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#1700 by jay343 on 15 Oct, 2016 16:05
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The onsite government personnel generally do not have classification authority. Also, they need to be under contract with the government for that before any new information could be treated as classified. If the drive was already classified, and Boeing employees knew about it they wouldn't have needed the original contract with Shawyer to get information about it.
Also, Boeing would still have wanted a follow on contract with Shawyer if they didn't give up on it. Even if it somehow got classified, they can't steal his IP. Please drop this conspiracy theory now, unless you have actual evidence to support your case.
Agreed !Quote from: Aviation Week & Space Technology... Shawyer's government funding has ended. Boeing's Phantom Works, which has previously explored exotic forms of space propulsion, was said to be looking into it some years ago. Such work has evidently ceased. “Phantom Works is not working with Mr. Shawyer,” a Boeing representative says, adding that the company is no longer pursuing this avenue.
http://aviationweek.com/awin/propellentless-space-propulsion-research-continues
David Hambling | Aviation Week & Space Technology
Nov 5, 2012
Notice that Hambling disclosed not only that “Phantom Works is not working with Mr. Shawyer” but even more final , they added that "the company (Boeing) is no longer pursuing this avenue." From what Hambling wrote, it reads to me that Boeing Phantom Works is no longer pursuing this (EM Drive) avenue. Furthermore, given a previous arrangement between Boeing and Shawyer, it would be highly non-standard for Boeing to continue working on Shawyer's technology without an Intellectual Property arrangement with Shawyer that would allow this. And, if Boeing would have conducted their own independent microwave cavity EM Drive R&D work prior to the arrangement with Shawyer, it would not make Intellectual Property sense that Boeing would have entered into an arrangement with Shawyer, as large companies usually refuse to discuss inventions with outside inventors because such discussions and arrangements create Intellectual Property issues (Ford Motor Company has an old classic legal case on such IP questions, involving intermittent windshield wipers (*)). Since Boeing is a very large public company with an extensive IP department, this implies that indeed "they are no longer pursuing this (EM Drive) avenue" at all, and if, hypothetically, Boeing were to conduct such development in the future, it would have to be a different (for Intellectual Property purposes) engineering design than Shawyer's EM Drive.
In other words, if Boeing would be engaged in any such development, Boeing would not be developing "his (Shawyer's) technology" but Boeing's separate, unique, different (for IP purposes) technology.
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(*) See: https://en.wikipedia.org/wiki/Robert_KearnsQuoteKearns won one of the best known patent infringement cases against Ford Motor Company (1978–1990) and a case against Chrysler Corporation (1982–1992). Having invented and patented the intermittent windshield wiper mechanism, which was useful in light rain or mist, he tried to interest the "Big Three" auto makers in licensing the technology. They all rejected his proposal, yet began to install intermittent wipers in their cars, beginning in 1969
First of all, let me make it clear that this is pure speculation on my part, and I have no intention to create yet another conspiracy theory, and I don't want to waste any bandwidth here.
You may have mistaken my implication. I'm suggesting that Roger is keeping us in the dark, not Boeing. I suspect that this in no mystery at all to Roger, and I suspect that Roger's IP is perfectly safe with Boeing, but maybe not so with the Chinese.
Aviation Week and Space Technology is an awesome publication, but their track record on classified programs is hit or miss. It's a frustratingly amusing fact of which some of you are aware, I'm sure. -
#1701 by aero on 15 Oct, 2016 16:53
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Here is the details as given in the meep manual, maybe that makes a difference for you.
you can also compute force spectra: forces on an object as a function of frequency, computed by Fourier transforming the fields and integrating the vacuum Maxwell stress tensor
\sigma_{ij} = E_i^*E_j + H_i^*H_j - \frac{1}{2} \delta_{ij} \left( |\mathbf{E}|^2 + |\mathbf{H}|^2 \right)
over a surface S via \mathbf{F} = \int_S \sigma d\mathbf{A}. We recommend that you normally only evaluate the stress tensor over a surface lying in vacuum, as the interpretation and definition of the stress tensor in arbitrary media is often problematic (the subject of extensive and controversial literature.
