Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1873879 times)

Offline birchoff

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It is amazing that Yang achieves record thrust force and record thrust force/powerInput by doing the complete opposite of common wisdom:

* lowest Q of any recorded test   (common wisdom: highest Q the better)
* smallest cone angle, closest to cylinder  (common wisdom: highest cone angle the better)
* longer cavity than Shawyer's Demo at same small diameter
* smaller big diameter than Shawyer's Demo at same small diameter  (common wisdom: the larger the big diameter the better)
...

1. Q represents energy stored. If ALL the energy is stored, it doesn't do any "work". A lower Q does not imply more waste, it implies more work is being done. It can be due to thrust or heat.
...

Todd

Assuming your perspective on Q, wouldn't you eventually want to build a cavity where the energy lots it heat is drastically reduced. So that there is more energy to do work via attenuation? If so would simply building a super conducting frustum do the trick, Or would super-conduction also reduce attenuation?

Offline WarpTech

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1. Q represents energy stored. If ALL the energy is stored, it doesn't do any "work". A lower Q does not imply more waste, it implies more work is being done. It can be due to thrust or heat.
...

Todd

Assuming your perspective on Q, wouldn't you eventually want to build a cavity where the energy lots it heat is drastically reduced. So that there is more energy to do work via attenuation? If so would simply building a super conducting frustum do the trick, Or would super-conduction also reduce attenuation?

IMO a superconducting frustum, that is longer with a much lower cone angle should exhibit more thrust and less loss to heat. The attenuation, I've learned is caused by the phase differences causing pressure imbalance, not resistive "heat" losses.

Consider this a game of tug-of-war. Both teams (pressures) are pushing equally in both directions, but when one team lets go of the rope, the other team experiences "thrust". In this case, the team that lets go, is the destructive interference at one end, and constructive interference at the other end gets to hold onto the rope, er, I mean frustum. :)
Todd

Offline aero

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I'm still fooling with meep, trying to get some 3D images and yesterday I had an interesting accident with the code. Using what I thought was a sealed "Bradycone" cavity in 3D, I continued to see extensive RF energy outside the cone. Of course that can not be as we understand things. I made a Force/Power run and meep measured 10+ times ideal photon rocket.

On further investigation i discovered that I had modeled the dielectric shifted in the Y direction - sideways - instead of axially. (Meep uses different coordinates in 3D that in 2D, it seems) So with the dielectric penetrating the center of one of the sidewalls of the cone (still oriented with dielectric and cavity ends parallel) F/P was 10 times higher than typical. OH, and this model won't resonate, that's where my investigation started.

I don't know if the above is meaningful but I found it interesting. I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.
Retired, working interesting problems

Offline Dortex

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  Thus, the geometrical attenuation theory is in accord with experiments also in this respect, as the experiments show that the highest thrust forces have been produced without dielectric inserts.

I'm starting to get the impression EW did everything possible not to get thrust.

Offline dustinthewind

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...Hmm this gets me thinking about the standing or non-standing waves in the cavity.  Let us say, that for some reason attenuation of light is happening at the top end so reflected light is weaker than light from the bottom.  As a result the wave is no longer a standing wave and is instead a traveling wave or semi (standing-traveling wave).  We all know if you hold a magnet near an aluminum plate and move it it will drag the plate because of the resistance to change in magnetic field.  So maybe the semi-traveling waves could do the same to the cavity and drag it along?  Does that sound like a possibility?
You can move an object by using a magnet from the outside.

However you cannot accelerate the center of mass of an object by moving a magnet  inside it.

An Astronaut with a powerful magnet inside the ISS can move the magnet all she/he wants,  and still will not be able to accelerate the center of mass of the ISS.


The traveling waves appear at the injection port then travel to the attenuated end then disappear I would think.  Isn't that sort of like a magnet just appears at the big end and then travels towards the narrow end then just suddenly disappears?  I'm not sure where the back reaction would come from to counter the drag from the changing magnetic field.  In the ISS, it would be the astronaut pushing against the hull to move the magnet, I would suspect.  Pressure from a changing magnetic field should be much larger than the pressure from light. 
« Last Edit: 06/12/2015 04:58 PM by dustinthewind »

Offline Star One

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  Thus, the geometrical attenuation theory is in accord with experiments also in this respect, as the experiments show that the highest thrust forces have been produced without dielectric inserts.

