Possible Overunity DC to DC Transformer

[Smudge]

Since my age will prohibit me from getting any financial gains from this idea, I offer it to all here for consideration and possibly experimentation.

Smudge

[Verpies]

Smudge,

I cannot open this attachment.  It appears truncated to 570kB.
Please check on your local hard drive what its length is supposed to be.

[Smudge]

My computer says size 570 KB (584,490 bytes) and size on disk 572 KB (585,728 bytes).

[Smudge]

I have downloaded the file from OUR and it has the same properties and opens OK.

[Smudge]

I note that we had over 1800 people on line today!!!!

[Verpies]

I note that we had over 1800 people on line today!!!!

I wouldn't exactly call them "people".

[Magluvin]

i dont get anything with the download. blank download page only.  tried earlier also.
can you try to uplaid again in a new post?

mags

[Thaelin]

I got a 4 page write up on core memory and femm display of the mag density. Is 4 pages all there is?
thay

[Smudge]

Yes it is only 4 pages.  Here is the file again but this time I saved it on a memory stick first.  Earlier it was saved in the cloud.

[PhysicsProf]

Thank you, Smudge - for the intriguing 4-page PDF.  I'm reading it ...

[Verpies]

I don't get anything with the download. blank download page only.  tried earlier also.
Can you try to uploid again in a new post?

Mags,

Try downloading it again.  I think it was problem with compression by the new forum software and I have rectified it.
Please report your success / failure.

[Magluvin]

ok.  yes. it works now
thanks

mags

[F6FLT]

@Smudge
Since the electrons are free, they rearrange their distribution within the iron conductor to equalize the potential. The Coulomb force resulting from the redistribution of free electrons counteracts the force associated with the magnetic gradient.
One might draw a parallel with the Hall effect, where a magnetic field transverse to a conductive strip forces the current-carrying electrons to move across the conductor, allowing a transverse voltage to be measured.
However, in the Hall effect, the Lorentz force acts on the electrons from one edge of the conductive strip to the other, while no force acts on the transverse circuit outside the strip; consequently, the net force on the electrons in the transverse circuit is non-zero, and a current flows.
In the first case, however, if electrons were to leave the high-magnetic-intensity zone to follow a transverse path, they would experience a force equal and opposite to the one that brought them there. In other words, connecting the top and bottom of the iron circuit shown in Fig. 4 via an external circuit would not generate any current, because the existence of the same "magnetic potential difference" between these two points means they are at the same electric potential.

[Verpies]

Since the electrons are free, they rearrange their distribution within the iron conductor to equalize the potential.

So why don't they all immediately repel themselves to the surface of any conductor ?

[Smudge]

@Smudge
Since the electrons are free, they rearrange their distribution within the iron conductor to equalize the potential. The Coulomb force resulting from the redistribution of free electrons counteracts the force associated with the magnetic gradient.

Yes that is true if there is no electrical connection, no load current to flow.  Same thing happens in a non magnetic conductor that has an induced electric field.  That rearranged distribution makes one side have an excess of electrons while the other has a deficiency.  There will be a voltage that can be measured.

One might draw a parallel with the Hall effect, where a magnetic field transverse to a conductive strip forces the current-carrying electrons to move across the conductor, allowing a transverse voltage to be measured.
However, in the Hall effect, the Lorentz force acts on the electrons from one edge of the conductive strip to the other, while no force acts on the transverse circuit outside the strip; consequently, the net force on the electrons in the transverse circuit is non-zero, and a current flows.

Yes

In the first case, however, if electrons were to leave the high-magnetic-intensity zone to follow a transverse path, they would experience a force equal and opposite to the one that brought them there.

Why are we talking about transverse paths?  The magnetic force is along the field lines, it is longitudinal.  And that path where the electrons have left the iron is not within a magnetic field so there is no equal and opposite force.

In other words, connecting the top and bottom of the iron circuit shown in Fig. 4 via an external circuit would not generate any current, because the existence of the same "magnetic potential difference" between these two points means they are at the same electric potential.

Is that a new law that you have discovered?  The redistribution of the electrons will alter the "magnetic potential difference" so it can't remain the same.  I don't follow your reasoning here.

Smudge

[F6FLT]

So why don't they all immediately repel themselves to the surface of any conductor ?

...
Is that a new law that you have discovered?  The redistribution of the electrons will alter the "magnetic potential difference" so it can't remain the same.  I don't follow your reasoning here.

