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#2380 by rfmwguy on 14 May, 2016 18:47
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Not the only way...you might appreciate my solution when I reveal the frustum.
To put this in context:See here rfmyguy's spectrum analysis of one of his magnetrons. It starts very close to 2.45Ghz and then drifts to 2.445Ghz as the magnetron heats. I do not think the frequencies emitted by a magnetron are close enough to 2.4575Ghz to excite TM212, even when the dirty magnetron signal is accounted for.
This is the exact problem that plagues all of us. You have to be spot on with resonance. Like Doc said, its a shame we didn't have exact dimensions from original builders but suppose that's to be expected from a commercial entity. The original EW dimensions certainly were off 2450 even if an insert was introduced, which lowers resonance frequency.
I built a "dummy" frustum out of aluminum sheets and PC board to get my final dimensions and hopefully the old magnetron and the new frustum connect somewhere...my hope anyway.
It is very important the end plates are very highly parallel, like to better than 100um or better still 25um. You may not see much improvement on VNA numbers but it only tells a small part of the story. Tuning for max thrust, under power, tells a very different story, which I learned from experience.
100 um = 0.004 inches
25 um = 0.001 inches (one thousands of an inch)
some builders are using only 1 mm copper, some are using thicker
1 mm = 1000 um
1 mm = 0.039 inch
so this specification for parallelism is asking to be parallel within 1/10 or 1/40 of that thickness of the copper walls ! !
Since it is easy to deform a plate to a fraction of its thickness, the only way to ensure this parallelism is to have much thicker walls than 1 mm -
#2381 by Rodal on 14 May, 2016 18:50
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I meant a 1mm plate by itself: in other words the only way without using stiffeners, sandwiched plates, or other materials to stiffen the flat end plates (since bending stiffness is related to the separation between the fascia). Stiffeners are common in aerospace, also in civil engineering (I-beams to increase bending moment of inertia), and so are sandwich plates, etc....
Not the only way...you might appreciate my solution when I reveal the frustum.
Since it is easy to deform a plate to a fraction of its thickness, the only way to ensure this parallelism is to have much thicker walls than 1 mm
Stiffeners and sandwich construction is how rockets and airplanes are constructed to save weight.
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#2382 by TheTraveller on 14 May, 2016 18:53
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so this specification for parallelism is asking to be parallel within 1/10 or 1/40 of that thickness of the copper walls ! !
Roger's spec is +- 10x skin depth (~13um), which my next frustum will achieve.
Based on my experimental experience +-100um should generate some thrust, while +-25um is better and +-13um is ideal.
As the travelling wave will bounce back & forth over 10, 000 times, work out the phase distortion introduced into the travelling wave from the non parallelism and the 2nd effect of the travelling waves being forced to bunch up at the side wall point of widest separation.
What I now know is building good EmDrives is not quick, cheap nor simple. Extremely high manufacturing accuracy & precision are needed. Add to that the polish requirements and avoiding oxidation during the life of the EmDrive. Another high hurtle to get over. -
#2383 by Rodal on 14 May, 2016 18:55
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...
Let's hope the thrust force is with you.
Quote from: Qui-Gon Jinn (The Phantom Menace)Remember: Your focus determines your reality
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#2384 by TheTraveller on 14 May, 2016 18:59
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I meant a 1mm plate by itself: in other words the only way without using stiffeners, sandwiched plates, or other materials to stiffen the flat end plates. Stiffeners are common in aerospace, also in civil engineering (I-beams to increase bending moment of inertia), and so are sandwich plates, etc....
Not the only way...you might appreciate my solution when I reveal the frustum.
Since it is easy to deform a plate to a fraction of its thickness, the only way to ensure this parallelism is to have much thicker walls than 1 mm
That's how rockets and airplanes are constructed to save weight.
My professionally built commercial quality EmDrive will have 6mm thick side walls and min 10mm thick spherical end plates. Will look very much like the final Flight Thruster, just a bit larger.
Saving weight is not really an option when max build and operational dimensional change from ideal is +- 10x skin depth. -
#2385 by meberbs on 14 May, 2016 19:08
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Don't believe everything FEKO tells you. Like other analysis packages it gets cutoff wrong. I know as I found out the hard way. Have verified this with Roger.
There has been exactly zero experimental evidence to support your statement. All evidence has been that all of the analysis programs used in these threads have correctly predicted resonance when used correctly (unit mistakes and such have caused issues, but that is user error). I do not believe there has been a single case where the prediction was not accurate to within parameters such as the accuracy of the build geometry.
