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#1560 by X_RaY on 10 Jan, 2016 13:42
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What are the dimensions of the dielectric insert and how its implemented in the spreadsheet?2) reports of experiments showing what happens to the anomalous force when the EM Drive has a small diameter that is below the cut-off condition. On the contrary, all the NASA tests are for an EM Drive that has a small diameter that is below the cut-off condition (as pointed out by TT), and on top of that they use a dielectric insert which lowers the natural frequency even further. Yet, NASA reports an anomalous thrust. Now, somebody could answer "well that's why NASA reports thrust orders of magnitude lower than Shawyer and Yang", but there are problems with that explanation:
The use of a dielectric in the small end lowers the cutoff freq such that, assuming a dielectric constant of 2.2, the small end on the EW copper frustum is operating above cutoff and thus there is no apparent small end cutoff affecting the EW measured thrust.
I can write the opposite. Rather than argue with words, let's be constructive. Please show us the mathematical equation and the source of the equation you are using to support your conclusion!
!
The equations are standard microwave engineering equations to calculate the cutoff and guide wavelength in a circular waveguide. But I think you already do know that.
Next you will say those equations don't apply as this is not a constant diameter waveguide.
I will reply saying that is your unproven theory and we will go around and round.
Bottom line is the DIY EmDrive builders are here because of Roger Shawyer and his invention. His advice is that if you calc the small end cutoff, as you would a constant diameter waveguide and it is at or below cutoff, the small end needs to be enlarged so it is operating above cutoff or you will not generate any thrust. That advise is backed up if you use Roger's Df equation and calc the Df of the frustum as attached. Maybe you should try it and be sure to use a cutoff equation that includes the Dielectric Constant when calculating a small end cutoff with a dielectric. Hint: the cutoff freq drops when you add in the dielectric.
My spreadsheet has been updated to support dielectrics inside the frustum. Attached see runs with no dielectric (small end in cutoff) and a dielectric (small end not in cutoff).
I ask this because the shift of the wavelength inside a closed cavity depends on the given waveform/mode, its not simple λ/√μ_r*ε_r like in an open wave guide, it depends on the local strength of the E-field. The boundary conditions(*) of the end plate is important since we are dealing with standing waves and may be there is no full half wavelength inside of the dielectric material. Therefore the calculations for a dielectric plate, not equal to n*lambda/4 inside the cavity and with respect to the mode and boundary conditions will much more complicated. At that point its better to use 3D FDTD or FEM. IMHO
(*)I think of a dielectric placed at the surface of one of the end plates
http://forum.nasaspaceflight.com/index.php?topic=38577.msg1455560#msg1455560
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#1561 by ludkokanta on 10 Jan, 2016 13:49
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Here I analyse the possibility of a propellentless thrust along with a statistical mechanics analysis of why it "would" work. Loosely related to "EM Drive".
http://www.failed-experiments.com/viewtopic.php?f=7&t=4 -
#1562 by glennfish on 10 Jan, 2016 14:19
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Regardless of whether the frustum is oriented East or West the mid-points of pendulum oscillation appear to follow the same pattern during the run (attached).
Those oscillations are pretty nasty and swamping all but the strongest signals. If you can figure out how to dampen them, that would be strongly advisable. The 2nd attachment shows a best fit model of the oscillations overlayed on your channel 1 data.
Here's an approximation of your oscillation function for BigE-West-2A-Battery-Run2
oscillation = sin(millisecond_value*.001860761 - 1999.89)*0.063842732 - 0.21635
Also attached is the residual mapping with channel 2 & channel 3 overlayed
On the residuals scale you can see that I couldn't take out all of the oscillation, the convergence values went to underflow before they converged. I need a better convergence algorithm.
The random variations appear to be > 30% of the residual range, and over a large # of runs, a signal, if present, might emerge, however, the oscillation is really nasty. It's difficult to to determine the SD on the sample channel if the oscillation were totally damped.
With the oscillation in there swamping everything, I frankly couldn't confirm or deny that you have anything or nothing other than a noisy oscillator that responds well when you apply the high-voltage.
