What relative magnetic permeability did you input into MEEP for copper? Did you input a value less than one?

Quote What relative magnetic permeability did you input into MEEP for copper? Did you input a value less than one?Unfortunately I don't have a Drude model for copper at 2 GHz. We are cautioned in the meep literature that meep becomes unstable for magnetic permeability less than, or very much less than 1. I'm still using perfect metal and thick sheets at that. It's the computer resolution issue still, not enough memory and not enough CPU. Thickness of 0.002 * height, or 0.002 * 9 inches = 0.018 inches is the best I can do in under a day of CPU time.And FYI, I went ahead and calculated the probability of an electron tunnelling through the thin copper (I used Maxima) and found the probability to be 0.0. I guess no one is surprised by that result. That result could be off by a few orders of magnitude, but then what is 0E10?

Quote from: aero on 02/27/2015 04:37 amIt's to bad that we can't find a way that one of the little known or unknown solutions to Maxwell's equations can cause a momentum. There is obviously momentum in electromagnetic waves. There are also some little known electromagnetic effects that create torque.

It's to bad that we can't find a way that one of the little known or unknown solutions to Maxwell's equations can cause a momentum.

Unless of course it is surface electrons excited by the high power resonant RF, tunnelling through the 35 micron copper ends.

Yes, and many other things may happen were not tested nor even proposed. I wonder how people can know that a lot of energy is pumped in this device and imagine nothing will get out. At the very least thermal effects should happen. Testing it in (near) vacuum doesn't eliminate the thermal hypothesis. Even Pioneer's acceleration that was due to thermal effects after all: http://spectrum.ieee.org/aerospace/astrophysics/finding-the-source-of-the-pioneer-anomalyAnother thing that strikes me is that people search for a unique cause explaining everything, which is a bit unlikely.One last thought: If a simulator shows results, build this device and publish results in a mainstream conference. Interesting things may happen ;-)

Is Maxima a descendant of Macsyma? Are they competing with Mathematica and Maple ?(I remember using Macsyma at MIT in the late 1970's )

Stainless Steel 304L (the material of the vacuum chamber) is weakly paramagnetic (the opposite of copper). What relative magnetic permeability did you input into MEEP for the StSt 304L for the chamber? Did you input a value greater than one? Did you report to us that value? I don't recall.

...I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.Been working with the supplier with a machine shop I posted about way back:http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669 ...

Quote from: Mulletron on 02/27/2015 09:00 pm...I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.Been working with the supplier with a machine shop I posted about way back:http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669 ...Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?Somebody should do so!Such an experiment would disprove all the theories with formulas advanced so far. All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.

Notsosureofit:Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)But your formula does not explicitly include a dielectric.QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ? (With no nonlinearities)

Quote from: Rodal on 02/27/2015 09:41 pmQuote from: Mulletron on 02/27/2015 09:00 pm...I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.Been working with the supplier with a machine shop I posted about way back:http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669 ...Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?Somebody should do so!Such an experiment would disprove all the theories with formulas advanced so far. All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.Yes, a cylinder experiment absolutely has to be done.

Quote from: Rodal on 02/27/2015 10:11 pmNotsosureofit:Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)But your formula does not explicitly include a dielectric.QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ? (With no nonlinearities)Presumably no, as long as you have frequency dispersion along the axis you should predict a force.Edit: But it adds complication to the formula in that the second term doesn't cancel out.

Are you planning to conduct an experiment with a dielectric inside a cylindrical cavity?Somebody should do so!Such an experiment would disprove all the theories with formulas advanced so far. All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.

Quote from: Notsosureofit on 02/27/2015 10:18 pmQuote from: Rodal on 02/27/2015 10:11 pmNotsosureofit:Your equation gives zero thrust for a cylinder (constant diameter along its length). Thus, according to your formula a cylinder will give no thrust, only a geometrical object with decreasing diameter will (ditto for Shawyer's and McCulloch's)But your formula does not explicitly include a dielectric.QUESTION: would your line of thinking also give zero thrust for a cylinder with an inserted dielectric at one end of the cavity with constant, homogeneous, isotropic dielectric properties ? (With no nonlinearities)Presumably no, as long as you have frequency dispersion along the axis you should predict a force.Edit: But it adds complication to the formula in that the second term doesn't cancel out.Since your existing formula:NT = P*Q*(1/(4*pi*L*f^3))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))gives NT=0 (zero thrust force) for Rs=Rb (a cylinder), just like Shawyer's and McCulloch's also go to zero for that case, then my understanding from your reply is that this formula is a simplification and that there are terms you neglected that do not go to zero for a cylinder.Do you know the 2nd order term that will still be finite for Rs=Rb (a cylinder) ? Or is the issue deriving an equation for the dispersion including the dielectric?

Since there is really no proven theory of how would an EM Drive generate thrust in space, thereby apparently violating the law of conservation of momentum, and we are still debating whether the experimental measurements are an artifact, the best way to cut to the chase is to test an EM Drive made of Aluminum, and to also test an iron (or a material coated with an interior thin film of Metglas) for the big flat end.

Quote from: Rodal on 02/27/2015 09:41 pmAre you planning to conduct an experiment with a dielectric inside a cylindrical cavity?Somebody should do so!Such an experiment would disprove all the theories with formulas advanced so far. All the formulas (Shawyer's, McCulloch and Notsosureofit...) give zero thrust if both diameters are the same.Before removing the internal dielectric, Shawyer considered a cylindrical EmDrive, in which a conical dielectric would be mandatory for thrust. See his 1990 UK Patent Application GB2229865 "Electrical Propulsion Unit for Spacecraft", attached below.

Seems like wherever we go with this project, Shawyer has already been there - and like as not, built a test model. Given that, is there any way to bring his reasoning/math in line with conventional physics, or do they remain fundamentally irreconcilable? I keep getting the impression he's on to something major, but consistently fails to communicate that 'something' in a coherent manner.

http://www.ebay.com/itm/AERCOM-Microwave-RF-Isolator-Circulator-2-4GHz-20dB-isolation-Low-I-L-TESTED-/281549538390?ssPageName=ADME:L:OU:US:1120Picked up one of these puppies on Ebay to protect my amp. Another example of broken time reversal symmetry in action.Got about an oz of very expensive liquid metal from here:http://www.amazon.com/Gallium-Indium-Eutectic-GaInSn-68-5%25/dp/B00KN92MWW/ref=sr_1_3?ie=UTF8&qid=1425074693&sr=8-3&keywords=galinstanSo back to the copper from way back: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326742#msg1326742I've determined that both the 10 and 16 mil are too thin to serve as frustum walls, thus they will become end caps.Been working with the supplier with a machine shop I posted about way back:http://forum.nasaspaceflight.com/index.php?topic=36313.msg1326669#msg1326669 I'm going that route. The quote I got is: price: $120.00 layout + $51.63 for part + freight. So I have to pay the layout, then anyone else who wants one of these: but built in 16oz copper, with a smooth butt seam inside, and 1/4" flange around edges, can get one for about 50 bucks plus shipping. If all this works out, it'll fulfill my goal of making a replication by DIYers easier. For me, paying the layout plus price about breaks even with buying the sheet myself and fumblefuddeling around trying to solder up a cone at home. So I'm happy. I'll get back with more later, when the items are at home.Again thanks to @Paul March for providing the dimensions for his DUT over at Eagleworks:http://forum.nasaspaceflight.com/index.php?topic=36313.msg1327467#msg1327467