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Author Topic: partzmans board ATL  (Read 36174 times)
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Itsu,

I ran a test comparison with a 4uf film cap for Cs between having no shield and having a near complete grounded shield.  The results were identical with a Ccs of 3.42uf.

I did searches for "Can aether energy be shielded?" including ChatGPT and no answer was really forthcoming.  In fact, most references do not even acknowledge an aether.

Pm
   

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PM, you say:

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As I reviewed your test, I'm not sure if you had the copper shield connected to ground or not.  It would be interesting if you connected the shield to ground and re-tested!

I did mention that i did ground the copper tape when i wrote:

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When i isolate Cs with a copper tape like shown in the below picture and I GROUND THIS COPPER TAPE, it (Cs) should be shielded significantly from any E-Field IMO.

I think it is consensus among scientists that indeed aether does not exist, so that would explain IMO why so little reference is available.

Itsu
   
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PM, you say:

I did mention that i did ground the copper tape when i wrote:

I think it is consensus among scientists that indeed aether does not exist, so that would explain IMO why so little reference is available.

Itsu

Itsu,

Yes, I missed your comments as I assumed too much from your pix!

Pm
   
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The following test is very informative IMO.  All previous tests showing dielectric induction or charge separation in a capacitor placed in an E-Field, have been done with fast rise time pulses on the primary.  This test uses a relatively slow ramp voltage as the L1 primary source for induction.

The schematic is shown first.  The voltage ramp generator in this case is a Crown XLi-2500 Class D power amplifier.  Resistor R1 is used to stabilize the internal feedback loop in the output stages of the amp as it has a problem otherwise directly driving the L1 inductive load.  The pulse on CH1(yel) is simply a sync pulse to initiate the ramp in a Rigol DG4162 generator that supplies the Crown amp.  C1 is open circuited with a Tek TPP0500B 10Meg, 3.9pf probe connected to Vc1.

The scope pix RoP1 shows the basic waveforms with the Math(red) channel indicating the average Pin=1.445W over 18.56us for a Uin=1.445*18.56e-6=26.8uJ .  This is the energy required for C1 to reach a peak voltage of 2.953v which equates to a UC1=2.953^2*550e-6/2=2.4mJ .  If the energy in C1 was able to be harvested at this time, this would make the COP=2.4e-3/26.8e-6=89.4 .

Scope pix Rop2 shows the peak voltage of the ramp driving L1.

What is most revealing IMO of this experiment, is that there is no way that the high impedance CH3 scope probe has any part in creating the ramp voltage across C1 over the 18.5us time period.  It is simple showing the voltage measurement over time of the charge separation in C1.  BTW, let's calculate the required current to raise C1 to 2.953v over this time period.  So, di=dE*C/dt therefore di=2.953*550e-6/18.5e-6=87.8 amps!  I personally find no evidence in any of the circuit measurements of this amount of current.  So I again propose the question, where is this current or energy to charge C1 coming from?

The other revelation here is that fast rise times of the E-Field are not needed to create this charge separation! 

I know many don't believe the Aether exists but call it whatever you wish, some outside energy source is at work here!

Regards,
Pm



   
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Thanks PM,

If i understand this correctly ,a specific  delay on vc1  might allow c1 extraction at the right time ?
   
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Thanks PM,

If i understand this correctly ,a specific  delay on vc1  might allow c1 extraction at the right time ?

3D,

Therein lies the problem!  Vc1 tracks the voltage waveform on L1 precisely based on the coupling or K factor.  For example, if K=.95 and the V/t=3, a single Cs will reach a peak voltage of 3*.95=2.85v.

If no load is presented to Cs, the voltage on Cs will track the voltage on L1 at any rate such as sine, pulse, ramp, or complex.  IOW, the voltage across Cs can not be separated from the E-Field generated by L1 as far as I have been able to determine.  Once this is accomplished by means of delay or whatever, then the gates are opened!

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Thanks PM

Could another fine wire  winding just used to trigger a high speed fet connected to Cs work ?

It might be the same as triggering from vc1 but isolation might be important as the traveling magnetic flux in of itself is very fast.

I keep refereeing back to the first TPU videos which were never responded to in any communications except to say that they are not piezio stacks.



I'm not in a good headspace atm . Keep up the good cause!
   
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Thanks PM

Could another fine wire  winding just used to trigger a high speed fet connected to Cs work ?

