....I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM
Quote from: Mulletron on 04/03/2015 09:55 pm....I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnMAll the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?
Quote from: Rodal on 04/03/2015 11:32 pmQuote from: Mulletron on 04/03/2015 09:55 pm....I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnMAll the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?Yep. Empty. It wasn't on the balance. Just sitting on the floor.
Quote from: Mulletron on 04/03/2015 11:42 pmQuote from: Rodal on 04/03/2015 11:32 pmQuote from: Mulletron on 04/03/2015 09:55 pm....I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnMAll the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?Yep. Empty. It wasn't on the balance. Just sitting on the floor.The entire frustum was on the floor, or just the HDPE disk? Sorry, I'm a little confused about what test you conducted.
It appears my previous post got deleted.So I'll ask again. Has this psudoscience nonsense been found to be false yet? Have any proper scientists tested this contraption so that they can show that it doesn't work?
Have any proper scientists tested this contraption so that they can show that it doesn't work?
I found the following paper http://www-ssc.igpp.ucla.edu/personnel/russell/papers/skip_ed/node4.html, ( http://onlinelibrary.wiley.com/doi/10.1029/98JA02101/pdf ) for example, that confirms the simple result I discussed above, for standing waves:QuoteFirst, we may obtain some information from the simultaneous Poynting vectors as shown in Figures 5 and 6. If we consider a transverse wave causing field line oscillations, the Poynting vectors behave very differently depending on whether the wave is traveling or standing. Figure 9 is a diagram of the Poynting vectors for the two different schemes. Even though the wave amplitudes for both conditions are set to be the same and the magnitude of Poynting vector oscillations is consequently the same, the traveling wave propagates energy, while the standing wave produces no net energy flux. The Poynting vectors in Figures 5 and 6 more resemble the traveling wave pattern. Thus for the Pc3-4 wave activities in our observations the traveling wave component is stronger. We may also estimate the resonant condition by examining the phase difference between dE and dB [e.g., Singer et al., 1982]. If the phase difference is 90, the wave is standing and a resonant condition is reached. TRAVELLING WAVE STANDING WAVE (EM Drive) Poynting Vector time average is (+1/2) Poynting Vector time average is zero (Cos[ ω t])^2 =( 1+Cos[ 2 ω t]) /2 Sin[ ω t] Cos[ ω t] = Sin[ 2 ω t] /2 So it is as simply as this: to transfer energy or momentum from virtual particles in the Quantum Vacuum, as proposed by Dr. White, a traveling wave would be needed, but then one would have no resonance, and no Q.If one has a cavity EM Drive then resonance can take place, and hence a high Q, but that precludes the possibility of transferring energy or momentum, according to Maxwell's equations.
First, we may obtain some information from the simultaneous Poynting vectors as shown in Figures 5 and 6. If we consider a transverse wave causing field line oscillations, the Poynting vectors behave very differently depending on whether the wave is traveling or standing. Figure 9 is a diagram of the Poynting vectors for the two different schemes. Even though the wave amplitudes for both conditions are set to be the same and the magnitude of Poynting vector oscillations is consequently the same, the traveling wave propagates energy, while the standing wave produces no net energy flux. The Poynting vectors in Figures 5 and 6 more resemble the traveling wave pattern. Thus for the Pc3-4 wave activities in our observations the traveling wave component is stronger. We may also estimate the resonant condition by examining the phase difference between dE and dB [e.g., Singer et al., 1982]. If the phase difference is 90, the wave is standing and a resonant condition is reached.
When the proper frequency is used, the interference of the incident wave and the reflected wave occur in such a manner that there are specific points along the medium that appear to be standing still. Because the observed wave pattern is characterized by points that appear to be standing still, the pattern is often called a standing wave pattern... These points vibrate back and forth from a positive displacement to a negative displacement; the vibrations occur at regular time intervals such that the motion of the medium is regular and repeating.
...so that they can show that it doesn't work
...Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years. My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism. Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?...
