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#2000 by Bob012345 on 25 Oct, 2016 18:23
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emdrive fan here.
Here's my latest attempt to make sense of emdrive : Consider reflection vs absorption. If a plate reflects a photon, its will take twice as much force from the photon than if it absorbed the photon. Like if a person catches a ball, it takes twice as much force to reflect the ball back at the same speed than to catch it and stop it.
The EMDrive's 2 end plates may be reflecting and absorbing photons at different rates to each other, hence they will have different radiations pressure and different forces. In other words, one plate reflects photons more, while the other absorbs them more. Hence net force, hence thrust.
This suggests that the different sizes of endplates is not necessary. An alternative is a straight cavity, with different end plate materials than absorb and reflect at different rates to each other.
thanks for any replys
That's more or less where I've been trying to go in calculating the forces within the cavity, except using atmospheric air molecules instead of photons. Air particles have the major advantage of being far more massive than photons hence differences in directional acceleration/rebound would be far more pronounced.
But unfortunately irregardless of what happens within the cavity, the COM issue still exists. For that idea to work the air particles or photons would need to be only an intermediary for some other mechanism which allows/causes something massive to escape.
The location of the stored "potential" energy and the location where that energy is more readily dissipated forms a gradient inside the frustum. As long as the internal potential energy moves along this gradient to be dissipated, the frustum will move the other way. COM is not violated, regardless of the fact that nothing but heat is escaping. The heat is not a propellant, the internal gradient and the amount of energy flowing with that gradient is.
If that's the correct operating principle, can you think of an easier way to set up a proper gradient without fancy microwave equipment and cavities? -
#2001 by meems on 25 Oct, 2016 19:03
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Quote from: WarpTech
The location of the stored "potential" energy and the location where that energy is more readily dissipated forms a gradient inside the frustum.
In other words, since one end-plate is absorbing more while the other is reflecting more, the energy density of the cavity will be lower nearer one end : the absorbing end. Because an absorbing wall represents an energy sink.
Correct?
Any emdrive experimenters out there want to try building a straight sided cavity with different material endplates?
Am I right in thinking that emdrive experimenters have already noted a change in thrust by varying the end plate materials? -
#2002 by D_Dom on 25 Oct, 2016 19:39
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Seems almost too simple, I like it. I was awake last night and couldn't stop thinking a demonstration is all around us all the time. Much like gravity was exploited by mankind long before Newtons explanation. Still I would expect cylindrical cavities with varying endplates have been studied extensively.
A poor example off the top of my head, a common laser has partially silvered mirror at one end. Optical wavelengths reflected power are controlled with a "Brewster angle" at the junction between glass and air. Been far too long since I studied optics, have to add more to the reading list. -
#2003 by WarpTech on 25 Oct, 2016 19:52
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Quote from: WarpTech
The location of the stored "potential" energy and the location where that energy is more readily dissipated forms a gradient inside the frustum.
In other words, since one end-plate is absorbing more while the other is reflecting more, the energy density of the cavity will be lower nearer one end : the absorbing end. Because an absorbing wall represents an energy sink.
Correct?
Any emdrive experimenters out there want to try building a straight sided cavity with different material endplates?
Am I right in thinking that emdrive experimenters have already noted a change in thrust by varying the end plate materials?
Correct with a caveat. This sets up a gradient. But the resonance produces a large enough equivalent field "mass" to move along this gradient with enough force to move the frustum the other way. m1*a1 = m2*a2. Without a resonant cavity, the amount of power required would be enormous.
As far as I know, copper, HDPE, PTFE and Shawyer's super conductor, and spherical end plate designs, are the only experiments that have been done with different materials and/or shapes. I'm still trying to make heads or tails of the results as they pertain to my my model.
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#2004 by Rodal on 25 Oct, 2016 19:52
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Quote from: WarpTech
The location of the stored "potential" energy and the location where that energy is more readily dissipated forms a gradient inside the frustum.
In other words, since one end-plate is absorbing more while the other is reflecting more, the energy density of the cavity will be lower nearer one end : the absorbing end. Because an absorbing wall represents an energy sink.
Correct?
...
