Quote from: aero on 07/10/2015 04:18 amQuote from: Rodal on 07/10/2015 02:01 amI went outside for a walk and I found out that actually:(Group velocity*Electric Permitivity of Free space)*(Phase velocity*Magnetic Permeability of Free space) = 1(all these multiplied together give exactly one)How about that?great, but not the least bit surprising as vg * vp = c2 and eo * muo = 1/c2Of course, it was a contrived attempt at humor answering these posts concerning somebody posting several posts remaking the well known fact that it is not the least beat surprising that 2c is not the same thing as c^2 (unless c=2): Quote from: deltaMass on 07/10/2015 01:06 amHave you considered going outside and walking around a bit? LOLQuote from: deltaMass on 07/10/2015 12:24 amQuote from: TheTraveller on 07/10/2015 12:20 amQuote from: deltaMass on 07/10/2015 12:19 amQuote from: TheTraveller on 07/09/2015 10:38 pmPhase velocity + group velocity = 2c. WrongGroup velocity X phase velocity = 2c.Also wrong
Quote from: Rodal on 07/10/2015 02:01 amI went outside for a walk and I found out that actually:(Group velocity*Electric Permitivity of Free space)*(Phase velocity*Magnetic Permeability of Free space) = 1(all these multiplied together give exactly one)How about that?great, but not the least bit surprising as vg * vp = c2 and eo * muo = 1/c2
I went outside for a walk and I found out that actually:(Group velocity*Electric Permitivity of Free space)*(Phase velocity*Magnetic Permeability of Free space) = 1(all these multiplied together give exactly one)How about that?
Have you considered going outside and walking around a bit? LOL
Quote from: TheTraveller on 07/10/2015 12:20 amQuote from: deltaMass on 07/10/2015 12:19 amQuote from: TheTraveller on 07/09/2015 10:38 pmPhase velocity + group velocity = 2c. WrongGroup velocity X phase velocity = 2c.Also wrong
Quote from: deltaMass on 07/10/2015 12:19 amQuote from: TheTraveller on 07/09/2015 10:38 pmPhase velocity + group velocity = 2c. WrongGroup velocity X phase velocity = 2c.
Quote from: TheTraveller on 07/09/2015 10:38 pmPhase velocity + group velocity = 2c. Wrong
Phase velocity + group velocity = 2c.
The Q factor of the cavity is defined as the stored energy divided by the energy lost per cycle. Thus as stored energy is transferred to kinetic energy, the decrease in stored energy results in a decrease in Q factor. Thus as acceleration increases, Q decreases and thus thrust decreases. The performance of superconducting thrusters was predicted using this simple energy theory, but without identifying the actual mechanism. This paper corrects this situation by describing the Doppler shifts which cause a decrease in stored energy, but which, more importantly, cause the frequency of the propagating wave to move outside the narrow resonant bandwidth of the cavity.
Quote from: TheTraveller on 07/10/2015 12:12 pmShawyer's theory of how cavity acceleration generates asymmetric Force is attached as an image:Please review sections 1, 2 & 3 of the IAC 2013 paper as attached for more details....I'm curious as to why you are attaching images of a text page of the report that you also attach as a pdf. Attaching an image of the text page takes more of your time, and more bandwidth, while the text information in the image is already in the pdf you attach. If it is to call attention to a particular section, you already detail that in your message ("Please review sections 1, 2 & 3")
Shawyer's theory of how cavity acceleration generates asymmetric Force is attached as an image:Please review sections 1, 2 & 3 of the IAC 2013 paper as attached for more details....
..Basically cavity acceleration causes asymmetric resonant wave path length variation that causes either MOTOR mode or GENERATOR mode operation. No acceleration, no asymmetric resonant wave path length variation and the cavity is in IDLE mode.
