Does Energy Gain NEED To Go Back to the Positive?

[Jim Mac]

I have my opinion on this, but I wanted to present it to hear other thoughts on it.

Say be manage to get some energy gain within a circuit.  The common thought is that we need to send this extra energy back to the positive source lead, so either the gain can go back to the battery,  or recirculate in the circuit.  But does it really need to?

Isn't it true that when we put positive voltage into the negative terminal of another power source, the voltage we put into the negative comes right out the other source's positive?

Think of two 12V batteries in series..  We have 2 power sources..  The positive of 1 battery feeds directly into the negative of another battery.  And BOOM-  24V comes out of the 2nd battery's positive.  So sending positive voltage into the negative terminal passes the voltage right through the other battery and out the positive..

So lets think of Flyback for a moment..  A HUGE spike of Voltage comes out the negative side of the inductor and enters the source's negative terminal.  That voltage passes right through the source and then comes out the positive. And our electronics can not take that voltage amplitude, so they blow up.  

So back to the question..  Is sending extra voltage gain back to the negative even more advantageous?  The negative terminal should have absolutely no resistance and can probably suckup the entire amount and pass it right out.  But sending gain back to the positive, we face much resistance and impedance.

[Classic]

Yes, it needs to go back, but only what is not consumed by the load. Just think of Tesla switch and how many times more we can use the same energy from same batteries without charging them.

Any load that is a short circuit will consume all electricity provided to complete depletion of source until there will be no more fight between + and -

[floor]

Yes, it needs to go back, but only what is not consumed by the load. Just think of Tesla switch and how many times more we can use the same energy from same batteries without charging them.

Any load that is a short circuit will consume all electricity provided to complete depletion of source until there will be no more fight between + and -

I would think that the priority should be first to achieve that gain ?
Secondly, perhaps then to effectively / efficiently store it ?
Only then to work on looping it ?

[Jim Mac]

Storing it is one school of thought.

The other school of thought is using it right away. If you have a gain, you then have 2 power sources (or batteries) in the circuit. And I believe they can be placed in series (sending the gain to the negative), so the voltages sum. Now to run XX amount of current through the circuit will take less voltage from the supply.

But to utilize the 2nd method, the circuit must meet certain criteria. It must be calling for energy at the exact time it is producing the gain. Think in the lines of 2 pistons of an engine. as one is going down, the other is raising.  Or like 2 of Floor's twist drives, while 1 is in motion, the other is resetting.

[Classic]

Maybe this video could be of help https://youtu.be/ng-Xu4foAic?si=2UJiflo9H9urpV9H

Jim, indeed impedance changes, but is important where this change take place and when. So, if the battery receiving charge have a greater impedance, the load between 2 batteries become source of negative resistance and the battery getting charged is the load. Always the load must have a higher resistance.

Floor, gain mechanism is quite simple, just collapse the magnetic field. This is the disturbance in local environment where we trick the nature to think there is a huge leak from the other side of the gradient. If we can trigger this effect with energy 1 we get energy 2 from environment and we only need to spend less energy at 1 than 2.
At least this is how i see it.

[floor]

At this point in time, I'm going to reserve most of my opinions, as they
are too tenuous. I need to study up on my dynamics (magnetic, electric,
inductive, capacitive and others).

Some observations / comparisons though.

Permanent magnets although not conventionally considered to be
energy sources, are in fact energy sources during several,
(importantly) various kinds of magnet interactions.

4 variations or classes.

1. Pseudo Solid motor
2. Shear to direct pull / Kedron_EDEN_Project.PDF
3. Rack action 2 complete 3 pn .PDF / Newtons magnets pn 3.PDF /
Lumens innovation 2 - b.JPG / Amazing I to O ratio b. PDF
4. Twist Drive When also including the magnet through a coil additional action.

That energy source is more so, in essence, electron spin.
It is also probable that energy from electron orbit can be accessed.
But...
We are not dealing with permanent magnets here, although, still working
toward methods to access energy from electron spin ?