No, the location of the peak energy density is not due to any "absorption". See the excellent discussion by Greg Egan
http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html
for an exact solution. The location of the peak in energy density is due to the mode shape, and the mode shapes are a result of the eigenvalue problem which is dictated by the equations of motion and the boundary conditions, and not by dissipation.
Here you have the first 3 TE mode shapes in an electromagnetically resonant truncated cone cavity.
There is no change whatsoever in absorption at each end for these modes. The modes are a result of the eigenvalue problem. That's why they are called eigenmodes !
TE011
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TE012
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TE013
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You have mode shapes in any vibration problem, in a string for example, or in a horn instrument.
See any undergraduate textbooks in Physics for discussion of the eigenvalue problem: Halliday and Resnick, or the Feynman Lectures in Physics, etc.
The eigenmodes are not due to dissipation at one end, or with dissipation at a node.
You have mode shapes even with zero damping !
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#2005 by WarpTech on 25 Oct, 2016 20:04
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Seems almost too simple, I like it. I was awake last night and couldn't stop thinking a demonstration is all around us all the time. Much like gravity was exploited by mankind long before Newtons explanation. Still I would expect cylindrical cavities with varying endplates have been studied extensively.
A poor example off the top of my head, a common laser has partially silvered mirror at one end. Optical wavelengths reflected power are controlled with a "Brewster angle" at the junction between glass and air. Been far too long since I studied optics, have to add more to the reading list.
The problem with a laser is the wave velocity is too fast. ~c, so T ~ P/c. With the frustum, T ~ P/v, where;
1/v ~ -(1/w)*d[ln(Q)]/dr
Where w is 2pi*f. We can replace Q by tau for decay time, or replace Q by Reactive power stored over Real power lost per cycle, as a function of r from the apex of the cone. We could cut the frustum simulations from front to back into circular cross sections and look at how much energy is stored and how much energy is lost in each cross section to calculate the derivative of Q. We can also approximate this as;
1/v ~ -(1/w*L)*ln(Qb/Qs)
Where L is the length, and the Q's are at the big and small end, respectively.
In any case, this is the equation we need to maximize by design. The larger this value is, the more thrust we will get.
Todd
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#2006 by WarpTech on 25 Oct, 2016 20:14
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Quote from: WarpTech
The location of the stored "potential" energy and the location where that energy is more readily dissipated forms a gradient inside the frustum.
In other words, since one end-plate is absorbing more while the other is reflecting more, the energy density of the cavity will be lower nearer one end : the absorbing end. Because an absorbing wall represents an energy sink.
Correct?
...
No, the location of the peak energy density is not due to any "absorption". See the excellent discussion by Greg Egan
http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html
for an exact solution. The location of the peak in energy density is due to the mode shape, and the mode shapes are a result of the eigenvalue problem which is dictated by the equations of motion and the boundary conditions, and not by dissipation.
Here you have the first 3 TE mode shapes in an electromagnetically resonant truncated cone cavity.
There is no change whatsoever in absorption at each end for these modes. The modes are a result of the eigenvalue problem. That's why they are called eigenmodes !
TE011
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TE012
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TE013
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You have mode shapes in any vibration problem, in a string for example, or in a horn instrument.
The eigenmodes are not due to dissipation at one end, or with dissipation at a node.
You have mode shapes even with zero damping !
Thanks Jose, but I do not think he or I meant to imply that the mode shapes were due to absorption. Only that the energy lost will be asymmetrical. Greg Eagan's derivation does not take into consideration sources or sinks, both of which contribute to a divergence of decay time and energy (Volt-sec), that is not present in his derivation. He assumes all waves to be time symmetric and therefore average to zero, but that is not necessarily true when there are sources and sinks. Isn't that true?
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#2007 by Rodal on 25 Oct, 2016 20:24
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...Thanks Jose, but I do not think he or I meant to imply that the mode shapes were due to absorption. Only that the energy lost will be asymmetrical. Greg Eagan's derivation does not take into consideration sources or sinks, both of which contribute to a divergence of decay time and energy (Volt-sec), that is not present in his derivation. He assumes all waves to be time symmetric and therefore average to zero, but that is not necessarily true when there are sources and sinks. Isn't that true?
