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#1860 by Rodal on 17 Jan, 2016 23:02
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I figure this may be of interest to @Rodal (you dig Finite element analysis) and @Notsosureofit (your hypothesis featuring an accelerated frame of reference).
http://people.clarkson.edu/~nanosci/jse/B/inpress/vaish.pdf
Yes, that is exactly what the Notsosureofit hypothesis is based on.
They give an example with a very coarse finite element mesh, including a figure showing significant deviation of the eigenvalue with and without gravitational effects vs. aspect ratio of the waveguide.
That looks amazing and makes one very interested in the results.
However:QuoteAssume that the wave guide is rectangular and the gravitational potential
U is generated by a piece of point matter located at the point (R, 0). Then,
U(x, y) = − GM ((x − R)2 + y2)1/2
where G is the gravitational constant and M is mass of the earth. The
waveguide cross section is [0, A] × [0, B] where R >> A, B. For example,
we can take A = 3, B = 2 and R = 10 and formulate the finite element
technique for this problem.
Side Note: the values of 2, and 3 are not much smaller than 10; and hence their numerical example does NOT satisfy their conditions of R>>A,B (this does not bode well for their paper
)
So in other words, the spacecraft is at a distance of 10 from a point mass with the mass of the Earth, and the waveguide dimensions are 2 and 3?
The average distance to the center of the Earth is about 3,959 miles
Suppose a spacecraft 100 miles above the surface of the Earth, it would be at 4059 miles.
If that distance is "10" then "3" is 1200 miles and "2" is 800 miles (give or take a few miles, which doesn't matter when your waveguide is over 1,000 miles long on one side).
So they are modeling a waveguide that has a cross-section of 1200 by 800 miles, near the Earth.
Obviously then we have to chose a point much farther away but you get my point ...
Say that your waveguide is at a distance from the Earth like the Moon: 238,900 miles, that is "10"
then your waveguide has a cross-section of 71,700 miles by 47,800 miles. That is, ahem, an EM Drive that is larger than the EM Drives presently tested...
So the example sounds like nonsense, unless they are calculating the effect of gravitation from a black hole on a spacecraft having a waveguide near it.
Once again showing that the fields in an EM Drive with dimensions of barely a couple of feet in diameter are way too small to show any significant gravitational effects on an EM Drive at 2.45 GHz having barely 1 KW power into it... unless you happen to fly a really big EM Drive close to a black hole
Anyway, according to the paper, for General Relativity to have an effect on the electromagnetic field, and hence for GR to explain the anomalous forces claimed for the EM Drive: this example shows that you have to make your EM Drive with dimensions comparable to your distance to a large gravitational mass. Make your EM Drive as large as possible. -
#1861 by Notsosureofit on 18 Jan, 2016 00:27
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Remember that 10^-17 ?
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#1862 by Rodal on 18 Jan, 2016 00:30
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Remember that 10^-17 ?
G = 6.674×10^(−11) N⋅m^2/kg^2
We can make it much bigger by expressing it in terms of parsecs, km/s and solar masses:
The Sun has a Schwarzschild radius of approximately 3.0 km (1.9 mi), whereas Earth's is only about 9.0 mm -
#1863 by SeeShells on 18 Jan, 2016 00:52
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I'll just have to make sure I fix that so I can have something other than questionable models, this includes FEKO and meep....
I had little control over what was modeled or presented only was asked for the dimensions of the cavity in the FEKO model.
I've requested a version of FEKO to run and confirm my cavity actions. Waiting for a email from them to be able to download.
I stated I was after TE012.
As far as meep goes other than pulling out the CSV file and log files you have little idea of values.
Didn't you and aero come up with a agreed on X, Y, Z coordinate system where the frustum as viewed sitting flat on the large end plate? See attachment
According to the convention we had agreed, which is the same as the picture you show, then SEZ_vueEx should be the Electric field in the direction parallel to the end plate, and measured at the small end (SE).
But then, the SEZ_vueEx (and other) images shown in the Meep files you linked to do not make any sense. No mode shape can result in an electric field parallel to a metal, measured at the metal.
BOUNDARY CONDITION FOR AN ELECTRIC FIELD: The tangential Electric Field will always be zero on a metal surface. This is because the free charge will swim around and cancel it out, simply by the attractive nature of charge.
