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#1900 by Monomorphic on 21 Oct, 2016 20:11
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I formally challenge the EM drive to a race. CID on one torsion balance and the EM on another and you can only use 12V @ 3 amps. Lets see which one makes a rotation first.
Sincerely,
Harry Sprain
Looks like a simple gyroscope performing attitude control. Does it work on a linear air bearing? -
#1901 by X_RaY on 21 Oct, 2016 20:17
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The tanδ(ε“)of the dielectric material produces some losses at one end...How about I just re-write your result using the definition of skin depth... then the equation makes a lot more sense to me. Q increases as "resistivity" decreases, but the relationship to frequency and permeability would seem to suggest a small frustum with high permeability.
Consider these equations at constant frequency and constant dimensions. What two variables do we have left to play with, in terms of materials?
Again, "To continue this discussion both of us must stop writing Q ~ and instead write Q = , detailing what precise expression, in terms of what variables, we are talking about"
in your response, you continue to write Q~. You need to write Q= referring to something specific and precise
to be able to have a discussion.
EDIT: Basically, this is the issue, I realize that one has to be explicit as to what is the Q= expression also in your dQ/dr
because there are many ways to write Q~ that lead to completely different expressions depending on what multiplying factors one is referring to
Like this? It's non-linear.
Thinking more about your theory, please comment on whether (and if not, why not) your theory is effectively nullified by the test results of Zeller and Kraft, at California Polytechnic State University, San Luis Obispo, who performed experiments with a cylindrical cavity having a HDPE dielectric ( https://www.linkedin.com/in/kurtwadezeller ) (NSF member Zellerium https://forum.nasaspaceflight.com/index.php?action=profile;area=showposts;u=47993 ). The cylindrical metal cavity was symmetric, but the HDPE was placed asymmetrically in the cylindrical cavity. Their conclusion was that there was no thrust. This experiment by Zeller and Kraft is a falsification of a hypothesis that thrust is generated simply by the HDPE (whether by electrostriction or other means), when asymmetrically placed in a cylindrical cavity.
Their experiments at California Polytechnic State University, San Luis Obispo, showing no thrust when using HDPE dielectric asymmetrically placed in a cylindrical cavity, can be seen in the attachments on https://www.linkedin.com/in/kurtwadezeller under "Researcher" section on Kurt Zeller's LinkedIn profile.
A theory explaining thrust of the EM Drive based on HDPE dielectric would need to also explain Zeller's experiments showing no thrust.
I just read the paper on the cylindrical cavity with HDPE dielectric, and the results were "inconclusive", not "null". They did measure thrust, but the results were in either direction.
Primarily, it seems they excited a TM mode of the cylinder. The TM mode concentrates the E field on the end plates and the H field on the side walls. The E field does not cause nearly as much (if any) power dissipation. It is the H field can cause power dissipation, and dissipation on the side walls of a cylinder will have a symmetrical effect. In order to have power dissipation, the material must have some conductivity. The resistivity of HDPE is too high, so there is very little power dissipation there. Instead, I think it's practically invisible to the MW's.
If they construct a cylinder where one end is copper and the other end is steel, they might see a different result.
This effect acts on the E component of the field.
Zellerium's experiment did not confirm the first patent from Shawyer related to the emdrive:
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#1902 by jay343 on 21 Oct, 2016 20:18
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http://forum.nasaspaceflight.com/index.php?topic=41475.0
Doesn't the Coriolis effect produce a torque on the gyroscope on the end of the balance? -
#1903 by Rodal on 21 Oct, 2016 20:19
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I formally challenge the EM drive to a race. CID on one torsion balance and the EM on another and you can only use 12V @ 3 amps. Lets see which one makes a rotation first.
to have any relevance whatsoever to the subject of this thread (space flight applications) the contest should be performed in a vacuum chamber under at least 10^(-6) Torr, and the drives should be powered by a battery , self-integrated in the same body being accelerated.