Quote from: jmossman on 04/04/2015 01:42 am...Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years. My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism. Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?...Hi James,No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q. Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.One can also consider a non-ideal resonator using an exact solution. One must use complex variables. The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". See for example: http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDFMaxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied. See Equation 12.34 in the above link.The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity. Similarly, real materials have a magnetic susceptibility which is a complex quantity. An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.
Quote from: Rodal on 04/04/2015 03:11 amQuote from: jmossman on 04/04/2015 01:42 am...Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years. My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism. Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?...Hi James,No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q. Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.One can also consider a non-ideal resonator using an exact solution. One must use complex variables. The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". See for example: http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDFMaxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied. See Equation 12.34 in the above link.The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity. Similarly, real materials have a magnetic susceptibility which is a complex quantity. An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.Which part of Maxwell's equations account for the observed thermal losses due to the induced currents in the frustum's large base?
BIG CORRECTIONIt has bothered me that, if the Poynting vector (ExB) would be a quadratic function of the harmonic function so that it never changes sign and therefore does not change orientation with time, even for an AC field, as for example discussed in http://en.wikipedia.org/wiki/Poynting_vector#Time-averaged_Poynting_vector, then the operation of the EM Drive would not be a subject of so much controversy.For the Poynting vector to vary like (Cos[ ω t + phaseAngle])^2 one needs that the E and B fields to be in phase with each other, as shown in the following image for example:But then it dawned on me, that the E and B fields cannot be in-phase with each other (therefore the above image is only true for a travelling wave and it is inappropriate for an EM Drive cavity which instead has standing waves), because Maxwell's equation states that they must 90 degrees out of phase with each other:One of Maxwell's equations (http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction#Maxwell.E2.80.93Faraday_equation) states that:Curl E = - d B / dtso, for example if the magnetic field B varies as Cos[ ω t], then the electric field must vary as its time derivative: - d(Cos[ ω t])/dt = ω Sin[ ω t] , and therefore the Poynting vector ExB should vary as Sin[ ω t] Cos[ ω t] = Sin[ 2 ω t] /2, which oscillates at twice the frequency of the electromagnetic fields and has a time average value of zero.Since the Poynting vector has a time average of zero, there cannot be any net energy flow out of the EM Drive.This is due to the fact that the waves inside the EM Drive are standing waves. Therefore the Poynting vector is just describing how energy is transferred between the electric and magnetic fields.Also this means that there cannot be momentum outflow either, due to the Poynting vector, if the electromagnetic fields are harmonic functions of time.Imagine, for discussion's sake, that it could indeed be possible that virtual electron-positron pairs would materialize out of the Quantum Vacuum, and that when such a pair materializes the Poynting vector is pointing towards the big base of the truncated cone EM Drive. Then the electron-positron pair would be transported by the Poynting vector field towards the big base of the truncated cone, and shortly during that transport the electron-positron would cease to exist, returning back to the vacuum. Then (as shown by Einstein himself in a though-model he proposed a long time ago concerning light particles being transported within a friction-less railroad car) the truncated cone would experience a recoil -simultaneous with the transport of the electron-positron pair-, which would result in a net force towards the small base of the truncated cone. If the Poynting vector would always be pointing towards the big base, this would function as proposed by Dr. White.Unfortunately, the standing waves within an EM Drive cavity are such that the E and B fields must be 90 degrees out of phase with each other (due to Maxwell's equations), and this dictates that the Poynting vector is changing direction at a frequency twice as high as the frequency of the electromagnetic fields. Therefore, if electron-positron pairs would materialize such as in the thought-model discussed above resulting in a recoil of the EM Drive towards the small base, it would occur just as often that electron-positron pairs would be transported in the completely opposite direction and the EM Drive would experience a force in the opposite direction. Therefore what would be expected (out of the Quantum Vacuum model) is to have forces in the EM Drive pointing towards the small base just as often as having forces pointed in the opposite direction towards the big base, and this would result in no net transport of the EM Drive over a period of such oscillations.I will need to correct some of my previous postings concerning the Poynting vector for the EM Drive: for a cavity like the EM Drive, the Poynting vector oscillates with time as Sin[ 2 ω t] /2: therefore the time average of the Poynting vector must be zero. This article in Wikipedia does not apply to a resonating cavity like the EM Drive:http://en.wikipedia.org/wiki/Poynting_vector#Time-averaged_Poynting_vectorbecause it does not obey Maxwell's equation Curl E = - d B / dt which must be obeyed for a resonating cavity.
