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

good luck at all

Laurent

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%   

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

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  . 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

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

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! 
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