....
Sincerely,
Harry Sprain
No power cords allowed, as:
*Prof. Juan Yang defenestrated all her prior tests by performing tests with a battery showing that in her previous tests the power cords suffered thermal expansion as a result of getting hot, which produced the anomalous force artifact. Yang reported no thrust when taking into account the relative error of her experiments.
*Samsonov is the only "citizen-scientist" that to this date has reported EM Drive tests performed with a battery.
Samsonov reported no thrust when taking into account the relative error of his experiments.
Are you suggesting that all reported results without internal batteries are spurious effects, Shawyer, Fetta, everyone building devices here?
I gave my opinion as to what the proper rules for a contest for a propellant-less space drive should be, to be able to judge a winner.
FACTS: In space you do not have air at 1 atmospheric pressure and you cannot run a practical space drive with power cords
No, your suggested interpretation of what I advised is a non-sequitur.
https://en.wikipedia.org/wiki/Non_sequitur_(logic)
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#1904 by rq3 on 21 Oct, 2016 21:29
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Take the exact same vehicle, put it in deep space and turn on the EmDrive, which immediately accelerates the 1,000kg mass at 1g for as long as the power is on.
Cool! You hit light speed in only about 354 days! -
#1905 by X_RaY on 21 Oct, 2016 21:34
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Let's consider the coupling factor:
If you eliminate R in the resonant circuit (superconductive resonator) there is still Rs, this is the source resistance. It can't be ignored, its part of the circuit.No, this high Q is NOT QLoaded it's QUnloaded=Q0!
This is not an accelerator cavity with a big hole in the middle so particles can be accelerated through the hole by the massive E field. The only losses are the eddy current heating and the momentum xfer to the accelerating mass.
It is loaded Q as the momentum loss from the EmWave is another loss factor to the cavity that drains stored cavity energy faster than just eddy current heating losses. The force curves published in the patent application are the result of the lost cavity energy and so they are in fact the loaded Q fill and decay time for the cavity with 5x TC being 1~ sec and 1x TC being ~0.2 sec.
From that you can calc the cavities effective loaded Q as the forces were generated.
Ql = 2 Pi Freq Tc. Assuming 2.45GHz excitement, the loaded Q is approx 3x10^9.
EDIT
I am sure we discussed this in the past. If my memory serves you has agreed with this at that time.
The antenna defines a specific coupling factor between the inner and the outer circuit. The same antenna what is used to couple the RF-energy into the cavity is able to couple the energy out, back into the source with the same effectivity. This is basic microwave 101 knowlage.
When it's just 1 it means the external Qext is as high as the Q0 of the resonator, the coupling is "critical", VSWR≈1. Now let's focus in a situation where Q0 of the cavity is the dominant part (1/Qeff=1/Q0+1/Qext´while K<<1, K=Q0/Qext),
the coupling factor(K) is low ...
Now we have a huge problem because the VSWR is huge, therefore the energy delivered into the resonator is small compared to the possibilities of the HF source generator.
Do you see the problem?? -
#1906 by WarpTech on 21 Oct, 2016 22:53
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,,,I just read the paper on the cylindrical cavity with HDPE dielectric, and the results were "inconclusive", not "null". They did measure thrust, but the results were in either direction. ...
Argumentative.
Zellerium himself reported his own experiments here:
http://emdrive.wiki/Experimental_Results
and he reported them very clearly as zero thrust.
As documented at NSF in several of his posts, Zellerium reported this at http://emdrive.wiki/Experimental_Results after much thought and deliberation.
The conclusions of the paper say "inconclusive". Regardless however, the electrical permittivity of HDPE is ~10^-11, and the dissipation factor ~10^-4, so that power dissipation will be proportional to 10^-15. Also, the HDPE is practically transparent to MW's, so the majority of the reflection I believe is happening at the surface of the copper behind the insert, not at the surface of the insert. In other words, HDPE insert will have at most a very tiny effect, due to very tiny losses, due to such a high electrical resistance.