I'm starting to get the impression EW did everything possible not to get thrust.

Curious statement, why would they move in that direction?

Offline aero

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  Thus, the geometrical attenuation theory is in accord with experiments also in this respect, as the experiments show that the highest thrust forces have been produced without dielectric inserts.

I'm starting to get the impression EW did everything possible not to get thrust.

Curious statement, why would they move in that direction?

If you look back at what we knew or even suspected when we started thread 1, you'll see that EW was in an almost totally unknown world at the time they started their work. That's why.
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Offline Rodal

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  Thus, the geometrical attenuation theory is in accord with experiments also in this respect, as the experiments show that the highest thrust forces have been produced without dielectric inserts.

I'm starting to get the impression EW did everything possible not to get thrust.

Curious statement, why would they move in that direction?

I doubt the comment was meant literally, but instead as how funny it is that things sometime turn out different than one imagined. 

Now, concerning NASA Eagleworks, we have posts in these threads (and in the media) stating that perhaps the researchers wanted "too much" to believe there is real thrust going on.  If now the feeling is that they did everything to diminish thrust, it is probable that the truth is somewhere in the middle, that the NASA team conducted the tests in an objective manner and sometimes the complexity and the unknown nature of the experiment makes it look like they are trying too hard to get thrust and sometimes makes it look like they are trying not to get thrust, while the truth is that the experimenter is doing its human best to be objective.

The real reason is that scientifically, one doesn't expect to get thrust out of a microwave cavity, and therefore one is in a situation similar to von Braun's description of R&D: 

" Basic research is what I'm doing when I don't know what I'm doing. "
-- Wernher von Braun
« Last Edit: 06/12/2015 05:14 PM by Rodal »

Offline SeeShells

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1. Q represents energy stored. If ALL the energy is stored, it doesn't do any "work". A lower Q does not imply more waste, it implies more work is being done. It can be due to thrust or heat.
...

Todd

Assuming your perspective on Q, wouldn't you eventually want to build a cavity where the energy lots it heat is drastically reduced. So that there is more energy to do work via attenuation? If so would simply building a super conducting frustum do the trick, Or would super-conduction also reduce attenuation?

IMO a superconducting frustum, that is longer with a much lower cone angle should exhibit more thrust and less loss to heat. The attenuation, I've learned is caused by the phase differences causing pressure imbalance, not resistive "heat" losses.

Consider this a game of tug-of-war. Both teams (pressures) are pushing equally in both directions, but when one team lets go of the rope, the other team experiences "thrust". In this case, the team that lets go, is the destructive interference at one end, and constructive interference at the other end gets to hold onto the rope, er, I mean frustum. :)
Todd
The harmonic wave collapse is towards the small end if I'm gleaning this right? Visualizing on the fly here so forgive me if I muck it up.  If I consider the internal Frustum structures of a stabilized harmonic wave pattern, it makes sense that the small end would collapse and decay first (simply less harmonic wave support) into effervescent waves and the large end would "rush" in to fill the void where the EM pattern lost structure in the small end. Whereas they also would also decay into effervescent waves. It seems to tie in with the Chinese using the TM mode as the collapsing harmonics at the large end of the Fructum structure would travel down the center of the Frustum towards the small end?

Offline SeeShells

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I'm still fooling with meep, trying to get some 3D images and yesterday I had an interesting accident with the code. Using what I thought was a sealed "Bradycone" cavity in 3D, I continued to see extensive RF energy outside the cone. Of course that can not be as we understand things. I made a Force/Power run and meep measured 10+ times ideal photon rocket.

On further investigation i discovered that I had modeled the dielectric shifted in the Y direction - sideways - instead of axially. (Meep uses different coordinates in 3D that in 2D, it seems) So with the dielectric penetrating the center of one of the sidewalls of the cone (still oriented with dielectric and cavity ends parallel) F/P was 10 times higher than typical. OH, and this model won't resonate, that's where my investigation started.

