Quote from: frobnicat on 05/03/2015 07:30 pm.../...For weak gravitational field the frequency ratio between top and bottom is ft/fb = (1 + Rs/2rt - Rs/2rb) where Rs is Schwarzschild radius, rt and rb distance from centre of body (earth centre). From there, sorry this is French wikipedia, I don't find a convenient English resource for the same formula.For Earth Rs is about 9mm, so lets say we have 0.3m altitude difference at earth surface (6.371e6m) => ft/fb = 1 - 3.3e-17This is one part in 3e16 redshift in frequency.Likewise any Doppler effect affecting the relative wavelengths (momentums) of photons between forward and backward plates of an accelerating frustum would indeed induce a non 0 net force : this force would always be opposite to the acceleration (ie. never a thrust) and in fact could be interpreted as the inertia of the mass equivalent of energy bouncing back and forth in the cavity (whatever its shape). The time constant of a photon in a Q=10000 about 0.3m across frustum would be like 10µs, at 100W pumped into the frustum there is then on the order of 1e-3 J EM energy content at any given time, that is equivalent to 1.1e-20 kg of mass, or an apparent added "force of inertia" of 1.1e-19N for a spacecraft accelerating at 1g, or equivalently an added weight of 1.1e-19N vertically for a resting frustum on earth.In summary, within classical frameworks, yes there can be non 0 net force of EM radiation in an accelerating cavity, but this will be vanishingly small forces, and always opposite to acceleration (aka "inertia").Is it correct ?Close, those forces are similar to the observed forces. Perhaps @Rodal has the page reference back to my calculation which is what I beleive you are stating.
.../...For weak gravitational field the frequency ratio between top and bottom is ft/fb = (1 + Rs/2rt - Rs/2rb) where Rs is Schwarzschild radius, rt and rb distance from centre of body (earth centre). From there, sorry this is French wikipedia, I don't find a convenient English resource for the same formula.For Earth Rs is about 9mm, so lets say we have 0.3m altitude difference at earth surface (6.371e6m) => ft/fb = 1 - 3.3e-17This is one part in 3e16 redshift in frequency.Likewise any Doppler effect affecting the relative wavelengths (momentums) of photons between forward and backward plates of an accelerating frustum would indeed induce a non 0 net force : this force would always be opposite to the acceleration (ie. never a thrust) and in fact could be interpreted as the inertia of the mass equivalent of energy bouncing back and forth in the cavity (whatever its shape). The time constant of a photon in a Q=10000 about 0.3m across frustum would be like 10µs, at 100W pumped into the frustum there is then on the order of 1e-3 J EM energy content at any given time, that is equivalent to 1.1e-20 kg of mass, or an apparent added "force of inertia" of 1.1e-19N for a spacecraft accelerating at 1g, or equivalently an added weight of 1.1e-19N vertically for a resting frustum on earth.In summary, within classical frameworks, yes there can be non 0 net force of EM radiation in an accelerating cavity, but this will be vanishingly small forces, and always opposite to acceleration (aka "inertia").Is it correct ?