If you have an example shape where Maxwell's equations predict resonance, but an experiment shows no resonance due to some "cutoff," please share this geometry since this would be an incredible and revolutionary discovery of new physics.
Edit: I am glad to hear that you are feeling better.
Thanks for the health mention. Yes it is good to be able to see a light at the end of a long tunnel.
BTW my 8mN is real. So something is happening that is outside currently accepted theory.
Roger advised me packages like FEKO & others do not properly model what happens in reality at the small end, near cutoff. Advised me to never go below diam in mtrs X 0.82 in TE013 expressed as drive freq despite what any analysis package indicated was not at cutoff.
As existing physics says the EmDrive should not work but it does work, just maybe the assumptions used in the modelers to determine small end cutoff in a tapered waveguide with reflecting end plates just might not be as accurate as you may assume.
I asked a simple (but excessively wordy) question in this post about your 8mN measurement, to get a better idea of the setup, and what magnitude of errors are associated it. I understand if you missed the post, since it is hard to keep up with threads on this site. If you can answer that, I think it would help everyone here get a better feel for your results.
There is a difference between "no thrust is generated when small end is in cutoff" and "no resonance exists when small end is in cutoff." The first one seems to be in line with what you learned from Shawyer, but you are claiming that the second one is therefore also true when you say not to trust FEKO. This logical leap is not justified. The available evidence (including X_RaY's real world experience) indicates that FEKO models resonance and cut off very well and therefore whatever changes to theory the EMDrive effect may require (if it is conclusively shown to exist) will not change predicted resonance significantly.
Note that I am not dismissing any results you have seen, but I do not have enough information on your experiment, so I can't use it as the basis for any conclusions. -
#2386 by X_RaY on 14 May, 2016 19:13
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There is a possibility to increase the stiffness of thin structures in general and spare a lot of weight.
I meant a 1mm plate by itself: in other words the only way without using stiffeners, sandwiched plates, or other materials to stiffen the flat end plates. Stiffeners are common in aerospace, also in civil engineering (I-beams to increase bending moment of inertia), and so are sandwich plates, etc....
Not the only way...you might appreciate my solution when I reveal the frustum.
Since it is easy to deform a plate to a fraction of its thickness, the only way to ensure this parallelism is to have much thicker walls than 1 mm
That's how rockets and airplanes are constructed to save weight.
My professionally built commercial quality EmDrive will have 6mm thick side walls and min 10mm thick spherical end plates. Will look very much like the final Flight Thruster, just a bit larger.
Saving weight is not really an option when max build and operational dimensional change from ideal is +- 10x skin depth.
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#2387 by A_M_Swallow on 14 May, 2016 19:25
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...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Why? there is a minimum size is a scientific investigation rather than an engineering exercise. To produce a working machine engineers have to stick to things that work. Where as weird behaviour on both sides of the limit may reveal to scientists something about the underlying physics.
edit:spelling -
#2388 by Rodal on 14 May, 2016 19:34
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Yes, but the problem with the EM Drive is that it has not been shown what "works" yet....Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Why there is a minimum size is a scientific investigation rather than an engineering exercise. To produce a working machine engineers have to stick to thinks that work. Weird behaviour on both sides of the limit may reveal to scientists something about the underlying physics.
Goddard, Von Braun and even Project Orion (explosively loaded rockets) were carrying out rocket experiments achieving lift off. Hence they could tell what "works" and what "does not work".
The Wright Brothers and the Human Powered Airplane were experimenting with flight powered vehicles: they could tell what "works" and what "does not work".
In great contrast, in decades of reporting experiments, Shawyer has not yet reported a single experiment that shows what works (not just the fact that there is no lift off and no flight like the Goddard, and the Wright Brothers) but, not conducting experiments as done in 50 years of experimenting with micro-thrusters in the Aerospace industry:
1) no torsion pendulum experiments
2) no vacuum chamber experiments
3) no battery-powered experiments
Hence whether the experiments show something that "works" or an experimental artifact is still a matter of contention. Look forward for somebody (NASA, etc.) to conduct battery-powered rotational experiment under partial vacuum to show that indeed something "works", at which point in time engineers will have something to follow that indeed is "shown to work", and then at that time, the EM Drive could develop in an engineering way.
Otherwise we have cases like Prof. Yang, who was a disciple of Shawyer, claiming the highest force/inputPower, with cord-powered experiments until she conducted them with battery power and nullified her previous claims.