For the reddit trolls, this is neither a positive nor a null result. It's a characterization of a test environment showing measurement concerns prior to a test.
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#1563 by TheTraveller on 10 Jan, 2016 14:52
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What are the dimensions of the dielectric insert and how its implemented in the spreadsheet?
I ask this because the shift of the wavelength inside a closed cavity depends on the given waveform/mode, its not simple λ*√με like in an open wave guide, it depends on the local strength of E-field. The boundary conditions(*) of the end plate is important since we are dealing with standing waves and may be there is no full half wavelength inside of the dielectric material. Therefore the calculations for a dielectric plate, not equal to n*lambda/4 inside the cavity and with respect to the mode and boundary conditions will much more complicated. At that point its better to use 3D FDTD or FEM. IMHO
(*)I think of a dielectric placed at the surface of one of the end plates
http://forum.nasaspaceflight.com/index.php?topic=38577.msg1455560#msg1455560
Hi XRay,
Thanks for the reply.
At this point, the entire frustum is filled with the dielectric, so no air to dielectric interface to deal with..
If you or anyone else can calc the non & dielectric small end cutoff & guide wavelengths for either the EW or RfPlumber frustum please do so. Would love to see the calculations and the data.
Please remember this discussion is about Roger's advised method to ensure the small end is not operating in cutoff. Roger never said his method calculated the exact cutoff data, more it was a simple method to keep a DIYer from building a frustum with too narrow a small end diameter.
Do you intend to share your spreadsheet?
Phil -
#1564 by Rodal on 10 Jan, 2016 14:59
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...Those oscillations are pretty nasty and swamping all but the strongest signals. ... this is neither a positive nor a null result. It's a characterization of a test environment showing measurement concerns prior to a test.
Sorry for the elipsis (if this is somehow putting the quotation out of context which is NOT my intention).
Rather I would like to agree on a common language and what and how it should be entered on the Exp Results wiki table.
Taking this definition from WikipediaQuoteIn science, a null result is a result without the expected content: that is, the proposed result is absent. It is an experimental outcome which does not show an otherwise expected effect.
I guess, that following Shawyer as expressed by TheTraveller, if no thrust was expected because it was below cut-off (*) one could say that the result was not unexpected and hence not null.
On the other hand, if by Null is meant no evidence of thrust , then if the oscillations swamp any signal, there is no evidence, and the result could be classified as Null, or inconclusive.
So the use of the word "Null" is not explicit enough as it implies some kind of expectation.
The Experimental Results wiki table avoids this (need to make the expectation explicit) by entering a number for thrust instead of categorizing results as positive or null.
The remaining issue however, is that sometimes an average result is entered for thrust without the experimental variation being entered (hence no measure of uncertainty) and the other issue is that certain DIY results are entered by the tester himself while others (Iulian Berca for example) are not, including institutional results. There is also controversy regarding how results should be categorized (some people think that certain results are due to Lorentz' force artifacts, or to thermal artifacts and others wrote that results like Berca's were due to effects on the digital scale he used).
What would you propose to enter in the Experimental Results wiki table for RFPlumber's present results ?
I had proposed that RFPlumber should enter his own results, as rfmwguy entered his own results...
________
(*) whatever that means for a closed resonant cavity (using cut-off equations that are known to be used only for open waveguides does not appear consistent to me) -
#1565 by glennfish on 10 Jan, 2016 15:10
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...Those oscillations are pretty nasty and swamping all but the strongest signals. ... this is neither a positive nor a null result. It's a characterization of a test environment showing measurement concerns prior to a test.
Sorry for the elipsis (if this is somehow putting the quotation out of context which is NOT my intention).
Rather I would like to agree on a common language and what and how it should be entered on the Exp Results wiki table.
Taking this definition from WikipediaQuoteIn science, a null result is a result without the expected content: that is, the proposed result is absent. It is an experimental outcome which does not show an otherwise expected effect.