Edit: I think this was it (have to check)"The proposition that dispersion caused by an accelerating frame of reference implied an accelerating frame of reference caused by a dispersive cavity resonator. (to 1st order using massless, perfectly conducting cavity)Starting with the expressions for the frequency of an RF cavity:f = (c/(2*Pi))*((X/R)^2+((p*Pi)/L)^2)^.5For TM modes, X = X[sub m,n] = the n-th zero of the m-th Bessel function.[1,1]=3.83, [0,1]=2.40, [0,2]=5.52 [1,2]=7.02, [2,1]=5.14, [2,2]=8.42, [1,3]=10.17, etc.and for TE modes, X = X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function.[0,1]=3.83, [1,1]=1.84, [2,1]=3.05, [0,2]=7.02, [1,2]=5.33, [1,3]=8.54, [0,3]=10.17, [2,2]=6.71, etc.Rotate the dispersion relation of the cavity into doppler frame to get the Doppler shifts, that is to say, look at the dispersion curve intersections of constant wave number instead of constant frequency.df = (1/(2*f))*(c/(2*Pi))^2*X^2*((1/Rs^2)-(1/Rb^2))and from there the expression for the acceleration g from:g = (c^2/L)*(df/f) such that:g = (c^2/(2*L*f^2))*(c/(2*Pi))^2*X^2*((1/Rs^2)-(1/Rb^2))Using the "weight" of the photon in the accelerated frame from:"W" = (h*f/c^2)*g => "W" = T = (h/L)*dfgives thrust per photon:T = (h/(2*L*f))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))If the number of photons is (P/hf)*(Q/2*pi) then:NT = P*Q*(1/(4*pi*L*f^3))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))"
Quote from: MazonDel on 05/04/2015 09:37 pmZellerium, I am uncertain about if your University allows for this sort of thing or not, but for the purposes of the nascent Wiki under development, would you be willing to provide (later once you have them) some Bills of Materials as well as instructions documenting how you assembled and set up your rig?...Dr. Rodal:Do you think the thrust signiture observed from the null Cannae drive could be due to air currents? I assume the frustum yielded similar thrusts while in a vacuum and in air, which shows how much the air is effecting thrust. Unless the shape of the Cannae drive led to a significant air current, it seems a symmetric cavity should produce thrust. Although we may be able to find the resonant frequencies for the frustum using software, changing parameters like the dielectric thickness will change the frequency required. And if we have to use a microwave magnetron then we have to change the frustum's size which is more challenging, but still possible.However, if we recieve enough money for the variable frequency test a tapered cavity would be a much better option.I just downloaded EM Pro recently, but I only have experience with 3-D design software (Creo) and Matlab. We may have licenses available for Multiphysics, I'll look into it. ...
Zellerium, I am uncertain about if your University allows for this sort of thing or not, but for the purposes of the nascent Wiki under development, would you be willing to provide (later once you have them) some Bills of Materials as well as instructions documenting how you assembled and set up your rig?
I thought this might be an alternative way of measuring small forces. Maybe it could be exploited to make a working model if some one thought it was easier to make.
Speaking of KM, is there anyone out there that has what it takes and is willing to volunteer and set up a wiki or something?http://en.wikipedia.org/wiki/MediaWikiI've set up a MediaWiki server before and it wasn't too bad. That is a good platform. I'm simply stretched too thin right now to try it again.
Quote from: Notsosureofit on 05/03/2015 08:12 pmQuote from: frobnicat on 05/03/2015 07:30 pm.../...For weak gravitational field the frequency ratio between top and bottom is ft/fb = (1 + Rs/2rt - Rs/2rb) where Rs is Schwarzschild radius, rt and rb distance from centre of body (earth centre). From there, sorry this is French wikipedia, I don't find a convenient English resource for the same formula.For Earth Rs is about 9mm, so lets say we have 0.3m altitude difference at earth surface (6.371e6m) => ft/fb = 1 - 3.3e-17This is one part in 3e16 redshift in frequency.Likewise any Doppler effect affecting the relative wavelengths (momentums) of photons between forward and backward plates of an accelerating frustum would indeed induce a non 0 net force : this force would always be opposite to the acceleration (ie. never a thrust) and in fact could be interpreted as the inertia of the mass equivalent of energy bouncing back and forth in the cavity (whatever its shape). The time constant of a photon in a Q=10000 about 