I Got a Speedup Effect..

Think you may be overlooking a key distinction. The reversal I’m describing on departure isn’t caused by resistive decay of the induced current,

That is exactly it. You understood me ...but you don't believe me. That's OK - we will get there.

...it’s caused by the geometry of induction itself. Even in a superconducting coil, the induced current is not free to “just persist” in one polarity.

In a shorted superconducting coil that is exactly what happens! The induced current persists indefinitely. There is a superconducting coil on Earth in which the current has been persisting for over 20 years !

As the rotor’s pole sweeps past the coil face, the flux linkage changes sign.

Ah, but the magnetic flux that penetrates a shorted ideal coil cannot change (by penetration I mean a loop integral of the flux). That is the beauty of it - it is the key to understanding the entire electromagnetic induction phenomenon.
The current induced in the shorted loop is such that the internal magnetic flux generated by that current cancels any change that the external flux is trying to make, thus keeping the total flux penetrating the coil constant and preventing any flux lines from cutting the coil (i.e. changing your flux linkage). This simulation shows this principle beautifully.

Oh, ...and and one more thing: The magnitude of current induced in an ideal shorted coil by an external flux source DOES NOT DEPEND on the speed with which that external flux changes. Nanosecond or a year - the induced current ens up the same magnitude.

By Faraday’s law, the induced EMF must change sign as well, which forces the current to reverse.

Faraday's law (induced EMF ℰ by a changing external flux) breaks down with an ideal shorted coil because when R=0 and i=ℰ/R then this would means that i=∞ for any ℰ>0, which is ridiculous.
In my opinion analyzing coils through the prism of induced EMF (ℰ) and Ohm's Law: i=ℰ/R, is exactly the kind of wrong path that prevents people from fully grasping electromagnetic induction.

So the attraction I see on departure is not a symptom of copper losses

Oh, yes it is ! ...and the transfer of coil's stored energy to the load, too, of course,

—it’s the inevitable consequence of the flux derivative changing sign across the coil’s two walls.

The external flux's time derivative dΦ/dt has absolutely no influence on the magnitude of the current induced in a shorted ideal coil. Like I already wrote: The flux can change in a nanosecond or in a year - it makes no difference.
You are just so used to common coils leaking their stored energy/current all the time (like Mr. Faraday), that you think that the speed of the external flux changes makes a difference in the induced current (because the external flux represents energy replenishment to the leaky coil) ...but I got news for you - to an ideal shorted coil that speed (rate of Φ change) makes not an iota of difference.

Superconductivity would only eliminate resistive losses and extend the magnitude/duration of the induced current, but it would not prevent the polarity flip.

Not only - it also prevents the flux from cutting the coil and changing flux linkage which completely defeats the Faraday's Law. Not a small feat.
Also, what "polarity flip" ? The direction of induced current in a shorted ideal coil does not change direction by itself.

In other words:
Approach phase → flux increasing → induced current opposes increase → repulsion.

I agree with the proviso that when you write "flux increasing" you mean external flux, because the total flux (internal + external) that penetrates an ideal shorted coil stays constant.
Also, you skipped one arrow - it should have been:
Approach phase → external flux increasing → induced current generates an internal flux → internal flux opposes the external flux (or increase of total penetrating flux) → repulsion and the total penetrating flux stays constant.

Zero crossing → flux change = 0 → current crosses zero at coil center.
Departure phase → flux decreasing → induced current opposes decrease → attraction.

No, in an ideal shorted coil you do not start from 0 flux. The current induced during the approach phase still circulates in the coil and it generates internal flux that opposes any changes in total flux penetration attempted by the external flux. This means repulsion. If you were starting form zero internal flux, then you'd have attraction on departure ...but you don't because an ideal shorted coil has not lost its current (and the internal flux it generates) that has been induced during the approach phase.

That sequence is dictated by Faraday’s law and coil symmetry, not by resistance. Superconductivity strengthens the effect but doesn’t alter the fundamental reversal.
...
Superconductivity would only amplify the magnitude and persistence of those induced currents, but it wouldn’t eliminate the polarity flip dictated by rotational symmetry.

