The Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.

Quote from: TheTraveller on 05/14/2015 04:34 AM....The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.OKThanks for the answer Very clear answerThat's NOT what deltaMass had written (there is no N=1,2,3,4, in equation 5)That's exactly the same factor of 1.7 I derived, based on TE110Now I'm going to sleep.

....The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.

Quote from: deltaMass on 05/14/2015 04:26 AMThe Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.Shawyer always says the small end cutoff wavelength should be just above small end cutoff for best thrust generation. The higher the Df, the higher the resultant thrust from T = 2 Df Po Q / cSo the small end operating at just above cutoff is important to achieve a Df close to 1.

Quote from: TheTraveller on 05/14/2015 04:40 AMQuote from: deltaMass on 05/14/2015 04:26 AMThe Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.Shawyer always says the small end cutoff wavelength should be just above small end cutoff for best thrust generation. The higher the Df, the higher the resultant thrust from T = 2 Df Po Q / cSo the small end operating at just above cutoff is important to achieve a Df close to 1.Not if you take that Df at face value, no. Let's do some high school algebra. I use x to mean lambda, and I can't be bothered using suffices throughout.We start with the relation we know to be true: x2 >= x1 >= x0. We now quantify this with these definitions:x2 := a*x1 (a >= 1)x1 := b*x0 (b >= 1)The original Df expression: Df = x0*(x2 - x1) / (x1*x2 - x0^{2})becomes, using the above definitions:Df = b*(a - 1) / (a*b^{2} - 1)What's Df when x1 = x0?This is tantamount to setting b=1. Doing that, we getDf = (a-1) / (a-1) = 1. QED.i.e. it doesn't matter what the value of 'a' might be; i.e. it doesn't matter about cutoff.

I ought to mention that Shawyer does not specify how to compute v_{g} or lambda_{g}. What he does do is specify the functional relation between them - i.e. v(lambda) or lambda(v) - and that's all.Anyone know how to compute v_{g} without using lambda_{g}?Anyone know how to compute lambda_{g} without using v_{g}?

Wrong. (You need AND)

Quote from: deuteragenie on 05/13/2015 08:57 PMAny resemblance with existing or future device(s) is purely coincidental.Made with MEEP. Does not use the correct frequencies, material, etc. but looks cute.Is that a 2-Dimensional model with MEEP? Maxwell's equations in a flat 2-D surface?modeling the truncated cone as a FLAT trapezium ?Is the magnetic field (for TM modes) a point scalar (only able to have + or - sign but the direction is always perpendicular to the surface) instead of being a vector in the azimuthal (circumferential) direction ?What are we seeing out of the EM Drive? evanescent wave field?If the answers are yes, do you have enough memory to run a 3D model instead?Thanks

Any resemblance with existing or future device(s) is purely coincidental.Made with MEEP. Does not use the correct frequencies, material, etc. but looks cute.

Quote from: deuteragenie on 05/13/2015 08:57 PMAny resemblance with existing or future device(s) is purely coincidental.Made with MEEP. Does not use the correct frequencies, material, etc. but looks cute.You might want to increase your skin thickness or Meep resolution. Looks kind of like a Gaussian source with some of the shorter wavelengths bouncing around and longer wavelengths stepping over the boundary (numerically)?Yes, I noticed that and tried to improve the resolution with not much success so far. I'll keep playing.If you are using perfect metal for the skin, then the thickness won't change the result but will help to avoid the model numerically spanning the skin. If you are modelling copper in the GHz range, please tell me about your model as I have struggled for 6 months trying to find a Drude model for copper at 2 GHz.

Have modified my Shawyer Df calculator and best Df scanner as per the derived Shawyer Df equation, using cutoff wavelength and guide wavelength as per microwave industry supplied equations. I assume Shawyer did not supply these equations in his papers as they are equations that should be known to microwave industry individuals skilled in the art. Anyway they are now in the public record.The scanner still sweeps the frequency range 0Hz to 10GHz but reports the frequency that generates a Df as close to 1 as possible but not over.The attached results are very interesting as the frequency needed to get the Df to just below 1 is very close to the Rf driving frequency used to generate Lambda0 or free wavelength in the selected medium.While I'm still testing the spreadsheet, which meets both of Shawyers boundary conditions, the results for my Flight Thruster design are looking to be very close to what I could build. Bit of dimension tweaking should get the Df 1 frequency to the 3.85GHz Shawyer used.Will post the spreadsheet after a bit more testing.

Published 12 May 2015This represents the current state of the art.http://iopscience.iop.org/0953-8984/27/21/210301/articlehttp://iopscience.iop.org/0953-8984/27/21 (TOC)I'm pretty excited about what could come out of applying these materials too...maybe a way forward:http://www.lap.physik.uni-erlangen.de/lap/?page=research_krstic_chiral&language=en

Quote from: Mulletron on 05/14/2015 10:51 AMPublished 12 May 2015This represents the current state of the art.http://iopscience.iop.org/0953-8984/27/21/210301/articlehttp://iopscience.iop.org/0953-8984/27/21 (TOC)I'm pretty excited about what could come out of applying these materials too...maybe a way forward:http://www.lap.physik.uni-erlangen.de/lap/?page=research_krstic_chiral&language=enExcited? I'm more than excited. I think I've reached my limits of current understanding and applying these effects to the EM device. I need more time to read and digest because every article seems to have a little something that fits in the overall scheme of why. Mullutron I think you have a tiger by the tale here.Like this one.Transfer of linear momentum from the quantum vacuum to a magnetochiral moleculeM Donaire1, B A van Tiggelen2 and G L J A Rikken3Show affiliationsM Donaire et al 2015 J. Phys.: Condens. Matter 27 214002. doi:10.1088/0953-8984/27/21/214002Received 14 April 2014, accepted for publication 2 July 2014. Published 12 May 2015. © 2015 IOP Publishing LtdAbstractIn a recent publication [1] we have shown using a QED approach that, in the presence of a magnetic field, the quantum vacuum coupled to a chiral molecule provides a kinetic momentum directed along the magnetic field. Here we explain the physical mechanisms which operate in the transfer of momentum from the vacuum to the molecule. We show that the variation of the molecular kinetic energy originates from the magnetic energy associated with the vacuum correction to the magnetization of the molecule. We carry out a semiclassical calculation of the vacuum momentum and compare the result with the QED calculation.