0.3m across frustum would be like 10µs, at 100W pumped into the frustum there is then on the order of 1e-3 J EM energy content at any given time, that is equivalent to 1.1e-20 kg of mass, or an apparent added "force of inertia" of 1.1e-19N for a spacecraft accelerating at 1g, or equivalently an added weight of 1.1e-19N vertically for a resting frustum on earth.In summary, within classical frameworks, yes there can be non 0 net force of EM radiation in an accelerating cavity, but this will be vanishingly small forces, and always opposite to acceleration (aka "inertia").Is it correct ?Close, those forces are similar to the observed forces. Perhaps @Rodal has the page reference back to my calculation which is what I beleive you are stating.What do you mean by "similar to the observed forces" ? Quantitatively what my very crude estimations of effect of acceleration on EM net force imbalance are 14 orders of magnitude below observed forces at EW, and that is for 1g acceleration, which is far from acceleration of the frustum (horizontal wise), even taking into account recorded noise.I should really not put my feet in that but : if Doppler effects are to be given importance ( to the point of wanting to compensate for them ) and related influence of acceleration of device on resonance, then this should be quantitatively assessed. A 30cm journey from one end to the other takes a photon 1ns. In such a short time a 1g accelerating frustum would have added or subtracted only v=1e-8 m/s to the velocity seen by the photon on the "last bounce" (this is very crude !). What is the Doppler effect of such a tiny velocity difference ? f1/f0 = 1 +or- v/c that would amount to a ratio of 1 +or- 3.3e-17, in agreement with above value for gravitational redshift (equivalently, for the same 1g and 30cm distance). I fail to understand how even a superconducting cavity with a Q of a billion and a super narrow bandwidth would care about such a tiny Doppler shift... And the net force imbalance induced by this Doppler shift, for 100W pumped into frustum at Q=10000 would be 1.1e-19N, at Q=1e9 would be 1.1e-14N ... (and opposite to acceleration). Basically this is just the "inertial force" of the equivalent EM mass multiplied by acceleration...Beyond this crude photon model I must admit I don't get the more sophisticated equations, I know this was explained quite a bit. Notsosureofit would you care plugging the same parameters in your formulas and derive values, what a ~30cm frustum would experience as net EM force imbalance due to acceleration at 1g, given a feed of 100W at Q=10000 ? What is wrong with a crude photon model that don't show dependency on frequency that would be correct with your model where I see some f^3 but no acceleration ?I have : F = - acc * P*Q*L/c^3 F net EM radiation imbalance due to acceleration of frustumacc acceleration in m/s²P input power in WL length of cavity in mc speed of lightThis is not the levels of "thrust" (well, this is not a thrust at all !) experimentally recorded, but is it physically sound ? QuoteEdit: I think this was it (have to check)"The proposition that dispersion caused by an accelerating frame of reference implied an accelerating frame of reference caused by a dispersive cavity resonator. (to 1st order using massless, perfectly conducting cavity)Starting with the expressions for the frequency of an RF cavity:f = (c/(2*Pi))*((X/R)^2+((p*Pi)/L)^2)^.5For TM modes, X = X[sub m,n] = the n-th zero of the m-th Bessel function.[1,1]=3.83, [0,1]=2.40, [0,2]=5.52 [1,2]=7.02, [2,1]=5.14, [2,2]=8.42, [1,3]=10.17, etc.and for TE modes, X = X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function.[0,1]=3.83, [1,1]=1.84, [2,1]=3.05, [0,2]=7.02, [1,2]=5.33, [1,3]=8.54, [0,3]=10.17, [2,2]=6.71, etc.Rotate the dispersion relation of the cavity into doppler frame to get the Doppler shifts, that is to say, look at the dispersion curve intersections of constant wave number instead of constant frequency.df = (1/(2*f))*(c/(2*Pi))^2*X^2*((1/Rs^2)-(1/Rb^2))and from there the expression for the acceleration g from:g = (c^2/L)*(df/f) such that:g = (c^2/(2*L*f^2))*(c/(2*Pi))^2*X^2*((1/Rs^2)-(1/Rb^2))Using the "weight" of the photon in the accelerated frame from:"W" = (h*f/c^2)*g => "W" = T = (h/L)*dfgives thrust per photon:T = (h/(2*L*f))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))If the number of photons is (P/hf)*(Q/2*pi) then:NT = P*Q*(1/(4*pi*L*f^3))*(c/(2*pi))^2*X^2*((1/Rs^2)-(1/Rb^2))"
The easiest way to measure small forces is to make them bigger.
