If I did the calculation right, a volume change of ~125 litres would be needed to account for this apparent weight change, due to thermal ballooning of a sealed volume. That's far too high a change in volume that could be reasonably expected, so it can't be the whole story. Did I calculate this right?air density = 4*10^{-4} Kg/m^{3}

Quote from: deltaMass on 05/18/2015 03:25 AMIf I did the calculation right, a volume change of ~125 litres would be needed to account for this apparent weight change, due to thermal ballooning of a sealed volume. That's far too high a change in volume that could be reasonably expected, so it can't be the whole story. Did I calculate this right?air density = 4*10^{-4} Kg/m^{3}1 litre weights about 1.2 grams, or 12 mN . Heated up by 10 degrees Kelvin it expands by about 1/30 , resulting in buoyancy of approximately 12/30 = 0.4 mN . Definitely enough to make an experiment inaccurate (4mN from 10 litres), albeit probably less than thermal and electromagnetic effects in cabling (which is basically impossible to estimate).The only thing he can do with cabling as it is, is see if it replicates turned upside down, with some accuracy (~5% for example), without excessive fiddling. But it probably won't even if the effect is there because of all the classical forces.

Quote from: txdrive on 05/18/2015 12:39 PMQuote from: deltaMass on 05/18/2015 03:25 AMIf I did the calculation right, a volume change of ~125 litres would be needed to account for this apparent weight change, due to thermal ballooning of a sealed volume. That's far too high a change in volume that could be reasonably expected, so it can't be the whole story. Did I calculate this right?air density = 4*10^{-4} Kg/m^{3}1 litre weights about 1.2 grams, or 12 mN . Heated up by 10 degrees Kelvin it expands by about 1/30 , resulting in buoyancy of approximately 12/30 = 0.4 mN . Definitely enough to make an experiment inaccurate (4mN from 10 litres), albeit probably less than thermal and electromagnetic effects in cabling (which is basically impossible to estimate).The only thing he can do with cabling as it is, is see if it replicates turned upside down, with some accuracy (~5% for example), without excessive fiddling. But it probably won't even if the effect is there because of all the classical forces.Would you agree drilling 6 x 1 mm diameter equally spaced holes around the circumference at both ends of the frustum wall, say 5 mm away from the end plates, and the same around the frustum middle would eliminate any buoyancy or hot air jet false positives?

Quote from: TheTraveller on 05/18/2015 12:54 PMQuote from: txdrive on 05/18/2015 12:39 PMQuote from: deltaMass on 05/18/2015 03:25 AMIf I did the calculation right, a volume change of ~125 litres would be needed to account for this apparent weight change, due to thermal ballooning of a sealed volume. That's far too high a change in volume that could be reasonably expected, so it can't be the whole story. Did I calculate this right?air density = 4*10^{-4} Kg/m^{3}1 litre weights about 1.2 grams, or 12 mN . Heated up by 10 degrees Kelvin it expands by about 1/30 , resulting in buoyancy of approximately 12/30 = 0.4 mN . Definitely enough to make an experiment inaccurate (4mN from 10 litres), albeit probably less than thermal and electromagnetic effects in cabling (which is basically impossible to estimate).The only thing he can do with cabling as it is, is see if it replicates turned upside down, with some accuracy (~5% for example), without excessive fiddling. But it probably won't even if the effect is there because of all the classical forces.Would you agree drilling 6 x 1 mm diameter equally spaced holes around the circumference at both ends of the frustum wall, say 5 mm away from the end plates, and the same around the frustum middle would eliminate any buoyancy or hot air jet false positives?I wouldn't assume that IMHO, on the basis that the Ph.D. thesis that Shawyer uses as his only reference on radiation pressure measurements (Dr. Cullen's thesis) makes it clear that Cullen had to use a mesh as follows to eliminate air effects that have plagued these experiments since Maxwell's times:"drilling 6 x 1 mm diameter equally spaced holes around the circumference at both ends of the frustum wall, say 5 mm away from the end plates" would not be enoughIt is puzzling to a researcher's mind that Shawyer uses Cullen as his only reference for radiation pressure from closed frustum cavities, when Cullen only used open waveguides with constant cross section in his pressure experiments, while Shawyer apparently ignores Cullen's prescription on the need for a wired mesh to eliminate gas effects.

.....Don't see how this relates to what EW, Shawyer and the Chinese did. None of those used the Cullen grid you are suggesting, so why bash Shawyer for not using it and not the others?...

