The difference between the Metal and the Copper model is perfect metal and our model of copper material. The difference between the Copper and zCopper model is the orientation of the dipole antenna. Both (all 3) models are excited by an Ez source. The Copper model antenna is aligned parallel to the y-axis while the zCopper model antenna is aligned parallel to the z-axis. That is it. Performance wise, the zCopper model gave a meep calculated quality factor ~40 times higher than the already high quality factor of the copper model, and over 10 times higher than the Metal model.If the Poynting vector field is weaker, doesn't this tell us something? I wonder if the Metal model Poynting vector field is intermediate because the Medal model Q was intermediate.
Quote from: aero on 07/04/2015 03:33 amThe difference between the Metal and the Copper model is perfect metal and our model of copper material. The difference between the Copper and zCopper model is the orientation of the dipole antenna. Both (all 3) models are excited by an Ez source. The Copper model antenna is aligned parallel to the y-axis while the zCopper model antenna is aligned parallel to the z-axis. That is it. Performance wise, the zCopper model gave a meep calculated quality factor ~40 times higher than the already high quality factor of the copper model, and over 10 times higher than the Metal model.If the Poynting vector field is weaker, doesn't this tell us something? I wonder if the Metal model Poynting vector field is intermediate because the Medal model Q was intermediate.The Metal model and the Copper models give the same Poynting vector field, no difference.The zCopper model gives a weaker Poynting vector field that both the Copper and Metal models.1) So it looks like what is important is the orientation of the antenna. It is better to orient the antenna perpendicular (y) to the source current Ez than to align it in the same z directionQUESTION: Why ?_________________________________________2) The material model difference between Copper and Metal makes no difference to the Poynting vector field
You guys tickle me pink. It doesn't matter where you put the antenna if you are using Maxwell/Poynting to figure that out. That theory says that you get zero thrust. You're better off with a Hail Mary. It's logical.
Traveler - Not attacking, just pointing something out.In my book, Shawyer gets major points. Best I can tell, the conical Frustum EM Drive thing is his concept. He spent a lot of years trying to figure things out, building one experimental model after another. In some areas, he was clearly ahead of the curve - I used to make comments to this effect in prior versions of this thread.Lately, though, what I see are grand claims lacking substance - most of them coming from you. Grand claims, to be taken seriously, REQUIRE grand evidence. A new theory or model won't suffice here. To back these claims up, to have them taken seriously, Shawyer will need to cough up a truly impressive EM Demo model - one that can be replicated by others. Talking about something that can lift itself, or dang close to it.At this point, your relentless promotions are hurting Shawyer's case far more than you are helping it.
Quote from: ThinkerX on 07/04/2015 02:32 amTraveler - Not attacking, just pointing something out.In my book, Shawyer gets major points. Best I can tell, the conical Frustum EM Drive thing is his concept. He spent a lot of years trying to figure things out, building one experimental model after another. In some areas, he was clearly ahead of the curve - I used to make comments to this effect in prior versions of this thread.Lately, though, what I see are grand claims lacking substance - most of them coming from you. Grand claims, to be taken seriously, REQUIRE grand evidence. A new theory or model won't suffice here. To back these claims up, to have them taken seriously, Shawyer will need to cough up a truly impressive EM Demo model - one that can be replicated by others. Talking about something that can lift itself, or dang close to it.At this point, your relentless promotions are hurting Shawyer's case far more than you are helping it. Shawyer is in business to make money from his IP. Can't see him providing DIY kits.Building a total system that generates 9.8N/ kg of total mass is not easy. Way beyong DIY.Any statements I make are based on data, images & documents in my files that I recently shared.I have showed how Shawyers use of Cullen's equation 15 is valid and those that claim otherwise don't understand what guide wavelength is and that it will not vary if you turn a 1 end closed waveguide into a 2 end closed waveguide.In the near future I will go through each of the 16 equations and 4 diagrams in the Shawyer Theory paper, invite comments, put the negative comments to Shawyer and post back his comments.You see it is easy to say Shawyers theory is wrong but maybe not so easy to go equation by equation through the theory paper and make such claims when Shawyer will respong.So let's see how willing those who attach Shawyer when he is not here to defend his theory will be to have to defend their claims when Shawyer will defend them.If I was a betting man, I would bet the critics don't score one goal.
