Quote from: StrongGR on 05/20/2015 01:53 pmThanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect....Even if the thrust was not macroscopic, ......

Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.

Quote from: WarpTech on 05/20/2015 02:01 pmBut what happens when a wave is attenuated in a perfectly conducting circular waveguide? That energy is not lost as "heat" because there is no resistance to dissipate it.How will the wave attenuate (lose energy) if no wall losses?

But what happens when a wave is attenuated in a perfectly conducting circular waveguide? That energy is not lost as "heat" because there is no resistance to dissipate it.

I've been reading for several days, I have a question regarding the Maxwell Equations simulations of the Frustrum. I recently read a paper by Tuval and Yahalom (2013). They note that Maxwell's equations are conservative of momentum if the action of forces is taken to be instantaneous. However, if the forces are taken to be propagating at c they are not, in general, for particles. [The momentum of the fields is needed to make conservation of momentum work.]So, the question is: do the simulations of the frustrum account for the retarded forces and fields on and created by the moving charges in the walls/end caps?R.

Quote from: hhexo on 05/20/2015 03:05 pmQuote from: StrongGR on 05/20/2015 01:53 pmThanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect....Even if the thrust was not macroscopic, ......Ok, I've run some back-of-the-envelope calculations in pure engineer style, i.e. just looking at the orders of magnitude.I am handwaving constants as follows:pi^2 = 10^1G = 10^-11c^4 = 81 * 10^32 = 10^34mu0^2 = 16 * pi^2 * 10^-14 = 10^-12h = 1 (this is the height of the frustum, right? Not Planck!)and I get that the geometry related effects have to counter-balance a term which is 10^-32 * U0^4.At this point we are left with a term (1 / ln2(r2/r1)) * (r2^4 * XXX - (r2^6 / r1^2) * YYY), where XXX and YYY are those complicated expressions for the fourth power of the magnetic field that I can't rewrite here. But, out of the two terms the second one is clearly dominant if r2 >> r1.So we're left with 10^-32 * U0^4 * YYY * (r2^6 / r1^2) * (1 / ln2(r2/r1)).I can't quite follow what the order of magnitude of U0 and YYY would be. Can anybody help me? Are they going to be significant powers of ten?Anyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:r2 = 10^2.5 = several hundred metersr1 = 10^-3 = one millimetergiven a frustum height of just one meter... Things get better if h is small, but not too much better.Not very encouraging in terms of practicality. And if U0 or YYY are going to be negative powers of ten, it's going to get even worse.Damn gravity being such a weak force!

Anyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:r2 = 10^2.5 = several hundred metersr1 = 10^-3 = one millimetergiven a frustum height of just one meter...

Quote from: hhexo on 05/20/2015 05:16 pmAnyway, if we just want to balance that 10^-32, we need something like r2/r1 = 10^5.5, for example:r2 = 10^2.5 = several hundred metersr1 = 10^-3 = one millimetergiven a frustum height of just one meter...Gah! Sorry, big glaring error there in my last passage. Because it's r2^6 / r1^2, I can't just consider r2/r1 and scale the ratio.r2=10^5.5r1=1is a correct example.

Quote from: SeeShells on 05/20/2015 03:49 pm...Reading it I had a question (more than one but..). And this is for everyone, why did Eagle Works observe no thrust in a EM device with no HDPE insert? Makes me wonder what effect achiral materials like this that can induce chirality would have with relativity and electromagnetic fields in your equations? Thanks nice work!Excellent question that has been puzzling me for a long time as well. To examine this, first let's recapitulate the statement in Brady et.al.' report:QuoteThere appears to be a clear dependency between thrust magnitude and the presence of some sort of dielectric RF resonator in the thrust chamber. The geometry, location, and material properties of this resonator must be evaluated using numerous COMSOLŪ iterations to arrive at a viable thruster solution. We performed some very early evaluations without the dielectric resonator (TE012 mode at 2168 MHz, with power levels up to ~30 watts) and measured no significant net thrust My exact solution for the truncated cone, given the dimensions reported by Paul March and also used in NASA's COMSOL FEA analysis, for mode shape TE012, gives me a natural frequency of:2.2024 GHzSo they operated at frequency of 2.168 GHz which is 1.59% away from the frequency given by the exact solution. Perhaps the reason was that that was the frequency given by the Finite Element analysis for mode TE012 (let's recall that the Finite Element solution converges from below to the correct eigensolution, and only for an infinite Finite Element mesh one can theoretically converge to the eigensolution). So, they were looking in the right frequency range for TE012.My recollection is that Paul March thought that the reason maybe that to generate thrust without a dielectric one needs to provide Amplitude, Frequency and Phase Modulation, and that at the time of the experiments detailed in the NASA report ( http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdf ), they were not providing the needed AM, FM and PM modulation. I was not able to find the exact quotation (I wonder why, given the fantastic search function we have at this site ). I was only able to find this quotation from Paul March, in answer to this question:Quote from: Star-Drive(Shawyer and the Chinese used the magnetron excited TE012 mode in their frustum cavities without dielectrics being present.)