I don't know if the above is meaningful but I found it interesting. I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.
I'm just digging into it so I need some more time, it's not easy to learn. When I feel I can offer some intelligent bread crumbs I'll speak up.
But your right that is a weird result.
Shell

Offline aero

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I'm still fooling with meep, trying to get some 3D images and yesterday I had an interesting accident with the code. Using what I thought was a sealed "Bradycone" cavity in 3D, I continued to see extensive RF energy outside the cone. Of course that can not be as we understand things. I made a Force/Power run and meep measured 10+ times ideal photon rocket.

On further investigation i discovered that I had modeled the dielectric shifted in the Y direction - sideways - instead of axially. (Meep uses different coordinates in 3D that in 2D, it seems) So with the dielectric penetrating the center of one of the sidewalls of the cone (still oriented with dielectric and cavity ends parallel) F/P was 10 times higher than typical. OH, and this model won't resonate, that's where my investigation started.

I don't know if the above is meaningful but I found it interesting. I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.
I'm just digging into it so I need some more time, it's not easy to learn. When I feel I can offer some intelligent bread crumbs I'll speak up.
But your right that is a weird result.
Shell

Oh thank you, thank you!

It took me about 2 months in a vacuum to get my first valid Force/Power run working, but then I've had more time to forget my education than you have. About 5 years more, if my gleaning from your remarks is correct. That, and side trips into Maxima and ParaView made my progress a little slow.

I hope you won't need to study Meep in a vacuum though. Ask, or PM me if you have questions about Meep that you'd rather not share with the forum.
Retired, working interesting problems

Offline SeeShells

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1. Q represents energy stored. If ALL the energy is stored, it doesn't do any "work". A lower Q does not imply more waste, it implies more work is being done. It can be due to thrust or heat.
...

Todd

Assuming your perspective on Q, wouldn't you eventually want to build a cavity where the energy lots it heat is drastically reduced. So that there is more energy to do work via attenuation? If so would simply building a super conducting frustum do the trick, Or would super-conduction also reduce attenuation?

IMO a superconducting frustum, that is longer with a much lower cone angle should exhibit more thrust and less loss to heat. The attenuation, I've learned is caused by the phase differences causing pressure imbalance, not resistive "heat" losses.

Consider this a game of tug-of-war. Both teams (pressures) are pushing equally in both directions, but when one team lets go of the rope, the other team experiences "thrust". In this case, the team that lets go, is the destructive interference at one end, and constructive interference at the other end gets to hold onto the rope, er, I mean frustum. :)
Todd
The harmonic wave collapse is towards the small end if I'm gleaning this right? Visualizing on the fly here so forgive me if I muck it up.  If I consider the internal Frustum structures of a stabilized harmonic wave pattern, it makes sense that the small end would collapse and decay first (simply less harmonic wave support) into effervescent waves and the large end would "rush" in to fill the void where the EM pattern lost structure in the small end. Whereas they also would also decay into effervescent waves. It seems to tie in with the Chinese using the TM mode as the collapsing harmonics at the large end of the Fructum structure would travel down the center of the Frustum towards the small end?
Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?


Offline Rodal

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Hence, why I said the frustum needs to be longer, so we can have something "closer" to Zeng & Fan's waveguide for traveling waves. If we are confined to 0-length past the cut-off diameter, attenuation is minimized, reflection and Q are higher. If we extend it out a full wavelength, we may attenuate 66% of the energy. The other 33% will be reflected with a larger phase shift than what Shawyer's design allows.

Also, the standing waves in a damped cavity will also frequency shift due to the damping. That's what gives Shawyer's design "some" thrust, but as I said, it is the rate of attenuation that will exert a higher force. So a longer front end to give the waves some traveling room to be attenuated faster, is what I believe is needed.

In other words, "design" the thruster more like Zeng and Fan and less like Shawyer.
Todd

Prof. Yang's EM Drive is significantly longer than Shawyer's

Description Mode Shape  Length (m)   Db (m) Ds (m)   Frequency (GHz)  Q  Force / PowerInput (mN/kW)
Shawyer Demo  TE012    0.187           0.28    0.14921    2.45             45000   80-243
Yang                  TE012    0.24            0.201  0.1492       2.45               1531   1070

Both have the same frequency, same mode shape, same Small Diameter

Yang achieves 10 to 5 times greater force/input power by operating with 29 times lower Q with a 28% longer EM Drive and 39% smaller big diameter.  All the opposite of what Shawyer recommends.