The "g" in the formulas is the amount of acceleration needed to make the tapered cavity "look like" a cylindrical one, ie. compensated
Thermal convection forces have plagued radiation pressure measurements since the time of Maxwell, for about 139 years and counting. There is a rich history showing this.The shape of the Cannae device maximizes the effect of thermal convection effects.Please notice that the thrust/PowerInput obtained by NASA Eagleworks in a vacuum is only a fraction of the one that they measured in air, which shows quantitatively the huge problem with conducting tests under ambient conditions like Shawyer in the UK and Yang in China have done. But, again, if you don't have a vacuum chamber, there are proven ways to minimize this effect that neither Shawyer nor Yang appear to have utilized.I strongly advise against performing any Finite Element Analysis calculations unless you have an analyst available that has taken University courses in Finite Element Analysis and has practical experience with such Finite Element packages. If you want to perform FEA you must add to your team such an analyst.EDIT: If you use the same geometrical dimensions and materials used by NASA, couldn't you use NASA's COMSOL calculations (available in this thread) to assess the frequencies and mode shapes ? (and thus avoid the need to perform any numerical calculations on your own)Best regards,
Quote from: Rodal on 05/04/2015 11:36 pmThermal convection forces have plagued radiation pressure measurements since the time of Maxwell, for about 139 years and counting. There is a rich history showing this.The shape of the Cannae device maximizes the effect of thermal convection effects.Please notice that the thrust/PowerInput obtained by NASA Eagleworks in a vacuum is only a fraction of the one that they measured in air, which shows quantitatively the huge problem with conducting tests under ambient conditions like Shawyer in the UK and Yang in China have done. But, again, if you don't have a vacuum chamber, there are proven ways to minimize this effect that neither Shawyer nor Yang appear to have utilized.I strongly advise against performing any Finite Element Analysis calculations unless you have an analyst available that has taken University courses in Finite Element Analysis and has practical experience with such Finite Element packages. If you want to perform FEA you must add to your team such an analyst.EDIT: If you use the same geometrical dimensions and materials used by NASA, couldn't you use NASA's COMSOL calculations (available in this thread) to assess the frequencies and mode shapes ? (and thus avoid the need to perform any numerical calculations on your own)Best regards,Oh! I hadn't found the data from the vacuum test, I assume it is buried in the forum somewhere? If there isn't strong evidence that a symmetric cavity will work, we will build a frustum. We could use EW's dimensions if we recieve funding for the variable frequency equipment. However, if we have to use a microwave magnetron we will have to do the FEA. I might be able find someone willing to help. We don't have a vacuum chamber large enough so we will look into minimizing the affects of convection. Thank you for the response!
....The frustum is a waveguide that has a strong gradient in the group velocity, near the cut-off modes, relative to the microwave photons moving inside it....
Quote from: WarpTech on 05/05/2015 01:54 am....The frustum is a waveguide that has a strong gradient in the group velocity, near the cut-off modes, relative to the microwave photons moving inside it....As the initial photons travel towards one of the ends, and it hits the end-plate, yes it functions as a travelling wave in a waveguide. However, once the wave hits the end it gets reflected (almost perfectly since the skin depth is extremely small compared to the wavelength and hence the losses are negligible). At that point we have standing waves. The high Q of the cavity is a result of these standing waves producing resonance. So instead of awaveguide with travelling waves having a non-zero Poynting vector transmitting energy from one end to the other, what we have upon reflection is a closed cavity with standing waves having a self-cancelling zero mean value (over a period) Poynting vector and there is no transmission of energy So, the question is, do you see your mechanism as resulting into an acceleration vs. time that gives constant acceleration at constant power input, for ever and ever (which involves an energy paradox) or do you see your mechanism as just resulting into a one-time short-impulse or Dirac delta-function spike in acceleration only (due to the initial photons hitting the end) upon energizing the cavity ?Regards,JR
...I do not think it will have constant acceleration for constant power input over time, because the matter of the frustum and it's source will experience relativistic effects, on mass, time and length. It takes more than a frustum to make an actual warp drive....
Quote from: WarpTech on 05/05/2015 03:40 am...I do not think it will have constant acceleration for constant power input over time, because the matter of the frustum and it's source will experience relativistic effects, on mass, time and length. It takes more than a frustum to make an actual warp drive....Question: under your interpretation, the Emdrive would have a maximum speed always less than c, because acceleration would eventually decrease until maybe becoming zero due to relativistic effects on the frustum at some speed. Is that correct?If so, what speed are we talking about? something close to c or much lower?This topic of the Emdrive's maximum speed and the potentially diminishing acceleration is something that has appeared repeatedly in the discussions, without a clear answer yet because there is no experimental data backing it or disproving it yet.This notion has also been rebuffed by some people, because assuming a "maximum speed" also assumes a privileged reference frame, which is a big no no in current theories.I feel there could be a GR explanation, related to the fact that we do have an absolute speed limit: the speed of light, which is the same on all reference frames, including that of the microwaves inside the frustum.
Want to make sure it isn't forgotten that:Shawyer said to use narrow band source for cavity with shaped ends. Wideband is for cavity with flat ends. Thus, Eagleworks is using the wrong type of signal source. Can't say we didn't tell them.