[...You made statements Shawyer ignores the length. That statement is not correct.You made statements Shawyer has not fully disclosed data. That is his right.

the latest interpretation of Shawyer's Design Factor, completely ignores the cavity length.

Quote from: Rodal on 05/15/2015 12:21 pm3D Plot of Shawyer's Design Factor vs. frequency and vs. small diameter; for same big diameter as Flight Thruster, but with the small diameter ranging from zero to same size as big diameter.Remember: according to Shawyer the Design Factor multiplies the Power Input and the Q. The higher the Design Factor, the higher the thrust of the EM Drive, the smaller the Design Factor, the smaller the thrust.Observe that at high frequency, the Design Factor changes almost linearly with small diameter, such that the Design Factor goes to zero as the small diameter approaches the big diameter.The Design Factor approaches 1 for the small diameter approaching zero.As the small diameter approaches zero, the cut-off frequency clips the Design Factor, such that to be able to have a smaller small diameter one has to operate at higher frequency (in order to avoid cut-off).A very nice feature of Shawyer's Design Factor (as opposed to McCulloch's formula) is that Shawyer's Design Factor incorporates the cut-off frequency and hence it prevents consideration of a pointy cone, as the cut-off prevents too small of a small diameter to be considered.The highest value of the Design Factor is reached at frequencies just a little over the cut-off frequency for the small end:Cut-Off frequency for small end= cM/(cst*sD) wheresD= small end diameter (m)cst=1.7062895542683174cM = light speed in selected medium (m/s) = 299705000 (m/s) (speed of light in air) = 299792458 (m/s) (speed of light in vacuum) The Design Factor has little dependence on frequency, except near the cut-off frequency. The Design Factor of Shawyer asymptotically approaches this value for high frequencies (it becomes practically independent of frequency)Limit[DesignFactor, f -> Infinity] = (bD^2 - sD^2)/(bD^2 + sD^2)wherebD = big end diameter (m)sD= small end diameter (m)Whether Shawyer's Design Factor is correct, remains to be proven. For example, Shawyer's Design Factor predicts that the smaller the small diameter the better (hence larger cone angles, for constant frustum length), in contrast with Todd's conjecture that the highest attenuation the better (which leads to small cone angles ~7.5 degrees as the optimal design).Reference: formula for Design Factor here: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1374110#msg1374110designFactor = (bD^2 - sD^2)/( (bD^2)*Sqrt[1 - (cM/(bD*cst*f))^2] + (sD ^2)*Sqrt[1 - (cM/(cst*f*sD))^2] )bD = big end diameter (m)sD= small end diameter (m)f = applied frequency (Hz)cst=1.7062895542683174cM = light speed in selected medium (m/s) = 299705000 (m/s) (speed of light in air) = 299792458 (m/s) (speed of light in vacuum)Very interesting plot. Was working Excel to do the same, but why reinvent the wheel? Nice data.One element we have yet to nut out is external applied Rf cavity resonance between the 2 end plates. Is vital for this to happen, as without resonance to applied Rf frequency, there will be no thrust developed as Q will be very low.Simple to calc wavelength 1/2 wavelength between and call it done but with constantly variable guide wavelength between the end plates, my gut says end plate resonance is not 1/2 of Lambda0 (free) as Lambda0 only exists outside the cavity, nor Lambda1 (big end) guide wavelength nor Lambda2 (small end) guide wavelength. Additionally guide wavelength varies continuously, at each point of diameter change, from one end of the cavity to the other.So how to calc the external Rf frequency required to bring the cavity into end plate to end plate resonance and allow Q to grow very large?Of course all this assumes any external waveguide / coax is tuned (SWR 1:1) to supply Rf energy with minimal loss.It is important as Shawyer says in the 1st attachment.In the 2nd attachment there is notes an interesting object built into the side of the cavity.It is used as in the 3rd attachment to assist the cavity getting into resonance?