Not a new law, just an application of existing ones.
When the medium is neutral, the electrons have no reason to move. When the magnetic field gradient tends to concentrate them on one side and reduce their number on the other, the internal Coulomb forces between them counteracts this, and a new equilibrium is reached.
In order to observe a voltage between the thick and thin regions of an asymmetrical ring, the circuit must be closed externally using a voltmeter. The current flowing through it is what I call the transverse current, which is supposed to be summed and measured across the output+ / output- terminals in Fig. 5.
However, as in the case of induction in a loop, there is no guarantee that the EMF along the output circuit will be equal to the potential difference calculated across the diameter. We must consider the forces acting on all the electrons in the output circuit, particularly near the junction points where the magnetic field gradient will tend to have the same effect as in the ring. This case would be equivalent to connecting the diametrical potential difference of the ring to the same potential difference in the external circuit. If the positive and negative terminals of a battery are connected, respectively, to the positive and negative terminals of another battery of the same voltage, no current flows.

[Verpies]

When the medium is neutral, the electrons have no reason to move.

So what happened to the Coulomb forces between the electrons in the "medium" ?

...the internal Coulomb forces between them counteracts this, and a new equilibrium is reached.

[Smudge]

Not a new law, just an application of existing ones.
When the medium is neutral, the electrons have no reason to move. When the magnetic field gradient tends to concentrate them on one side and reduce their number on the other, the internal Coulomb forces between them counteracts this, and a new equilibrium is reached.
In order to observe a voltage between the thick and thin regions of an asymmetrical ring, the circuit must be closed externally using a voltmeter. The current flowing through it is what I call the transverse current, which is supposed to be summed and measured across the output+ / output- terminals in Fig. 5.

OK now I understand what you are saying.

However, as in the case of induction in a loop, there is no guarantee that the EMF along the output circuit will be equal to the potential difference calculated across the diameter. We must consider the forces acting on all the electrons in the output circuit, particularly near the junction points where the magnetic field gradient will tend to have the same effect as in the ring.

I agree with this considerations, and I must point out that the two potentials associated with the two magnetic field gradients in the copper conductor at the connection points are of opposite polarity with respect to the current flow, so they cancel out. 
Edit. My apologies but I took F6's reply s applying to something I am currently working on but haven't yet posted.  Yes the electrons move through magnetic field gradients at the connection points and the two gradients would negate the internal induced potential if they acted in the same manner as that within the Fe material.  But the external connections are not Fe, and the electron behaviour there is different. Within Fe the magnetization at any point is established in the local atom's magnetic dipole polarisation and the conduction electrons take on that polarisation.  Thus the tendency for the electrons to drift according to the field gradient.  That is not the case in non-magnetic conductors like copper. There the atoms and electrons remain unpolarised so zero drift even in a magnetic field gradient.

This case would be equivalent to connecting the diametrical potential difference of the ring to the same potential difference in the external circuit. If the positive and negative terminals of a battery are connected, respectively, to the positive and negative terminals of another battery of the same voltage, no current flows.

Clearly I disagree for the reason stated.  I was expecting this scheme to be challenged as I am not an expert in spintronics, but from what I have learned about electron spin interaction it involves a spin trnsfet torque in which angular momentum is conserved.  Conservation of angular momentum leads to conservation of energy, and AI has already put that forward as a reason my (edit other unpublished scheme) won't work.  But the driver for spin transfer torque is not the electron's angular momentum, it is the electron's magnetic dipole moment.  I don't see a conservation law for dipole moment so maybe there is something here to build on.

Smudge

[F6FLT]

...Yes the electrons move through magnetic field gradients at the connection points and the two gradients would negate the internal induced potential if they acted in the same manner as that within the Fe material.  But the external connections are not Fe, and the electron behaviour there is different. Within Fe the magnetization at any point is established in the local atom's magnetic dipole polarisation and the conduction electrons take on that polarisation.  Thus the tendency for the electrons to drift according to the field gradient.  That is not the case in non-magnetic conductors like copper.
...

I disagree on this point. The effect referred to in the principle is the magnetic gradient. Regardless of the medium through which the electrons flow in the circuit, they will have to pass through the same magnetic gradients, both positive and negative. The net effect is zero.

[F6FLT]

So what happened to the Coulomb forces between the electrons in the "medium" ?

The question isn't clear; please be more specific. By "Coulomb forces," I meant the Coulomb forces between charges, even though we know that a Coulomb force can also arise from an external field.
When no current flows, all the forces acting on the electrons are in equilibrium; if a force is not due solely to the charges themselves—for instance, due to an electric field resulting from a magnetic gradient—then the Coulomb forces balance out following the rearrangement of electrons within the volume.

[Smudge]

I disagree on this point. The effect referred to in the principle is the magnetic gradient.