Quote from: TheTraveller on 07/10/2015 12:12 pm..Basically cavity acceleration causes asymmetric resonant wave path length variation that causes either MOTOR mode or GENERATOR mode operation. No acceleration, no asymmetric resonant wave path length variation and the cavity is in IDLE mode.So you are saying that the EM Drive needs to be accelerated by other means, in order to itself engage in "motor" mode and produce any acceleration.But the EM Drive in Shawyer's and other researcher's experiments is located on a rotating Earth that is experiencing centripetal acceleration: 0.034 m/s^2 near the Equator , (and a centripetal acceleration around the Sun of 0.005928 m/s^2) Why isn't the centripetal acceleration of the Earth enough to cause the "motor" mode to engage?How much acceleration is required for the "motor" mode to engage and what is this acceleration value based on, or due to?
Neither a Coke can, nor the fluid inside it have a "motor" mode or a "generator" mode that are triggered by acceleration. Nobody to my knowledge has posited to use a Coke can as a means of space propulsion.The questions:1) Why isn't the centripetal acceleration of the Earth enough to cause the conjectured "motor" mode of an EM Drive to engage?2) How much threshold acceleration is required for the "motor" mode to engage and what is this acceleration threshold value based on, or due to?are not answered by substituting an analogy of the EM Drive to a Coke can with a fluid inside it.I guess that there is no answer to the above questions at the moment.
Quote from: zen-in on 07/10/2015 08:38 amThe extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong. I have already showed they are not calculating Q correctly. The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data. These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters. A 145 MHz cavity typically has a Q = 350. Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000. The skin effect and other factors increase the losses at higher frequencies.So easy to claim the Chinese data has been made up, despite having no proof. Along with silly claims that would also say Eagleworks doesn't know how to measure Q either as per attached measured Q of 50,995.I'll not bother to send you my data as you will claim it is also made up.
The extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong. I have already showed they are not calculating Q correctly. The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data. These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters. A 145 MHz cavity typically has a Q = 350. Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000. The skin effect and other factors increase the losses at higher frequencies.
Last night I read:Wimmer, M., Regensburger, A., Bersch, C., Miri, M., Batz, S., Onishchukov, G., & ... Peschel, U. (2013). Optical diametric drive acceleration through action-reaction symmetry breaking. Nature Physics, 9(12), 780-784. doi:10.1038/nphys2777Two sentence summary: Photon-Photon interactions in clever devices are predicted and observed to break Newton's third law Symmetry. I.e. S1 collides with S2 and both end up traveling in the same direction and velocity as S1's original vector.If S2 is trapped bouncing off of the end plate of your resonator and a steady stream of input energy S1, then it isn't difficult to imagine this broken symmetry leading to a violation of Conservation of Momentum.If your library has ebscohost you can get the paper free. If you don't have access, PM me.Are there other examples of broken symmetries?
Quote from: WarpTech on 07/10/2015 04:59 amQuote from: deltaMass on 07/10/2015 04:32 amQuote from: WarpTech on 07/10/2015 03:16 amIn a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.You must have been reading someone else's posts, because I never wrote any of that (and nor do I plan to)! In my simple Newtonian analysis, the value of 'k' is not specified at all. I have no clue where you get this stuff!You said,Quote from: deltaMass on 07/09/2015 07:55 pm...1. Writing 'u' for the phase velocity, you get k = u/c2 Newton/Watt = 1/c when u=c,or in other words a pure photon rocket. But experimental evidence suggests a much higher value for k, and so if your formula is correct, it is predicting a superluminal phase velocity.Is that your intent? Do you think that this observation is important?...I used 1/c as an example to solve for a particular case where break even was < c. But then I had an "AH HA Moment". LOL! Here is the Newtonian version too, in this case, the velocity goes to infinity rather than c, when 100% of the initial rest-energy has been spent. I hope you realize what this is saying. That for whatever energy is available in the battery to use for thrust, there will be a limiting velocity because the battery will go dead. It will not suddenly start to recharge when the speed exceeds some limit. ToddDude. Your Newtonian expression for kinetic energy is not 0.5*m*v2, so how can I take this seriously?
