I've been going over the figures here againhttp://emdrive.wiki/Experimental_ResultsYang is ahead of the pack by a very decent margin on both thrust and specific thrust (N/W).Consider:- Yang is at a Chinese establishment with ties to the military.- Yang does not respond to questions about her experiment.- It does not seem unreasonable to imagine that disinformation is at work here.- A perfect conspiratorial excuse for any such data is "military secrecy".So I'll just state flat out that I don't give Yang's results any credence.I also know that nobody here can prove me wrong unless they reproduce her results.

Quote from: deltaMass on 08/06/2015 06:53 PM@Rodal: Well, because the specific heat of copper isn't zero, it takes a finite time to incur a given temperature rise for a given power input. And when there's an impedance mismatch, the dissipated power will fall and so the corresponding temperature changes will be less than in the matched condition.On the other hand, measurable mismatch changes are sensed at the ports with response times on order roughly the cavity dimensions divided by the speed of light. This is far faster than measurable temperature changes.However, if the EmDrive is actually generating thrust, the statement about the equivalence of high- and low-Q dissipation may not be true. That's really the reason I'm pointing this out.There seems to be some assumption of reasonably monotonic behavior such that thermal latency will not be a problem...While the assumption of reasonable behavior applies to other known systems, the EM Drive output (non-repeatable, different by orders of magnitude between researchers) seems to not abide by that assumptionFor example, the issue of vibrations of unknown magnitude and frequency being required to "engage" the drive, etc...

@Rodal: Well, because the specific heat of copper isn't zero, it takes a finite time to incur a given temperature rise for a given power input. And when there's an impedance mismatch, the dissipated power will fall and so the corresponding temperature changes will be less than in the matched condition.On the other hand, measurable mismatch changes are sensed at the ports with response times on order roughly the cavity dimensions divided by the speed of light. This is far faster than measurable temperature changes.However, if the EmDrive is actually generating thrust, the statement about the equivalence of high- and low-Q dissipation may not be true. That's really the reason I'm pointing this out.

Quote from: deltaMass on 08/06/2015 07:27 PMI've been going over the figures here againhttp://emdrive.wiki/Experimental_ResultsYang is ahead of the pack by a very decent margin on both thrust and specific thrust (N/W).Consider:- Yang is at a Chinese establishment with ties to the military.- Yang does not respond to questions about her experiment.- It does not seem unreasonable to imagine that disinformation is at work here.- A perfect conspiratorial excuse for any such data is "military secrecy".So I'll just state flat out that I don't give Yang's results any credence.I also know that nobody here can prove me wrong unless they reproduce her results.So that means that you give Shawyer the most credence is that right? as you say that he has the coolest experimental setup?

...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.

About cavity energy.When an optimal impedance match is achieved between the RF source and its load (the cavity), all the forward power gets dissipated in the cavity walls as real ohmic heat. Thus maximum heating corresponds to maximum energy within the cavity, the condition that is being sought. Notionally then it would be possible to keep the impedance match optimal over temperature changes by simply monitoring the cavity temperature (I'm assuming that the mode does not switch). The latency of thermal feedback is slower than a 1-port or 2-port VSWR measurement, it's true, but it should nevertheless work well. And it can be made contactless.Because there's no free lunch, a perfectly matched high-Q cavity and a perfectly matched low-Q cavity will dissipate exactly the same amount of power in ohmic heating for the same input power. At least, that's the theory.

Indeed, one can actually build a self-accelerating system. But it has limited utility because of its limited lifetime. It works like this.Take a railroad car and inside place another smaller car or "puck". The puck starts out full of sand, but continually leaks sand straight out the bottom via a slit, contactlessly. The puck bounces between the two walls of the cart, elastically. You give the system an initial push and you'll see it self-accelerate until the sand is all gone.That's because it always hits the forward wall with more momentum than it will hit the back wall

Quote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.Todd

Quote from: WarpTech on 08/06/2015 09:13 PMQuote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.ToddIf F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart. Specifically, as the size of the waveguide decreases Vp increases. An increase in Vp decreases F. That would make the force on the little end --smaller-- than the force on the big end.Is that right?