I guess, that following Shawyer as expressed by TheTraveller, if no thrust was expected because it was below cut-off one could say that the result was not unexpected and hence not null.
On the other hand, if by Null is meant no evidence of thrust , then if the oscillations swamp any signal, there is no evidence, and the result would be Null.
So the use of the word "Null" is not explicit enough as it implies some kind of expectation.
The Experimental Results wiki table avoids this by entering a number for thrust instead of categorizing results as positive or null.
The remaining issue however, is that sometimes an average result is entered for thrust without the experimental variation being entered (hence no measure of uncertainty) and the other issue is that certain DIY results are entered by the tester himself while others (Iulian Berca for example) are not, including institutional results.
What would you propose to enter in the Experimental Results wiki table for the RFPlumber's present results ?
LOL I accept your definitions, with the following caveat, can a test result in a null result if the test environment does not support any result?
In this case I'd be curious to know if this constitutes a test of EM drive, or if it constitutes a test of a test environment?
If it's a test of the EM drive, then I accept it as a null result.
If it's a test of a test environment, then I'd characterize it as a resounding success of showing measurement issues to resolve prior to testing an EM drive.
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#1566 by TheTraveller on 10 Jan, 2016 15:24
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Roger's 0.82 cutoff rule
It is real simple to calc your TE01x frustum small end cutoff lower freq from Roger's cutoff rule:
Small end TE01x cutoff freq = c / (0.82 * small end diam in mtrs)
No spreadsheet required for this one. Only works for TE01x mode excitation. -
#1567 by X_RaY on 10 Jan, 2016 16:18
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OK, If the whole cavity is filled with the dielectric it's nearly the same situation as to use a larger resonator. In this case I agree to use simple λ/√μ_r*ε_r.What are the dimensions of the dielectric insert and how its implemented in the spreadsheet?
I ask this because the shift of the wavelength inside a closed cavity depends on the given waveform/mode, its not simple λ/√μ_r*ε_r like in an open wave guide, it depends on the local strength of E-field. The boundary conditions(*) of the end plate is important since we are dealing with standing waves and may be there is no full half wavelength inside of the dielectric material. Therefore the calculations for a dielectric plate, not equal to n*lambda/4 inside the cavity and with respect to the mode and boundary conditions will much more complicated. At that point its better to use 3D FDTD or FEM. IMHO
(*)I think of a dielectric placed at the surface of one of the end plates
http://forum.nasaspaceflight.com/index.php?topic=38577.msg1455560#msg1455560
Hi XRay,
Thanks for the reply.
At this point, the entire frustum is filled with the dielectric, so no air to dielectric interface to deal with..
If you or anyone else can calc the non & dielectric small end cutoff & guide wavelengths for either the EW or RfPlumber frustum please do so. Would love to see the calculations and the data.
Please remember this discussion is about Roger's advised method to ensure the small end is not operating in cutoff. Roger never said his method calculated the exact cutoff data, more it was a simple method to keep a DIYer from building a frustum with too narrow a small end diameter.
Do you intend to share your spreadsheet?
Phil
As I wrote the situation is more complicated if the cavity is partial filled with a dielectric. One can find approximations for a spreadsheet for any mode deriving formula from the Maxwell equations (which makes me some headache for a while last year) and I came to the conclusion its better to use field calculation programs for such a situation. Of course I could do this using EMPro or Comsol, but at the moment I don't have the time for this simulation (maybe sometime in the future).
At the moment I don't like to publish the spreadsheet.
For the cut off frequency I use:
fc = Xmn Λ X'mn / 2πa√με
Wherein the radius of the waveguide is "a", "fc" is the cutoff frequency and "μ" & "ε" are the natural not the relative values (permeability and permittivity in vacuum) ! For dielectrics, one have to add the relative factor under the square root.
"Xmx Λ Xmn" is the mode dependent Hankel/Bessel value.
But again this is a formula for cylindrical wave guides not for conical cavity resonators! -
#1568 by Rodal on 10 Jan, 2016 16:35
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OK, If the whole cavity is filled with the dielectric it's nearly the same situation as to use a larger resonator. In this case I agree to use simple λ*√με.What are the dimensions of the dielectric insert and how its implemented in the spreadsheet?
