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#1680 by WarpTech on 15 Oct, 2016 02:22
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Since when is the "integral of the Poynting vector over equal sized squares" equal to the Energy density...What is shown is the integral of the Poynting vector takes over equal sized squares at each data point. I believe that to be the energy density.
aero
I would agree....
Can you explain that?
Hahaha, probably not, but I thought I knew what he meant. Poynting vector is W/m^2, which is Energy density times velocity, and we know what that is.
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#1681 by Rodal on 15 Oct, 2016 02:26
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Since when is the "integral of the Poynting vector over equal sized squares" equal to the Energy density...What is shown is the integral of the Poynting vector takes over equal sized squares at each data point. I believe that to be the energy density.
aero
I would agree....
Can you explain that?
Hahaha, probably not, but I thought I knew what he meant. Poynting vector is W/m^2, which is Energy density times velocity, and we know what that is.
The statement <<the integral of the Poynting vector takes over equal sized squares at each data point. I believe that to be the energy density>> is not even dimensionally correct,
where Poynting vector S = E x H
and
energy density
A correct statement would be, for example, that the boundary integral of the Poynting vector over the surface boundary of a volume V, equals the negative of the power (not the power density), if the volume integral of the current density J and the electric field E is zero, integrated over the volume V. -
#1682 by WarpTech on 15 Oct, 2016 02:53
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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.
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#1683 by Rodal on 15 Oct, 2016 02:56
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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.
You can simplify further by noticing that
(1/tau)dtau/dr = d(Ln tau)/dr
so the thrust depends on the rate of change of the natural Logarithm of tau with respect to r
(times - P, divided by 2*omegao)
Here is a picture of the natural log (the rate of change changes a lot for small tau, but changes little for large tau, but one has to know the dependence of tau on r to make quantitative statements ...):
Notice that the relationship for Q is also the rate of a log of Q with respect to r -
#1684 by ThinkerX on 15 Oct, 2016 03:16
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Quote
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.
Thrust of the levels reported in the experiments? (Greater than a photon rocket)
CoE / CoM issues? Or are we still tapping into forces outside the cavity, be it gravity or the quantum vacuum?
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#1685 by WarpTech on 15 Oct, 2016 03:22
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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.
You can simplify further by noticing that
(1/tau)dtau/dr = d(Ln tau)/dr
so the thrust depends on the rate of change of the natural Logarithm of tau with respect to r
(times - P, divided by 2*omegao)
Here is a picture of the natural log (the rate of change changes a lot for small tau, but changes little for large tau, but one has to know the dependence of tau on r ...):
Notice that the relationship for Q is also the rate of a log of Q with respect to r
True, but the way I have it written, it is more obvious that the "force" is dependent on P/2v, where v = -dr/dtau. -
#1686 by aero on 15 Oct, 2016 03:25
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You agree that the "integral of the Poynting vector SA= E x H...What is shown is the integral of the Poynting vector takes over equal sized squares at each data point. I believe that to be the energy density.
aero
I would agree....
over equal sized squares" is equal to the Energy density
Can you explain that?
I miss quoted - it 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.
aero -
#1687 by WarpTech on 15 Oct, 2016 03:32
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Quote
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.
Thrust of the levels reported in the experiments? (Greater than a photon rocket)
CoE / CoM issues? Or are we still tapping into forces outside the cavity, be it gravity or the quantum vacuum?
As Dr. Rodal is pointing out, we do not know the dependence of tau on r. If we control that, then the thrust levels will be just an engineering problem.
Yes, it will be greater than a photon rocket, because F = P/2v, where v = -dr/dtau, and we will "control" this derivative by design. (I hope.)
No CoE or CoM issues. It works like gravity. The energy input has a potential energy, which is m*g*L. Energy escapes the frustum into the metal through heat dissipation. As it is dissipated, the CM of the internal field moves from the small end toward the big end, and the CM of the frustum moves the other way. If the energy could not escape, it would be a 1 shot deal, but since energy can escape as heat dissipation, the cycle can be repeated. The frustum can't move forward, unless the energy inside moves backwards and is lost as heat, almost the same as if it escaped out the back, but the velocity is much, much slower, boosting the thrust.