Quote from: deltaMass on 07/10/2015 04:32 amQuote from: WarpTech on 07/10/2015 03:16 amIn a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.You must have been reading someone else's posts, because I never wrote any of that (and nor do I plan to)! In my simple Newtonian analysis, the value of 'k' is not specified at all. I have no clue where you get this stuff!You said,Quote from: deltaMass on 07/09/2015 07:55 pm...1. Writing 'u' for the phase velocity, you get k = u/c2 Newton/Watt = 1/c when u=c,or in other words a pure photon rocket. But experimental evidence suggests a much higher value for k, and so if your formula is correct, it is predicting a superluminal phase velocity.Is that your intent? Do you think that this observation is important?...I used 1/c as an example to solve for a particular case where break even was < c. But then I had an "AH HA Moment". LOL! Here is the Newtonian version too, in this case, the velocity goes to infinity rather than c, when 100% of the initial rest-energy has been spent. I hope you realize what this is saying. That for whatever energy is available in the battery to use for thrust, there will be a limiting velocity because the battery will go dead. It will not suddenly start to recharge when the speed exceeds some limit. Todd
Quote from: WarpTech on 07/10/2015 03:16 amIn a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.You must have been reading someone else's posts, because I never wrote any of that (and nor do I plan to)! In my simple Newtonian analysis, the value of 'k' is not specified at all. I have no clue where you get this stuff!
In a way, you are correct. In the Newtonian approximation the break even happens for k=1/c when 2/3 of the initial rest mass has been converted into kinetic energy. That is a significant speed where relativity is no longer negligible.
...1. Writing 'u' for the phase velocity, you get k = u/c2 Newton/Watt = 1/c when u=c,or in other words a pure photon rocket. But experimental evidence suggests a much higher value for k, and so if your formula is correct, it is predicting a superluminal phase velocity.Is that your intent? Do you think that this observation is important?...
...]Dude. Your Newtonian expression for kinetic energy is not 0.5*m*v2, so how can I take this seriously?
... you ignore where the energy is coming from by assuming that what comes out of the battery has negligible mass, then claim over unity by counting the energy gained from that expenditure as gravy. Paradox resolved, case closed. My job is not to convince you, but to make sure others don't buy into more of this paradox fantasy.Todd
Quote from: TheTraveller on 07/10/2015 09:39 amQuote from: zen-in on 07/10/2015 08:38 amThe extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong. I have already showed they are not calculating Q correctly. The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data. These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters. A 145 MHz cavity typically has a Q = 350. Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000. The skin effect and other factors increase the losses at higher frequencies.So easy to claim the Chinese data has been made up, despite having no proof. Along with silly claims that would also say Eagleworks doesn't know how to measure Q either as per attached measured Q of 50,995.I'll not bother to send you my data as you will claim it is also made up.You guys are in my world when you talk Q, and yes, I've used $100K network analyzers and handhelds. Here's the deal...A Q of 10K is theoretically possible but highly unlikely. A 2.45 Ghz bandpass cavity would have to have a 3dB (half power BW) of 245 Khz...I've never seen such a beast. Here's the problem...center frequency drift of both the cavity and the source. You'd be chasing your tail trying to keep it centered.I'm for projecting Qs of this design between 1 and 5K, meaning between about 2 & 5 Mhz 3dB BW. Even at this reduced Q, there will be drift concerns due to heating.Bottom line, Qs are not 5 digits. On paper, yes...real world, no.