Nothing of the sort. Faraday's law is not applicable to ideal coils because when R=0 then Ohm's Law i=ℰ/R dictates that i=∞ which is ridiculous.
The Quantitative Lenz Effect (QLE) nips the Faraday's Law in the bud before the bastard has a chance to act, because QLE does not allow for the change of the flux linkage through an ideal shorted coil.
His little brother (the qualitative Lenz Effect) which you all know and love, defines only the sign of the induced EMF in the parent law. The qualitative Lenz Effect is the minus sign in the Faraday's parent Law: ℰ= -dΦ/dt. Both of them say noting about the quantity of the induced current. The big brother (QLE) - does ...and it slaughters the brother and the father.

floodrod

Re: I Got a Speedup Effect.. « Reply #62 on: 2025-12-03, 18:28:36 »

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The “20‑year persistence” example refers to a static trapped current in a superconducting loop — a DC bias that remains when you induce it once and then leave the coil isolated.
That’s a very different case from a coil under a continuously alternating or rotating field.

When the external flux keeps changing polarity, the coil must adjust its current to conserve the fluxoid.

If the external flux excursion crosses the conserved value, the induced current polarity flips. In other words, under AC drive the coil does not sit there producing one‑direction DC forever — it responds dynamically, and the output polarity changes with the inducing field.

So:

Persistent DC for decades = trapped current in an isolated superconducting loop.
Polarity reversal = inevitable when the external field swings through the conserved fluxoid value.
Mixing those two cases is what causes the confusion.

As the rotor’s field sweeps, the coil generates currents that oppose flux changes.

Even though there are no resistive losses, the rotor still feels drag because it must do mechanical work to maintain fluxoid conservation.

That work is stored as magnetic energy in the coil’s persistent current.

If the rotor keeps rotating, the coil’s current adjusts continuously — and if the external flux crosses the conserved value, the current polarity flips. Each adjustment resists the rotor’s motion, manifesting as drag.

---------------------------

Jim Mac - Owner of Mooker.Com Opensource Free Energy Forum

Magluvin

Re: I Got a Speedup Effect.. « Reply #63 on: 2025-12-03, 18:58:04 »

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im with flood on this one. saying that the magnet approaching the face of a super coil begins somthing that cant be stopped is silly. if the approaching magnet sees a lenz repell effect on approach, and the coils condition cannot be changed from there, then id say the magnet could never make it past or to tdc.. besides, why are we shorting the super coil??? why would we not use that possitive and then negative output from the magnet pass?? special useless case i suppose.

like saying the cap to cap losses are because of resistance. another useless circuit. but that 50% loss is not due to resistive loss. ideal, same loss.

mags

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Re: I Got a Speedup Effect.. « Reply #64 on: 2025-12-03, 19:02:58 »

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The “20‑year persistence” example refers to a static trapped current in a superconducting loop — a DC bias that remains when you induce it once and then leave the coil isolated.
That’s a very different case from a coil under a continuously alternating or rotating field.

The behavior is the same - complete and total penetrating flux preservation.
What the current does, to keep the flux constant is a different issue

When the external flux keeps changing polarity, the coil must adjust its current to conserve the fluxoid.
If the external flux excursion crosses the conserved value, the induced current polarity flips. In other words, under AC drive the coil does not sit there producing one‑direction DC forever — it responds dynamically, and the output polarity changes with the inducing field.

I agree, but an approaching and departing pole of a motor is not an AC drive.

So:
Persistent DC for decades = trapped current in an isolated superconducting loop.

It's more of a "trapped flux".
Current is only reactionary to keep the trapped flux constant.

Polarity reversal = inevitable when the external field swings through the conserved fluxoid value.

No flux nor current polarity reversal happens when a pole approaches and departs an ideal coil.
If another pole comes along and is of a different polarity, then it can reverse the current ...but not the flux through the ideal coil.

As the rotor’s field sweeps, the coil generates currents that oppose flux changes.

Yes.
It is important to answer the question "changes measured from what datum ?". In case of an ideal shorted coil the datum is always the state of the coil the moment it was frozen (when it became superconductive). In case of resistive coils, the datum is always moving in time but stays behind us just a short time. What "time" ? The RL time, of course.

Even though there are no resistive losses, the rotor still feels drag because it must do mechanical work to maintain fluxoid conservation.

No because with an ideal coil there is no drag. There is only repulsion on approach and departure ...and the magnitudes of these repulsions are the same.

That work is stored as magnetic energy in the coil’s persistent current.

The work done by an approaching pole is stored, yes. In an ideal case - indefinitely.
Work is energy. The same stored energy repels the pole when it is departing.

If the rotor keeps rotating, the coil’s current adjusts continuously.

Yes and if there is no rotor pole in the vicinity to bother it, then it defaults to the current which was flowing through it the moment it was frozen with Liquid Nitrogen or Helium (I think you call it the "conserved value" of flux).