Yes that in itself would work, but we still haven't disconnected Cs from the E-Field.

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It might be the same as triggering from vc1 but isolation might be important as the traveling magnetic flux in of itself is very fast.

Other windings can be placed on the core and operated independently.  The voltage on Cs will then reflect the simple sum of each winding's E-Field.  What needs to happen is this- An E_Field is applied to the core thus charge separating Cs.  Then, the E-Field is somehow removed leaving the charge in Cs.  The gain of this operation would be large thus yielding OU.  This is way easier said than done!

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I keep refereeing back to the first TPU videos which were never responded to in any communications except to say that they are not piezio stacks.

I missed that comment!  I have refrained from making any similarities between this concept and the TPU but I do see the possibilities.

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I'm not in a good headspace atm . Keep up the good cause!

Take care!

Pm
   
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...
What needs to happen is this- An E_Field is applied to the core thus charge separating Cs.  Then, the E-Field is somehow removed leaving the charge in Cs.  The gain of this operation would be large thus yielding OU.
...


Hi PM,

So you mean switching off the input current (i.e. the source of the E-field in this case) during applying a switched load across Cs, right?
So Cs should see the load whenever the input current is disconnected.
However, Cs should be discharged to a certain degree only to avoid the high input current demand for recharging an empty Cs.
It might be the high COP would still remain higher than 1.

Thanks for sharing your efforts.

Gyula
   
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Hi PM,

So you mean switching off the input current (i.e. the source of the E-field in this case) during applying a switched load across Cs, right?
So Cs should see the load whenever the input current is disconnected.

Yes, you could do this as you describe.  So when you release the input current from L1 or the primary, the voltage across the primary and thus the E-Field will reverse to a voltage level that depends on the what the collapsing current "sees" as a limit.  The load resistor will "see" this reversed voltage and will produce power.  This sudden load connected to Cs at this point in time will create a decrease in the collapsing current of L1 to a degree that the energy to power the primary will exceed the output energy across the load.

Quote
However, Cs should be discharged to a certain degree only to avoid the high input current demand for recharging an empty Cs.
It might be the high COP would still remain higher than 1.

Thanks for sharing your efforts.

I created a device called "Simple OU' previously in my post #487 on page 20 in this thread that used a load resistor driven by the bipolar output voltage of Cs with a 4 cycle square wave applied to the primary.  Alas, it turn out COP<1.

Pm

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Hi Jon,
This is above my pay grade so I asked the new o1 model from OpenAI. I posted your comment along with the diagram and scope shots.

I'd be interested in others thoughts.

---------------
Below is a summary of what is actually going on in this “slow‐ramp” test, why your capacitor ends up with more apparent stored energy than your scope math says you put in, and why there is no need to invoke “the Aether” (or any other unseen energy source) to explain it. In short, the measured “COP ≈ 90” arises from incomplete measurement of the true input power/energy and from overlooking how the amplifier, coil, and parasitic coupling really behave in the time domain.

1. Why the capacitor’s voltage rises

Even with a “slow” or relatively low‐dV/dt ramp, there is still an electric‐field coupling from the primary inductor  to the open‐circuited capacitor . This coupling can be broken down into:
   1.   Capacitive coupling from the coil’s windings and/or leads to .
   2.   Displacement current (sometimes called “displacement current through parasitic capacitances”) from the amplifier’s output node and coil windings into the capacitor node.

Because of these parasitic coupling paths, a small displacement current can flow into  even though it appears “open‐circuited.” On a scope plot, it can easily look like “charge separation with no conduction current,” but in reality the coil, wiring, test leads, and the capacitor physically create stray/pickup paths.

2. The huge “87.8 A” calculation

You computed I = C\,\Delta V/\Delta t \approx 88\,\mathrm{A}  based on
I \;=\; \frac{(550\,\mu\mathrm{F})(2.953\,\mathrm{V})}{18.5\,\mu\mathrm{s}} \;\approx\; 88\,\mathrm{A}.