3D Plot of Shawyer's Design Factor vs. frequency and vs. small diameter; for same big diameter as Flight Thruster, but with the small diameter ranging from zero to same size as big diameter.Remember: according to Shawyer the Design Factor multiplies the Power Input and the Q. The higher the Design Factor, the higher the thrust of the EM Drive, the smaller the Design Factor, the smaller the thrust.Observe that at high frequency, the Design Factor changes almost linearly with small diameter, such that the Design Factor goes to zero as the small diameter approaches the big diameter.The Design Factor approaches 1 for the small diameter approaching zero.As the small diameter approaches zero, the cut-off frequency clips the Design Factor, such that to be able to have a smaller small diameter one has to operate at higher frequency (in order to avoid cut-off).A very nice feature of Shawyer's Design Factor (as opposed to McCulloch's formula) is that Shawyer's Design Factor incorporates the cut-off frequency and hence it prevents consideration of a pointy cone, as the cut-off prevents too small of a small diameter to be considered.The highest value of the Design Factor is reached at frequencies just a little over the cut-off frequency for the small end:Cut-Off frequency for small end= cM/(cst*sD) wheresD= small end diameter (m)cst=1.7062895542683174cM = light speed in selected medium (m/s) = 299705000 (m/s) (speed of light in air) = 299792458 (m/s) (speed of light in vacuum) The Design Factor has little dependence on frequency, except near the cut-off frequency. The Design Factor of Shawyer asymptotically approaches this value for high frequencies (it becomes practically independent of frequency)Limit[DesignFactor, f -> Infinity] = (bD^2 - sD^2)/(bD^2 + sD^2)wherebD = big end diameter (m)sD= small end diameter (m)Whether Shawyer's Design Factor is correct, remains to be proven. For example, Shawyer's Design Factor predicts that the smaller the small diameter the better (hence larger cone angles, for constant frustum length), in contrast with Todd's conjecture that the highest attenuation the better (which leads to small cone angles ~7.5 degrees as the optimal design).Reference: formula for Design Factor here: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1374110#msg1374110designFactor = (bD^2 - sD^2)/( (bD^2)*Sqrt[1 - (cM/(bD*cst*f))^2] + (sD ^2)*Sqrt[1 - (cM/(cst*f*sD))^2] )bD = big end diameter (m)sD= small end diameter (m)f = applied frequency (Hz)cst=1.7062895542683174cM = light speed in selected medium (m/s) = 299705000 (m/s) (speed of light in air) = 299792458 (m/s) (speed of light in vacuum)

Quote from: TheTraveller on 05/20/2015 12:59 pm[...You made statements Shawyer ignores the length. That statement is not correct.You made statements Shawyer has not fully disclosed data. That is his right.Incorrect. Read again. I stated that your interpretation of Shawyer's Design Factor ignores the cavity length.I wrote: Shawyer's Design Factor, completely ignores the cavity length.

While such a Design Factor, and hence such a formula for thrust force, that completely ignores the cavity length

...T = (2 * Po * Df * Q) / c seems to ignore the requirement that the applied Rf must cause the frustum to operate in electrical length resonance. But in reality it is there.Po does imply the frustum is operating at resonance as otherwise little energy would be transferred from the Rf generator to inside the frustum. Additionally the frustum load impedance must be matched to that of the Rf generator's output impedance or energy will be lost in the xfer.Likewise for Q as using a Rf driving frequency that was not at frustum resonance would generate a low Q.Which says both Shawyers Po and Q imply the frustum is impedance matched to the Rf generator and operating at length resonance with the driving Rf generator.

As promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.

Quote from: TheTraveller on 05/20/2015 01:17 pm...T = (2 * Po * Df * Q) / c seems to ignore the requirement that the applied Rf must cause the frustum to operate in electrical length resonance. But in reality it is there.Po does imply the frustum is operating at resonance as otherwise little energy would be transferred from the Rf generator to inside the frustum. Additionally the frustum load impedance must be matched to that of the Rf generator's output impedance or energy will be lost in the xfer.Likewise for Q as using a Rf driving frequency that was not at frustum resonance would generate a low Q.Which says both Shawyers Po and Q imply the frustum is impedance matched to the Rf generator and operating at length resonance with the driving Rf generator.It is known that the criitcal parameter in the solution of a conical cavity is the cone angle. It is obvious that Shawyer's Design Factor does not take into account the cone angle. As I stated in the note to my original post, since the initial cone angles of the experiments by Shawyer (the Experimental. Demonstration and Flight Thruster) involved small cone angles, according to the published photographs, it is apparent that Shawyer's thrust formula, and hence his Design Factor must be based on a small cone angle approximation.