...My current thoughts still hinge on the evanescent waves through the gaps in the structure as being key to the operation of this device. That is, evanescent waves are unidirectional, they do not average to zero but do collapse to zero at some distance from the cavity. This distance is considerably greater than one third or even one full wavelength.My current thought is that perhaps, as the evanescent waves pass through the gaps, they apply force to the virtual particles in one direction only and then collapse as we know they do. However, the virtual particles have already dissipated back into the QV so only the momentum of the cavity remains. I hope to run some cases testing this in the near future but a Mathematica model would be much more desirable, and telling, than a Meep test result. Fortunately, Meep does provide a QV model in 2-D - it was prepared in order to measure Casimir forces but should be useful for our situation, too. Unfortunately, only 1-D and 2-D models were implemented. Once I figure out how to run the model, and what it means, I will be ready to attempt some results.
Quote from: jmossman on 04/04/2015 03:53 amQuote from: Rodal on 04/04/2015 03:11 amQuote from: jmossman on 04/04/2015 01:42 am...Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years. My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism. Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?...Hi James,No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q. Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.One can also consider a non-ideal resonator using an exact solution. One must use complex variables. The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". See for example: http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDFMaxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied. See Equation 12.34 in the above link.The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity. Similarly, real materials have a magnetic susceptibility which is a complex quantity. An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.Which part of Maxwell's equations account for the observed thermal losses due to the induced currents in the frustum's large base? In order to solve differential equations one must provide boundary conditions. The thermal losses due to the induced eddy-currects in the copper are a result of the imaginary part of the material properties appearing in the solution of the boundary conditions to solve Maxwell's differential equations.That is how COMSOL's Finite Element Analysis provided the solution for the thermal losses and the predicted temperature for NASA's EM Drive truncated cone.The frequencies obtained by COMSOL's FEA were obtained by solving an eigenvalue problem.Damping is responsible for the finite amplitude of the response, but the presence of damping does not preclude standing waves, waves which have fixed nodes and anti-nodes. The boundary conditions due to the end plates do not disappear due to heat production. For example, one of the boundary conditions is that the tangent electric field must be continuous at the material interface. This boundary condition (and therefore the node produced in that component of the electric field) does not cease to exist due to heat generation. That only affects the amplitude. The only way to remove that boundary condition would be to remove the end plate, and if you would do that, the EM Drive would no longer be a completely enclosed cavity.
Quote from: mlindner on 04/04/2015 01:08 amIt appears my previous post got deleted.So I'll ask again. Has this psudoscience nonsense been found to be false yet? Have any proper scientists tested this contraption so that they can show that it doesn't work?I'm sure some "proper scientists" will come round for a look eventually. In the meantime, history has a lot to teach us about these sorts of things:http://amasci.com/weird/vindac.htmlhttp://www.lifehack.org/articles/lifestyle/6-world-changing-ideas-that-were-originally-rejected.htmlhttp://www.cracked.com/article_18822_5-famous-scientists-dismissed-as-morons-in-their-time.html
Quote...so that they can show that it doesn't workJust curious, what makes you so certain that it doesn't work? I mean, I have no idea if it works or not. The only certain thing I know here is that I don't know for certain if it works or doesn't.There is a growing body of evidence which suggest that it does work. It hasn't been proven by anybody that it doesn't work.