I would also like to remind you that EW test of the Cannae drive showed no thrust "except" when the dielectric insert was present. So doesn't that confound @Zellurium's results as "inconclusive", just as he said in the paper?
Also, my model is based on shedding (divergence) of (scalar) magnetic flux, as @notsosureofit realized, from my conservation of flux equation. If that is the case, then using a TM mode in the cylinder concentrates the E field on the end plates, not the B field. So any B field shedding is happening on the side walls, which will be divergence-less. Therefore, no thrust is predicted by my present model.
Another thing I've been looking at is, that with a TM mode, the current flows along the length of the cavity, and resistance will go "UP" at the small end where the width of the conductor is narrower. With a TE mode, the current flows around the circumference of the cavity, and the resistance will go "UP" at the big end where the length of the conductor is longer. So there may be a change in direction based on TM vs TE modes, and how they apply. IMO, the TE mode will have thrust in a cylinder if one end has higher power dissipation than the other end, and that resistor is not transparent, allowing reflection off an identical copper end plate. In Zellurium's experiment, these conditions were not satisfied due to the relatively small losses of HDPE and the use of a TM mode shape.
That's my story and I'm satisfied with it. I prefer to work with magnetic fields, which should cause a much larger power dissipation, due to higher current flow.
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#1907 by Rodal on 21 Oct, 2016 23:48
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...Another thing I've been looking at is, that with a TM mode, the current flows along the length of the cavity, and resistance will go "UP" at the small end where the width of the conductor is narrower. With a TE mode, the current flows around the circumference of the cavity,...
?
There are no currents on the surfaces of a perfect metal conductor in a TE mode in an electromagnetically resonant cavity. The boundary condition is that any electrical component parallel to a perfect electrical conductor is zero.
http://www.antenna-theory.com/tutorial/electromagnetics/electric-field-boundary-conditions.php
For an imperfect conductor, the magnitude of such currents is extremely small (see Jackson). -
#1908 by dustinthewind on 22 Oct, 2016 00:10
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That is odd because I thought there were currents even for a superconductor. The reduced skin depth because the current has less resistance and can more effectively stop the light from passing through. That is the incoming light with its associated electric field encounter the cavity wall. The cavity wall currents are stimulated to generate a counter-current which negates/reduces the magnetic/electric field from existing beyond the wall. As a result the electric field generated from the accelerating currents makes an electric field inside that is equal and opposite to the incident electric field but the magnetic field inside (with opposite velocity v x E = B) constructively interferes. Hence the magnetic field parallel to the metal plate and the non-existent parallel electric field at resonant equilibrium....Another thing I've been looking at is, that with a TM mode, the current flows along the length of the cavity, and resistance will go "UP" at the small end where the width of the conductor is narrower. With a TE mode, the current flows around the circumference of the cavity,...
?
There are no currents on the surfaces of a perfect metal conductor in a TE mode in an electromagnetically resonant cavity. The boundary condition is that any electrical component parallel to a perfect electrical conductor is zero.
http://www.antenna-theory.com/tutorial/electromagnetics/electric-field-boundary-conditions.php
For an imperfect conductor, the magnitude of such currents is extremely small (see Jackson).
Well take for instance, "TE012 square of norm of magnetic field.png" at this link: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1352878#msg1352878 where the square of the magnetic field should be from currents moving in a circle at the base. Edit: Well, half from the light and the other half from the currents.
http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html#CURRENTSQuoteCurrents, heating and Q factors...
...The Ampère-Maxwell Law then tells us that the current enclosed by the loop, running through the wall perpendicular to the loop, will be proportional to the magnetic field just inside the cavity.
...In the TE modes, the currents run around the axis. -
#1909 by Rodal on 22 Oct, 2016 01:05
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...That is odd because I thought there were currents even for a superconductor. ...