Greg Egan solved the eigenvalue problem under well-stated assumptions.
Whenever Monomorphic or X_Ray have shown a FEKO solution, or Paul March or Samsonov (RFPlumber) have shown a COMSOL solution, they have solved the eigenvalue problem too, just like Egan, but done it numerically, instead of analytically.
All those solutions compare very well with experiments, concerning eigenmodes and eigenfrequencies.
The difference is that those solutions (COMSOL, FEKO, Egan) cannot calculate any thrust, since Maxwell's equations will show no thrust for a closed cavity, acting as a closed system. Even if you allow for dissipation of heat and release of heat into space as heat radiation one can readily show that the thrust would be less than that of a perfectly collimated photon rocket, and hence much less than what is claimed for the EM Drive.
Aren't you justifying the EM Drive thrust as due to a polarized quantum vacuum theory?
(not due to heat dissipation and heat radiating into Space)? -
#2008 by Monomorphic on 25 Oct, 2016 20:56
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...Thanks Jose, but I do not think he or I meant to imply that the mode shapes were due to absorption. Only that the energy lost will be asymmetrical. Greg Eagan's derivation does not take into consideration sources or sinks, both of which contribute to a divergence of decay time and energy (Volt-sec), that is not present in his derivation. He assumes all waves to be time symmetric and therefore average to zero, but that is not necessarily true when there are sources and sinks. Isn't that true?
Greg Egan solved the eigenvalue problem under well-stated assumptions.
Whenever Monomorphic or X_Ray have shown a FEKO solution, or Paul March or Samsonov (RFPlumber) have shown a COMSOL solution, they have solved the eigenvalue problem too, just like Egan, but done it numerically, instead of analytically.
All those solutions compare very well with experiments, concerning eigenmodes and eigenfrequencies.
The difference is that those solutions (COMSOL, FEKO, Egan) cannot calculate any thrust, since Maxwell's equations will show no thrust for a closed cavity. Even if you allow for dissipation of heat and release of heat into space as heat radiation one can readily show that the thrust would be less than that of a perfectly collimated photon rocket, and hence much less than what is claimed for the EM Drive.
Aren't you justifying the EM Drive thrust as due to a polarized quantum vacuum theory?
(not due to heat dissipation and heat radiating into Space)?
Dr. Rodal, I have been looking at other EM software suites and noticed that Opera Simulation Software is the preferred package used by many scientists at CERN. http://operafea.com/product/
They have a number of application modules that are listed on that page. Many seem like they could be of benefit for our purposes. For example, the stress analysis module allows external forces and constraints to be applied globally or as boundary conditions on the surface of the model. There is a HF EM module, a dynamic EM module, and even a sputtering module!
This software is probably extremely expensive, but it looks like exactly what is needed here. I was hoping you would take a quick look through the modules and comment. Thanks! -
#2009 by meems on 25 Oct, 2016 21:03
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No, the location of the peak energy density is not due to any "absorption". See the excellent discussion by Greg Egan
I looked at the link. I am right in thinking Greg is arguing against emdrive?
> The eigenmodes are not due to dissipation at one end, or with dissipation at a node.
By that logic eigenmodes would be unaffected even without end plates on emdrive, since 0% or 100% dissipation makes no difference, and an open ended emdrive is effectively 100% dissipation. Which is nonsense.
I think the reason dissipation doesn't factor in Greg's analysis is that he simply doesn't consider an emdrive with a high dissipation ( absorption ) end plate.
Anyway, I consider energy distribution within the cavity to be of secondary importance. I don't know why on this forum you seem to focus on it. Roger says the thrust is generated by reaction forces at the end plates, and that seems sensible to me. -
#2010 by Rodal on 25 Oct, 2016 21:20
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...
The solution to the eigenvalue problem includes the boundary conditions.
> The eigenmodes are not due to dissipation at one end, or with dissipation at a node.
By that logic eigenmodes would be unaffected even without end plates on emdrive, since 0% or 100% dissipation makes no difference, and an open ended emdrive is effectively 100% dissipation. Which is nonsense.
...