As I said, a force is the integral of a stress tensor component over a surface area.
The statement in the Meep manual you are quoting now is simply a statement about calculating forces, and not about calculating energy density.
As I said in my previous post, stress and energy density have the same units (which is not a coincidence). A force does not have the same units as energy. Energy has units of force times displacement. Energy density has units of force times displacement over volume, or force over area.
Your statementQuoteit is not the Poynting vector, rather the curve shows the integral of the Maxwell Stress Tensor taken over equal sized (one side only) square detectors, perpendicular to the z axis of rotation. And that has the same units as Energy Density.
that the integral of the stress tensor has the same units as energy density is incorrect, and not at all supported by the Meep manual.
Sorry.
So please explain to me how this statement is either true or false.
"The pressure exerted by an electromagnetic field on neutral matter is simply equal to its energy density (the two quantities have the same units), and is uniformly exerted in all directions orthogonal to the lines of magnetic flux."
From: http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html#FORCE
Thanks
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#1702 by zen-in on 15 Oct, 2016 16:58
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I would like to suggest, in the interest of readability, that people refrain from posting speculation, conspiracy theories and other unsupported conclusions. It would be more interesting to read the results of actual experiments, see photos of the apparatus, etc, etc. We have visited these conspiracy theories about EM-Drive "going black" many times before.
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#1703 by rfmwguy on 15 Oct, 2016 16:59
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For all we know, SPR's flight thrusters may have had significant weight, volume, service life, unexpected signature characteristics, or some other factor that made them operationally unfavorable over ion drive based solar electric station keeping systems. In the end, it's all speculation, and doesn't really matter. The onus is still on EM Drive proponents to demonstrate an indisputably functional system.
Speculation on Boeing being a Saint or Sinner on this issue is all the same, speculative. Everyone has seen my honest, open efforts on the 18.4 mN of observed torsion beam displacement. They can take it or leave it. Its all the same to me...onus is not relevant as I have nothing to prove to anyone but myself, thereby no duty nor obligation to anyone except to share the journey...which I did. -
#1704 by Rodal on 15 Oct, 2016 17:01
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Here is the details as given in the meep manual, maybe that makes a difference for you.
you can also compute force spectra: forces on an object as a function of frequency, computed by Fourier transforming the fields and integrating the vacuum Maxwell stress tensor
\sigma_{ij} = E_i^*E_j + H_i^*H_j - \frac{1}{2} \delta_{ij} \left( |\mathbf{E}|^2 + |\mathbf{H}|^2 \right)
over a surface S via \mathbf{F} = \int_S \sigma d\mathbf{A}. We recommend that you normally only evaluate the stress tensor over a surface lying in vacuum, as the interpretation and definition of the stress tensor in arbitrary media is often problematic (the subject of extensive and controversial literature.
As I said, a force is the integral of a stress tensor component over a surface area.
The statement in the Meep manual you are quoting now is simply a statement about calculating forces, and not about calculating energy density.
As I said in my previous post, stress and energy density have the same units (which is not a coincidence). A force does not have the same units as energy. Energy has units of force times displacement. Energy density has units of force times displacement over volume, or force over area.
Your statementQuoteit is not the Poynting vector, rather the curve shows the integral of the Maxwell Stress Tensor taken over equal sized (one side only) square detectors, perpendicular to the z axis of rotation. And that has the same units as Energy Density.
that the integral of the stress tensor has the same units as energy density is incorrect, and not at all supported by the Meep manual.
Sorry.
So please explain to me how this statement is either true or false.
"The pressure exerted by an electromagnetic field on neutral matter is simply equal to its energy density (the two quantities have the same units), and is uniformly exerted in all directions orthogonal to the lines of magnetic flux."
