Quote from: Ricvil on 07/09/2015 08:34 pmQuote from: Rodal on 07/09/2015 07:26 pmQuote from: WarpTech on 07/09/2015 07:22 pmQuote from: aero on 07/09/2015 06:59 pm.........Just another point of attention.The resulting electromagnetic force, with the fields confined inside cavity, calculated from the poyting vector, is F=-(1/c^2).d(integral_vol(S))/dt, where S=ExHSo a constant, at principle, make no difference to resulting force on whole cavity....I think you are overlooking the point I brought up earlier, that the DC magnetic flux can pass through copper. There is no skin effect for the DC offset, so it makes this an open system. Magnetic flux can carry momentum and energy "out" of the frustum and allow it to escape. For DC currents flowing in the copper, the frustum is wide open.Todd
Quote from: Rodal on 07/09/2015 07:26 pmQuote from: WarpTech on 07/09/2015 07:22 pmQuote from: aero on 07/09/2015 06:59 pm.........Just another point of attention.The resulting electromagnetic force, with the fields confined inside cavity, calculated from the poyting vector, is F=-(1/c^2).d(integral_vol(S))/dt, where S=ExHSo a constant, at principle, make no difference to resulting force on whole cavity....
Quote from: WarpTech on 07/09/2015 07:22 pmQuote from: aero on 07/09/2015 06:59 pm.........
Quote from: aero on 07/09/2015 06:59 pm......
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Quote from: flux_capacitor on 07/09/2015 08:34 pmQuote from: rfmwguy on 07/09/2015 08:04 pmShawyers frustum appeared to have an insulated larger frustum plate, maybe both. In effect, these plates might have been charged capacitors exhibiting the "corona wind".Interesting. This raises two questions:- Does the process of resonating microwaves inside a cavity is naturally known to produce an electric potential difference between two conductive plates (both isolated from the frustum and each other by a dielectric gasket) ?- If so, for EmDrive designs with plates electrically isolated from the frustum, could we just connect the two plates together with a ground wire (outside of the cavity) to prevent any voltage and thus any ion wind around the cavity?So, I've been SLOWLY working on my own theory at the 40K ft level. Assymetrical (end) plates, insulated from one another are capacitive plates with a dielectric (air) medium. An even better capacitor would be created by putting in a dielectric puck, HDFE, which is what EW did. But don't think thats why they did it.RF radiation striking the end plates impart Ev and a potential builds up ONLY if the plates are insulated from one another. The EW frustum does not appear to have insulated end-plates and I would not call it a capacitor, nor do I think it is exhibiting the corona/ion effect. Outgassing? Maybe.Regardless, I've set my design up to be able to insulate the end plates, so I can test the capacitor/corona theory. My first test will be with uninsulated end plates, test #2 will be with insulated end plates...as it stands now.
Quote from: rfmwguy on 07/09/2015 08:04 pmShawyers frustum appeared to have an insulated larger frustum plate, maybe both. In effect, these plates might have been charged capacitors exhibiting the "corona wind".Interesting. This raises two questions:- Does the process of resonating microwaves inside a cavity is naturally known to produce an electric potential difference between two conductive plates (both isolated from the frustum and each other by a dielectric gasket) ?- If so, for EmDrive designs with plates electrically isolated from the frustum, could we just connect the two plates together with a ground wire (outside of the cavity) to prevent any voltage and thus any ion wind around the cavity?
Shawyers frustum appeared to have an insulated larger frustum plate, maybe both. In effect, these plates might have been charged capacitors exhibiting the "corona wind".
More info is in the paper, which is in Chinese. Attached.I don't think the drawing is dimensionally correct.
Quote from: frobnicat on 07/10/2015 01:22 am...We are not going to see an EMdrive effect unless some other "teeming streaming background field" is put in that picture of energy_repartition_end minus energy_repartition_start in a closed system.But equally so, the "teeming streaming background field" will do you no good if you don't have an electromagnetic stress acting on it, so until the "teeming streaming background field" is found we might as well calculate the electromagnetic stress stress energy tensor components and at least verify whether they are pointing in the right direction.
...We are not going to see an EMdrive effect unless some other "teeming streaming background field" is put in that picture of energy_repartition_end minus energy_repartition_start in a closed system.
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?
Quote from: WarpTech on 07/10/2015 03:16 amQuote from: deltaMass on 07/10/2015 12:04 amQuote from: WarpTech on 07/09/2015 09:12 pm[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important It's a good thing I'm going to bring this debate to a close. The answer is embarrassingly simple. It embarrasses me, you and everyone else who's been gawking at it and writing about this paradox. Not to forget to mention everyone of us who "thinks" we know rocket science! 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.So then I decided to look at it relativistically rather than solve for an incorrect speed... That's when I said "OUCH!" This is going to hurt.... but you said yourself, you're power source is a battery on board the moving vehicle. Convinced now?ToddSorry, but I've no idea how you got the final line, and I certainly don't accept the expression for Eout without some explanation.
Quote from: deltaMass on 07/10/2015 12:04 amQuote from: WarpTech on 07/09/2015 09:12 pm[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important It's a good thing I'm going to bring this debate to a close. The answer is embarrassingly simple. It embarrasses me, you and everyone else who's been gawking at it and writing about this paradox. Not to forget to mention everyone of us who "thinks" we know rocket science! 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.So then I decided to look at it relativistically rather than solve for an incorrect speed... That's when I said "OUCH!" This is going to hurt.... but you said yourself, you're power source is a battery on board the moving vehicle. Convinced now?Todd
Quote from: WarpTech on 07/09/2015 09:12 pm[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important
[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]
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.