Quote from: TheTraveller on 07/04/2015 11:58 amQuote from: ThinkerX on 07/04/2015 02:32 amTraveler - Not attacking, just pointing something out.In my book, Shawyer gets major points. Best I can tell, the conical Frustum EM Drive thing is his concept. He spent a lot of years trying to figure things out, building one experimental model after another. In some areas, he was clearly ahead of the curve - I used to make comments to this effect in prior versions of this thread.Lately, though, what I see are grand claims lacking substance - most of them coming from you. Grand claims, to be taken seriously, REQUIRE grand evidence. A new theory or model won't suffice here. To back these claims up, to have them taken seriously, Shawyer will need to cough up a truly impressive EM Demo model - one that can be replicated by others. Talking about something that can lift itself, or dang close to it.At this point, your relentless promotions are hurting Shawyer's case far more than you are helping it. Shawyer is in business to make money from his IP. Can't see him providing DIY kits.Building a total system that generates 9.8N/ kg of total mass is not easy. Way beyong DIY.Any statements I make are based on data, images & documents in my files that I recently shared.I have showed how Shawyers use of Cullen's equation 15 is valid and those that claim otherwise don't understand what guide wavelength is and that it will not vary if you turn a 1 end closed waveguide into a 2 end closed waveguide.In the near future I will go through each of the 16 equations and 4 diagrams in the Shawyer Theory paper, invite comments, put the negative comments to Shawyer and post back his comments.You see it is easy to say Shawyers theory is wrong but maybe not so easy to go equation by equation through the theory paper and make such claims when Shawyer will respong.So let's see how willing those who attach Shawyer when he is not here to defend his theory will be to have to defend their claims when Shawyer will defend them.If I was a betting man, I would bet the critics don't score one goal.That would be interesting. I try as an researcher and engineer to question everything. Even you and RS's theories as there exists holes in his theories that have been pointed out by people better than me. I wrote you on the Reddit blog after I read RS's application for a patent and I'll repost here and maybe you'll reply.QUOTE: Good luck with trying to invalidate Newton's second law by this overly simplistic explanation. On one of my patents invoking the fracture mechanics of a silicon wafer with a wafer carrier I played hell with the reviewers. Fracture mechanics was and still is poorly understood and they beat the crap out of me pushing me to explain what couldn't be explained. It breaks cleanly just didn't do it.I see SPR having the same issues talking about thrust with this over simplistic explanation. If you are having a tough time convincing some of the more knowledgeable in physics of this working how does SPR think they can convince the reviewers for patents? If you are a friend to RS and a good engineer please tell him it just may not fly because he is seemingly violating one of the most inviolate laws we have and his explanation is severely lacking.END QUOTE
I'm an enemy of exposition. I feel there's no need to overstate.
Quote from: SeeShells on 07/04/2015 12:27 pmQuote from: TheTraveller on 07/04/2015 11:58 amQuote from: ThinkerX on 07/04/2015 02:32 amThat would be interesting. I try as an researcher and engineer to question everything. Even you and RS's theories as there exists holes in his theories that have been pointed out by people better than me. I wrote you on the Reddit blog after I read RS's application for a patent and I'll repost here and maybe you'll reply.QUOTE: Good luck with trying to invalidate Newton's second law by this overly simplistic explanation. On one of my patents invoking the fracture mechanics of a silicon wafer with a wafer carrier I played hell with the reviewers. Fracture mechanics was and still is poorly understood and they beat the crap out of me pushing me to explain what couldn't be explained. It breaks cleanly just didn't do it.I see SPR having the same issues talking about thrust with this over simplistic explanation. If you are having a tough time convincing some of the more knowledgeable in physics of this working how does SPR think they can convince the reviewers for patents? If you are a friend to RS and a good engineer please tell him it just may not fly because he is seemingly violating one of the most inviolate laws we have and his explanation is severely lacking.END QUOTEI'm online maybe 2 hours a day every 3 days or so. Some of my meds literally take me off line. Very confused dreams. Weird for me as I normally can almost control my dreans.BTW I have no theory.Shawyer recently added more to his theory. Read how he believes Force generation is built up, bounce by bounce, as the EMDrive initially accelerates http://www.emdrive.com/IAC13paper17254.v2.pdfI really do wish you luck in your build efforts but believe you will be lucky to see anything.Had a talk with Shawyer and now understand why scales are a very bad way to try to measure Force. I have changed my build to a rotary test rig as that is really the only way to allow Force to develope as the EMDrive, being free to move, starts to accelerate.