Either the Meep model is not correctly modeling the boundary conditions, in which case the whole Meep model is complete nonsense, or SEZ_vueEx has a value very close to zero. If SEZ_vueEx is showing a very small number, there is no way to tell what the images mean: they have no numerical field values and there is no way to tell which images are significant. From your answer << have little idea of values.>> you cannot tell either, since the Meep files do not show the numerical fields.
So, there is no way to tell from these Meep images which ones make any sense, except to tell that SEZ_vueEx for example does not make any sense and therefore there is no way to tell what is going on, much less to tell what mode shape is present.
Shell -
#1864 by Rodal on 18 Jan, 2016 01:13
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I snagged this image from the reference:
Please take into account that the internal height is adjusted internally with a screw prior to testing, so the internal height at resonance can be much smaller than the actual exterior height of the cavity
https://tu-dresden.de/die_tu_dresden/fakultaeten/fakultaet_maschinenwesen/ilr/rfs/forschung/folder.2007-08-21.5231434330/ag_raumfahrtantriebe/JPC%20-%20Direct%20Thrust%20Measurements%20of%20an%20EM%20Drive%20and%20Evaluation%20of%20Possible%20Side-Effects.pdf
and counted pixels using a tool new to me, so the count was not very repeatable, but these are the numbers:
Small diameter = 134 px
big diameter = 184 - 187 px
slant height = 142 - 147 px
Using small diameter, Ds = 77 mm and calculating by ratio, gives big diameter ~ 105.7 mm to 107.5 mm and slant height ~ 81.6 mm to 84.5 mm
I don't know what that proves, but as a sanity check it doesn't seem all that close to anyone's numbers.
Yes, consider that it can be shorter. But it can't be longer which kind of rules out axialLength = 2*(0.0686 - 0.003) meter (since the wall thickness is 3 mm)
Which means that
The dimensions I calculated a long time ago here: https://forum.nasaspaceflight.com/index.php?topic=39004.msg1477474#msg1477474
Axial Length = 0.100842 m = 0.735*2*0.0686 m
are very much in play !
Welll... The next thing I did was to snag the COMSOL image from the reference, attached. It seems to me that this drawing should be to scale and should show the interior dimensions. Using the pixel measuring method, I find:
px px ref. 77 77
sd 244 241.9 77 77 ave
bd 341.1 339.2 107.6422131148 107.0426229508 107.3424180328
shi 211.4 206.8 66.712295082 65.2606557377 65.9864754098
shi 212.5 208.1 67.0594262295 65.6709016393 66.3651639344
66.1758196721
That is, with small diameter = 0.77 mm, by ratio the big diameter = 107.34 mm and the height = 66.176 mm. But at least that is close to one of the dimensions given previously.
OK, hold your horses
Your persistence in looking at these dimensions was a good motivator to see what could be going on.
When I looked at the height being 0.0686 m instead of 2*0.0686 I was looking at mode TE111 instead of TM010 (see https://forum.nasaspaceflight.com/index.php?topic=37642.msg1410336#msg1410336).
It turns out that TM010 is fairly insensitive to height (which makes sense because it is the analog of the mode for a cylinder that is constant in the height direction).
Bottom line: the natural frequency does not change much for TM010 whether the height is 2*0.0686m or 0.0686 m. It is around the magnetron frequency (2.45 GHz) even if the height is changed by a factor of 2!!
Which makes the Tajmar EM Drive all the more weird, because if this is the mode he excited, what was the point of the screw-driven mechanism then ?
Details coming maybe tomorrow.
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#1865 by TheTraveller on 18 Jan, 2016 01:16
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I'll just have to make sure I fix that so I can have something other than questionable models, this includes FEKO and meep.
Shell
Nicely said.
TE01x mode has a unique end plate characteristic. The end plate currents do not cross over to the side walls & there is only 1 annular current ring. See attached.
No other TE & TM mode does this. Should be very easy to see this on any sim.
Have also read building in a thin but totally insulating gap between the end plates to the side walls suppresses all modes that need end plate to side wall currents to exits (so the mode can form). This should stop the degenerative but equal guide wavelength TM11x mode from trying to form / forming.
Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.
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#1866 by Rodal on 18 Jan, 2016 01:29
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I'll just have to make sure I fix that so I can have something other than questionable models, this includes FEKO and meep.
Shell
Nicely said.
TE01x mode has a unique end plate characteristic. The end plate currents do not cross over to the side walls & there is only 1 annular current ring. See attached.
No other TE & TM mode does this. Should be very easy to see this on any sim.