I ask this because the shift of the wavelength inside a closed cavity depends on the given waveform/mode, its not simple λ*√με like in an open wave guide, it depends on the local strength of E-field. The boundary conditions(*) of the end plate is important since we are dealing with standing waves and may be there is no full half wavelength inside of the dielectric material. Therefore the calculations for a dielectric plate, not equal to n*lambda/4 inside the cavity and with respect to the mode and boundary conditions will much more complicated. At that point its better to use 3D FDTD or FEM. IMHO
(*)I think of a dielectric placed at the surface of one of the end plates
http://forum.nasaspaceflight.com/index.php?topic=38577.msg1455560#msg1455560
Hi XRay,
Thanks for the reply.
At this point, the entire frustum is filled with the dielectric, so no air to dielectric interface to deal with..
If you or anyone else can calc the non & dielectric small end cutoff & guide wavelengths for either the EW or RfPlumber frustum please do so. Would love to see the calculations and the data.
Please remember this discussion is about Roger's advised method to ensure the small end is not operating in cutoff. Roger never said his method calculated the exact cutoff data, more it was a simple method to keep a DIYer from building a frustum with too narrow a small end diameter.
Do you intend to share your spreadsheet?
Phil
As I wrote the situation is more complicated if the cavity is partial filled with a dielectric. One can find approximations for a spreadsheet for any mode solve Maxwell equations (which makes me some headache for a while last year) and I came to the conclusion its better to use field calculation programs for such a situation. Of course I could do this using EMPro or Comsol, but at the moment I don't have the time for this simulation (maybe sometime in the future).
At the moment I don't like to publish the spreadsheet.
For the cut off frequency I use:
fc = Xmn Λ X'mn / 2πa√με
But again this is a formula for cylindrical waveguides not for conical cavity resonators.
So someone is using a cut-off formula which is known to apply only for an open waveguide, whereby the waveguide is completely filled with a dielectric medium with relative permittivity 2.2 (going from memory here) opening into what? What happens at the opening? How is this related to a closed resonant cavity, with copper walls, whereby there are two HDPE disks inserted at the small end, and there is no opening?
What happened to the difference we learnt at school between an open waveguide and a closed cavity?
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#1569 by aero on 10 Jan, 2016 17:23
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...
I have a machine at work dedicated to running big/long engineering simulations - a quad core XEON with 16GB RAM. I just installed VirtualBox on it and have the Ubuntu VM with meep on an SD card to take in and install on it. It can run as long as we like - I don't care if it runs a meep run for a few weeks, and nobody but me uses it
This is a case where more computer, or longer running times could be helpful. Perhaps Tajmar's cavity model will give an opportunity to test higher resolution meep/Harminv runs compared to published data. . Let me know what you'd like to run on it.
I am not sure if this is correct... but I don't think you're ever going to get great cpu performance out of virtualbox. As I understand virtualbox doesn't really even talk to the hardware of your computer, but its cpu is simulated.
For simulation purposes I'm almost positive you want to install a native copy of Linux, not just boot up on a stick or any such thing.
I have made timing runs with different installs of meep running on my computer. Using a native copy of Linux (Ubuntu) with meep compiled from source as the reference, I find that:
Meep running on VMBox takes about 3% overhead
Meep running on a pre-compiled binary download takes over 20% overhead.
Yes, there is overhead associated with VMBox, but it is a surprisingly small overhead. -
#1570 by rfmwguy on 10 Jan, 2016 19:33
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Ethan Bearman - KBYR emdrve podcast 11/4/15
http://podcast.ethanbearman.com/podpress_trac/web/2942/0/KBYR2015-11-04.mp3 -
#1571 by X_RaY on 10 Jan, 2016 20:24
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Roger's 0.82 cutoff rule
@TT if you uses the geometric average "BD+SD/2" instead of the smallest diameter for the cut off calculation you'll see that the cutoff frequency is always below the resonant frequency (propagation condition of a waveguide made of such open ended cavities) as long as one of the ends is still above the cut off diameter of a circular wave guide. (That does not mean that the wave propagates like in a waveguide with the middle diameter over large distances, I have tried this please see here https://forum.nasaspaceflight.com/index.php?topic=37642.msg1403569#msg1403569 , the same is true if every secound cavity ,what forms the waveguide is turned by 180 degree.)