Thus following what Prof. Yang was doing, is by her latest results, not following something that "works".
Both NASA and Hackaday Aachen teams show that what works best is to use a polymer insert, and Cannae also when using normal conducting metals like copper.
For the time being "what works" varies with the eyes of the beholder. -
#2389 by Monomorphic on 14 May, 2016 19:40
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...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value? -
#2390 by A_M_Swallow on 14 May, 2016 19:42
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Although the same man can do both in practice pure research is performed at universities and science laboratories.
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#2391 by X_RaY on 14 May, 2016 19:48
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http://forum.nasaspaceflight.com/index.php?topic=39004.msg1472751#msg1472751...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value?Quote from: TheTravellerRoger's 0.82 cutoff rule
Just a crude approximation/ hand rule to calculate the cut off diameter of a cylindrical waveguide for TE01 mode.
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) -
#2392 by Monomorphic on 14 May, 2016 19:57
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http://forum.nasaspaceflight.com/index.php?topic=39004.msg1472751#msg1472751...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value?Quote from: TheTravellerRoger's 0.82 cutoff rule
Just a crude approximation/ hand rule to calculate the cut off diameter for a cylindrical waveguide for TE01 mode.
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)
Thanks! I wonder how this cutoff rule works with the wedge geometry. -
#2393 by Rodal on 14 May, 2016 20:04
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http://forum.nasaspaceflight.com/index.php?topic=39004.msg1472751#msg1472751...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value?Quote from: TheTravellerRoger's 0.82 cutoff rule
Just a crude approximation/ hand rule to calculate the cut off diameter for a cylindrical waveguide for TE01 mode.
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)
Thanks! I wonder how this cutoff rule works with the wedge geometry.
Roger's 0.82 cutoff rule
This is just the standard cut-off rule for open waveguides (no metal ends) for a cylindrical waveguide having constant cross-sectional dimensions equal to the small end diameter:
1/0.82 = 3.83170597020751/Pi
=BesselRoot X'01/Pi (see http://wwwal.kuicr.kyoto-u.ac.jp/www/accelerator/a4/besselroot.htmlx for value of the Root of Derivative of Bessel function 01: BesselRoot X'01 )
2 Pi fc = c BesselRoot X'01/r
hence
fc = c BesselRoot X'01/(Diameter*Pi)
= c (3.83170597020751/Pi )/Diameter
where Shawyer uses the small end diameter, as the diameter in the above equation
For rectangular waveguide use (as shown below by X-Ray)
ωc = 2 Pi fc
where a, b are the small end dimensions of the wedge EM Drive
and where n,m are the mode shape numbers in TEmn, so for TE01 use n=1,m=0
In general, an excitation of the guide at a cross-section (a location along the longitudinal axis) excites all waveguide modes. The modes with cutoff frequencies higher than the frequency of excitation decay away from the source. Only the dominant mode has a sinusoidal dependence on the longitudinal axis and thus possesses fields that are periodic in the longitudinal axis and "dominate" the field pattern far away from the source, at distances larger than the transverse dimensions of the waveguide. -
#2394 by X_RaY on 14 May, 2016 20:18
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http://forum.nasaspaceflight.com/index.php?topic=39004.msg1472751#msg1472751...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value?Quote from: TheTravellerRoger's 0.82 cutoff rule
Just a crude approximation/ hand rule to calculate the cut off diameter for a cylindrical waveguide for TE01 mode.
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)
Thanks! I wonder how this cutoff rule works with the wedge geometry.
This is the related equation for rectangular waveguides.
"grenz" is german for border =cutoff (λc; fc)
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#2395 by X_RaY on 14 May, 2016 20:27
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Therefore the term with "b" inside is negligible for TE01 since m=0 for this mode.
http://forum.nasaspaceflight.com/index.php?topic=39004.msg1472751#msg1472751...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value?Quote from: TheTravellerRoger's 0.82 cutoff rule
Just a crude approximation/ hand rule to calculate the cut off diameter for a cylindrical waveguide for TE01 mode.
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)
Thanks! I wonder how this cutoff rule works with the wedge geometry.