Have also read building in a thin but totally insulating gap between the end plates to the side walls suppresses all modes that need end plate to side wall currents to exits (so the mode can form). This should stop the degenerative but equal guide wavelength TM11x mode from trying to form / forming.
Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.
<<Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.>>
No, that can not be an electric current ring at the end plate of an EM Drive, that would only be true for an open waveguide without any metallic ends !!!
Again, the tangential electric field, parallel to a metal, must be zero at the metal end plates.
The only field that can show up at the end plates of the EM Drive having metal end plates for mode TE01p is a magnetic field.
People, we learned this at school: BOUNDARY CONDITIONS for Electric Fields, remember
NASA Eagleworks does not have this misunderstanding. NASA got the fields correct, including the boundary conditions from the very first COMSOL analysis run by Frank Davis.
The EM Drive metallic ends cannot be ignored. The EM Drive is not a waveguide with open ends !!!!
There are boundary conditions at the ends due to the metal !!!
The electric field, and hence the electric current at the metal end plates in mode TE012 is a big zero (see plot below)
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#1867 by TheTraveller on 18 Jan, 2016 02:00
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No, that can not be an electric current ring at the end plate of an EM Drive, that would only be true for an open waveguide without any metallic ends !!!
Again, the tangential electric field, parallel to a metal, must be zero at the metal end plates.
It is for a resonant cavity end plate. Think H field.
See
http://www.ganino.com/BSTJ/images/Vol26/bstj26-1-31.pdf
As per the attached, Blue = H field, the NASA & Bell lab data for the currents in the end plate of a resonant cavity align.
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#1868 by SeeShells on 18 Jan, 2016 02:23
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This was just posted on the other site. Things are now making a little more sense.I'll just have to make sure I fix that so I can have something other than questionable models, this includes FEKO and meep.
Shell
Nicely said.
TE01x mode has a unique end plate characteristic. The end plate currents do not cross over to the side walls & there is only 1 annular current ring. See attached.
No other TE & TM mode does this. Should be very easy to see this on any sim.
Have also read building in a thin but totally insulating gap between the end plates to the side walls suppresses all modes that need end plate to side wall currents to exits (so the mode can form). This should stop the degenerative but equal guide wavelength TM11x mode from trying to form / forming.
Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.
<<Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.>>
No, that can not be an electric current ring at the end plate of an EM Drive, that would only be true for an open waveguide without any metallic ends !!!
Again, the tangential electric field, parallel to a metal, must be zero at the metal end plates.
The only field that can show up at the end plates of the EM Drive having metal end plates for mode TE01p is a magnetic field.
People, we learned this at school: BOUNDARY CONDITIONS for Electric Fields, remember
NASA Eagleworks does not have this misunderstanding. NASA got the fields correct, including the boundary conditions from the very first COMSOL analysis run by Frank Davis.
The EM Drive metallic ends cannot be ignored. The EM Drive is not a waveguide with open ends !!!!
There are boundary conditions at the ends due to the metal !!!
The electric field at the metal end plates in mode TE012 is a big zero (see plot below)
Quote
Some quick notes.
The movies are mislabelled.
The TE01 mode should be TE10. It is not any reference to the frustum, but the dominant mode of the waveguides that are excited by the RF source. I should have not have mentioned the rectangular waveguide mode, it is confusing.
The Feko solver I'm using is method of moments.
The E-field magnitudes shown are not taken on the surface of the frustum (Where they are zero) but 2mm inside.
Feko calculates the standing waves. The animation in these movies are made my changing the phase of the RF sources so affecting the instantaneous E-field magnitude displayed. At 2.47 Ghz, this would be pretty fast!
The frustum walls are perfect conductors.
Note that the E-field scale is logarithmic. This can be misleading or helpful, I'm not sure. Will maybe try a linear scale next time.
I'll do an update with an improved model using copper and S-port measurements soon.
End Quote -
#1869 by Rodal on 18 Jan, 2016 02:26
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...
This was just posted on the other site. Things are now making a little more sense.
Quote
Some quick notes.
The movies are mislabelled.
The TE01 mode should be TE10. It is not any reference to the frustum, but the dominant mode of the waveguides that are excited by the RF source. I should have not have mentioned the rectangular waveguide mode, it is confusing.
The Feko solver I'm using is method of moments.
The E-field magnitudes shown are not taken on the surface of the frustum (Where they are zero) but 2mm inside.
