Team SPR has clever people working on the superconducting EMDrive.In the latest patent application it is revealed the SC EMDrive is only driven for 0.2 of the 1st TC as attached.In reference to the total amount of power that must flow into a TC driven device to fully charge it, the power input at 0.2 of the 1st TC is almost nothing as attached.This means the Rf energy does not spend much time bouncing from end plate to end plate and inducing thermal heating effects on the walls of the SC frustum. Instead the input Rf energy is quickly converted into kinetic.Seems pulsing the Rf energy input for 0.2 of 1 frustum TC maybe something to investigate further.

Quote from: WarpTech on 08/01/2015 05:55 PM@Notsosureofit,EDIT: Never mind. I answered my own dumb question. I shouldn't post before I finish my coffee. The bandwidth used for Q is not the same, but f is. Therefore, it results in;acceleration g = (c^{2}/L)*(delta_f/f), wheredelta_f = (2pi/f)*(fs^{2} - fb^{2})Q = f/delta_b (b for bandwidth)N*T = (P/2pi*L*f)*(delta_f/delta_b)This implies a lower frequency, large delta_f/L implies a short length, wide half-angle. Small delta_b implies narrow bandwidth. So a wide stubby frustum with both a smaller small end and a larger big end? Using a narrow band RF amplifier rather than a Magnetron. Just as @TT and Shawyer have said.Thank you.ToddThis should make what I'm saying a little clearer. Using @Notsosureofit's theory. From this, we can see how the design of the frustum could be maximized for thrust. What puzzles me is why we are working in microwaves when the equation clearly shows that lower frequency is better. Todd

@Notsosureofit,EDIT: Never mind. I answered my own dumb question. I shouldn't post before I finish my coffee. The bandwidth used for Q is not the same, but f is. Therefore, it results in;acceleration g = (c^{2}/L)*(delta_f/f), wheredelta_f = (2pi/f)*(fs^{2} - fb^{2})Q = f/delta_b (b for bandwidth)N*T = (P/2pi*L*f)*(delta_f/delta_b)This implies a lower frequency, large delta_f/L implies a short length, wide half-angle. Small delta_b implies narrow bandwidth. So a wide stubby frustum with both a smaller small end and a larger big end? Using a narrow band RF amplifier rather than a Magnetron. Just as @TT and Shawyer have said.Thank you.Todd

TT,I see you're opting for spherical end plates, how are you going to make the end plates?shape hammering or CNC milling from a larger block of copper/aluminum?Aluminum end plates might be easier to get and cheaper to mill, no?

Spin forming.... interesting.. had not seen that one yet, but as i see, it needs a pre-made template form to pull the sheet on... Unless you plan for mass production, isn't that an expensive way to produce a prototype?

FYI Yang's Sidewalls in degrees.Shell

Quote from: SeeShells on 08/03/2015 11:34 AMFYI Yang's Sidewalls in degrees.ShellMedian = 14.97Mean = 16.5257Max = 25.48Min = 11.45Skewness = 0.667531Kurtosis = 1.91077It is a very skewed distribution, hence the Median is a better measure of central tendency. This confirms that 15 degrees is the best estimate for the cone half-angle used by Yang, in agreement with my prior analysis

Quote from: Rodal on 08/03/2015 11:49 AMQuote from: SeeShells on 08/03/2015 11:34 AMFYI Yang's Sidewalls in degrees.ShellMedian = 14.97Mean = 16.5257Max = 25.48Min = 11.45Skewness = 0.667531Kurtosis = 1.91077It is a very skewed distribution, hence the Median is a better measure of central tendency. This confirms that 15 degrees is the best estimate for the cone half-angle used by Yang, in agreement with my prior analysisThose drawing are just for reference. They are not representative of actual frustum builds. We have no photographs and no real frustum dimensions. And thus Prof Yang apparently protects Chinese state secrets.

Quote from: TheTraveller on 08/03/2015 11:53 AMQuote from: Rodal on 08/03/2015 11:49 AMQuote from: SeeShells on 08/03/2015 11:34 AMFYI Yang's Sidewalls in degrees.ShellMedian = 14.97Mean = 16.5257Max = 25.48Min = 11.45Skewness = 0.667531Kurtosis = 1.91077It is a very skewed distribution, hence the Median is a better measure of central tendency. This confirms that 15 degrees is the best estimate for the cone half-angle used by Yang, in agreement with my prior analysisThose drawing are just for reference. They are not representative of actual frustum builds. We have no photographs and no real frustum dimensions. And thus Prof Yang apparently protects Chinese state secrets.Yeah, let's just ignore all those drawings and let's use VooDoo to come up with a Yang geometry, because the drawings don't agree with pre-conceived notions. Let's don't believe our eyes

So we have 7 drawings from the author's in the author's previously ballyhooed "peer-reviewed papers".The smallest angle in any of these drawings is practically TWICE as much as 6 degrees, the median is 15 degrees and the max is 25 degrees. Yet we are going to ignore all these drawings in peer-reviewed journals?The drawings are part of a big master conspiracy?The peer-reviewers are part of a master conspiracy or they just did not realize that the drawings did not match the text?Since we are at it, why not just say that Yang used a cylinder for her experiments?

...What I find interesting is it seems the stress values are located big end or small end and little shows up in the sidewalls which is what the current theory of Yang's proposes as to thrust. I find it interesting that we see little or no enhanced stress on the side wall which if we did would tend to go hand and hand with their theories. ....Shell

....This confirms that 15 degrees is the best estimate for the cone half-angle used by Yang, in agreement with my prior analysis. There is absolutely no basis in any of her drawings for a cone half-angle of 6 degrees