It is real simple to calc your TE01x frustum small end cutoff lower freq from Roger's cutoff rule:
Small end TE01x cutoff freq = c / (0.82 * small end diam in mtrs)
No spreadsheet required for this one. Only works for TE01x mode excitation.
This viewpoint matches the fact that the cavity still resonates although one end is clear under the cutoff diameter for a circular wave guide.
EDIT
I have got no idea whatever that means for any thrust generation in this situation, I wait for results from SeeShells as everyone else here.
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#1572 by SteveD on 10 Jan, 2016 20:56
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It is not a problem of duration of "effect", but of frame invariance of the effect. 1s of thrust at .01µN/kW is already too much not to apparently break CoE if such thrust/power is possible regardless of absolute velocity (relative to what ?). Said otherwise, the apparent problem is with a "frame agnostic" EMdrive : a device that with the same given power will have the same thrust (an accelerometer will record the same acceleration for the driven craft) whether or not it has previously been subject to some arbitrary deltaV by conventional mean. If I can send a .01µN/kW for 30s (as measured in the lab) EMdrive on a conventional rocket (say, a fission fragment rocket) at 200km/s, power it for 30s and still measure .01µN/kW onboard, then I have a CoE issue. I'd have a hard time actually extracting useful energy from that, but that doesn't prevent the problem within known frameworks. Happen to be, sun is orbiting galaxy at about 200km/s. On a 12 hours span, the same experiment (depending on latitude...) would experience a differential of up to 400km/s wrt. its orientation to some galactocentric frame. So in a sense we can say that EMdrives have already (are) tested at such velocities... even while standing almost still on the bench and doing their µm displacement against balances. Can you find a natural frame with lower relative velocities that still makes sense for deep space application ?
BTW, while a "frame agnostic" EMdrive can in principle (>3.33µN/kW) and in practice (approx. >1N/kW) reach break-even and be used as a generator, the same device can be used to "erase" energy, i.e. act as a net sink of power, that apparently (in waste heat) radiates less than spent power. See how ?
You're thinking too small. Your problem is that the EMDrive is mass invariant. If KE=1/2mv^2 then, at some point, if the efficiency is higher than that of a photon rocket, then you're going to need to lose some mass to keep the books balanced.
So what about a supernova? We've been thinking about this problem in terrestrial scales. I note that the acceleration formulas for photonic systems are as follows:
Reflection of photons (a photon rocket that reuses its photons) can allow for a constant -- mass invariant -- acceleration. So what happens when a star blows off some energy on a stellar scale and part of that energy gets stuck in a, temporarily, resonant system? Various stellar phenomena can certainly put out enough energy to accelerate an object (or the plasma that use to be the object) to the point that this becomes an issue in short order.
If you can build a scenario where a naturally occurring phenomena can cause this problem, then odds are it's happened somewhere, at some time, within the visible universe. That would suggest that either 1. the underlying equation is fundamentally wrong in its assumptions or 2. some form of relativistic force exists causing a virtual loss of mass, keeping the equations in balance. Once you accept either of these as true, it follows that odd effects from photons in a copper can should not be dismissed out of hand. -
#1573 by RotoSequence on 10 Jan, 2016 22:19
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On the hypothesis that EM drive test results are not the product of thrust, what interactions take place to distinguish the different degrees of apparent effect that have been seen in and out of resonance?
