Im ok with it.  I just cant see why we lost 50% woops. Something is up.

If you could do the 2 caps and resistor test, we will at least nailed down something that is on the list. I just cant imagine a 50% loss in heat here. Where did it go. We must be calculating something wrong.  Will think on it.

Well, this is more fun than titos puzzles.  No offense Teets.  ;D We just seem to have plain objectives to work with and interesting results.

Mags