Quote from: snow on 05/20/2015 09:19 amcan someone explain in normal people terms were is the recoil going from turning on the microwave?shouldn't the metal box push on the microwave just as much as the microwave pushes on the metal box?also, it gets really hot... could the recoil just be delayed?Shawyer's explanation is that the waves bounce back and forth across the device.The speed of the waves depends upon the shape of the container (the container is acting as a wave-guide, which reduces the propagation speed to be less than C). Shawyer asserts that the momentum change is thereby asymmetric, due to the differing effective impact speeds. Note that the wider physics community do not agree with this analysis, although I do not understand the details.However, there are other theories, in particular Dr.White at NASA thinks that the EM fields from the standing waves are interacting with quantum 'virtual particles', effectively pushing off these. There are problems to do with special relativity and conservation of energy that would seem to make this unlikely unless you go and revive some very old theories of the aether and absolute reference frames, which are very marginal (special relativity has been well tested and aether theories abandoned as unproductive).Another theory that allows breaking of local conservation of momentum and which _is consistent with observations is Woodward 'Mach' effects. These relate to accelerating bodies which are changing in internal energy levels. The momentum is effectively transferred to all other bodies in the universe, at the speed of light. Attempts to produce or measure Mach effects have thus far failed, but there are reasonable theoretic reasons to believe they might be possible (they help to explain inertia and provide a preferred reference frame which special relativity lacks). I don't think anyone is explaining the EM drive in terms of this, however.

can someone explain in normal people terms were is the recoil going from turning on the microwave?shouldn't the metal box push on the microwave just as much as the microwave pushes on the metal box?also, it gets really hot... could the recoil just be delayed?

Quote from: StrongGR on 05/20/2015 12:13 pmAs promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.Thank you for an interesting paper. What I think would be helpful for anchoring the reader in reality would be a small worked example for the thrust prediction.

Quote from: pogsquog on 05/20/2015 09:40 amQuote from: snow on 05/20/2015 09:19 amcan someone explain in normal people terms were is the recoil going from turning on the microwave?shouldn't the metal box push on the microwave just as much as the microwave pushes on the metal box?also, it gets really hot... could the recoil just be delayed?Shawyer's explanation is that the waves bounce back and forth across the device.The speed of the waves depends upon the shape of the container (the container is acting as a wave-guide, which reduces the propagation speed to be less than C). Shawyer asserts that the momentum change is thereby asymmetric, due to the differing effective impact speeds. Note that the wider physics community do not agree with this analysis, although I do not understand the details.However, there are other theories, in particular Dr.White at NASA thinks that the EM fields from the standing waves are interacting with quantum 'virtual particles', effectively pushing off these. There are problems to do with special relativity and conservation of energy that would seem to make this unlikely unless you go and revive some very old theories of the aether and absolute reference frames, which are very marginal (special relativity has been well tested and aether theories abandoned as unproductive).Another theory that allows breaking of local conservation of momentum and which _is consistent with observations is Woodward 'Mach' effects. These relate to accelerating bodies which are changing in internal energy levels. The momentum is effectively transferred to all other bodies in the universe, at the speed of light. Attempts to produce or measure Mach effects have thus far failed, but there are reasonable theoretic reasons to believe they might be possible (they help to explain inertia and provide a preferred reference frame which special relativity lacks). I don't think anyone is explaining the EM drive in terms of this, however.My personal preference goes to the "Todd conjecture"... (poke poke)This theory, developped by Warptech (aka Todd D) formulates that due to the shape of the cone, waves slow down towards the small end (compression?). As a consequence, the fall in speed is compensated by a change in frequency, causing them to fall out of resonance (between small en big plate). So they and up being attenuated or cut. The energy/momentum of that wave can't just vanish, but it is transferred into the frustum walls...dunno if I formulated it 100% correctly, but that's how I understood the idea.. more or less... sort of... I like this idea because it gives a very comprehensible way to explain where the momentum of the frustum comes from. Something I really miss in all the other theories. Quantum Vacuum field? dunno, but feels more like an artificial mathematical model that attempts to explain, then a valid theory to explain something real. QV has yet to be proven by experiments, btw...However, it would also mean that the pursuit of a higher Q, to achieve higher performances is a dead end.If Shawyer can prove with his supercooled rig that he is indeed getting considerable thrust improvements, then this theory goes down the drain...So far, little news has been brought on his nitrogen cooled device.. tbh, that worries me a little. I'll admit, I've never believed Shawyer's linear extrapolations that promise 1ton of thrust. I know very few devices that scale perfectly linear in power/thrust output.

Quote from: deltaMass on 05/20/2015 01:43 pmQuote from: StrongGR on 05/20/2015 12:13 pmAs promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.Thank you for an interesting paper. What I think would be helpful for anchoring the reader in reality would be a small worked example for the thrust prediction.Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.