As I have mentioned before, if anything is escaping the cavity, and if it's somehow real and not quantum, then it can only escape through gaps. Gasket gaps are good in that regard. But if something is escaping, shouldn't the real world direction of the escape be a consideration? It was back when I was looking at forces through gaps, trying to measure evanescent waves. Making the ends like corks in a bottle, instead of like a cap on a pipe made a big difference, percentage wise in the forces.
Quote from: deltaMass on 07/10/2015 03:36 amQuote from: WarpTech on 07/10/2015 03:16 amQuote from: deltaMass on 07/10/2015 12:04 amQuote from: WarpTech on 07/09/2015 09:12 pm[...]What can I say, except today I have a better handle on it than I did yesterday, or the day before... I'm learning too! Now, I understand that "break even" means that Pin = Pout, or 100% conversion efficiency. It doesn't mean anything more than that. [...]Oh, but indeed it does! It means that if the breakeven velocity is physically attainable (which implies at least that it's less than c), then you have free energy forever in a perpetual motion machine of the first kind. That's kinda important It's a good thing I'm going to bring this debate to a close. The answer is embarrassingly simple. It embarrasses me, you and everyone else who's been gawking at it and writing about this paradox. Not to forget to mention everyone of us who "thinks" we know rocket science! 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.So then I decided to look at it relativistically rather than solve for an incorrect speed... That's when I said "OUCH!" This is going to hurt.... but you said yourself, you're power source is a battery on board the moving vehicle. Convinced now?ToddSorry, but I've no idea how you got the final line, and I certainly don't accept the expression for Eout without some explanation.Your in denial! Sit down, have a pint and you'll see how it works. It's just algebra, set them equal and solve for gamma.Eout is self explanatory. You start with an initial total rest-energy of your vehicle plus charged battery. You expend battery energy Ein, in order to generate thrust, regardless of how it is used to generate force and acceleration, this is the energy from the battery. That energy spent is subtracted from the initial total rest-energy and converted to kinetic energy by whatever means. At t=0, v=0, gamma=1. Plug in the total energy available in your battery for Ein. Then you have the gamma value and the maximum attainable speed. It will never give you more than you started with. Please don't embarrass us further...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!
...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?...
Quote from: TheTraveller on 07/09/2015 12:00 pmMore info is in the paper, which is in Chinese. Attached.I don't think the drawing is dimensionally correct.I hunted throughout the Chinese Physical Society site (http://wulixb.iphy.ac.cn) but didn't find any English translations of the 2014 Chinese paper. So after going through various machine translation software, I've uploaded to the wiki a more readable English version, with most of the formatting and images intact; may be of help.http://emdrive.wiki/images/9/9c/2014_Yang_NWPU_Paper.docx
...Yes clever Chinese, getting Q of 117,500 without needing to go to spherical end plates.Anybody translate the tags pointing to the 2 flat to/from spherical converter sections? I think I know how to calc the min length but knowing what those tags say may help.
Quote from: TheTraveller on 07/09/2015 11:33 pm...Yes clever Chinese, getting Q of 117,500 without needing to go to spherical end plates.Anybody translate the tags pointing to the 2 flat to/from spherical converter sections? I think I know how to calc the min length but knowing what those tags say may help.Earlier in this thread there was a discussion on how the Chinese calculated Q. The conclusion was they are not calculating Q correctly. The actual Q of their cavity, based on network analyzer sweeps was closer to 3,000. I don't recall the exact number but it was 1000 < Q < 5000.
AbstractA microwave resonator system is made, which has a tapered resonant cavity, a microwave source, and a transmission device. Because of the electromagnetic pressure gradient on the tapered resonant cavity, a net electromagnetic force along the axis of the cavity may be observed, which is needed to verify experimentally the use of the independent microwave resonator system. It is also needed to keep the independent microwave resonator system in resonating state, which is the important procedure to demonstrate the possibility of net electromagnetic force. Thus, a low-signal resonating experiment on the tapered resonant cavity combined with resonating parts is completed to accurately find out the resonant frequency of 2.45 GHz and to analyze the influence of temperature on the resonant state. Experimental result shows that the resonant frequency and quality factor of the independent microwave resonator system are 2.44895 GHz and 117495.08 respectively. When the temperature of the tapered resonant cavity wall rises, the resonant frequency will be decreased and the quality factor changed separately.
Quote from: zen-in on 07/10/2015 05:30 amQuote from: TheTraveller on 07/09/2015 11:33 pm...Yes clever Chinese, getting Q of 117,500 without needing to go to spherical end plates.Anybody translate the tags pointing to the 2 flat to/from spherical converter sections? I think I know how to calc the min length but knowing what those tags say may help.Earlier in this thread there was a discussion on how the Chinese calculated Q. The conclusion was they are not calculating Q correctly. The actual Q of their cavity, based on network analyzer sweeps was closer to 3,000. I don't recall the exact number but it was 1000 < Q < 5000.The calcs in the latest paper are correct. Check it out.Suggest the earlier low Q was related to cavity bandwidth needed to accept their magnetron output bandwidth....Experimental result shows that the resonant frequency and quality factor of the independent microwave resonator system are 2.44895 GHz and 117495.08 respectively. When the temperature of the tapered resonant cavity wall rises, the resonant frequency will be decreased and the quality factor changed separately.
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
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.
The layman EMDrive thread has disappeared down the list, so could anyone give a brief (and simple) description of where you guys are at on this issue?Thanks.
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
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.