That equation is correct only if the capacitor current truly flows through a single well‐defined conduction path that you have measured. But in this experiment:
   1.   A Class‐D power amplifier can (and does) deliver short, high‐peak current pulses into reactive loads, especially if there is any pulse or ramp edge.
   2.   The circuit layout almost certainly has stray and distributed impedances that you are not measuring with your single “current sense” approach.  In particular:
   •   Your R1 or CH2 measurement might not register the fast spike(s).
   •   The amplifier itself may deliver pulses outside the bandwidth or sample window you are capturing with the scope’s math channel.
   •   Your scope probe on  is high‐impedance, but there is still stray conduction/current flow in the cables and amplifier output stage.

Hence the true current path responsible for charging  is simply “hiding” in plain sight within the distributed circuit geometry. It is not that “88 A mysteriously appears.” It is that short bursts of current do flow, but your scope’s measurement method is missing them.

3. Why the scope math suggests only 26.8 µJ in

You mention measuring

which integrates to

Yet the capacitor’s energy is
\[
U_{C_1} \;=\;\tfrac12\,C\,V^2
\;=\;\tfrac12\times 550\times 10^{-6}
\times (2.953)^2
\;\approx\; 2.4\,\mathrm{mJ}.
\]
And you note that , giving “COP ≈ 90.”

Missing or mis‐measured input

What is really happening is that your scope’s math trace is almost certainly not capturing the real (and sometimes very brief) current/voltage pulses that the Class‐D amplifier is generating. A Class‐D stage can switch large current in tens of nanoseconds or microseconds, and typical “Power = Voltage × Current” scope math may only be capturing the lower‐amplitude portion or the time‐averaged portion.

So your actual energy input is being under‐reported by orders of magnitude. Once you sum up all the energy delivered—including the brief spikes or transitions that your math trace might be missing—the real input to the circuit comes out comparable to (or greater than) the  stored on . This resolves the “Where did all that energy come from?” question by conventional circuit theory.

4. Slow‐ramp vs. fast‐pulse

You note that a slower ramp also leads to the capacitor being charged. That is not surprising. Even when dV/dt is smaller, any finite “ramp” across an inductor + capacitor arrangement (with a parasitic coupling path) can result in net charge buildup. Whether the waveform edges are “fast” or “slow,” the principle is the same: the amplifier and coil geometry provide a path for displacement current into . The main difference is that with a slower ramp, the absolute current peaks can be lower (though still possibly short spikes on transitions), and the total time is longer.

5. There is no need for an “Aether” hypothesis

From standard electromagnetics:
   1.   Displacement currents flow whenever there is a time‐varying electric field. Even in “open” circuits, you can get net charge on a conductor or capacitor by virtue of parasitic capacitances.
   2.   A Class‐D amplifier plus an inductive load can have high instantaneous current pulses that do not show up if you are only looking at a lower‐bandwidth or time‐averaged trace.
   3.   Apparent “over‐unity” or “COP > 1” arises routinely when the experimenter’s measurement method does not capture all real input power/energy. This is not a condemnation—it is easy to make these errors with modern power scopes that average or sample in ways that miss fast edges.

Everything we see in these waveforms is consistent with normal conservation of energy once the full circuit parasitic paths and the high‐speed amplifier switching dynamics are properly accounted for. Put differently, the amplifier really is delivering the current/energy that shows up as voltage on . It just does so in short bursts and via stray coupling paths that are subtle to measure.

Bottom line

   1.   The capacitor is being charged from the amplifier/coil system via normal electromagnetic coupling (not from “outside” or “the Aether”).
   2.   The “88 A” is simply the instantaneous current demanded by ; short bursts of current do flow, but are missed by your scope’s power measurement.
   3.   The “COP ≈ 90” is an artifact of under‐measuring the true input energy.

No new physics is required—just the usual Maxwell‐based explanation of parasitic coupling, displacement currents, and the pitfalls of incomplete power/energy measurements in pulse or Class‐D driven circuits.
------------------------- edit added screenshots for the Tex format calculations
   

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In my opinion that AI generated response is a good example of the dangers of relying on AI.  PM's post #628 assumed the reader had been following this thread where the capacitor is placed within the central hole of a ring core.  The circuit in that post did not show that, it merely showed the capacitor near an inductor.  AI then concluded that stray capacitive coupling was the only connection.  It then decided that there were two paths, (a) capacitive coupling and (b) displacement current, as though these were two separate entities which they are not.  Irrespective of that blunder, AI then gave its long winded answer that you posted.