...Reading it I had a question (more than one but..). And this is for everyone, why did Eagle Works observe no thrust in a EM device with no HDPE insert? Makes me wonder what effect achiral materials like this that can induce chirality would have with relativity and electromagnetic fields in your equations? Thanks nice work!

There appears to be a clear dependency between thrust magnitude and the presence of some sort of dielectric RF resonator in the thrust chamber. The geometry, location, and material properties of this resonator must be evaluated using numerous COMSOLŪ iterations to arrive at a viable thruster solution. We performed some very early evaluations without the dielectric resonator (TE012 mode at 2168 MHz, with power levels up to ~30 watts) and measured no significant net thrust

(Shawyer and the Chinese used the magnetron excited TE012 mode in their frustum cavities without dielectrics being present.)

...I honestly think we have a "physical" system in the EM Drive if we use magneto-chiral matter (HDPE is a weak one in the way the carbon atoms are layered). REF: http://www.nature.com/nmat/journal/v7/n9/full/nmat2256.html

Note here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero

As promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.

QuoteNote here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero Doesn't that imply zero net thrust?

Quote from: hhexo on 05/20/2015 05:42 pmr2=10^5.5r1=1is a correct example.It is my approximation: r2 increasingly large and r1 decreasing toward zero otherwise one should change the final formula. So, in your example it would be better to have r2=10 m and r1=10^-3 m and so on. This is a cone.

r2=10^5.5r1=1is a correct example.

Another way around, as suggested above, is to fill with some material the cavity. The formula goes like mur^-2 and mur much smaller than 1.

Quote from: deltaMass on 05/20/2015 06:14 pmQuoteNote here that we are choosing to work with the momentum density associated with the canonical energy momentum tensor rather than the Poynting vector; the latter is expected to integrate to zero Doesn't that imply zero net thrust?Certainly yes for a linear isotropic system, but apparently not for an anisotropic nonlinear system as the one they are considering. They are considering a 4th order term.

Quote from: StrongGR on 05/20/2015 05:46 pmQuote from: hhexo on 05/20/2015 05:42 pmr2=10^5.5r1=1is a correct example.It is my approximation: r2 increasingly large and r1 decreasing toward zero otherwise one should change the final formula. So, in your example it would be better to have r2=10 m and r1=10^-3 m and so on. This is a cone. ...What if we considered other shapes? We started with the truncated cone because of the EmDrive, but what if we considered a pillbox cavity (which is basically a rounded cone...)? All the equations would be different, for sure, but it might be worth exploring. Maybe in another thread. ...

I'm incredibly appreciate of all the work that went into the dimensional estimates of the various drives but I'm having some difficulty understanding where the dimensional values came from regarding the attached image:Shawyer reported a large plate diameter of .28m - so using that as a given.Going straight off the image (not taking into account perspective) at the junction of the cone and the cylinder I get about .169m for the small plate diameter. This is a difference of 40mm with what was originally estimated and about 73mm from Dr. Rodal's calculation (96.13) of the small plate diameter. This is an enormous difference and I can't believe it. I'm going to model this and lay it on top of the photo in perspective so I can find out what the numbers are closer to.Also the cone length was estimated to be .345m - which is larger than the plate diameter. Between which two points is this measurement for?