Has anyone seen this paper yet. I am wondering if this is the paper @Traveller was talking about some pages back.http://www.sciencedirect.com/science/article/pii/S0094576515002726
Quote from: rfmwguy on 07/10/2015 01:45 pmQuote from: TheTraveller on 07/10/2015 09:39 amQuote from: zen-in on 07/10/2015 08:38 amThe extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong. I have already showed they are not calculating Q correctly. The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data. These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters. A 145 MHz cavity typically has a Q = 350. Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000. The skin effect and other factors increase the losses at higher frequencies.So easy to claim the Chinese data has been made up, despite having no proof. Along with silly claims that would also say Eagleworks doesn't know how to measure Q either as per attached measured Q of 50,995.I'll not bother to send you my data as you will claim it is also made up.You guys are in my world when you talk Q, and yes, I've used $100K network analyzers and handhelds. Here's the deal...A Q of 10K is theoretically possible but highly unlikely. A 2.45 Ghz bandpass cavity would have to have a 3dB (half power BW) of 245 Khz...I've never seen such a beast. Here's the problem...center frequency drift of both the cavity and the source. You'd be chasing your tail trying to keep it centered.I'm for projecting Qs of this design between 1 and 5K, meaning between about 2 & 5 Mhz 3dB BW. Even at this reduced Q, there will be drift concerns due to heating.Bottom line, Qs are not 5 digits. On paper, yes...real world, no.Maybe you need to actually measure a EMDrive cavity, as Eagleworks did as per attached.With the low power from the NA, there will not be any significant heating of the cavity, so the resonant frequency will not be shifting around. You will need to sweep very slowly as it will take time for the cavity to fully fill with energy.
Quote from: birchoff on 07/10/2015 01:46 pmHas anyone seen this paper yet. I am wondering if this is the paper @Traveller was talking about some pages back.http://www.sciencedirect.com/science/article/pii/S0094576515002726That is the one.Which makes 6 peer reviewed EMDrive papers, published in 5 different journals / publications.
Judging by the tuning mechanisms used, looks like they used a rectangular waveguide. In my opinion, I don't think they would go through the effort of drawing their picture to scale.
Quote from: TheTraveller on 07/10/2015 02:10 pmQuote from: rfmwguy on 07/10/2015 01:45 pmQuote from: TheTraveller on 07/10/2015 09:39 amQuote from: zen-in on 07/10/2015 08:38 amThe extremely high Q claimed for the Chinese and other em-drive cavities is completely wrong. I have already showed they are not calculating Q correctly. The graph shown is not a photo taken from a network analyzer and so I believe it is just made up data. These cavities are similar in most respects to cavity filters used for VHF and UHF repeaters. A 145 MHz cavity typically has a Q = 350. Scaling this up to 2.5 GHz and the Q may be as high as 1,000 - 2,000. The skin effect and other factors increase the losses at higher frequencies.So easy to claim the Chinese data has been made up, despite having no proof. Along with silly claims that would also say Eagleworks doesn't know how to measure Q either as per attached measured Q of 50,995.I'll not bother to send you my data as you will claim it is also made up.You guys are in my world when you talk Q, and yes, I've used $100K network analyzers and handhelds. Here's the deal...A Q of 10K is theoretically possible but highly unlikely. A 2.45 Ghz bandpass cavity would have to have a 3dB (half power BW) of 245 Khz...I've never seen such a beast. Here's the problem...center frequency drift of both the cavity and the source. You'd be chasing your tail trying to keep it centered.I'm for projecting Qs of this design between 1 and 5K, meaning between about 2 & 5 Mhz 3dB BW. Even at this reduced Q, there will be drift concerns due to heating.Bottom line, Qs are not 5 digits. On paper, yes...real world, no.Maybe you need to actually measure a EMDrive cavity, as Eagleworks did as per attached.With the low power from the NA, there will not be any significant heating of the cavity, so the resonant frequency will not be shifting around. You will need to sweep very slowly as it will take time for the cavity to fully fill with energy.The shape of the emdrive cavity, just due to its nature, will have a lower Q than a cylindrical cavity. Any EM professional (or even just grad student) has done material characterization in a cylindrical cavity of high Q in a strict measurement environment. This is because it's a required course at most places. Large Q is tricky to get, and can vary wildly with tiny temperature variations. The higher the q, the larger the variations. Why you find it prudent to argue against multiple experts with a Chinese journal paper (notoriously low quality) will forever be beyond me.