Nothing odd about that, since superconductors are not perfect conductors:
http://www.differencebetween.com/difference-between-superconductor-and-vs-perfect-conductor/
https://en.wikipedia.org/wiki/Perfect_conductor
I never commingled the two in my post. -
#1910 by Rodal on 22 Oct, 2016 01:09
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...Well take for instance, "TE012 square of norm of magnetic field.png" at this link: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1352878#msg1352878 where the square of the magnetic field should be from currents moving in a circle at the base. Edit: Well, half from the light and the other half from the currents....
Not at all. The magnetic field being shown for TE012 is due to the continuity that must be satisfied at the interface boundary condition due to the magnetic field parallel to the surface. It is not at all due to the electric field! I do know because that came from my own exact solution , and I know what is being plotted and what BC are satisfied in my solution !
See Jackson, or any standard reference: http://home.sandiego.edu/~ekim/e171f00/lectures/boundary.pdf -
#1911 by Rodal on 22 Oct, 2016 01:13
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...
Greg Egan states clearly, in the section titled "Currents, heating and Q factors...", at the outset:
http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html#CURRENTSQuoteCurrents, heating and Q factors...
...The Ampère-Maxwell Law then tells us that the current enclosed by the loop, running through the wall perpendicular to the loop, will be proportional to the magnetic field just inside the cavity.
...In the TE modes, the currents run around the axis.Quote from: Greg EganSo far, we have treated the cavity walls as being made of a perfect conductor. In reality, of course, this will not be true, and the currents induced in the walls by the electromagnetic field will meet resistance
which is in agreement with my statement, and with Jackson:Quote from: RodalFor an imperfect conductor, the magnitude of such currents is extremely small (see Jackson).
From the same reference: http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.htmlQuote from: Greg EganTE modes
For the TE modes there is no electric field at the wall
which is exactly what I stated.
I carefully stated "perfect conductor" in my statements.
I carefully chose those words on purpose.
I never commingled perfect conductor with superconductor. i never commingled perfect conductor with imperfect conductor. I was careful to make separate statements about them. -
#1912 by Rodal on 22 Oct, 2016 01:25
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...That's my story and I'm satisfied with it. I prefer to work with magnetic fields, which should cause a much larger power dissipation, due to higher current flow.
Glad that you are satisfied with it.
My conclusion is that it is better to wait until your theory is developed to the point that it makes quantitative predictions, with a formula like Shawyer's, McCulloch's or Notsosureofit, that one can compare numerically with experimental results, otherwise the discussion becomes argumentative.
Looking forward to some numbers compared with experiment...
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#1913 by dustinthewind on 22 Oct, 2016 02:04
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The rest of the quote states that the current flow even with restive loss, and in equilibrium with the antenna is similar to a lossless material. So there should be just as much current as in a perfect conductor right?...
Greg Egan states clearly, in the section titled "Currents, heating and Q factors...", at the outset:
http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html#CURRENTSQuoteCurrents, heating and Q factors...
...The Ampère-Maxwell Law then tells us that the current enclosed by the loop, running through the wall perpendicular to the loop, will be proportional to the magnetic field just inside the cavity.
...In the TE modes, the currents run around the axis.Quote from: Greg EganSo far, we have treated the cavity walls as being made of a perfect conductor. In reality, of course, this will not be true, and the currents induced in the walls by the electromagnetic field will meet resistance
which is in agreement with my statement, and with Jackson:Quote from: RodalFor an imperfect conductor, the magnitude of such currents is extremely small (see Jackson).
From the same reference: http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.htmlQuote from: Greg EganTE modes
For the TE modes there is no electric field at the wall
which is exactly what I stated.