The wavelength is 299700000 m/s /(2.45*10^9 1/s) = 0.122 m

So Yang's EM Drive (which has the same small diameter as Shawyer's Demo) has a length (0.24 - 0.187) = 0.053 m

Yang's EM Drive is therefore about 1/2 wavelength longer than Shawyer's EM Drive truncated cone length


The most important parameter to Zeng & Fan is the cone half-angle.  It is drastically different between them:

Shawyer   (180/Pi) ArcTan[(0.28 - 0.14921)/(2*0.187)] = 19.275 degrees

Yang        (180/Pi) ArcTan[(0.201 - 0.1492)/(2*0.24)] =    6.159 degrees

Shawyer's EM Drive has a cone half-angle more than 3 times greater.

Yang's EM Drive is closer to a cylinder, which according to Zeng & Fan results in much greater attenuation.

Thank you for validating everything I just said!  8)

I have now added the dimensions of the EM Drive cavities used in experiments, figured out in spherical coordinates, which is the natural system to use for the microwaves inside the cavity (they are spherical waves), to the wiki page for Experimental Results:

http://emdrive.wiki/Experimental_Results

Of these spherical geometry parameters, Zeng and Fan single out the cone half angle as the most important, as the geometrical attenuation is a very strong function of the cone half angle.

Zeng and Fan show data for 7.5 degrees, 15 degrees, up to 90 degrees in 7.5 degree increments.

They show that the largest attenuation by far occurs at 7.5 degrees.

It is fascinating, again that Yang achieves 10 to 5 times greater force/input power by operating with 29 times lower Q with a cone half angle of 6 degrees, while Shawyer's EM Drive has a cone half-angle more than 3 times greater.

Offline SeeShells

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I'm still fooling with meep, trying to get some 3D images and yesterday I had an interesting accident with the code. Using what I thought was a sealed "Bradycone" cavity in 3D, I continued to see extensive RF energy outside the cone. Of course that can not be as we understand things. I made a Force/Power run and meep measured 10+ times ideal photon rocket.

On further investigation i discovered that I had modeled the dielectric shifted in the Y direction - sideways - instead of axially. (Meep uses different coordinates in 3D that in 2D, it seems) So with the dielectric penetrating the center of one of the sidewalls of the cone (still oriented with dielectric and cavity ends parallel) F/P was 10 times higher than typical. OH, and this model won't resonate, that's where my investigation started.

I don't know if the above is meaningful but I found it interesting. I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.
I'm just digging into it so I need some more time, it's not easy to learn. When I feel I can offer some intelligent bread crumbs I'll speak up.
But your right that is a weird result.
Shell

Oh thank you, thank you!

It took me about 2 months in a vacuum to get my first valid Force/Power run working, but then I've had more time to forget my education than you have. About 5 years more, if my gleaning from your remarks is correct. That, and side trips into Maxima and ParaView made my progress a little slow.

I hope you won't need to study Meep in a vacuum though. Ask, or PM me if you have questions about Meep that you'd rather not share with the forum.
I started to just play around with it and my first thought was from Star Trek as Scotty said "They be whales Capt'n". You're right it's not for the light of heart. Thanks for the help as I have somethings I really want to model and see if what I "see" using my mind's eye agrees. When I get stuck or have a dumb question that stumps me I'll pm you. Thanks again.
Shell

Offline Rodal

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Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?
OK, I need your help here Shell.
I understand Zeng and Fan's paper analytically, that the smaller the half-angle the greater the attenuation, but I don't have a physical understanding of this to explain it in simple terms.  Which means that when it comes down to it, I don't understand really understand it   :) .

This is my problem in understanding this:

1) A perfect cylinder has no geometrical attenuation: it has no evanescent waves, no progressive cut-off of modes.  The modes that are cut-off are the ones that have a bigger diameter than the cylinder, that's it.