The effect is not just the magnetic gradient, it is also the electron spin polarisation.  With zero polarisation (as in non-magnetic conductors) there is zero effect from the magnetic gradient.  The situation at the connection points is far more complex than my or your simple statements on this subject.  The magnetic simulations show a reduction from the internal field to almost zero external field over zero distance, but that is too simplistic. At the Fe to (say Cu or solder) junction the surfaces are not smooth so there will be atomic or molecular mixing through the magnetic gradient.  The spin polarisation over that distance will not be the same as that which occurs in pure Fe, so I challenge your view that the gradient-induced voltage there will produce a net zero effect.  I think an experiment could settle the dispute.

Having recently returned to doing experimental work I have obtained some 99.5% purity Fe sheet 0.5mm thick from Advent Research Materials http://www.advent-rm.com.  The sheets are 10mm square and cost about £20 each.  I have found a local company that does laser cutting who have cut me the shapes I want at a cost of £30 each.  I could easily produce the flux-compressing ring-core shape required for this experiment but I do not have equipment to measure in microvolts.  Many years ago Graham Gunderson sent me an amplifier that would do the trick but that is now in the posession of Itsu.  It could measure the Seebeck effect of the temperature from your fingers when connecting a croc clip to some circuitry.  I could send a shaped ring-core to Itsu if he agrees to do the experiment.

Smudge     

[Itsu]

Smudge, no problem.

I also gladly will send you the amplifier, so you can do the experiment yourself in your own time, no problem either.
Just send me a PM with the address, and you probably have it end of this week.

Itsu 

[F6FLT]

The effect is not just the magnetic gradient, it is also the electron spin polarisation.  With zero polarisation (as in non-magnetic conductors) there is zero effect from the magnetic gradient.  The situation at the connection points is far more complex than my or your simple statements on this subject.  The magnetic simulations show a reduction from the internal field to almost zero external field over zero distance, but that is too simplistic. At the Fe to (say Cu or solder) junction the surfaces are not smooth so there will be atomic or molecular mixing through the magnetic gradient.  The spin polarisation over that distance will not be the same as that which occurs in pure Fe, so I challenge your view that the gradient-induced voltage there will produce a net zero effect.  I think an experiment could settle the dispute.

Having recently returned to doing experimental work I have obtained some 99.5% purity Fe sheet 0.5mm thick from Advent Research Materials http://www.advent-rm.com.  The sheets are 10mm square and cost about £20 each.  I have found a local company that does laser cutting who have cut me the shapes I want at a cost of £30 each.  I could easily produce the flux-compressing ring-core shape required for this experiment but I do not have equipment to measure in microvolts.  Many years ago Graham Gunderson sent me an amplifier that would do the trick but that is now in the posession of Itsu.  It could measure the Seebeck effect of the temperature from your fingers when connecting a croc clip to some circuitry.  I could send a shaped ring-core to Itsu if he agrees to do the experiment.

Smudge     

To measure microvolts, there are some 'vintage' instruments. I have an HP3468A, which is very reliable. But I see it still costs around £100 even today (https://www.ebay.fr/itm/267638249097).

I recognise that the issue of interconnection is complex, particularly with regard to spin polarisation. The devil is in the details, which are difficult for us to grasp from a theoretical perspective. Experimentation is necessary. It's brave to undertake this, as it's not easy and requires an investment, but the idea is worth testing. I'll be following this with interest.

[panyuming]

Dear Mr. Smudge! Thank you for sharing this idea!
I did a simple experiment with your basic design the other day.
I didn't see any obvious results, so I didn't report it to you.

Today, it seems I got some good results, so I'm reporting to you.
I understand your idea,
which can actually be considered as power generation from a static magnetic field.
Then I modified your design by placing the Fe wire on a permanent magnet.
To observe it more clearly,
I connected 5 Fe wires in series and soldered copper wires to both ends of the irons.
All my measurements were recorded after things stabilized,
estimating that the thermoelectric effect due to temperature difference was negligible.

Multiple experiments gave roughly these results:
with no magnet, the multimeter read 0.000 mV;
placing the Fe wire on an N35 magnet, the reading was 0.025 mV.

I can only say that there was a relatively clear voltage appearing.
My measurement setup is really too ordinary, just a very common 4-3/4 digit multimeter.
So my measured results can't serve as proof.
I hope that you or other experts can conduct further experiments.
Sending you my sincere respect! Thank you, Mr. Smudge!

[Smudge]

Thank you for those results, they are significant.  I offered this experiment here https://www.overunityresearch.com/index.php?msg=118188 and the following posts show Itsu doing this with a single iron wire.  He got about 4 microvolts before losing the results and not being able to repeat them.  You got 5 microvolts per iron wire so similar.  Is your experiment repeatable? Please note that in this experiment with a magnet the B field passing from the iron into the solder connection is continuous, meaning there is a B field within part of the non-magnetic wire, whereas in the ring-core that is not the case.  That could offer a better result for the ring-core flux concentrator.

Smudge