What you write about <<By that logic eigenmodes would be unaffected even without end plates on emdrive>> is incorrect, because when you have no end plates, you change the boundary conditions.
Actually without end plates, you no longer have standing waves !
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#2011 by Rodal on 25 Oct, 2016 21:33
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...Dr. Rodal, I have been looking at other EM software suites and noticed that Opera Simulation Software is the preferred package used by many scientists at CERN. http://operafea.com/product/
Sorry, I am not familiar with Opera FE, and I could not find a technical paper describing what solution techniques would set it apart from other packages. I am familiar with other general solution packages like Abaqus, ANSYS, and ADINA.
Different people at CERN use different software, how do you determine which one is preferred ?. -
#2012 by meems on 25 Oct, 2016 22:12
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Seems almost too simple, I like it. I was awake last night and couldn't stop thinking a demonstration is all around us all the time. Much like gravity was exploited by mankind long before Newtons explanation. Still I would expect cylindrical cavities with varying endplates have been studied extensively.
A poor example off the top of my head, a common laser has partially silvered mirror at one end. Optical wavelengths reflected power are controlled with a "Brewster angle" at the junction between glass and air. Been far too long since I studied optics, have to add more to the reading list.
Here's some theory I think is central to emdrive :
Why doesn't a 100 Watt light torch give a reaction force ( kick back ) similar to a mechanical device with a power of 100 Watts?
An example of mechanical 100 Watt thruster is : A man throwing 2kg mass projectiles at 10 m/s once per second, every second. Such action would be plenty to accelerate the man in the opposite direction of his projectiles at m/s.
But the kickback from 100 W light torch would be effectively zero by comparison.
Why? The answer is : only a tiny fraction of the energy in light is as mechanical energy \ momentum. Most of it is stored in a EM wave which is lateral to its motion. In other words, a photon is like a thrown grenade, its kinetic energy is insignificant in comparison to its stored potential energy.
The key then, to making a thruster that uses light, is somehow exploiting the lateral EM energy : by converting it into parallel kinetic energy ( parallel to the motion of the thruster ).
We already know its possible to convert light to motion with near 100% efficiency : shine a light on a black body and the light will be converted to heat which is electron motion. But this is unorganised ( random ) motion. We need coherent motion.
How does a photon impart its energy into electron motion?
This question is a bit theoretical. When a photon hits an electron, the electron is exposed to the lateral electric and magnetic waves of the photon, which will push the electron laterally to the incident of the photon.
This might explain why shining a light torch does not give a significant kickback, and why an object being exposed to a light doesn't experience a significant kickback. Energy conversion between mechanical and light energy is lateral to the direction of travel of the light.
How to go about applying this line of theory to gain thrust
The photons must be incoming at a lateral angle to the intended direction of thrust. Then the lateral EM wave will act along the axis of intended thrust.
But it which direction along the axis of thrust?
If the light is incoherent, in both directions. But with a coherent laser, and with a material with precisely engineered layers of absorbing material layered with transparent material ( positioned where the EM wave would give a kick in the wrong direction ), then the magnetic and electric waves of the light will always be in the right direction to kick the thruster in the intended direction of thrust.
Well that's the end of this line of thought. Its not specific to emdrive. Its just a theory on how to convert light into motion, without using a propellent. It maybe that emdrive is somehow using this theory. -
#2013 by meems on 25 Oct, 2016 22:29
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...
The solution to the eigenvalue problem includes the boundary conditions.
> The eigenmodes are not due to dissipation at one end, or with dissipation at a node.
By that logic eigenmodes would be unaffected even without end plates on emdrive, since 0% or 100% dissipation makes no difference, and an open ended emdrive is effectively 100% dissipation. Which is nonsense.
...
What you write about <<By that logic eigenmodes would be unaffected even without end plates on emdrive>> is incorrect, because when you have no end plates, you change the boundary conditions.
Actually without end plates, you no longer have standing waves !
To understand this, look at undergraduate textbooks in Physics for discussion of the eigenvalue problem like the Feynman Lectures in Physics, and also at a book on partial differential equations, for the role of boundary conditions.