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#1688 by Rodal on 15 Oct, 2016 03:37
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You agree that the "integral of the Poynting vector SA= E x H...What is shown is the integral of the Poynting vector takes over equal sized squares at each data point. I believe that to be the energy density.
aero
I would agree....
over equal sized squares" is equal to the Energy density
Can you explain that?
I miss quoted - it 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.
aero
That doesn't make sense either.
Dimensionally, the units of Energy Density are energy per unit volume Energy/Volume or (Force*Length)/Volume, while the units of stress are force per unit area Force/Area. Since Volume = Area * Length, it is obvious therefore that stress and energy density have exactly the same units. (Which is not a coincidence).
So, if you integrate the stress, now you don't even have the same units as energy density.
If you integrate stress over an area, you end up with a unit of force.
If you integrate stress over a curved line, you end up with a unit of force per unit length.
Have to go ........think about it. I'll be back tomorrow
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#1689 by aero on 15 Oct, 2016 04:15
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You agree that the "integral of the Poynting vector SA= E x H...What is shown is the integral of the Poynting vector takes over equal sized squares at each data point. I believe that to be the energy density.
aero
I would agree....
over equal sized squares" is equal to the Energy density
Can you explain that?
I miss quoted - it 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.
aero
That doesn't make sense either.
Dimensionally, the units of Energy Density are energy per unit volume Energy/Volume or (Force*Length)/Volume, while the units of stress are force per unit area Force/Area. Since Volume = Area * Length, it is obvious therefore that stress and energy density have exactly the same units. (Which is not a coincidence).
So, if you integrate the stress, now you don't even have the same units as energy density.
If you integrate stress over an area, you end up with a unit of force.
If you integrate stress over a curved line, you end up with a unit of force per unit length.
Have to go ........think about it. I'll be back tomorrow
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. -
#1690 by birchoff on 15 Oct, 2016 04:35
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Quote
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.
Thrust of the levels reported in the experiments? (Greater than a photon rocket)
CoE / CoM issues? Or are we still tapping into forces outside the cavity, be it gravity or the quantum vacuum?
As Dr. Rodal is pointing out, we do not know the dependence of tau on r. If we control that, then the thrust levels will be just an engineering problem.
Yes, it will be greater than a photon rocket, because F = P/2v, where v = -dr/dtau, and we will "control" this derivative by design. (I hope.)
No CoE or CoM issues. It works like gravity. The energy input has a potential energy, which is m*g*L. Energy escapes the frustum into the metal through heat dissipation. As it is dissipated, the CM of the internal field moves from the small end toward the big end, and the CM of the frustum moves the other way. If the energy could not escape, it would be a 1 shot deal, but since energy can escape as heat dissipation, the cycle can be repeated. The frustum can't move forward, unless the energy inside moves backwards and is lost as heat, almost the same as if it escaped out the back, but the velocity is much, much slower, boosting the thrust.
How does a dielectric affect the idea your proposing? -
#1691 by jay343 on 15 Oct, 2016 05:17
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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.
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. -
#1692 by meberbs on 15 Oct, 2016 05:43
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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. -
#1693 by jay343 on 15 Oct, 2016 06:24
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Also, they can't[\b] just classify something. Only the government can classify something, and if you had the experience you just claimed, you should know that.
I wasn't saying that Boeing classified it, however, if any part of the technology was already classified, then indeed a Boeing employee could classify the information based on association. That's what "classifiers" do. That said, all government contractors like Boeing have government personnel on-site, representing all of their customers. Getting something entirely new classified because it is recognized as sensitive, could be done very quickly. For that matter, Boeing could also declare it as proprietary information which for all intents and purposes has the same effect as far as Roger's ability to share the information.
No, I think that if the prototype didn't work, Boeing would have said that very clearly. Unless it was already classified ;-) -
#1694 by Gilbertdrive on 15 Oct, 2016 08:11
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While there are situations here on earth that kinetic energy is an important component, it is not always necessary when exploring CoM and CoE.
Yes on the fact that, for many calculations, you do not need to use Kinetic energy.
Also, for deep space travel, the speed is important in the departure and the arrival reference frame, because it is important to know when we are at destination !