Quote from: StrongGR on 05/20/2015 01:53 pmQuote from: deltaMass on 05/20/2015 01:43 pmQuote from: StrongGR on 05/20/2015 12:13 pmAs promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.Thank you for an interesting paper. What I think would be helpful for anchoring the reader in reality would be a small worked example for the thrust prediction.Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.It would be interesting to see which cylindrical cavity mode gave the greatest integrated field along a given laser beam path. As far as detection goes, there has been great progress lately on separating out small sidebands from an intense laser line. (i'll look into that)

QuoteWarptechIf the thruster had 2 compartments, a cylinder, where resonance was easy to establish at high Q, and a long frustum designed for maximum attenuation connected at one end of the cylinder. Between the two, there is a "shutter" that can rapidly open and close. When closed, the cylinder resonates as a cylinder. When open, energy expands into the frustum chamber where it is attenuated. After the shutter closes again, the energy in frustum attenuates and energy in cylinder recharges.... repeat. I keep looking for ways to decouple the resonant amplifier from the attenuator.Firstly I need to ask what sort of timeframes you are looking at for connecting/ disconnecting cycle, micro_sec, milli_sec, seconds?. How long do you think the coupling will need to be in place to create resonance inside the thruster cavity...{or are you thinking the resonance is not even needed in that chamber at all, just force fed from the attached cylinder in burst mode}. I think I see where you are going with this but more info may help clarify the desired method, and help refine a model im working on for mechanical distribution of em waves.

WarptechIf the thruster had 2 compartments, a cylinder, where resonance was easy to establish at high Q, and a long frustum designed for maximum attenuation connected at one end of the cylinder. Between the two, there is a "shutter" that can rapidly open and close. When closed, the cylinder resonates as a cylinder. When open, energy expands into the frustum chamber where it is attenuated. After the shutter closes again, the energy in frustum attenuates and energy in cylinder recharges.... repeat. I keep looking for ways to decouple the resonant amplifier from the attenuator.

But what happens when a wave is attenuated in a perfectly conducting circular waveguide? That energy is not lost as "heat" because there is no resistance to dissipate it.

Quote from: Notsosureofit on 05/20/2015 01:58 pmQuote from: StrongGR on 05/20/2015 01:53 pmQuote from: deltaMass on 05/20/2015 01:43 pmQuote from: StrongGR on 05/20/2015 12:13 pmAs promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.Thank you for an interesting paper. What I think would be helpful for anchoring the reader in reality would be a small worked example for the thrust prediction.Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.It would be interesting to see which cylindrical cavity mode gave the greatest integrated field along a given laser beam path. As far as detection goes, there has been great progress lately on separating out small sidebands from an intense laser line. (i'll look into that)Do you mean shooting a laser beam into a cylindrical cavity or into a frustum?

Quote from: StrongGR on 05/20/2015 02:00 pmQuote from: Notsosureofit on 05/20/2015 01:58 pmQuote from: StrongGR on 05/20/2015 01:53 pmQuote from: deltaMass on 05/20/2015 01:43 pmQuote from: StrongGR on 05/20/2015 12:13 pmAs promised I am posting the latest draft about general relativity and electromagnetic field. The relevant conclusion is that there is thrust. Thanks to the comments by Jose Rodal, it can be shown that this can be meaningful and the best geometry is that of the frustum tending to a cone. There is no violation of conservation law due to the presence of the gravity that can escape the device producing a reaction.I post it here for your comments that are very welcome as usual. You can find the final equation at page 12 for your evaluations. Later on, I will post a version with a somewhat different presentation to arxiv.Thank you for an interesting paper. What I think would be helpful for anchoring the reader in reality would be a small worked example for the thrust prediction.Thanks. Yes, you are right. It is the next step to fill with numbers and put some plots to see the orders of magnitude. I am not sure yet that I am able to account for a macroscopic effect.It would be interesting to see which cylindrical cavity mode gave the greatest integrated field along a given laser beam path. As far as detection goes, there has been great progress lately on separating out small sidebands from an intense laser line. (i'll look into that)Do you mean shooting a laser beam into a cylindrical cavity or into a frustum? Just cylindrical at this point.

"Bessel function - magnetron - (SRF) superconducting RF cavity - Fermi Labs"Sound interesting? I thought so: http://www.fnal.gov/pub/today/archive/archive_2014/today14-11-03.htmlShould the emdrive become a reality...Fermi might be on the verge of helping to develop a massive, cost-effective RF source: http://www.fnal.gov/pub/today/archive/archive_2014/images/magnetron.jpg

The reason I'm not retired is that I want to build this prototype," Pasquinelli said. "It's a solution to a real-world problem, and it will be a lot of fun to build the first one.

Quote from: WarpTech on 05/20/2015 02:01 pmBut what happens when a wave is attenuated in a perfectly conducting circular waveguide? That energy is not lost as "heat" because there is no resistance to dissipate it.How will the wave attenuate (lose energy) if no wall losses?