Correct, and further, there would also be no standing waves if the boundary condition was 100% dissipation \ absorption of incident light energy. Therefore your assertion that absorption is not a factor in the energy distribution is wrong. -
#2014 by dustinthewind on 25 Oct, 2016 22:43
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...Thanks Jose, but I do not think he or I meant to imply that the mode shapes were due to absorption. Only that the energy lost will be asymmetrical. Greg Eagan's derivation does not take into consideration sources or sinks, both of which contribute to a divergence of decay time and energy (Volt-sec), that is not present in his derivation. He assumes all waves to be time symmetric and therefore average to zero, but that is not necessarily true when there are sources and sinks. Isn't that true?
Greg Egan solved the eigenvalue problem under well-stated assumptions.
Whenever Monomorphic or X_Ray have shown a FEKO solution, or Paul March or Samsonov (RFPlumber) have shown a COMSOL solution, they have solved the eigenvalue problem too, just like Egan, but done it numerically, instead of analytically.
All those solutions compare very well with experiments, concerning eigenmodes and eigenfrequencies.
The difference is that those solutions (COMSOL, FEKO, Egan) cannot calculate any thrust, since Maxwell's equations will show no thrust for a closed cavity, acting as a closed system. Even if you allow for dissipation of heat and release of heat into space as heat radiation one can readily show that the thrust would be less than that of a perfectly collimated photon rocket, and hence much less than what is claimed for the EM Drive.
Aren't you justifying the EM Drive thrust as due to a polarized quantum vacuum theory?
(not due to heat dissipation and heat radiating into Space)?
Speaking of the polarizeable vacuum, traveling waves as opposed to standing waves and EM propulsion has brought back an old idea I was tinkering with a while ago. The image is attached below and comes from this forum thread here: https://forum.nasaspaceflight.com/index.php?topic=39464.0 . The idea is, haven't they confirmed that the earths rotation drags on space time with gravity probe b? It reminds me of super-fluid viscosity in the youtube video I posted here: https://forum.nasaspaceflight.com/index.php?topic=41462.msg1600285#msg1600285
So if we know matter has a viscous drag on space-time then the question is, "does light also have viscous drag on the quantum vacuum?" Properties of light trapped in a reflective cavity gains rest mass so there are some parallels between light and mass.
The idea was to use a superconductive wall as in the image below of the toroid with a part of the wall open and shoot light in. The light goes in and is trapped inside, going around the superconductive walls indefinitely. Eventually, if light has drag on space time then energy will be transferred from the light to the vacuum inducing a vortex. The Earth is likely to have more mass, but the velocity of light might make a larger difference. If a vortex is set up around the toroid, then the toroid wouldn't experience any net drag it appears, but if you put a ship in the middle, it should experience an unidirectional frame drag.I can't remember the paper but there was one paper that suggested some one who had passed away had come up with derivations that light could have viscous drag on the vacuum over long distance and they wanted to explore if some of the red shift of distant light was the light losing energy to the vacuum. I'll have to see if I can find that article again.Here is the article: http://link.springer.com/article/10.1007/s10509-012-1260-x , I would think such an experiment as the light injected into the toroid may possibly explore that question.
Energy should be lost from the light to the vacuum as the vacuum accelerates (change in frame of the light), red-shifting the light and if the vacuum effects the ship in the middle then energy should be transferred to the ship. It should set up an effect where such a vortex would induce a back flow outside the toroid vortex on the outer vacuum also. It is a bit different than the current EM drive, but is a type of EM drive working off the frame drag effect.
The odd thing about the cavity is that it would have inside a traveling wave, instead of a standing wave and there may be concerns such as feeding in light in phase to constructively interfere. (or maybe destructive interference would be desired?) There is some energy loss gradients if we consider space as viscous which reminds me of the impedance Todd speaks of which may be in a sense an exchange with the vacuum.
Also, maybe the viscous drag could increase as a non-linear function of the electric field strength of the light at some point. Instead of a toroid a cross cut of the toroid, as dual cylinders may be an easier concept to test.
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#2015 by Rodal on 25 Oct, 2016 22:55
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If you are going to quote me, please use an actual quote and do it in context....