Had PM's post shown that the capacitor was within the alternating magnetic vector potential field within the ring core, resulting in it also enduring an alternating electric field there, then AI might have given an entirely different answer.  It would be interesting to give AI the challenging task of evaluating PM's "dielectric induction or charge separation" for that situation.

Smudge
   

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Accurately measuring the average input power is notoriously difficult, especially with high crest-factor i & v waveforms.

Measuring the average output power is relatively easy.
A Wattbox based on an incandescent light bulb with an uncoiled filament is accurate and cheap for this purpose, ...regardless of the waveform.
   
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In my opinion that AI generated response is a good example of the dangers of relying on AI.  PM's post #628 assumed the reader had been following this thread where the capacitor is placed within the central hole of a ring core.  The circuit in that post did not show that, it merely showed the capacitor near an inductor.  AI then concluded that stray capacitive coupling was the only connection.  It then decided that there were two paths, (a) capacitive coupling and (b) displacement current, as though these were two separate entities which they are not.  Irrespective of that blunder, AI then gave its long winded answer that you posted.

Had PM's post shown that the capacitor was within the alternating magnetic vector potential field within the ring core, resulting in it also enduring an alternating electric field there, then AI might have given an entirely different answer.  It would be interesting to give AI the challenging task of evaluating PM's "dielectric induction or charge separation" for that situation.

Smudge

I absolutely agree!

Pm

   
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Accurately measuring the average input power is notoriously difficult, especially with high crest-factor i & v waveforms.

Measuring the average output power is relatively easy.
A Wattbox based on an incandescent light bulb with an uncoiled filament is accurate and cheap for this purpose, ...regardless of the waveform.

Again, I absolutely agree!

Pm
   

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FWIW .
The o1 model is meant to excel at these sorts of problems. I'd be interested if I've missed more context.
   
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FWIW .
The o1 model is meant to excel at these sorts of problems. I'd be interested if I've missed more context.

Jim,

Thank you for taking the time to input this info into the AI models.  I will address each and every point made by this 01 model over the next few posts.

Regards,
Pm
   
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Jim and All,

This is regarding the first screenshot of AI model 01.

1.  Under-Measurement of Input Energy

In the RoP1 scope pix in my post #628, there is no evidence of of any spikes in the input voltage or current waveforms.  With the sample rate of 1.25GS/s, there are 23,200 samples taken over the 18.56us measurement period for the Math(red) channel to calculate the results of Pin=1.445 watts.  IMO, there is no error in these input measurements! 

Keep in mind that a ramped waveform is the one of easiest to perform math calculations on since the average or mean area under the waveform is simply b*h/2 .  So, we can make a quick calculation here with Eavg=37.94/2=18.97 and Iavg=.16/2=.08 for a resulting Pavg=18.97*.08=1.5176 watts.  This is higher than the actual Pin measured by the scope Math because the actual current ramp is slightly concave rather than linear.

2.  Conservation of Energy

This is a standard reply that is expected from any AI since this is all it has been taught!  The Pin is fully explained above and it is what it is!!!

3.  The role of the ring core

It is known that the E-Field is generated by E=dA/dt where A is the magnetic vector potential or the A-Field.  It is also known that the unit of the magnetic vector potential is Ampere-m (A-m) in the SI system.  However, if one looks at any of my many posts where the primary is driven by a fast rising square wave, one will see the the voltage across the Cs capacitor placed in the core window rises in 100s of nanoseconds before the current has risen any appreciable amount.  Therefore, the A-field has very little influence on generating the near instantaneous voltage on Cs but rather the experiment shows the E-Field (V/t) to be the influence.  This really becomes apparent when relatively large amounts of capacitance (>2000uf) are used for Cs!

Therefore, there remains no explainable source of energy to account for the voltages reached across Cs.  That is except the aether!

Conclusion

All points made here are addressed and refuted above.

Regards,
Pm 



   
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Jim and All,

This is regarding the second screenshot of the AI Model 01.

The Ring Core and Its Fields

Points #1 and #2 both rely on the use of the A-Field to generate any voltage or displacement current in Cs.  As shown previously, the A-Field has no influence on the voltage generated across Cs.  I would be happy to be shown or proven otherwise!

Point #3 creates voltages across Cs via stray capacitance or coupling leading to "non-negligible electric fields in the region of the capacitor".  This might be applicable if Cs were in the nano-farad range but when in the 100's or 1000's of micro-farads, no way would this be applicable IMO!