— and if the external flux crosses the conserved value, the current polarity flips. Each adjustment resists the rotor’s motion, manifesting as drag.

You need to analyze this point more precisely.
Let's say that the "conserved value" of flux is 1 Weber.
And the undisturbed coil needs 10kA to maintain it.

Then if an external flux source comes around and tries to push -1 Webers through the ideal coil (tries to reverse the flux), then the coil responds with 20kA of internal current (no current reversal !), which generates 2 Webers of internal flux and the total flux stays constant at 1 Weber again because 2 + (-1) = 1. This means repulsion.

However when an external flux source comes around and tries to push 2 Webers through the ideal coil (flux in the same direction), then the coil responds with -10kA of internal current (current reverses), which generates -1 Weber of internal flux and the total flux stays constant at 1 Weber because (-1) + 2 = 1. This means repulsion, also. An ideal coil always repels external flux. If it didn't, then the flux linkage would change.

For the sake of easier analysis it is better to start with a "conserved value" of flux equal to zero and the undisturbed coil current also zero.
Then an approaching pole of any polarity will induce nonzero current (reaching i_MAX at TDC) and will be repelled from the coil during the entire approach. When the same pole starts departing, it will also be repelled from the coil as the coil reduces its current from i_MAX to zero.
When the pole gets away, the coil returns to its undisturbed state, i.e. conserved value of flux equal to zero and the coil's current also zero. That is why I cal the QLE a "net zero" effect.

Verpies

Re: I Got a Speedup Effect.. « Reply #65 on: 2025-12-03, 19:08:07 »

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...besides, why are we shorting the super coil???

Because a shorted coil is a happy coil

Shorting a coil preserves its energy.
Shorting a capacitor dumps its energy.
Shorting a battery dumps its energy.

Opening a coil dumps its energy.
Opening a capacitor preserves its energy.
Opening a battery preserves its energy.

like saying the cap to cap losses are because of resistance. another useless circuit. but that 50% loss is not due to resistive loss. ideal, same loss.

Coils and capacitors do not act the same way. In fact they act in opposite ways.
Resistance is a killer of coil's stored energy and a saviour of capacitor's stored energy.
cap-to-cap energy transfers and coil-to-coil energy transfers are both 50% efficient because one meets its own match.
cap-to-coil energy transfers are almost 100% efficient because of their complementarity.

why would we not use that possitive and then negative output from the magnet pass??

Because coils are current sources (while capacitors are voltage sources).
The current in an ideal shorted coil does not reverse as a pole approaches it and departs it. It only decreases to its original value before the approach.

Saying that the magnet approaching the face of a super coil begins something that cant be stopped is silly.

It can be stopped. Just move the magnet away and the current will stop ....or add resistance. Resistance dissipates energy stored in a coil. The higher this resistance the faster the dissipation.
I find it amazing that you don't find it silly that a once-charged capacitor retains its charge, by the same token.

Quote from: floodrod on 2025-12-03, 19:36:45

if the approaching magnet sees a Lenz repel effect on approach, and the coils condition cannot be changed from there, then id say the magnet could never make it past or to TDC..

It couldn't if the magnet had its OD equal to the ID of the ideal shorted coil and its flux could not make a detour and was forced down the throat of the coil ...but when the magnet's flux can make a detour, then it does and the magnet passes (albeit with repulsion), while the total flux through an ideal shorted coil is preserved. The detour usually looks like the flux is getting bent/distorted. Look at this simulation to observe it.
You probably wonder, whether the magnet could be forced through that superconducting ring. The answer is yes as long as the magnet is smaller than the ring and sufficiently heavy or pushed. If the loop became superconductive while the magnet was inside the coil (or nearby) then after the passage of that magnet through the ring and moving it far away, the current induced in the would circulate indefinitely until the magnet (or another one like it) comes back and threads the ring again. See this.

floodrod

Re: I Got a Speedup Effect.. « Reply #66 on: 2025-12-03, 19:36:45 »

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Quote- "I agree, but an approaching and departing pole of a motor is not an AC drive."
"No flux nor current polarity reversal happens when a pole approaches and departs an ideal coil."

Rotational induced EMF is governed by motion of the field relative to conductor orientation.