I carefully stated "perfect conductor" in my statements. I never commingled perfect conductor with superconductor. i never commingled perfect conductor with imperfect conductor. I was careful to make separate statements about them.Quote from: Greg EganSo far, we have treated the cavity walls as being made of a perfect conductor. In reality, of course, this will not be true, and the currents induced in the walls by the electromagnetic field will meet resistance, and dissipate energy as heat. If this rate of loss is not too great, and the energy in the cavity is constantly being replenished by a source of microwaves at the resonant frequency, then it’s reasonable to assume that the field geometry will closely resemble that of the lossless mode.
...
The resistance of a given area of the wall will be constant, so the power dissipated as heat per unit area will be proportional to the square of the current, and hence proportional to the square of the magnetic field strength at the wall.
I am sure there are differences between superconductors and perfect conductors but as far as resistance to current flow at low temperatures I am not seeing a significant difference.
Below quote from: http://www.differencebetween.com/difference-between-superconductor-and-vs-perfect-conductor/QuoteWhen the temperature of the material is decreased pass the critical temperature the resistance of the material abruptly drops to zero.
So very little resistance to maintaining an induced ac current in equilibrium right? Isn't a superconductor at low temperature w.r.t. resistance a conductor with out resistance, as is a perfect conductor with out resistance. I am not sure I am understanding what the important difference is beyond this.
...
What is the difference between Superconductor and Perfect Conductor?
• Superconductivity is a phenomenon occurring in real life, while perfect conductivity is an assumption made to ease the calculations.
• Perfect Conductors can have any temperature, but superconductors only exist below the critical temperature of the material.
I had difficulty finding the quote from Jackson at that particular link. -
#1914 by WarpTech on 22 Oct, 2016 02:07
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...Another thing I've been looking at is, that with a TM mode, the current flows along the length of the cavity, and resistance will go "UP" at the small end where the width of the conductor is narrower. With a TE mode, the current flows around the circumference of the cavity,...
?
There are no currents on the surfaces of a perfect metal conductor in a TE mode in an electromagnetically resonant cavity. The boundary condition is that any electrical component parallel to a perfect electrical conductor is zero.
http://www.antenna-theory.com/tutorial/electromagnetics/electric-field-boundary-conditions.php
For an imperfect conductor, the magnitude of such currents is extremely small (see Jackson).
This is a TE012 mode, and it would be about the same for TE013.
The H2 field will cause currents to flow in the skin depth, that's what prevents the field from passing through the copper. If there were no current induced in the copper, the magnetic field would escape to the outside. A superconductor will prevent the magnetic field from entering because that current flows as a phase shift of the single quantum wave function that describes the motion of the charge, due to the Meissner effect.
My expectation would be, superconductor on one end of the cylinder, and a high resistivity material like carbon steel on the other end, and make the cylinder as short as possible and still support a TE012 or TE013 mode.
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#1915 by Rodal on 22 Oct, 2016 02:15
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.I am not sure I am understanding what the important difference is beyond this.
I had difficulty finding the quote from Jackson at that particular link.
As stated in Wikipedia (https://en.wikipedia.org/wiki/Perfect_conductor) a perfect conductor is an idealized material.
This means that a perfect conductor is a material that does not exist in Nature.
Whether to use the BC for a perfect conductor to model a superconductor or an imperfect conductor is a matter of numerical approximation. Most closed-form solutions to electromagnetism problems employ the assumption of a perfect conductor.
If you don't use perfect conductor BC's many problems become unsolvable in closed form.
They can be solved numerically, but then nobody is sure about the validity of such numerical solutions, as to uniqueness ,convergence, etc. -
#1916 by Rodal on 22 Oct, 2016 02:20
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[The H2 field will cause currents to flow in the skin depth, that's what prevents the field from passing through the copper. If there were no current induced in the copper, the magnetic field would escape to the outside....
I carefully separated my statements between BC for perfect conductor and imperfect conductor.
You are stating << If there were no current induced in the copper,>> which means that you either failed to read my statements or you are confusing one with another.
You are commingling the two, and moreover not addressing the numerical magnitude of the currents, since I stated that in an imperfect conductor like copper the currents in TE modes are very small, and you make no numerical statement whatsoever.