From: http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html#FORCE
Thanks
aero
Yes, as I have explained in my last several posts, stress and energy density have the same units, in agreement with what Greg Egan states.
I have further explained that the fact that stress and energy density have the same units is not a coincidence.
You, first postulated that the energy density is the integral of the Poynting vector. Then you changed this to state that energy density is the integral of the stress tensor.
Sorry, both of your above statements are incorrect. Even dimensionally, they are both incorrect and disagree with the quotation you are quoting from Greg Egan.
As I have previously formally proved:
STRESS AT ALL INTERNAL SURFACES FOR TE0np MODES
* all shear stress components are zero, and hence the stress field is a principal stress
* the stress is compressive
* the stress varies from zero to its maximum compressive value, never reversing sign, at a frequency which is twice as high as the frequency of the electromagnetic fields.
* the electric vector field is zero at all internal surfaces
* Coulomb's pressure and the electrostatic pressure are zero
* the stress is entirely due to the energy density
* the stress is entirely due to the magnetic component parallel to the surface
Again, the above -that the stress is entirely due to the energy density- is true, and only true for TE0np MODES, assuming perfectly conductive surfaces
(Transverse electric modes with quantum number in the azimuthal direction equal to zero)
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#1705 by Bob012345 on 15 Oct, 2016 18:08
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When I first read that Boeing received a working prototype of the EMDrive, and then didn't follow through with a "lucrative licensing agreement", it made perfect sense to me. Having worked 30+ years at an aerospace company even larger than Boeing, I can tell you that if the prototype demonstrated any thrust whatsoever, it would have have been classified immediately, and Roger would not be allowed to confirm nor deny anything further about that specific device nor any contracts nor agreements related to it.
So you are saying that you worked at Lockheed Martin, since they are the only one comparable to Boeing. Northrop is next, but they are definitely smaller than Boeing.
I remember some disappointment expressed earlier about Roger's companies being focused on terrestrial applications, but I think he made it very clear in the IBT interview that he expects that the general public will become familiar with all of this when autonomous flying cars become available, but that the "much more serious" applications will be related to cheap access to space, especially space-based solar power, which he sees as the answer to energy and environmental problems worldwide.
Also, they can't just classify something. Only the government can classify something, and if you had the experience you just claimed, you should know that.
My first reaction when reading that they didn't follow through was also not surprise, because it makes perfect sense that they would check it out, find out it doesn't work as advertised and drop the contract. Claiming "it went dark" and such basically amounts to conspiracy theory unless you have some real evidence to show.
Shawyer plainly says they verified the data. If they refuted Shawyer's data he would not have said that. -
#1706 by aero on 15 Oct, 2016 18:09
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Dr. Rodal,
So what is true about the meep force curves related to energy density? Is the Maxwell Stress Tensor at a point equal to the energy density at that point? This run was for a copper cavity. Could that be why it shows negative forces just inside the end boundaries?
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#1707 by Rodal on 15 Oct, 2016 18:16
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Dr. Rodal,
For a TE0np mode, for example, for TE012 or for TE013, assuming perfect conductors, the stress and the energy density are the same thing. You should not be integrating the stress to obtain the energy density.
So what is true about the meep force curves related to energy density? Is the Maxwell Stress Tensor at a point equal to the energy density at that point? This run was for a copper cavity. Could that be why it shows negative forces just inside the end boundaries?
aero
Please show the same graph for a TE0np mode, without integrating the stress. Do not integrate the stress.
For non-perfect conductors there are really tiny electric fields within the skin depth but they are usually ignored (ref. Jackson) for TE0np modes, and don't much affect what is stated above. -
#1708 by WarpTech on 15 Oct, 2016 18:18
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Missed a few minus signs.
It's still mind blowing when you realize that a smaller Q and a lower frequency should work better. Note, this is not assuming a cylinder. It is using the radial coordinate from the apex of the cone and gives the basic design relationships.
Thrust - With constant frequency, constant velocity and constant wavelength inside the cavity.
