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"Veritasium" (YT) - "The Big Misconception About Electricity" ?

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adx:

--- Quote from: aetherist on February 17, 2022, 11:00:30 pm ---
<big snip>


--- Quote from: adx on February 17, 2022, 12:48:29 pm ---
--- Quote from: aetherist on February 16, 2022, 09:43:00 pm ---Can any members here use old electricity to explain the traces for the AlphaPhoenix X pt1 & (later) pt2?
--- End quote ---
Once again, this explanation is in this thread way back - the answer is yes. The Maxwell simulation (or even all of them) replicates the features seen in the measurement I think better than expected given the problems with 'X' technique. The only thing I found 'interesting' is the "subtle lift", in both. I didn't quite go to town on the scope screenshot to the degree you have, but I did pore over it for a time not to treat it as some kind of smorgasbord of  Dunning-Krugeresque intrigue but because I use scopes and know what to look for. You are ignoring the fact pointed out in one of my first replies to you that the result of the measurement matches the Maxwellian simulator's output, confirming the theory for that particular case, which is what you question, resulting in the answer "yes" which is a simple word with a stable meaning and unlikely to be confusing unlike this unnecessarily long sentence which you have no problem understanding. Ask your rational core, it asked the question.
--- End quote ---
I am still not happy with lumped element TL models. And i admit that they can replicate the initial 0.2 V that AlphaPhoenix (Brian) got in his white trace for V across his bulb.

--- End quote ---
As I posted in my first reply on page 42, lumped TL models don't replicate the 0.2V, they predict higher, because they are an incomplete subset of conventional theory not intended for antennas. You are right to not like them in this particular application - they are not intended to be accurate for the job.

What does match experiment quite consistently including shape of the pulses, is the collection of field solver simulations based on Maxwell's theory. You ran through them in your second post, and they show conventional electricity theory matching measurement for the white trace.


--- Quote ---But i should have made it clear that i was referring to his green trace for the voltage across the resistor near his positive terminal.

<snipperoo>

--- End quote ---
Yes, that is what I meant too. I can see why you have a problem with it now, and needed more detail...


--- Quote from: adx on December 18, 2021, 02:19:09 pm ---If the scope were truly isolated (or ground lifted, depending on where EMC caps go) then the green trace should rise sharply more like the yellow.

--- End quote ---
and

--- Quote from: adx on February 15, 2022, 01:38:27 am ---I explained some of the problems with AlphaPhoenix's result many pages back, one of the main ones which distorts the send waveform I think is common mode coupling. I explained what I think it should look like, if it is measured with a better technique. Others did too, and went into quite some detail.

--- End quote ---

At 7:27 in the video is a diagram of the setup. The probe "reference GND" is the ground clip of the scope. This is tying one side of the pulse generator to Earth, loosely via extension cords and perhaps an inverter from the cars (described in discussions here at the time). The green probe, which is on the other side of the resistor, can thus not see the step directly from the step generator, because it is shorted to ground at the send end (by the ground clip). In essence it can only see voltage due to current getting around the circuit the long way, and a slow change of the GND voltage (which we can't directly see, because there is no probe measuring the voltage between this scope's GND and Earth under the desk).

This is not the way it's meant to be, but surprisingly the experiment still works. It's not necessarily an error if the person doing the test knows that taking this shortcut will still work. Again, I agree the green trace is "wrong", and this does represent the current in that resistor, and hence the current sent into that leg of the 'apparatus'. The other leg should be taking the balance, so it should be seeing nearly all the initial pulse missing from the green side (because that is shorted to ground).

From the clean white trace I can infer that the differential send current is probably fairly rectangular. But subtract the green trace and add the generator step, and that's going to make for a pretty messy voltage on the far side of the unprobed resistor, possibly best not to think about because it is guaranteed to confuse.

The situation would be the same but inverted traces (voltages) if the polarity of the generator is changed - other than that there is no difference and I likely would not test to confirm if I were doing the experiment.

BTW all this isn't so much from theory, as from experience. It just helps explain what is seen. The result is in the white trace, and matches simulation as you already know and I can now see you never really had a problem with. I rarely think about theory when doing engineering stuff, but I sometimes calculate things, and sometimes put it in a circuit simulator if I have really turned my brain inside out. By "Trevor's theorem" I try not to think through tricky situations to arrive at an answer, especially if it is about something that is inverted a number of times - he says you're most likely to get that out by one so would be better off flipping a coin, so save yourself the bother and guess, test, and then swap it round if it's wrong (sort of thing). The important thing is that anyone can be wrong at any time so don't put too much trust in thoughts. Or scopes. In either case trying to bulldoze through a problem with your mind is asking for trouble. It's not about intelligence, but experience in the biz.