The solution to the eigenvalue problem includes the boundary conditions.
> The eigenmodes are not due to dissipation at one end, or with dissipation at a node.
By that logic eigenmodes would be unaffected even without end plates on emdrive, since 0% or 100% dissipation makes no difference, and an open ended emdrive is effectively 100% dissipation. Which is nonsense.
...
What you write about <<By that logic eigenmodes would be unaffected even without end plates on emdrive>> is incorrect, because when you have no end plates, you change the boundary conditions.
Actually without end plates, you no longer have standing waves !
To understand this, look at undergraduate textbooks in Physics for discussion of the eigenvalue problem like the Feynman Lectures in Physics, and also at a book on partial differential equations, for the role of boundary conditions.
Correct, and further, there would also be no standing waves if the boundary condition was 100% dissipation \ absorption of incident light energy. Therefore your assertion that absorption is not a factor in the energy distribution is wrong.
Please read what I wrote.
What I stated was that << The eigenmodes are not due to dissipation at one end, or with dissipation at a node.>>.
Which is a correct statement.
You posted a question, which I took as stating that you were unsure, as you put Correct? in question marks:...In other words, since one end-plate is absorbing more while the other is reflecting more, the energy density of the cavity will be lower nearer one end : the absorbing end. Because an absorbing wall represents an energy sink.
Correct?
...
Which I answered showing that the position of the max energy density is a function of the mode shapes. If you don't like to get an answer, you should not be posting questions.
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#2016 by meems on 25 Oct, 2016 23:38
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Ok Rodal, whatever you say.
So, am I right in thinking you don't think emdrive can work? -
#2017 by Rodal on 25 Oct, 2016 23:56
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Ok Rodal, whatever you say.
I am not so sure of it.
So, am I right in thinking you don't think emdrive can work?
PS: Now I understand that your point (and Todd's) was that due to absorption the max energy density position would change. I expect this to be a very small effect for the EM Drive, due to the high Q in the experiments.
Also there does not seem to be an appreciable difference between experiments involving dissipation and calculations based on COMSOL, FEKO, etc, that do not involve dissipation and simply calculate the electromagnetic field, mode shapes and frequency based on Maxwell's equations and boundary conditions. -
#2018 by Rodal on 26 Oct, 2016 01:09
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Seems almost too simple, I like it. I was awake last night and couldn't stop thinking a demonstration is all around us all the time. Much like gravity was exploited by mankind long before Newtons explanation. Still I would expect cylindrical cavities with varying endplates have been studied extensively.
A poor example off the top of my head, a common laser has partially silvered mirror at one end. Optical wavelengths reflected power are controlled with a "Brewster angle" at the junction between glass and air. Been far too long since I studied optics, have to add more to the reading list.
The problem with a laser is the wave velocity is too fast. ~c, so T ~ P/c. With the frustum, T ~ P/v, where;
1/v ~ -(1/w)*d[ln(Q)]/dr
Where w is 2pi*f. We can replace Q by tau for decay time, or replace Q by Reactive power stored over Real power lost per cycle, as a function of r from the apex of the cone. We could cut the frustum simulations from front to back into circular cross sections and look at how much energy is stored and how much energy is lost in each cross section to calculate the derivative of Q. We can also approximate this as;
1/v ~ -(1/w*L)*ln(Qb/Qs)
Where L is the length, and the Q's are at the big and small end, respectively.
In any case, this is the equation we need to maximize by design. The larger this value is, the more thrust we will get.
Todd
NASA found best results with PTFE and HDPE dielectrics at the small end, which increases tan delta (dissipation) at the small end, hence lower Qs at the small end
1/v ~ -(1/w*L)*ln(Qb/Qs)
But this would argue for materials having much larger dissipation losses than HDPE or PTFE be better ?
I thought that they tried materials with more dissipation and they found worse results, not better. But have to check...to be sure
Also this prescription runs contrary to Shawyer, who maintains that it is better not to use any dielectrics (which were involved in his initial patents) because they decrease Q at one end and he thinks that's bad... -
#2019 by RotoSequence on 26 Oct, 2016 01:25
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Still too many low level questions and too little data.