With more definitions, since nobody seems to understand that thrust inversely proportional to Q is very different from anything done prior.
In these equations;
tau(r) is the decay time of the Q as a function of r, measured from the apex of the cone.
Zeta(r) is the damping factor as a function of r.
m(r) is the variation in equivalent mass of the MW photons inside the frustum, as a function of r.
g(r) is the acceleration due to the gradient in Zeta, wrt. r.
T is the thrust, written as a function of Q, or a function of tau.
Chi is (scalar) magnetic flux.
When there is current flowing through the copper, copper has resistance and resistance creates a voltage drop. Once that voltage drop has been created, magnetic flux can escape by the amount equal to the V-sec around the loop. So magnetic flux is escaping where there is heat being generated and power is dissipated. Preferably at the big end. This asymmetry provides the equation for force, the same as a photon rocket, except;
F ~ P/2v, where v = -1/(dtau/dr) << c
Hence, thrust can be generated much greater than a photon rocket, because the velocity at which magnetic flux is escaping is so very slow.
Edit: Pesky negative signs!
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#1709 by Bob012345 on 15 Oct, 2016 18:28
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Missed a few minus signs.
It's still mind blowing when you realize that a smaller Q and a lower frequency should work better. Note, this is not assuming a cylinder. It is using the radial coordinate from the apex of the cone and gives the basic design relationships.
Thrust - With constant frequency, constant velocity and constant wavelength inside the cavity.
With more definitions, since nobody seems to understand that thrust inversely proportional to Q is very different from anything done prior.
In these equations;
tau(r) is the decay time of the Q as a function of r, measured from the apex of the cone.
Zeta(r) is the damping factor as a function of r.
m(r) is the variation in equivalent mass of the MW photons inside the frustum, as a function of r.
g(r) is the acceleration due to the gradient in Zeta, wrt. r.
T is the thrust, written as a function of Q, or a function of tau.
Chi is (scalar) magnetic flux.
When there is current flowing through the copper, copper has resistance and resistance creates a voltage drop. Once that voltage drop has been created, magnetic flux can escape by the amount equal to the V-sec around the loop. So magnetic flux is escaping where there is heat being generated and power is dissipated. Preferably at the big end. This asymmetry provides the equation for force, the same as a photon rocket, except;
F ~ P/2v, where v = -1/(dtau/dr) << c
Hence, thrust can be generated much greater than a photon rocket, because the velocity at which magnetic flux is escaping is so very slow.
In effect, the speed of light is slow so Power/c is much greater than light? I always speculated that if light can be made slow, its momentum could be larger but I thought Todd's recent paper showed if the dielectric constants of space change, the force is invariant. Why is force P/2v and not P/v or 2P/v?
Also, when and how did thrust become inversely proportional to Q instead of proportional to Q? Physically, that doesn't seem to resonate with me.
Thanks.
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#1710 by Rodal on 15 Oct, 2016 18:29
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With more definitions, since nobody seems to understand that thrust inversely proportional to Q is very different from anything done prior.
In these equations;
tau(r) is the decay time of the Q as a function of r, measured from the apex of the cone.
Zeta(r) is the damping factor as a function of r.
m(r) is the variation in equivalent mass of the MW photons inside the frustum, as a function of r.
g(r) is the acceleration due to the gradient in Zeta, wrt. r.
T is the thrust, written as a function of Q, or a function of tau.
Chi is (scalar) magnetic flux.
When there is current flowing through the copper, copper has resistance and resistance creates a voltage drop. Once that voltage drop has been created, magnetic flux can escape by the amount equal to the V-sec around the loop. So magnetic flux is escaping where there is heat being generated and power is dissipated. Preferably at the big end. This asymmetry provides the equation for force, the same as a photon rocket, except;
F ~ P/2v, where v = -1/(dtau/dr) << c
Hence, thrust can be generated much greater than a photon rocket, because the velocity at which magnetic flux is escaping is so very slow.