Quote from: Rodal on 08/06/2015 06:57 PMQuote from: deltaMass on 08/06/2015 06:53 PM@Rodal: Well, because the specific heat of copper isn't zero, it takes a finite time to incur a given temperature rise for a given power input. And when there's an impedance mismatch, the dissipated power will fall and so the corresponding temperature changes will be less than in the matched condition.On the other hand, measurable mismatch changes are sensed at the ports with response times on order roughly the cavity dimensions divided by the speed of light. This is far faster than measurable temperature changes.However, if the EmDrive is actually generating thrust, the statement about the equivalence of high- and low-Q dissipation may not be true. That's really the reason I'm pointing this out.There seems to be some assumption of reasonably monotonic behavior such that thermal latency will not be a problem...While the assumption of reasonable behavior applies to other known systems, the EM Drive output (non-repeatable, different by orders of magnitude between researchers) seems to not abide by that assumptionFor example, the issue of vibrations of unknown magnitude and frequency being required to "engage" the drive, etc...As a real-world example, fine-grained CPU power management (i.e. fast decisions) has moved away from using thermal diodes since their response time is too slow. (fast decisions need microsecond responses vs 10's or 100's of milliseconds from thermal diodes) Since the EM drive cavity charge/discharge times are on the order of microseconds, I'm rather doubtful that the thermal data would be useful in a true control loop feedback for any EM drive experiment.However, perhaps the temperature might be useful to post-process and evaluate how much power was actually received/dissipated by the copper in the frustum during an experimental run. (i.e. temperature = integration of ohmic losses over time) Unfortunately, the possible effects of heating moisture within the frustum along with any number of other variables will introduce noise into the temperature reading (cooling from turbulent air around exterior of frustum, humidity at moment of test, etc). Still worth a look to see how well frustum temperature correlates with impedance matching... but as an input into a real-time EM drive feedback loop, I'd put my money on VSWR measurement as being far more useful.

As to how you can reconcile that the force being larger on the Big End results in motion towards the Small End

Quote from: Rodal on 08/06/2015 10:18 PMAs to how you can reconcile that the force being larger on the Big End results in motion towards the Small EndThat's why I prefer McCulloch's theory: according to MiHsC, as they travel towards the big end, photons gain momentum. When they bounce back towards the small end, they loose momentum. So to obey CoM, the cavity has to move from the big end towards the small end.

I tried to model TheTraveller's EmDrive Mark 2 in 3D, which was a bit difficult since the only dimensions TT provided were end diameters D_{s} and D_{b} and the length (R_{b} - R_{s})^{*}D_{s} = 159 mmD_{b} = 400 mmR_{b}-R_{s} = 240.7 mmMoreover the dimensions in his original drawing are not proportionally on scale, maybe due to a phenomenon known as the Yang Effect But with trial & error I finally found his "magical proportions": Mr. T inscribed his spherical cone (cone + spherical base) within a cube! Hence:R_{s} = D_{s} = 159 mmR_{b} = D_{b} = 400 mmhalf-cone angle = 30°@TheTraveller: Very clever! The more I look at your design, the more I like it.@Rodal: can you please measure mode and resonant frequency with COMSOL for such a cavity? This is important because this cavity will be monolithically fixed (no tunable small end).@SeeShells: It should please you in your quest for those kinds of mathematical relations, following your love with the golden ratio * Big and small radius from the cone apex

I tried to model TheTraveller's EmDrive Mark 2 in 3D, which was a bit difficult since the only dimensions TT provided were end diameters D_{s} and D_{b} and the length (R_{b} - R_{s})^{*}D_{s} = 159 mmD_{b} = 400 mmR_{b}-R_{s} = 240.7 mmR_{s} = D_{s} = 159 mmR_{b} = D_{b} = 400 mmhalf-cone angle = 30°

I'll be the first to point out that the folks here have probably had quite enough of me discussing CoE and over-unity. But to Bae...I had the same thought. The situation is indeed similar. He is asking us to believe that thrust is proportional to the number of bounces, which is directly related to the Q. So he also has a P*Q factor in his thrust equation. And the obvious question is: how can it be that a cavity which is being pumped steady-state with input power P can yield a force that depends on P*Q? How can that be sustainable for seconds and minutes? It appears that one's extracting more than goes in. I confess to being uncertain about this. I saw his experiment where he got thrust that was indeed 2*P*Q/c. The saving grace might well be that photon mirrors only become efficient as the mirror velocity with respect to the light source approaches c. By this logic, he is only sipping at each beam, and despite the fact that multiple reflections are simultaneously in play (which intensifies the beam of course) he is only taking a small fraction out of each individual beam.But that's an unsatisfactory explanation too, because one could engineer around it. Probably only a mathematical argument could settle it for me.

The Q factor of a resonator depends on the optical frequency ν0 (nu0), the fractional power loss l per round trip, and the round-trip time Trt. (assuming that l << 1)

For a resonator consisting of two mirrors with air (or vacuum) in between, the Q factor rises as the resonator length is increased, because this decreases the energy loss per optical cycle. However, extremely high Q values (see below) are often achieved not by using very long resonators, but rather by strongly reducing the losses per round trip.

Quote from: flux_capacitor on 08/06/2015 10:58 PMI tried to model TheTraveller's EmDrive Mark 2 in 3D, which was a bit difficult since the only dimensions TT provided were end diameters D_{s} and D_{b} and the length (R_{b} - R_{s})^{*}D_{s} = 159 mmD_{b} = 400 mmR_{b}-R_{s} = 240.7 mmR_{s} = D_{s} = 159 mmR_{b} = D_{b} = 400 mmhalf-cone angle = 30°I get slightly different dimensions, on the order of tenths of a millimeter. Which brings up the question, how tight are the tolerances on the manufacturing process, and exactly how tight they need to be to get good results?

You think if you achieve cavity resonance (charge/discharge) in microseconds (say less than 20microseconds) that you would see the EM-Drive/Q-thruster thrust phenomena in microseconds?