Also small apology that I didn't say "subtle lift" originally, or if I did I edited that to "some sort of frequency dependent tilt". This is the white trace before the first reflection arrives. I'm not very interested in why, just noticed it (it appears in the simulations too).

aetherist:

--- Quote from: adx on February 19, 2022, 08:33:23 am ---From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described.

So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense.

There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign.

There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed.
--- End quote ---
Margan explains the stepped charging of a capacitor on page 14 of his paper. I think i am happy with his explanation of steps.
Erik Margan repeated Wakefield's X.
http://www.ivorcatt.co.uk/x726.pdf

However, Margan makes a mistake on page 15.  He says that the battery continues to supply photons (ie energy) after the capacitor is full.
As the line is filled, the battery supplies the current (less after each reflection at V1), until several µs later the current has dropped practically to zero, and the voltage is practically equal to the battery voltage. But the important question to ask is this: Did the battery stop supplying energy to the circuit?
The correct answer is: no! Photons cannot stand still! The battery continues to supply the energy to the system, but the system now reflects all the energy back to the battery.
The effective current is zero, so no power is dissipated, but the energy continues to flow back and forth throughout the system. Another point to make is the following: if we put back the 50 k resistor as H VB and charge again the cable, we will not see any stairsteps. This is because the impedance mismatch is very high, 50 to 50 k , so the steps are less than 9 mV, too H H small to see (at the input sensitivity of 2V division). Because of the much smaller Î steps a much larger number of reflections is needed to charge up the line, leading to the classical capacitor charging equation e . It is important to…..

I don’t agree with Margan. I reckon that once the system is fully saturated with electons then the negative lead plate of the negative terminal of the lead acid battery can't feed any more electons out of the battery fluid onto the lead plate (or onto the lead strap or onto the lead terminal). The chemical process stops. Margan reckons in effect that the chemical process continues, but in effect he reckons that electons are both coming out of the fluid & going into the fluid. No.

U mentioned that i reckoned that the discharge of a capacitor must have a very weak additional long term discharge current due to the redistribution of the induced surface charge (electrons) on the positive plate.
And u said that i said that i reckoned that there will be a corresponding very weak additional long term charge current, ie a mirror image of the discharge.
Yes & No.
No i never said that.
And yes, i do believe that there is a mirror image effect during charging.
If the speed of surface electrons is say c/10,000 then their charge/discharge will be very weak compared to the c/1 speed of electons. But i am not sure how far the surface electrons need to move. If they move from the surface of the wire or plate to just under the surface then that distance might be only 1.0 nm. Or they might have to move 1.0 nm along the surface. I don’t think that any electrons have to go all of the way to the battery. Still thinking.

aetherist:

--- Quote from: SandyCox on February 19, 2022, 02:11:41 pm ---I will attempt to address your misconceptions in the note I am writing.
In the meantime, let me set a challenge: Let's change the termination resistor in Catt's paper from 75 Ohm to 47 Ohm. How, according to "new electricity", will the measured pulses look if we do that?
--- End quote ---
There will be stepped reflexions of current & voltage. I have never worked on that kind of stuff.
But i see that Wakefield already has some traces for 75 Ohm coax with a 40 Ohm termination.
Shown on page 42 & 43 of Forrest Bishop's paper re Reforming Electromagnetic Units…
http://www.naturalphilosophy.org//pdf//abstracts/abstracts_6554.pdf


One thing i can explain, that i think no-one else could explain.
See how the bottom trace jumps up a few V too high, & then falls down to the correct 4V.
That jump is due to the 40 Ohm resistor being saturated with my electons. When the switch is closed electons in that short wire near the switch already heading for the switch will instead of doing their usual u-turn at the open switch they will propagate through the closed switch & onto the core wire of the coax, followed by electons from the resistor. The short wire was saturated with electons too, but the resistor holds a lot more electons per m length than the wire. Hence the brief spike of over-voltage.
U can see the same spike of over-voltage in some of the other traces, but there it is a negative spike of under-voltage i suppose u could call it.