Why the Capacitor Charges

Point #1 summarizes on the idea that the magnetic vector potential of the A-Field generates the voltage across Cs.  I say it does not!

Point #2 summarizes that stray couplings, etc, create the voltage across Cs.  Again I say they do not!

Point #3 states that the source for the voltage across Cs comes from unseen and mis-measured transients in the input.  AI states that this is very possible with class D type amplifiers and yet no such transients are seen with a 350MHz scope using a 1.25GHz sample rate.

Let's say that the input voltage ramp was generated via an analog emitter or source follower.  Then the input energy would be doubled plus an addition 10% for loss in efficiency.  This would make the new Uin'~60uJ which would result in a new COP'=2.4e-3/60e-6=40 .

Regards,
Pm
   
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Another potential source to explain the charge separation could be the Poynting vector or S=EXH, where S is the Poynting vector (measured in watts per square meter, W/m²), which points in the direction of energy flow, E is the electric field vector (measured in volts per meter, V/m), and H is the magnetic field intensity vector (measured in amperes per meter, A/m).

However, the H-Field is in part defined by the MMF which is N*I or number of turns times coil current so again, with very little current flow within the first 150-200ns, it is logical to conclude that the Poynting vector is not responsible for the charge separation and the resultant energy contained in the larger valued electrolytics.

Regards,
Pm
   

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IMO, there is no error in these input measurements! 
Yes there is even if the Math channel properly multiplies the current and voltage and averages the products afterwards (I can't be certain that it does) because your probe positions are wrong for measuring the supply voltage AFTER the current sensing resistor (which is what the DUT sees).  This causes the Pin to be overestimated.
   
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Yes there is even if the Math channel properly multiplies the current and voltage and averages the products afterwards (I can't be certain that it does) because your probe positions are wrong for measuring the supply voltage AFTER the current sensing resistor (which is what the DUT sees).  This causes the Pin to be overestimated.

Verpies,

I chose that frame of reference as the most accurate for the Pin measurement because otherwise, the input current to L1 would begin at a negative value with 100ms between cycles.  This is due to the low frequency instability of the Class D amplifier output filter/feedback network trying to drive an inductive output.

The "overestimated" Pin even gets better due to the fact that with the inductance of L1 is 1.2mH and the input current peaks at 160ma, the energy in L1 reaches ~15uJ at the end of the 18.56us measurement window.

Pm
   
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FWIW, here is an example of charge separation or dielectric induction used in an example of electrolysis.  In this case, the sample is tap water in a very crude pill bottle with copper plates.  No electrolyte was added but our tap water is far from pure!

The basic schematic is shown and the purpose of diode D2 is to shift the output pulse in the cell to be mostly positive throughout the cycle in order for the cell to "see" an average positive voltage referenced to the cathode.

The pix shown is rather poor focus but I hope you can see the hydrogen bubbles being generated on the cathode plate.  With this rather crude cell it is impossible to quantify the amount of hydrogen produced at this time.

The interesting thing is the amount of energy drawn from the input to produce this electrolysis.

The first scope pix P1 shows the Pin to the primary L1 being 8.245W over 12.82us for an input energy of Uin=8.245*12.82e-6=105.7uJ.  CH2(blu) is the power supply voltage at Vs, CH3(pnk) is the Vcell output voltage, and CH4(grn) is the input current to the primary L1.

The second scope pix P2 shows the Pret or power returned to the power supply as 6.797W over 12.38us for a returned energy of Uret=6.797*12.38e-6=84.1uJ leaving a net input energy of (105.7e-6)-(84.1e-6)=21.6uJ.

Now, the third scope pix P3 shows the comparison between the current in the stored reference R1(wht) which was the input current with the cell inserted in the toroid core and operating, with the current shown in CH4(grn) withe cell removed from the toroid.  I separated the two traces so one can see they are identical and they superimpose perfectly. 

IOW, no energy is being consumed by the electrolysis but rather all the input energy is consumed by the magnetizing current of the core.

This disclosure was not meant to be a quantitative analysis but rather to show the concept of using charge separation for electrolysis.

Regards,
Pm   
   
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  Clever innovation, Jon! 

I hope to learn more about this.
   
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  Clever innovation, Jon! 

I hope to learn more about this.

Thank you Steve!

Jon
   
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