2) So, how can it be that the biggest geometrical attenuation occurs for the smallest half-angle ? in other words for the geometry that is closest to a cylinder?

3) What happens in the limit with the geometrical attenuation as the cone becomes a cylinder (as the half angle becomes zero)? How does one go from the largest geometrical attenuation for a cone to NO geometrical attenuation for a cylinder?
« Last Edit: 06/12/2015 06:06 PM by Rodal »

Offline SeeShells

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Hence, why I said the frustum needs to be longer, so we can have something "closer" to Zeng & Fan's waveguide for traveling waves. If we are confined to 0-length past the cut-off diameter, attenuation is minimized, reflection and Q are higher. If we extend it out a full wavelength, we may attenuate 66% of the energy. The other 33% will be reflected with a larger phase shift than what Shawyer's design allows.

Also, the standing waves in a damped cavity will also frequency shift due to the damping. That's what gives Shawyer's design "some" thrust, but as I said, it is the rate of attenuation that will exert a higher force. So a longer front end to give the waves some traveling room to be attenuated faster, is what I believe is needed.

In other words, "design" the thruster more like Zeng and Fan and less like Shawyer.
Todd

Prof. Yang's EM Drive is significantly longer than Shawyer's

Description Mode Shape  Length (m)   Db (m) Ds (m)   Frequency (GHz)  Q  Force / PowerInput (mN/kW)
Shawyer Demo  TE012    0.187           0.28    0.14921    2.45             45000   80-243
Yang                  TE012    0.24            0.201  0.1492       2.45               1531   1070

Both have the same frequency, same mode shape, same Small Diameter

Yang achieves 10 to 5 times greater force/input power by operating with 29 times lower Q with a 28% longer EM Drive and 39% smaller big diameter.  All the opposite of what Shawyer recommends.

The wavelength is 299700000 m/s /(2.45*10^9 1/s) = 0.122 m

So Yang's EM Drive (which has the same small diameter as Shawyer's Demo) has a length (0.24 - 0.187) = 0.053 m

Yang's EM Drive is therefore about 1/2 wavelength longer than Shawyer's EM Drive truncated cone length


The most important parameter to Zeng & Fan is the cone half-angle.  It is drastically different between them:

Shawyer   (180/Pi) ArcTan[(0.28 - 0.14921)/(2*0.187)] = 19.275 degrees

Yang        (180/Pi) ArcTan[(0.201 - 0.1492)/(2*0.24)] =    6.159 degrees

Shawyer's EM Drive has a cone half-angle more than 3 times greater.

Yang's EM Drive is closer to a cylinder, which according to Zeng & Fan results in much greater attenuation.

Thank you for validating everything I just said!  8)

I have now added the dimensions of the EM Drive cavities used in experiments, figured out in spherical coordinates, which is the natural system to use for the microwaves inside the cavity (they are spherical waves), to the wiki page for Experimental Results:

http://emdrive.wiki/Experimental_Results

Of these spherical geometry parameters, Zeng and Fan single out the cone half angle as the most important, as the geometrical attenuation is a very strong function of the cone half angle.

Zeng and Fan show data for 7.5 degrees, 15 degrees, up to 90 degrees in 7.5 degree increments.

They show that the largest attenuation by far occurs at 7.5 degrees.

It is fascinating, again that Yang achieves 10 to 5 times greater force/input power by operating with 29 times lower Q with a cone half angle of 6 degrees, while Shawyer's EM Drive has a cone half-angle more than 3 times greater.
It makes sense unless I'm off my meds. ;) I think the Frustum is a lot like the charging of a capacitor and pushing up the Q can lead to not that much more stored energy to release. See graph.

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Curious statement, why would they move in that direction?

To clarify: I wasn't implying they deliberately fudged results to arrive at an answer. I meant to say that now it seems that everything they did (dielectric, wall angles, high Q, etc.) just so happens to reduce thrust if the new guesses happen to be right.


Offline WarpTech

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How do we know that the truncated cone entropy distribution is maximized in an accelerated frame or reference?