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#1574 by SeeShells on 10 Jan, 2016 23:48
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Roger's 0.82 cutoff rule
@TT if you uses the geometric average "BD+SD/2" instead of the smallest diameter for the cut off calculation you'll see that the cutoff frequency is always below the resonant frequency (propagation condition of a waveguide made of such open ended cavities) as long as one of the ends is still above the cut off diameter of a circular wave guide. (That does not mean that the wave propagates like in a waveguide with the middle diameter over large distances, I have tried this please see here https://forum.nasaspaceflight.com/index.php?topic=37642.msg1403569#msg1403569 , the same is true if every secound cavity ,what forms the waveguide is turned by 180 degree.)
It is real simple to calc your TE01x frustum small end cutoff lower freq from Roger's cutoff rule:
Small end TE01x cutoff freq = c / (0.82 * small end diam in mtrs)
No spreadsheet required for this one. Only works for TE01x mode excitation.
This viewpoint matches the fact that the cavity still resonates although one end is clear under the cutoff diameter for a circular wave guide.
EDIT
I have got no idea whatever that means for any thrust generation in this situation, I wait for results from SeeShells as everyone else here.
Working hard at it to do just that.
Shell
PS: I've always like that idea of stacking. -
#1575 by TheTraveller on 11 Jan, 2016 00:03
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Roger's 0.82 cutoff rule
@TT if you uses the geometric average "BD+SD/2" instead of the smallest diameter for the cut off calculation you'll see that the cutoff frequency is always below the resonant frequency (propagation condition of a waveguide made of such open ended cavities) as long as one of the ends is still above the cut off diameter of a circular wave guide. (That does not mean that the wave propagates like in a waveguide with the middle diameter over large distances, I have tried this please see here https://forum.nasaspaceflight.com/index.php?topic=37642.msg1403569#msg1403569 , the same is true if every secound cavity ,what forms the waveguide is turned by 180 degree.)
It is real simple to calc your TE01x frustum small end cutoff lower freq from Roger's cutoff rule:
Small end TE01x cutoff freq = c / (0.82 * small end diam in mtrs)
No spreadsheet required for this one. Only works for TE01x mode excitation.
This viewpoint matches the fact that the cavity still resonates although one end is clear under the cutoff diameter for a circular wave guide.
EDIT
I have got no idea whatever that means for any thrust generation in this situation, I wait for results from SeeShells as everyone else here.
As I said this is a Roger quick tip, rule of thumb, to calc the external minimum freq that will safely operate the small end above cutoff and allow thrust to develop.
All my frustum designs comply with it. -
#1576 by TheTraveller on 11 Jan, 2016 00:10
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]So someone is using a cut-off formula which is known to apply only for an open waveguide, whereby the waveguide is completely filled with a dielectric medium with relative permittivity 2.2 (going from memory here) opening into what? What happens at the opening? How is this related to a closed resonant cavity, with copper walls, whereby there are two HDPE disks inserted at the small end, and there is no opening?
What happened to the difference we learnt at school between an open waveguide and a closed cavity?
Waiting for your cutoff & guide wavelength small end calcs for either the EW copper frustum or that of RfPlumber with and without a dielectric.
Serious, I'm interested to see your results and how you obtained them. You can do this? -
#1577 by oliverio on 11 Jan, 2016 00:21
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@rodal:
Would you agree with the following? If you consider the whole photonic field to be oscillating, it is sort of the case that the field will over one half of a period pass through the closed waveguide in a similar fashion to what the same mode might look like in an open waveguide for only that small timeslice.
I recognize the obvious fact that mode degeneration will, in every case, lead to resonating emdrive energy in the smaller side of a frustrum that is smaller than a cutoff diameter for an open waveguide.
But, if for tiny periods, the same effect of an open waveguide is seen by the oscillating field, one could understand why there might be some sort of pseudo cutoff diameter.
That said I could be very wrong that a closed waveguide appears to an oscillating field to be open for half its period. -
#1578 by Rodal on 11 Jan, 2016 00:31
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]So someone is using a cut-off formula which is known to apply only for an open waveguide, whereby the waveguide is completely filled with a dielectric medium with relative permittivity 2.2 (going from memory here) opening into what? What happens at the opening? How is this related to a closed resonant cavity, with copper walls, whereby there are two HDPE disks inserted at the small end, and there is no opening?