BTW in the Egan paper, could you please show me where he factors in the constantly changing cutoff wavelength, guide wavelength & group velocity as the EM wave bounces from end to end in a frustum with constantly varying diameter? Really can't follow how he calcs the resonate frequency for his frustum.

Quote from: TheTraveller on 05/18/2015 02:31 PM.....Don't see how this relates to what EW, Shawyer and the Chinese did. None of those used the Cullen grid you are suggesting, so why bash Shawyer for not using it and not the others?...As I said, because Shawyer is the only one that uses Cullen as his only reference on pressure radiation for the unorthodox pressure radiation calculations that Shawyer uses. Neither Prof. Yang nor NASA Eagleworks use Cullen as a reference. Pointing out the huge inconsistency in Shawyer using as his only reference a work that explicitly does NOT use cavities but instead uses open waveguides is not bashing somebody, it is pointing out a technical inconsistency that should bother any engineer. Engineers care about accuracy. Engineers that make bridges care about accuracy and aerospace engineers care even more about accuracy (because of the weight constraints in aerospace engineering, the safety factor in aerospace engineering cannot be as high as in civil engineering, hence accuracy is paramount in aerospace engineering). It is something that referees in peer-reviewed journals point out whenever engineering papers are reviewed for publication approval in a peer-reviewed journal. Ditto for Shawyer using Cullen as a reference for what Cullen does not do and ignoring Cullen's prescription on what Cullen recommends to do (to use a mesh).Quote from: TheTraveller on 05/18/2015 02:31 PMBTW in the Egan paper, could you please show me where he factors in the constantly changing cutoff wavelength, guide wavelength & group velocity as the EM wave bounces from end to end in a frustum with constantly varying diameter? Really can't follow how he calcs the resonate frequency for his frustum.I can't help making it any more clear than what Greg Egan did. Greg Egan's paper is extremely clear to me, as he uses textbook material. Concerning the cut-off wavelength and associated cut-off frequencies, they are automatically built-in in the two eigenvalue problems that Egan addresses.When one has an exact solution like Egan, one does not have to artificially impose a cut-off condition as a side-condition. The reason why Shawyer has to impose the cut-off condition as a side-condition is because Shawyer's solution is obviously not an exact solution. Shawyer's solution does not satisfy the Boundary Conditions of the problem as it has been pointed out in the forum previously.The fact that Greg Egan's solution automatically includes the cut-off condition (and not as a side-condition) in the eigenvalue problem is evident in the examples given by Egan: notice that the frequency and mode shape associated with the quantum number p=0, constant in the longitudinal direction of the truncated cone, is automatically cut-off by the eigensolution shown by Egan. The first natural frequency in the examples shown by Egan have p>0.

...And his exact solution to calc the resonant frequency of a frustum with spherical end plates is?

Quote from: TheTraveller on 05/18/2015 03:04 PM...And his exact solution to calc the resonant frequency of a frustum with spherical end plates is?It is clearly given by Egan in here http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html (for modes having the azimuthal quantum number m=0, but arbitrary n and p)Egan even walks you through, step by step, examples of how to solve the eigenproblems for both the Legendre associated function and for the spherical Bessel function (there are textbooks that don't bother to walk the student through on how to solve the eigenproblem).

I really have a problem with this Egan's explanation, attached below, which assumes all the wavelengths inside the cavity are the same length, which we know is not the case. ...

Looks like a photo of Shawyer's 1st knife edge balance beam test rig with the 1st EM Drive inside a sealed Faraday Cage.http://www.shelleys.demon.co.uk/fdec02em.htmScrews around the box seem to align up with the holes in the various mounting plates.

Quote from: TheTraveller on 05/18/2015 03:01 PMI really have a problem with this Egan's explanation, attached below, which assumes all the wavelengths inside the cavity are the same length, which we know is not the case. ...The solution shown in http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html has an infinite number of eigenwavelengths.I don't understand why one would assume that in a practical situation only one eigenwavelength (or equivalently only one eigenfrequency) would take place. I don't think that Egan implies that. On the contrary, Cullen (the reference used by Shawyer to justify his unorthodox calculations) explicitly shows how hard it was for him in his experimental Ph.D. thesis to excite the waveguide at just one eigenfrequency, as many eigenmodes are close to each other. Cullen shows that it was practically impossible, the best he could do was to suppress the magnitude of other eigenfrequencies to a relatively low magnitude.The reality is the contrary of what you apparently assume, Egan in http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html discusses a solution to an eigenvalue problem having an infinite number of eigenwavelengths and in reality it is very difficult to have the cavity resonate at a single pure eigenwavelength, as many eigenmodes are close to each other. This is particularly so when one uses a magnetron !