Congratulations on the work so far. Now we need some calculations to compare with NASA's and TU Dresden (Tajmar's) experimental results... -
#1711 by Rodal on 15 Oct, 2016 18:58
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I think you are shortchanging
With more definitions, since nobody seems to understand that thrust inversely proportional to Q is very different from anything done prior.
... the power of your theory.
Your theory does not say that thrust is inversely proportional to Q.
Your theory states that thrust is directly proportional to the negative rate of change of Log[Q] with r
, (times P, and divided by 2 times omegao)
That could mean almost no dependence or a high dependence on Q depending on the relationship between tau and r. This function remains "under wraps" and needs to be uncovered. It could be a nonlinear dependence
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#1712 by aero on 15 Oct, 2016 19:11
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Dr. Rodal,
For a TE0np mode, for example, for TE012 or for TE013, assuming perfect conductors, the stress and the energy density are the same thing. You should not be integrating the stress to obtain the energy density.
So what is true about the meep force curves related to energy density? Is the Maxwell Stress Tensor at a point equal to the energy density at that point? This run was for a copper cavity. Could that be why it shows negative forces just inside the end boundaries?
aero
Please show the same graph for a TE0np mode, without integrating the stress. Do not integrate the stress.
For non-perfect conductors there are really tiny electric fields within the skin depth but they are usually ignored (ref. Jackson) for TE0np modes, and don't much affect what is stated above.
The image that I attached, linked here, was the Maxwell Stress Tensor integrated over a detector 0.0354 square meep units. It was my consideration that I could recover the average stress on the detector by dividing the integral by the detector area. The cavity is resonating in the TE013 mode at high Q ~ 100,000. I'm using a meep built-in function to generate the curve above and to "not integrate" would require that I calculate the stresses from the fields as I know you have done, but doing so would result in so many data points that they would be difficult to deal with as I am looking for the internal values, not the values at the boundary.
I'm not sure what causes the curve to go negative though. Negative values aren't really at the ends, particularly the one at -1.5. The internal surface of the ends are at exactly -0.4 and +0.4 while the two end detectors are one grid point inside the ends. -
#1713 by SeeShells on 15 Oct, 2016 19:13
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Missed a few minus signs.
It's still mind blowing when you realize that a smaller Q and a lower frequency should work better. Note, this is not assuming a cylinder. It is using the radial coordinate from the apex of the cone and gives the basic design relationships.
Thrust - With constant frequency, constant velocity and constant wavelength inside the cavity.
With more definitions, since nobody seems to understand that thrust inversely proportional to Q is very different from anything done prior.
In these equations;
tau(r) is the decay time of the Q as a function of r, measured from the apex of the cone.
Zeta(r) is the damping factor as a function of r.
m(r) is the variation in equivalent mass of the MW photons inside the frustum, as a function of r.
g(r) is the acceleration due to the gradient in Zeta, wrt. r.
T is the thrust, written as a function of Q, or a function of tau.
Chi is (scalar) magnetic flux.
When there is current flowing through the copper, copper has resistance and resistance creates a voltage drop. Once that voltage drop has been created, magnetic flux can escape by the amount equal to the V-sec around the loop. So magnetic flux is escaping where there is heat being generated and power is dissipated. Preferably at the big end. This asymmetry provides the equation for force, the same as a photon rocket, except;
F ~ P/2v, where v = -1/(dtau/dr) << c
Hence, thrust can be generated much greater than a photon rocket, because the velocity at which magnetic flux is escaping is so very slow.
Fields outside of a frustum.
Shell -
#1714 by Rodal on 15 Oct, 2016 19:16
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The image that I attached, linked here, was the Maxwell Stress Tensor integrated over a detector 0.0354 square meep units. It was my consideration that I could recover the average stress on the detector by dividing the integral by the detector area. The cavity is resonating in the TE013 mode at high Q ~ 100,000. I'm using a meep built-in function to generate the curve above and to "not integrate" would require that I calculate the stresses from the fields as I know you have done, but doing so would result in so many data points that they would be difficult to deal with as I am looking for the internal values, not the values at the boundary.