Feko calculates the standing waves. The animation in these movies are made my changing the phase of the RF sources so affecting the instantaneous E-field magnitude displayed. At 2.47 Ghz, this would be pretty fast!
The frustum walls are perfect conductors.
Note that the E-field scale is logarithmic. This can be misleading or helpful, I'm not sure. Will maybe try a linear scale next time.
I'll do an update with an improved model using copper and S-port measurements soon.
End Quote
No, it that does not make any sense
There is no such thing as mode shape TE10
n in TEmnp can never be zero. n can only be 1,2,3, etc. (*)
I already showed in https://forum.nasaspaceflight.com/index.php?topic=39004.msg1477698#msg1477698 that the mode shape looks like TM112. (**)
1) The tangential electric field parallel to a metal wall must be zero, it can never be non-zero
2) The quantum number "n" in modes TEmnp or TMmnp can never be zero, it must be equal to or greater than 1. There is no such thing as a mode TE10 !!!!!
______________________
(*) TE10 would mean that you have a wave-pattern in the circumferential direction but constant field in the radial direction, which is an obvious impossibility.
(**) It doesn't look like TE11p either. (At least there is such a thing as mode shape TE11p)
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#1870 by SeeShells on 18 Jan, 2016 02:29
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This was just posted on the other site. Things are now making a little more sense.I'll just have to make sure I fix that so I can have something other than questionable models, this includes FEKO and meep.
Shell
Nicely said.
TE01x mode has a unique end plate characteristic. The end plate currents do not cross over to the side walls & there is only 1 annular current ring. See attached.
No other TE & TM mode does this. Should be very easy to see this on any sim.
Have also read building in a thin but totally insulating gap between the end plates to the side walls suppresses all modes that need end plate to side wall currents to exits (so the mode can form). This should stop the degenerative but equal guide wavelength TM11x mode from trying to form / forming.
Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.
<<Anyway just look at the end plate for a single annular current ring. If it is not there, it is not TE01x mode.>>
No, that can not be an electric current ring at the end plate of an EM Drive, that would only be true for an open waveguide without any metallic ends !!!
Again, the tangential electric field, parallel to a metal, must be zero at the metal end plates.
The only field that can show up at the end plates of the EM Drive having metal end plates for mode TE01p is a magnetic field.
People, we learned this at school: BOUNDARY CONDITIONS for Electric Fields, remember
NASA Eagleworks does not have this misunderstanding. NASA got the fields correct, including the boundary conditions from the very first COMSOL analysis run by Frank Davis.
The EM Drive metallic ends cannot be ignored. The EM Drive is not a waveguide with open ends !!!!
There are boundary conditions at the ends due to the metal !!!
The electric field at the metal end plates in mode TE012 is a big zero (see plot below)
Quote
Some quick notes.
The movies are mislabelled.
The TE01 mode should be TE10. It is not any reference to the frustum, but the dominant mode of the waveguides that are excited by the RF source. I should have not have mentioned the rectangular waveguide mode, it is confusing.
The Feko solver I'm using is method of moments.
The E-field magnitudes shown are not taken on the surface of the frustum (Where they are zero) but 2mm inside.
Feko calculates the standing waves. The animation in these movies are made my changing the phase of the RF sources so affecting the instantaneous E-field magnitude displayed. At 2.47 Ghz, this would be pretty fast!
The frustum walls are perfect conductors.
Note that the E-field scale is logarithmic. This can be misleading or helpful, I'm not sure. Will maybe try a linear scale next time.
I'll do an update with an improved model using copper and S-port measurements soon.
End Quote
No, that does not make any sense whatsoever
There is no such thing as mode shape TE10 !!!!!!!!!!!!!!!!!!!!
n in TEmnp can never be zero. n can only be 1,2,3, etc.
This is why I'm going to take the time to download and learn FEKO. This just messes with me and the work I've done.
Thanks Dr. Rodal for your time and insight in this. -
#1871 by glennfish on 18 Jan, 2016 02:42
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This is why I'm going to take the time to download and learn FEKO. This just messes with me and the work I've done.
Thanks Dr. Rodal for your time and insight in this.
Geeze Shells, do you always have to do it right?
I haven't had a good data analysis problem in months. I'm beginning to forget how things work...
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#1872 by TheTraveller on 18 Jan, 2016 02:55
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No, that can not be an electric current ring at the end plate of an EM Drive, that would only be true for an open waveguide without any metallic ends !!!