[Jim Mac]

I have done extensive and exhaustive tests and documenting on "splitting the positives" and came to the conclusion that there is no gain to be had there. A load placed between 2 positives only runs on the potential difference between the positives. The only possible advantage is it "could" be used to help re-capture some of the energy that is wasted during the charging process. But even that is a mute point, because there is a better way to do that anyway.

Regarding flyback, I am iffy on the subject.  But what I do know is that routing a flyback spike to the positive of a battery will waste almost all the flyback's possible advantage.

Any voltage you put into a batteries positive that is higher than the standing potential of the battery itself, comes right out the negative of the battery.


A simple circuit could conserve it though..  You can watch my simple demo here

[floor]

A scenario...  A starting point
Is there excess energy ?

An induction coil.
A charged capacitor.

We know the value of the joules of energy stored in the cap.
We charge the field in the coil using a very short pulse from the capacitor.

We then measure the back spike from the collapsing of the coils field. Its discharge.
(average voltage and duration of spike in time)  from 0 v to peak v and back to 0 v.

1 volt will drive 1 Coulomb of electrons through a resistance of
1 ohm in one second of time. This is an electric current flow of
1 ampere.

Do we know the ohmic resistance of the coil ?
Its impedance
Its total resistance ?

Amps x volts = watts. 1 watt of power is a transference of 1 joule
of energy for a duration of 1 second of time.

We calculate the energy and power that was present in the back spike.

We measure the energy content of the cap, once it is has done its
partial discharge through the coil.

The difference between the caps initial charge state and its partially
discharged state was stored in the coils electromagnetic field ?

Is that difference equal to, less than or greater than the energy present in the back spike ?

IF it is greater, then we have something to feed back into the positive... or the negative.

---

Or maybe we can ignore the characteristics of the coil.

Discharge it through a know resistance value, into a cap
of know capacitance, measure the voltage present on the
cap, and then calculate the joules in that cap.

Or

Simply charge a cap and   measure  the joules in it.

[Jim Mac]

A scenario...  A starting point
Is there excess energy ?

An induction coil.
A charged capacitor.

We know the value of the joules of energy stored in the cap.
We charge the field in the coil using a very short pulse from the capacitor.

We then measure the back spike from the collapsing of the coils field. Its discharge.
(average voltage and duration of spike in time)  from 0 v to peak v and back to 0 v.

1 volt will drive 1 Coulomb of electrons through a resistance of
1 ohm in one second of time. This is an electric current flow of
1 ampere.

Do we know the ohmic resistance of the coil ?
Its impedance
Its total resistance ?

Amps x volts = watts. 1 watt of power is a transference of 1 joule
of energy for a duration of 1 second of time.

We calculate the energy and power that was present in the back spike.

We measure the energy content of the cap, once it is has done its
partial discharge through the coil.

The difference between the caps initial charge state and its partially
discharged state was stored in the coils electromagnetic field ?

Is that difference equal to, less than or greater than the energy present in the back spike ?

IF it is greater, then we have something to feed back into the positive... or the negative.

---

Or maybe we can ignore the characteristics of the coil.

Discharge it through a know resistance value, into a cap
of know capacitance, measure the voltage present on the
cap, and then calculate the joules in that cap.

Or

Simply charge a cap and   measure  the joules in it.

Yes, a controlled test that can be repeated by others is what's needed.  Any wavy logic and assumptions in the test should not exist. Like if the conclusion says " since a cap loses 50% of it's power during transfer, this means the actual power transferred blah blah blah"..   No No No..  

The proof MUST clearly show that joules were created and caught, ready for re-use.  Too many claims rely on "Comparative Logic" to argue the claim, without achieving verifiable solid results that prove the claim.

Cadman posted his tests where I think it is claimed more was caught.  https://www.mooker.com/thread-29.html

This one is out of my lane of study, but I am watching if anyone can reproduce the results.

[floor]

Thanks for the link.