What happened to the difference we learnt at school between an open waveguide and a closed cavity?
Waiting for your cutoff & guide wavelength small end calcs for either the EW copper frustum or that of RfPlumber with and without a dielectric.
Serious, I'm interested to see your results and how you obtained them. You can do this?
Are you telling us that Shawyer is asking us to believe these things that not only go against what we learned at school, but that are moreover mutually self-contradictory ?:
1) That there is no radiation pressure on the lateral conical walls of the frustum of a cone cavity, and that only the forces on the end plates count ?
and
2) That the lateral walls of the closed cavity don't count for cut-off of resonance, and that therefore you can use for a closed cavity the cut-off condition for open waveguides with no end plates ?
Not only these two things go against what we learned at school, and against the experiments we performed at school, but they are self-contradictory: according to #1 he is asking us to ignore the lateral walls and only count the end-plates, but according to #2 he is asking us to ignore the end-plates and consider the resonant cavity as having only lateral walls.
I have not seen any experimental data from Shawyer supporting either of these two things, I'll continue to use the equations I learned at school and have worked so well in countless of experiments, including huge cavities and waveguides used at particle accelerators worldwide. Others are free to believe that all the past textbooks, Nobel Prizes (several of them worked on resonant cavities) and experiments are wrong, if they wish to do so.
___________________
Now seriously, regarding calculations with dielectric inserts: this of course can be done numerically, with Finite Element Analysis, Finite Difference, Boundary Element Method, etc. But also there are some closed form solutions for particular cases (some of them can be found in textbooks like Collin's).
With Notsosureofit we worked out a solution for dispersion for a cylindrical waveguide with a dielectric asymmetrically placed inside the cavity, regarding the Notsosureofit hypothesis, in previous threads.
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#1579 by TheTraveller on 11 Jan, 2016 00:37
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]So someone is using a cut-off formula which is known to apply only for an open waveguide, whereby the waveguide is completely filled with a dielectric medium with relative permittivity 2.2 (going from memory here) opening into what? What happens at the opening? How is this related to a closed resonant cavity, with copper walls, whereby there are two HDPE disks inserted at the small end, and there is no opening?
What happened to the difference we learnt at school between an open waveguide and a closed cavity?
Waiting for your cutoff & guide wavelength small end calcs for either the EW copper frustum or that of RfPlumber with and without a dielectric.
Serious, I'm interested to see your results and how you obtained them. You can do this?
Are you telling us that Shawyer is asking us to believe these things that not only go against what we learned at school, but that are moreover mutually self-contradictory ?:
1) That there is no radiation pressure on the lateral conical walls of the frustum of a cone cavity, and that only the forces on the end plates count ?
and
2) That the lateral walls of the closed cavity don't count, and that therefore you can use the cut-off condition for open waveguides with no end plates ?
Not only these two things go against what we learned at school, and against the experiments we performed at school, but they are self-contradictory: according to #1 he is asking us to ignore the lateral walls and only count the end-plates, but according to #2 he is asking us to ignore the end-plates and consider the resonant cavity as having only lateral walls.
Since I have not seen any experimental data from Shawyer supporting either of these two things, I'll continue to trust what I learnt at school and has worked so well in countless of experiments, including huge cavities and waveguides used at particle accelerators worldwide. Others are free to believe that all the past textbooks and experiments are wrong, if they wish.
___________________
Now seriously, regarding calculations with dielectric inserts: this of course can be done numerically. But also there are some closed form solutions for particular cases (some of them can be found in textbooks like Collin's).
With Notsosureofit we worked out a solution for dispersion for a cylindrical waveguide with a dielectric asymmetrically placed in the cavity.
And the small end cutoff & guide wavelengths for either the EW or RfPlumber frustums with & without dielectrics are?
If you ever do actually build a frustum, your choice to follow Roger's advise or not. Please let us know how that goes.
In this case I'd be curious to know if this constitutes a test of EM drive, or if it constitutes a test of a test environment?