I'm not sure what causes the curve to go negative though. Negative values aren't really at the ends, particularly the one at -1.5. The internal surface of the ends are at exactly -0.4 and +0.4 while the two end detectors are one grid point inside the ends.
What about this MEEP function that apparently is meant to output the total electric and magnetic energy density ?
http://ab-initio.mit.edu/wiki/index.php/Meep_Reference#Output_functions
output-tot-pwr
Output the total electric and magnetic energy density. Note that you might want to wrap this step function in synchronized-magnetic to compute it more accurately; see Synchronizing the magnetic and electric fields. -
#1715 by Notsosureofit on 15 Oct, 2016 19:26
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If the thrust mechanism is magnetic field "shedding", and I see no reason that that would be impossible, I would still think it would increase with Q. Power in would still equal the total dissipation and the coherence length of the surrounding field should increase with Q. I would think the "effective coherent volume" might represent a degree of coupling to the vacuum.
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#1716 by SeeShells on 15 Oct, 2016 19:31
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The point being that there are fields outside of the frustum and as to the field strengths they need to be derived from meep, this one was a TM mode. I'd recommend to aero that's another meep run he should do with his current TE013 frustum, so we can keep the current playing field....
Fields outside of a frustum.
Shell
What is the magnitude of:
1) the calculated fields outside the frustum of a cone
2) the calculated field inside the frustum of a cone
Need to understand the numerical accuracy of these calculations to understand what is being plotted for "outside fields"
Shell -
#1717 by X_RaY on 15 Oct, 2016 19:36
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If the thrust mechanism is magnetic field "shedding", and I see no reason that that would be impossible, I would still think it would increase with Q. Power in would still equal the total dissipation and the coherence length of the surrounding field should increase with Q. I would think the "effective coherent volume" might represent a degree of coupling to the vacuum.
Due to the skineffect* at the conductive walls the field is many many orders of magnitude greater inside the cavity than at the outside(when the source is still inside). Therefore its more likely that possible interactions with the QV should be orders of magnitude larger inside the cavity itself.
*and thick walls compared to the penetration depth for a given frequency and material conductivity of the wall -
#1718 by Rodal on 15 Oct, 2016 19:37
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If the thrust mechanism is magnetic field "shedding", and I see no reason that that would be impossible, I would still think it would increase with Q. Power in would still equal the total dissipation and the coherence length of the surrounding field should increase with Q. I would think the "effective coherent volume" might represent a degree of coupling to the vacuum.
Todd forgot the negative signs multiplying the rate of change of Log[Q] with r.
Todd's formula should show thrust proportional to the negative rate of change of Log[Q] with r.
EDIT: Todd corrected his posted formula, now reads OK -
#1719 by Rodal on 15 Oct, 2016 19:39
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The point being that there are fields outside of the frustum and as to the field strengths they need to be derived from meep, this one was a TM mode. I'd recommend to aero that's another meep run he should do with his current TE013 frustum, so we can keep the current playing field....
Fields outside of a frustum.
Shell
What is the magnitude of:
1) the calculated fields outside the frustum of a cone
2) the calculated field inside the frustum of a cone
Need to understand the numerical accuracy of these calculations to understand what is being plotted for "outside fields"
Shell
I think that those fields outside the frustum are due to well-known numerical lack of precision in the Finite Difference method (which cannot satisfy the Neumann boundary conditions all along the boundary, but only at the nodes). Take a look at their numerical magnitude in comparison with the fields inside the frustum.
Also should compare FDTD domain results with exact solutions to ascertain the veracity of the numerical output.
It's still mind blowing when you realize that a smaller Q and a lower frequency should work better. Note, this is not assuming a cylinder. It is using the radial coordinate from the apex of the cone and gives the basic design relationships.
Thanks.