But i have to have a think about what happens to electons inside resistors. Are they annihilated. Do they looz energy. Do they convert to infrared photons.
Anyhow, if there are lots of voids or porous bits or interface surfaces then there must be lots of electons on thems surfaces, & the electons would be doing lots of u-turns.
And they would be jostling lots of surface electrons, which would jostle atoms, & produce heating & resistance & would we know produce a voltage drop.
Its the voltage drop that has me worried. I am glad that SandyCox did not ask me re how exactly do electons produce a voltage drop across a resistor. Still thinking.

bsfeechannel:

--- Quote from: aetherist on February 19, 2022, 10:54:43 pm ---Shown on page 42 & 43 of Forrest Bishop's paper re Reforming Electromagnetic Units…
http://www.naturalphilosophy.org//pdf//abstracts/abstracts_6554.pdf

--- End quote ---

https://naturalphilosophy.org/about/


--- Quote ---The John Chappell Natural Philosophy Society (CNPS) provides an open forum for the study, debate, and presentation of serious scientific ideas, theories, philosophies, and experiments that are not commonly accepted in mainstream science.

--- End quote ---

Yeah. Right.

adx:
You misquote me, emphasis mine:


--- Quote from: aetherist on February 19, 2022, 10:00:28 pm ---...
U mentioned that i reckoned that the discharge of a capacitor must have a very weak additional long term discharge current due to the redistribution of the induced surface charge (electrons) on the positive plate.
And u said that i said that i reckoned that there will be a corresponding very weak additional long term charge current, ie a mirror image of the discharge.
Yes & No.
No i never said that.
And yes, i do believe that there is a mirror image effect during charging.
If the speed of surface electrons is say c/10,000 then their charge/discharge will be very weak compared to the c/1 speed of electons. But i am not sure how far the surface electrons need to move. If they move from the surface of the wire or plate to just under the surface then that distance might be only 1.0 nm. Or they might have to move 1.0 nm along the surface. I don’t think that any electrons have to go all of the way to the battery. Still thinking.

--- End quote ---

You said (in full for context, again emphasis mine):


--- Quote from: aetherist on February 19, 2022, 04:38:42 am ---
--- Quote from: SandyCox on February 18, 2022, 01:02:28 pm ---Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor?
--- End quote ---
I don’t know what the old (electron) electricity equation(s) is for discharge of a capacitor. But the new (electon) electricity equation(s) would be almost identical, except that it would have to show the correct steady half voltage for the correct double the distance, ie for double the time (at least it would for the case of zero reflexions).

And it would need an additional equation for the additional voltage from electons leaving the length of the wire from the capacitor to the resistor. This extra voltage would be for a doubled time, ie electons have to go the wrong way along that wire & later return along that wire, hence a doubled distance & a doubled time. For a giant capacitor & a short or thin wire this voltage might be insignificant.

And it would need an additional equation for the very small extra voltage happening for a very long time due to surface electrons gradually slowly entering the positive plate. But this voltage might be insignificant.

--- End quote ---

I then said (also in full for context, now emphasising):


--- Quote from: adx on February 19, 2022, 08:33:23 am ---From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described.

So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense.

There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign.

There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed.

--- End quote ---

You'll see I didn't say you reckoned anything. I merely quoted your words to show you how it is totally wrong in the context you used it. Your thoughts are irrelevant to the facts in this instance (where you are wrong). I didn't comment on what you think, only your wrong claims, using some of your exact words for clarity. I took no mental leap of assumption (like I did with your roo-tons, claiming that they were part of a planned deception scheme on your part - is that what you're upset with?).

You introduced the concept of a slow charge effect. We were both talking about a long-term charge retention and discharge mechanism, where the capacitor maintains a small voltage for longer than expected while discharging.

You are now starting to believe I have said things I didn't say. It seems to me you might be both externally and internally grasping at straws to ignore your "rational core". How can I know how you think if I am merely under your skin? To use your own device, I'm not saying I am your rational core, but what if I were?

Of course if you want to define capacitors to be transmission lines in either practical or theoretical effect (or both), then by your new definition then of course you go right ahead and show stair step changes in voltage, distance and that whole first line of nonsense. But I (or is it you?) think you should use the term "capacitor (transmission line)" to avoid ambiguity over your definition.

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