I was trying to help by seeking a mathematical solution to the boundary condition problem for your equations.
If you are not going to use perfect conductor BC's to solve your equations, please show me your mathematical solution using BC's for imperfect conductors, using your equations.
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#1917 by dustinthewind on 22 Oct, 2016 02:32
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But if very little current is induced in a restive conductor (bottom quote) then that would imply no current is induced in a perfect conductor, right? Is it that the current is zero so zero energy is dissipated through the walls, or is it the resistance is zero so zero energy is dissipated through the walls. Or is it the extremely small current as a relative term that is throwing me off. Or is resistance so great that current is reduced to extremely small but then I don't understand Greg Egan who is saying the below quote.[The H2 field will cause currents to flow in the skin depth, that's what prevents the field from passing through the copper. If there were no current induced in the copper, the magnetic field would escape to the outside....
I carefully separated my statements between BC for perfect conductor and imperfect conductor.
You are stating << If there were no current induced in the copper,>> which means that you either failed to read my statements or you are commingling one with another.
You are commingling the two, and moreover not addressing the numerical magnitude of the currents, since I stated that in an imperfect conductor like copper the currents in TE modes are very small, and you make no numerical statement whatsoever.Quote from: Greg EganSo far, we have treated the cavity walls as being made of a perfect conductor. In reality, of course, this will not be true, and the currents induced in the walls by the electromagnetic field will meet resistance, and dissipate energy as heat. If this rate of loss is not too great, and the energy in the cavity is constantly being replenished by a source of microwaves at the resonant frequency, then it’s reasonable to assume that the field geometry will closely resemble that of the lossless mode.
...
The resistance of a given area of the wall will be constant, so the power dissipated as heat per unit area will be proportional to the square of the current, and hence proportional to the square of the magnetic field strength at the wall....
There are no currents on the surfaces of a perfect metal conductor in a TE mode in an electromagnetically resonant cavity.
...
For an imperfect conductor, the magnitude of such currents is extremely small (see Jackson). -
#1918 by WarpTech on 22 Oct, 2016 02:35
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[The H2 field will cause currents to flow in the skin depth, that's what prevents the field from passing through the copper. If there were no current induced in the copper, the magnetic field would escape to the outside....
I carefully separated my statements between BC for perfect conductor and imperfect conductor.
You are stating << If there were no current induced in the copper,>> which means that you either failed to read my statements or you are confusing one with another.
You are commingling the two, and moreover not addressing the numerical magnitude of the currents, since I stated that in an imperfect conductor like copper the currents in TE modes are very small, and you make no numerical statement whatsoever.
I was trying to help by looking for a mathematical solution to the boundary condition problem for your equations.
If you are not going to use perfect conductor BC's to solve your equations, please show me your mathematical solution using BC's for imperfect conductors, using your equations.
Sorry, that's because in my mind, it is trivial to say that once the cylinder has reached it's maximum capacity at the loaded Q value;
Pin*Q
Then
Pd = Pin
It is how that power dissipation is mapped to the cylinder that is represented by my equations, and the properties of the material. So if the losses are "small", it just means it has a higher Q!
It's Friday night, I'll do math another time.
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#1919 by Rodal on 22 Oct, 2016 02:40
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...
I was trying to help by seeking a mathematical closed-form solution to his theory. This is only possible in most cases assuming BC's for perfect conductors, and not for imperfect conductors.
But if very little current is induced in a restive conductor (bottom quote) then that would imply no current is induced in a perfect conductor, right? Is it that the current is zero so zero energy is dissipated through the walls, or is it the resistance is zero so zero energy is dissipated through the walls. Or is it the very little as a relative term that is throwing me off.
...
Without a mathematical solution, the discussion becomes wordy, with shades of meaning, which I am too tired to pursue.
When Todd's theory makes numerical predictions that can be compared with experiments we can continue such a discussion.
to be able to have a discussion. 