Rotor induction (moving magnet/rotor): The reversal isn’t caused by resistive decay, and it’s not about a weakening field either — it’s built into the geometry of a rotor cutting the coil. A single‑polarity magnet sweeps across one side of the coil on approach and the opposite side on departure, and those conductors run in opposite directions around the loop. That change in which leg is being cut flips the terminal polarity, even though the magnet pole itself never changes. In a superconducting coil there are no resistive losses, but the induced output still reverses because the conductor orientation and motion dictate the sign — not decay.

Transformer induction (time‑varying flux in a core): In a transformer, reversal comes from the flux itself rising and falling in time. The primary winding drives a sinusoidal flux in the core, and the secondary sees that flux linkage increase, then decrease, so the induced voltage follows the derivative of that sinusoidal flux. In other words, transformer reversal is about the time profile of flux amplitude, while rotor reversal is about geometry — which coil wall is being cut and the winding orientation.

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Verpies

Re: I Got a Speedup Effect.. « Reply #67 on: 2025-12-03, 19:55:58 »

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Rotational induction polarity is cutting of conductor directions, not growth / decay.

But the external flux from the rotor's pole is unable to cut the windings of an ideal shorted coil. So all of your arguments that rely on flux cutting the coil go out of the window.
The only way an external flux can cut the coil is when the internally generated flux is weakened by resistance an unable to oppose the external flux completely.

P.S.
An ideal transformer is capable of transforming direct current

Magluvin

I Got a Speedup Effect.. « Reply #68 on: 2025-12-03, 20:43:03 »

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hey flood.
that vid is a perfect example of what i explained earlier. beautiful representation. but, with the coil setup of the tpu bottom ring coils cannot accomplish that. the arrangement of where the poles of the windings with only 2 phased inputs cannot even drive that motor presented earlier. that motor is an altered example that i cannot accept in this case of the tpu windings. that is my arguement.

and one more thing, the vid verp has presented of the magnet dropped above the super ring, also had another example showing a heavier magnet drop, of which the mag does fall through and the super ring does reverse its current. so his proofs shown, in sim mind you, does not hold water here.

https://www.youtube.com/watch?v=D-nAh9TMM30

mags

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Re: I Got a Speedup Effect.. « Reply #69 on: 2025-12-03, 21:20:07 »

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Quote from: Magluvin on 2025-12-03, 20:43:03

and one more thing, the vid verp has presented of the magnet dropped above the super ring, also had another example showing a heavier magnet drop, of which the mag does fall through and the super ring does reverse its current. so his proofs shown, in sim mind you, does not hold water here.

https://m.youtube.com/watch?v=yMg6j-UoUfo

Good catch, but that minuscule current movement after the exit was is not the way an ideal ring behaves.
I was corresponding about this with prof.Belcher and he acknowledged that this ring was simulated as very low resistance - not superconducting, (because his software could not handle dividing by zero), so the current in it has dissipated a smidge during the fall.

Like I wrote previously, if you allow the resistance to dissipate the current that was induced in the ring by the magnet on approach then it will start with zero current afterwards and upon departure of the magnet, the current will reverse, but that is ONLY because it was allowed to decay earlier.

floodrod

Re: I Got a Speedup Effect.. « Reply #70 on: 2025-12-03, 22:20:06 »

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Quote- But the external flux from the rotor's pole is unable to cut the windings of an ideal shorted coil.

I really don't have time to debate much more what a superconductive coil will do. From what I gather, you are now suggesting that flux cutting vanishes in an ideal coil. I think my point stands, in any normal setup "Lenz Drag" is an appripriate description where a magnets motion is opposed on the entry and exit while passing a coil.

This convo would have been better to place in it's own topic as it was initiated when Verps challenged what I consider a well established concep and has little to do with my Orion presentation.

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Jim Mac - Owner of Mooker.Com Opensource Free Energy Forum

Verpies

Re: I Got a Speedup Effect.. « Reply #71 on: 2025-12-03, 22:51:28 »

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From what I gather, you are now suggesting that flux cutting vanishes in an ideal coil.

Not only do I suggest it but I claim it.

I think my point stands, in any normal setup "Lenz Drag" is an appripriate description...

It isn't. The Lenz Effect is not responsible for the drag. The resistance is.

where a magnets motion is opposed on the entry and exit while passing a coil.

That is what happens in resistive coils. For 1 pole, drag is the difference of the force during approach minus the force during departure integrated over the two half cycles ...and it depends on resistance like this:

As you can see, the drag disappears with no resistance. Do you think that the Lenz Effects stops working when it is deprived of resistance ?

This convo would have been better to place in it's own topic as it was initiated when Verps challenged what I consider a well established concept

Rather a well established misconception.