Roger's 0.82 cutoff rule
This is just the standard cut-off rule for open waveguides (no metal ends) for a cylindrical waveguide having constant cross-sectional dimensions equal to the small end diameter:
1/0.82 = 3.83170597020751/Pi
=BesselRoot X'01/Pi (see http://wwwal.kuicr.kyoto-u.ac.jp/www/accelerator/a4/besselroot.htmlx for value of the Root of Derivative of Bessel function 01: BesselRoot X'01 )
2 Pi fc = c BesselRoot X'01/r
hence
fc = c BesselRoot X'01/(Diameter*Pi)
= c (3.83170597020751/Pi )/Diameter
where Shawyer uses the small end diameter, as the diameter in the above equation
For rectangular waveguide use (as shown below by X-Ray)
ωc = 2 Pi fc
where a, b are the small end dimensions of the wedge EM Drive
and where n,m are the mode shape numbers in TEmn, so for TE01 use n=1,m=0
In general, an excitation of the guide at a cross-section (a location along the longitudinal axis) excites all waveguide modes. The modes with cutoff frequencies higher than the frequency of excitation decay away from the source. Only the dominant mode has a sinusoidal dependence on the longitudinal axis and thus possesses fields that are periodic in the longitudinal axis and "dominate" the field pattern far away from the source, at distances larger than the transverse dimensions of the waveguide.
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#2396 by Rodal on 14 May, 2016 20:39
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http://forum.nasaspaceflight.com/index.php?topic=39004.msg1472751#msg1472751...Most of these packages do not have standard facility to plot the energy density, but as Monomorphic has shown one can visualize the energy density peak from the E and B fields. If one has the small diameter below the cutoff for an open waveguide, there maybe a region close to the small end without a significant intensity of the fields, which is not as good as having the energy density peak as close as possible to the small end.
Violate Roger's 0.82 rule and there will be no thrust. Don't need any software package to tell you that as none can.
Can someone better explain this 0.82 rule? 0.82 of what value?Quote from: TheTravellerRoger's 0.82 cutoff rule
Just a crude approximation/ hand rule to calculate the cut off diameter for a cylindrical waveguide for TE01 mode.
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)
Thanks! I wonder how this cutoff rule works with the wedge geometry.
This is the related equation for rectangular waveguides.
"grenz" is german for border =cutoff (λc; fc)
So for TE01
fc=c/(2a) where a is the longer dimension of the small end of the wedge.
The numbering of the modes is standardized. The dimension b is chosen as b < a, and the first index m gives the variation of the field along a. The TE10 mode then has the lowest cutoff frequency and is called the dominant mode. All other modes have higher cutoff frequencies (except, of course, in the case of the square cross-section for which TE01 has the same cutoff frequency). Guides are usually designed so that at the frequency of operation only the dominant mode is propagating, while all higher-order modes are "cutoff".
Rearranging this expression gives the normalized cutoff frequency as functions of the aspect ratio a/b of the guide (where w=b, and ω= 2 Pi f):
Graph (w = b)
Credit: http://web.mit.edu/6.013_book/www/chapter13/13.4.html -
#2397 by X_RaY on 14 May, 2016 20:49
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Dr Rodal is incredibly good in this field of physics therefore I "liked" most of his posts due to this fact not for personal reasons.
Thanks for the great advice!!
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#2398 by Monomorphic on 14 May, 2016 20:56
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fc=c/(2a) where a is the longer dimension of the small end of the wedge.
So a is Dt? Or should a be Wt? Dt = Wb and Db in Shawyer's wedges.
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#2399 by frobnicat on 14 May, 2016 21:13
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.../...
Also, the return to center (RTC) of the torsion wire is resolved. It was not thermal changes, it was a poor design for the top clamp. There was slight friction in a feed-through hole rather than a direct drop from the wire clamp. RTC is very stable, however, it is VERY slow...will need drag weights to "help it along" rather than a typical torsion spring labs use. Not a quick fix to install a return spring...weight will do just fine on the long beam assembly.
Trying (and failing) to visualize your design idea with "drag weights". When you say RTC is slow, is your setup already damped (no or not a lot of oscillations) and the RTC time is a long slow approach to equilibrium, or not already damped and RTC is long because of a long time for oscillations to dissipate (a lot of oscillations) ? Also are you talking of RTC relative to a clean perturbation mostly in torsion (the degree of freedom of interest when measuring thrust induced torque) or swings/tumbling around all axis ?
For an already damped torsion pendulum and considering only the main rotation along wire axis, for a given moment of inertia RTC time will be proportional to (mmh, square root of) mechanical sensitivity (angular deviation to torque ratio, that is inverse of stiffness). Shorter RTC time implies lower mechanical sensitivity, that is higher stiffness . Not saying this is necessarily bad since you have a high resolution LDS. But I can't see how adding weights can decrease RTC time in this specific case (already damped torsion pendulum and considering only the main rotation along wire axis).