Again, the tangential electric field, parallel to a metal, must be zero at the metal end plates.
It is for a resonant cavity end plate. Think H field.
See
http://www.ganino.com/BSTJ/images/Vol26/bstj26-1-31.pdf
As per the attached, Blue = H field, the NASA & Bell lab data for the currents in the end plate of a resonant cavity align.
Recent quote from an expert I respect. No not from Roger.QuoteThere are no end-plate to end-walls currents in the TE01X modes only if the symmetry in the cavity is perfect to within ~1/100th of wavelength of the driven frequency. For a 2.45 GHz drive signal that has a wavelength of 0.1223643 meters in-vacuum, that implies a required frustum build tolerance of +/-1.2mm.
So Shell YES, you can easily identify TE01x mode excitation by the end plate H field pattern & if available by the single annular current ring induced in the end plate by the H field and by the lack of end plate to side wall currents. -
#1873 by Rodal on 18 Jan, 2016 03:10
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No, that can not be an electric current ring at the end plate of an EM Drive, that would only be true for an open waveguide without any metallic ends !!!
Again, the tangential electric field, parallel to a metal, must be zero at the metal end plates.
It is for a resonant cavity end plate. Think H field.
See
http://www.ganino.com/BSTJ/images/Vol26/bstj26-1-31.pdf
As per the attached, Blue = H field, the NASA & Bell lab data for the currents in the end plate of a resonant cavity align.
Recent quote from an expert I respect. No not from Roger.QuoteThere are no end-plate to end-walls currents in the TE01X modes only if the symmetry in the cavity is perfect to within ~1/100th of wavelength of the driven frequency. For a 2.45 GHz drive signal that has a wavelength of 0.1223643 meters in-vacuum, that implies a required frustum build tolerance of +/-1.2mm.
So Shell YES, you can easily identify TE01x mode excitation by the end plate H field pattern & if available by the single annular current ring induced in the end plate by the H field and by the lack of end plate to side wall currents.
There are no electric currents on a metal produced by the electric field component tangential to it, period, no matter what the tolerance is.
This is true for flat ends and for spherical ends, or any end shape you like.
It has nothing to do with the geometrical tolerance, it has everything to do with the cancellation of the tangential field component. What matters is the direction of the electric field component. It is the tangential field that gets cancelled. The normal electric field, normal to the metal survives. The tangential field gets cancelled.
If you have a wavy surface, then it is the electric field component tangential to the surface waviness that gets cancelled.
For spherical ends, TE01p the electric field tangential to the spherical ends is zero.
For flat ends, TE01p the electric field tangential to the flat ends is zero.
What is common to both ends is that the tangential field gets cancelled and the electric field normal to the surface survives. Whatever the shape of the surface.
The tangential electric field on a conductive metal will be zero, no matter what the shape. This is an elementary boundary condition taught in school, and verified in all experiments.
The lack of electric currents on a conductor produced by the tangential component of the electric field has to do again with Boundary Conditions. Electric currents (charge) produced by the electric field should not be confused with eddy currents produced by the magnetic field.
Eddy currents produced by the magnetic field in mode shape TE01p will be present even with zero tolerance.
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#1874 by TheTraveller on 18 Jan, 2016 03:23
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Eddy currents produced by the magnetic field in mode shape TE01p will be present even with zero tolerance.[/b]
The data I posted was about H field induced eddy currents in the end plates of a TE01x resonant cavity. Image posted again so you can see the H field reference to the end plate annular current ring.
Point was TE01x produces a unique end plate H field and induced current signature that should easily be seeable in any simulation that correctly models TE01x mode excitation.
The build tolerance quote was about end plate H field induced annular eddy currents staying local to the end plate and not also becoming partly side wall currents.
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#1875 by Mulletron on 18 Jan, 2016 03:33
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You know you the got EmDrive fever when you wake up at 3am and stare at mode shapes for two hours playing "spot the quadrupole moment". Based on previous discussions about the possibility of photons becoming massive while confined within waveguide and resonators, assuming this to be true, then the deformation of these resonant modes by virtue of the cavity geometry should produce mass quadrupole moments all over the place.