This has been the cause of much discussions in these threads, for example @frobnicat, @deltaMass and @wallofwolfstreet insisting that there is no preferred frame of reference, that the photons don't accelerate, and that everything should be based on frame-indifference.  They insist on applying frame-indifference to the cavity.

To be fair, doesn't the bulk of experimental evidence outside the realm of EM Drives support frame indifference?

Here's the issue with frame invariance or "indifference" as you put it. In QED, the force on an atom in the quantum vacuum, EM field is proportional to (Milonni Appendix B);

F = R*v ~ [Rho(w) -  (w/3)*d(Rho(w))/dw]

Where Rho(w) is the spectral energy density of the vacuum EM field. We know that in the ZPF;

Rho(w) = (hbar*w3)/(2pi2*c3)

So when Rho(w) is plugged into the above equation, we get F = 0. THIS is the QM equivalent of an "Inertial Reference Frame". The force on the atom goes to zero, at "any" velocity. 

This equation however is degenerate. Rho, being a quantized by hbar and because photons are Bosons, Bose-Einstein statistics apply, then Rho(w) can be modified to be Rho(n*w), where n is an integer called the "photon number"

This NEW vacuum is also an Inertial frame, yet there are n X more photons in the field. In other words, the ZPF is a thermal field with a Gaussian distribution. There are variations in the thermal field that modify "n", to behave as a variable intensity of the ZPF. A higher "n" value, means a higher ZPF spectral energy density value, as a function of the general coordinates;

Rho(n*w) = (hbar*(n*w)3)/(2pi2*c3)

Current theories Normalize n=1 in the vacuum, because it cannot be measured directly.

This "PROVES" that all inertial reference frames are NOT alike. There exist an infinite number of inertial reference frames, each differs from the other by the density of photons in the spectrum of the thermal field. This is true for ANY spectral energy density proportional to w3!

THIS is why frame invariance is treated differently in SR than it is in GR, where the differences represent different gravitational potentials or different inertial mass content.

Todd

Offline WarpTech

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Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?
OK, I need your help here Shell.
I understand Zeng and Fan's paper analytically, that the smaller the half-angle the greater the attenuation, but I don't have a physical understanding of this to explain it in simple terms.  Which means that when it comes down to it, I don't understand really understand it   :) .

This is my problem in understanding this:

1) A perfect cylinder has no geometrical attenuation: it has no evanescent waves, no progressive cut-off of modes.  The modes that are cut-off are the ones that have a bigger diameter than the cylinder, that's it.

2) So, how can it be that the biggest geometrical attenuation occurs for the smallest half-angle ? in other words for the geometry that is closest to a cylinder?

3) What happens in the limit with the geometrical attenuation as the cone becomes a cylinder (as the half angle becomes zero)? How does one go from the largest geometrical attenuation for a cone to NO geometrical attenuation for a cylinder?

In Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;

alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta=0, there is no attenuation, that's a cylinder.

If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.

Todd
« Last Edit: 06/12/2015 07:30 PM by WarpTech »

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Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?
OK, I need your help here Shell.
I understand Zeng and Fan's paper analytically, that the smaller the half-angle the greater the attenuation, but I don't have a physical understanding of this to explain it in simple terms.  Which it comes down to,it  means that I don't understand it really   :) .

This is my problem in understanding this:

1) A perfect cylinder has no geometrical attenuation: it has no evanescent waves, no progressive cut-off of modes.  The modes that are cut-off are the ones that have a bigger diameter than the cylinder, that's it.

Yep, that's about it and very well said.

2) So, how can it be that the biggest geometrical attenuation occurs for the smallest half-angle ? in other words for the geometry that is closest to a cylinder?

The way I see it it it really doesn't matter after the harmonic pattern starts its collapse at the small end. The collapse of the wave functions is part of the Quantum collapse into effervescent waves and where the discussions of any harmonic actions are mute as there aren't any harmonic cavities to support a Quantum environment. All we are left with are the quantum force vectors... I think.
http://rationalwiki.org/wiki/Quantum_collapse


3) What happens in the limit with the geometrical attenuation as the cone becomes a cylinder (as the half angle becomes zero)? How does one go from the largest geometrical attenuation for a cone to NO geometrical attenuation for a cylinder?

I think number 2 answers that.

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