Quote from: TheTraveller on 05/18/2015 03:40 PMLooks like a photo of Shawyer's 1st knife edge balance beam test rig with the 1st EM Drive inside a sealed Faraday Cage.http://www.shelleys.demon.co.uk/fdec02em.htmScrews around the box seem to align up with the holes in the various mounting plates.I like this, because besides of eliminating hot air buoyancy as an explanation, by showing the device producing thrust in any direction, the logical next step for any DYI fan is to run the device fully enclosed in a Faraday cage, showing it producing thrust in any direction as well. Such setup would still be affordable for most DYI fans capable of building an Emdrive.And from there, what's needed are vacuum tests. But those are significantly more expensive for a regular DYI fan, and would be easier done by people in universities and labs with facilities like vacuum chambers.

Quote from: Rodal on 05/18/2015 01:39 PMQuote from: TheTraveller on 05/18/2015 12:54 PMQuote from: txdrive on 05/18/2015 12:39 PMQuote from: deltaMass on 05/18/2015 03:25 AMIf I did the calculation right, a volume change of ~125 litres would be needed to account for this apparent weight change, due to thermal ballooning of a sealed volume. That's far too high a change in volume that could be reasonably expected, so it can't be the whole story. Did I calculate this right?air density = 4*10^{-4} Kg/m^{3}1 litre weights about 1.2 grams, or 12 mN . Heated up by 10 degrees Kelvin it expands by about 1/30 , resulting in buoyancy of approximately 12/30 = 0.4 mN . Definitely enough to make an experiment inaccurate (4mN from 10 litres), albeit probably less than thermal and electromagnetic effects in cabling (which is basically impossible to estimate).The only thing he can do with cabling as it is, is see if it replicates turned upside down, with some accuracy (~5% for example), without excessive fiddling. But it probably won't even if the effect is there because of all the classical forces.Would you agree drilling 6 x 1 mm diameter equally spaced holes around the circumference at both ends of the frustum wall, say 5 mm away from the end plates, and the same around the frustum middle would eliminate any buoyancy or hot air jet false positives?I wouldn't assume that IMHO, on the basis that the Ph.D. thesis that Shawyer uses as his only reference on radiation pressure measurements (Dr. Cullen's thesis) makes it clear that Cullen had to use a mesh as follows to eliminate air effects that have plagued these experiments since Maxwell's times:"drilling 6 x 1 mm diameter equally spaced holes around the circumference at both ends of the frustum wall, say 5 mm away from the end plates" would not be enoughIt is puzzling to a researcher's mind that Shawyer uses Cullen as his only reference for radiation pressure from closed frustum cavities, when Cullen only used open waveguides with constant cross section in his pressure experiments, while Shawyer apparently ignores Cullen's prescription on the need for a wired mesh to eliminate gas effects.18 holes, 1mm diameter = 14.2mm^2 of total air hole area to allow the internally heat air to not cause buoyancy or hot air gas jet false positives. Much simpler than the Cullen grid. As engineer, the rule is KISS.Don't see how this relates to what EW, Shawyer and the Chinese did. None of those used the Cullen grid you are suggesting, so why bash Shawyer for not using it and not the others?BTW in the Egan paper, could you please show me where he factors in the constantly changing cutoff wavelength, guide wavelength & group velocity as the EM wave bounces from end to end in a frustum with constantly varying diameter? Really can't follow how he calcs the resonate frequency for his frustum.

Interesting version of the C band Flight Thruster and dimensions.www.slideshare.net/Stellvia/emdrive-presentation-at-space-08-conference-barbican-london-presentation

....This grid might account for the unusual thermal signature of the large end base plate. Pardon my "warping" of the perspective to get it as close to a circle as time would permit.This thermal pattern doesn't remind me of a concave surface pattern. It "could" be that the large base is copper-clad PC board with the "bulls eye" patern etched on the upper surface (connected to ground potential) and the bottom (outside) surface of the PC board completely copper. Perhaps this is the reason for the apparent thickness of the large base, a "suspended" pattern a few mm above the copper.Just speculation...hope there is someone who can connect the dots. Original thermal image: https://d253pvgap36xx8.cloudfront.net/editor_uploads/1277/2015/05/06/NASA_emdrive2.jpg

Quote from: TheTraveller on 05/18/2015 04:16 PMInteresting version of the C band Flight Thruster and dimensions.www.slideshare.net/Stellvia/emdrive-presentation-at-space-08-conference-barbican-london-presentationNever seen this before Not a truncated cone.What happened to the spherical waves? Is there an explanation to its flat sides and the departure from the conical shape ?