For instance, if I had a resonant sphere, I would have no quadrupole moment.QuoteThe quadrupole represents how stretched-out along some axis the mass is. A sphere has zero quadrupole. A rod has a quadrupole. A flat disk also has a quadrupole, with the opposite sign of the quadrupole of a rod pointing out from its flat sides. The rod is a sphere stretched along that axis and the disk is a sphere squashed along that axis. In general, objects can have quadrupole moments along three different axes at right angles to each other. (The quadrupole moment is something called a tensor.)
https://van.physics.illinois.edu/qa/listing.php?id=204
So yeah, I'm squashing massive spheres here at 5:25 in the morning and not really sure if this makes any sense. Is this the definition of pathological science?
Also thanks to @Rodal for offline discussion about TM212 vs TE012. -
#1876 by SteveD on 18 Jan, 2016 04:51
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Here’s to successfully modified frustum which now happily resonates at 2,331 MHz. I removed 8 mm from the small end, resulting in a shorter central length and a bigger diameter of the small end, thus pushing the small end cut-off frequency further down and away from resonance. New cut-off freq is 2,256 MHz which is 75 MHz below test frequency.
These resonance cavities turn out to be remarkably predictable. Simulating new dimensions (in COMSOL) was showing a freq shift of +20 Mhz (2323 MHz-> 2343 Mhz). Actual as-measured freq moved by +19 MHz (2312 MHz->2331 Mhz). 80(?) years-old technologies rule.
New dimensions:
D_big: 264 mm (as before)
D_small: 162 mm (+4 mm)
L_center: 196 mm (-8 mm)
TE012 freq (MHz): Simulated: 2343, Actual: 2331.
Df: 0.69
Small end cut-off: 2,256 MHz.
Q factor (at -3 dB S11): 2300 (went down from the original 3100, not sure if this is due to coupling mismatch under new dimensions, or oxidation from minor torch work, or just aging since the last time it was measured).
If anyone knows why this frustum should not be producing thrust then speak now or forever hold your peace.
Ladies and Gentleman, you can now make your bets. (I have already simulated mode frequency shifts for difference thicknesses of HDPE disks at the small end, so this gives a good hint as to what my own prediction for this upcoming test is…)
0.0019 Newtons (1900 uN). Looking forward to seeing your results.
Um that was for 100w. Based on my little spreadsheet I'm showing 0.000581N (581 uN). Good luck and keep safe.
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#1877 by SteveD on 18 Jan, 2016 04:55
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That Woodward's and White's formulations respect Relativity, including time dilation, does not, by itself, preclude the Alcubierre metric "warp drive". Rather, the Alcubierre "warp drive" problem is the need for "negative mass/energy", something that is not available (and hence considered by many physicists as a non-starter), and it has other important dificulties: https://en.wikipedia.org/wiki/Alcubierre_drive#Difficulties . The creator of the Alcubierre concept, Alcubierre himself is on record saying that the "Alcubierre drive" is not something that can be done:Quote from: Alcubierre"from my understanding there is no way it can be done, probably not for centuries if at all" https://twitter.com/malcubierre/status/362011821277839360
If negative energy existed, what properties would it have? Would it have an effect on measurable heat? -
#1878 by RERT on 18 Jan, 2016 11:26
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RFPlumber -
Just had a browse over the COMSOL Frustrum data you posted. What I'm reading in these plots is:
3.13*10^5 V/m peak Electric Field (circulating about a quarter way down from the small end)
1.62*10^-3 Tesla peak Magnetic Field (axial in about the same location), and something in the region of 0.6*10^-3 T at the wall near the peak fields, which is presumably what drives the induced current.
At least 1Kw/M^2 power density in a roughly 10cm ring round the frustrum around the location of the peak fields, and at least 350 A/m over the same 10cm ring.
The total power in the ring looks to me to be about 60W, which is a good match for the 100W input power, given there is power elsewhere. Seems to me the frustrum would be a few tens of degrees above ambient at that rate of heating.
More speculatively, one can turn 350 A/m into 35A by multiplying by the width of the current 'ring'.
If I've got the units or anything else wrong please let me know.
One last question, since one would expect a standing wave rather than static fields, it would be good to know whether your plots are a measurement at a specific point in time, or whether they are a cycle average of some kind.
Thanks again,
R.
[Modified to add missing word.] -
#1879 by SeeShells on 18 Jan, 2016 11:39
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On the discussion on propagation, fields or modes within a cavity or waveguide.
https://archive.org/stream/ClassicalElectrodynamics/Jackson-ClassicalElectrodynamics#page/n273/mode/2up
I've got to take my little Great Pyrenees in to be spade this morning and will be back on later.
Shell
)