And the speed is as much relative to a reference frame that the Kinetic Energy. Since the rest mass does not move, Kinetic Energy is entirely defined by the speed, and the speed is entirely desined by the Kinetic energy. If one is fiftitious, the other is also.
When you want to explore CoE, you have to do it from a point of view of an inertial reference frame, and you have to take into account the Kinetic energy in this reference frame. You can't do the economy of Kinetic energy when exploring CoE of anything having accelerated from it's original reference frame.
But what is true, and maybe it is what you intended to exprime, is the fact that you do not need to evaluate CoE when actually travelling. In fact, you need to verify CoE essentially when you are unsure of the performance of a device. For example, if one tell me that his LH2/LOX motor send burnt gaz at 20 000m/s, I can calculate easily with CoE that it is not possible, using Kinetic energy of the gaz, and comparing it to the chemical energy of LH2/LOX.
It is simpler than using pressure calculus and expansion in a nozzle...
But, once you know really how your motor perform, you do not need to evaluate CoE anymore. I think I agree with you on that point.
Once we know how the emdrive works, how it is interacting with fields, in what way it performs, it may become useless to verify CoE.
Since we do not know how it works, it is at the opposite important to explore CoE, using Kinetic Energy, because it will give us tracks about theories.
That is why any verified experiment showing more force than P/V would dismiss many possible theories. And we know that because we explorate CoE.
I shall give an example of theory.
If the drive is pushing against the earth by using a new distant interaction, still working from orbit. Doing the same thing that a maglev train on earth, but at bigger distances.
This theory would be dismissed by exploration of CoE if thrust bigger than P/V is measured.
That is why CoE is so important to evaluate in the context of creating new theories. At the opposite, CoE evaluation is not needed for calculating the trajectory of a classical rocket.
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#1695 by Rodal on 15 Oct, 2016 12:07
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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. -
#1696 by meberbs on 15 Oct, 2016 13:33
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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, 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.
I wasn't saying that Boeing classified it, however, if any part of the technology was already classified, then indeed a Boeing employee could classify the information based on association. That's what "classifiers" do. That said, all government contractors like Boeing have government personnel on-site, representing all of their customers. Getting something entirely new classified because it is recognized as sensitive, could be done very quickly. For that matter, Boeing could also declare it as proprietary information which for all intents and purposes has the same effect as far as Roger's ability to share the information.
No, I think that if the prototype didn't work, Boeing would have said that very clearly. Unless it was already classified ;-)
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. -
#1697 by rfmwguy on 15 Oct, 2016 13:40
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Been out of this industry for about a decade, but what you say is mainly true. General export regs are what companies tend to refer to first for guidance. If materials, power, frequency, processing speed, et al, is mentioned, its the company's duty to control these items in the marketplace, which then involves disclosure to the government, which then can result in classification. Not an expert here, but have been on that path before and relaying what I picked up over the years.Also, they can't[\b] just classify something. Only the government can classify something, and if you had the experience you just claimed, you should know that.
I wasn't saying that Boeing classified it, however, if any part of the technology was already classified, then indeed a Boeing employee could classify the information based on association. That's what "classifiers" do. That said, all government contractors like Boeing have government personnel on-site, representing all of their customers. Getting something entirely new classified because it is recognized as sensitive, could be done very quickly. For that matter, Boeing could also declare it as proprietary information which for all intents and purposes has the same effect as far as Roger's ability to share the information.
No, I think that if the prototype didn't work, Boeing would have said that very clearly. Unless it was already classified ;-)
In pure speculation, I sense Shawyers IP was buried for R&D, possibly partnered with a government entity, thereby Boeing was not, within itself, working on it as "Boeing". Semantics have a way of answering questions they was not intended to answer. But thats just one person's opinion...no evidence to back it up.
The disagreement I have is your comment that companies cannot just steal IP. Stealing and enhancement/development are a fine line which has been crossed many times over many years. Its a rough and tumble world out there. -
#1698 by Rodal on 15 Oct, 2016 14:10
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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
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#1699 by RotoSequence on 15 Oct, 2016 16:03
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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.
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.