Quote from: TheTraveller on 07/04/2015 11:58 amQuote from: ThinkerX on 07/04/2015 02:32 amThat would be interesting. I try as an researcher and engineer to question everything. Even you and RS's theories as there exists holes in his theories that have been pointed out by people better than me. I wrote you on the Reddit blog after I read RS's application for a patent and I'll repost here and maybe you'll reply.QUOTE: Good luck with trying to invalidate Newton's second law by this overly simplistic explanation. On one of my patents invoking the fracture mechanics of a silicon wafer with a wafer carrier I played hell with the reviewers. Fracture mechanics was and still is poorly understood and they beat the crap out of me pushing me to explain what couldn't be explained. It breaks cleanly just didn't do it.I see SPR having the same issues talking about thrust with this over simplistic explanation. If you are having a tough time convincing some of the more knowledgeable in physics of this working how does SPR think they can convince the reviewers for patents? If you are a friend to RS and a good engineer please tell him it just may not fly because he is seemingly violating one of the most inviolate laws we have and his explanation is severely lacking.END QUOTE
Quote from: ThinkerX on 07/04/2015 02:32 am
Quote from: deltaMass on 07/04/2015 06:46 amYou guys tickle me pink. It doesn't matter where you put the antenna if you are using Maxwell/Poynting to figure that out. That theory says that you get zero thrust. You're better off with a Hail Mary. It's logical.You need to look at the equation I posted yesterday. I shows there is a huge thrust, many orders larger than a photon rocket, going like 1/r^2 from the apex of the cone.http://forum.nasaspaceflight.com/index.php?topic=37642.msg1399882#msg1399882Todd
for a tapered rectangular waveguide of width and height;x*A*tanθ and x*B*tanθ
thrust... , going like 1/r^2 from the apex of the cone.
Quote from: Rodal on 07/03/2015 07:58 pmQuote from: aero on 07/03/2015 04:52 pmDr. Rodal- Time slices 3 thru 7 have been added to the csv folder on Google Drive. Same place as the other ones, I just changed the name to reflect 3 thru 13.@aero:two important questions to investigate this further:1) How do you impose boundary conditions? What are your boundary conditions and how do you actually implement them in Meep (I read that boundary conditions can be imposed such that the problem becomes nonlinear, which would also partly explain the results)2) TS013 : does this mean that you only marched the FD solution through 13 time steps total? If so, this is way insufficient to make sure that this is not just a transient, if so we would need to investigate marching forwards many more time steps to investigate the time evolutionIt appears from what we're seeing in the gif movies and what I've calculated for a tapered waveguide, that the two match. For a very long frustum waveguide, starting at the small end with a k vector perpendicular to the x-axis, the axis of the frustum;k2 = 0 + ky2 + kz2, initially.This is the rocket equation (per photon);dpx = (E(w)/(c2*px))*[dE(w) + (c/p)*hbar2*(Xmn2/x^3)*dx]Xmn2 = [(m*pi/A*tanθ)2 + (n*pi/B*tanθ)2], for a tapered rectangular waveguide of width and height;x*A*tanθ and x*B*tanθMultiple reflections from the side wall will rotate k into kx, thereby transferring all of the momentum that was perpendicular to the axis, to be parallel to the axis. That rotation occurs a little bit, each time it reflects from the side walls. You can see this clearly in the Hx-z images from @aero's work. (Thank you @aero and @VAXheadroom) This is WHY a smaller cone angle and a longer frustum is better. "If it were long enough", there would be so many bounces that no energy would reach the big end, it would all be converted into thrust.I retract my previous statement that I've given up on the microwaves, thanks to all the work done here. Theory and simulation seem to be merging. Todd
Quote from: aero on 07/03/2015 04:52 pmDr. Rodal- Time slices 3 thru 7 have been added to the csv folder on Google Drive. Same place as the other ones, I just changed the name to reflect 3 thru 13.@aero:two important questions to investigate this further:1) How do you impose boundary conditions? What are your boundary conditions and how do you actually implement them in Meep (I read that boundary conditions can be imposed such that the problem becomes nonlinear, which would also partly explain the results)2) TS013 : does this mean that you only marched the FD solution through 13 time steps total? If so, this is way insufficient to make sure that this is not just a transient, if so we would need to investigate marching forwards many more time steps to investigate the time evolution
Dr. Rodal- Time slices 3 thru 7 have been added to the csv folder on Google Drive. Same place as the other ones, I just changed the name to reflect 3 thru 13.
...I've just fried my brain on reading a great paper from MIT even considering my math skills felt like pidgin math. after. Good paper and I highly recommend it. http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDF...
Quote from: deltaMass on 07/04/2015 06:46 amYou guys tickle me pink. It doesn't matter where you put the antenna if you are using Maxwell/Poynting to figure that out. That theory says that you get zero thrust. You're better off with a Hail Mary. It's logical.You have overstated your case. There is nothing in Maxwell's or Poynting's theory saying that one should get zero thrust out of anything. Actually, Maxwell was the first scientist to derive the equations predicting that electromagnetic radiation can produce stresses, and the stress-energy tensor carries his name to honor that achievement.Case in point: if there is a net Poynting vector due to energy that gets dissipated into heat asymetrically, there maybe asymmetric heat transfer (by convection and/or radiation) resulting in asymmetric forces: "thrust". That is fully consistent with Maxwell's and Poynting's equations as well as consistent with Newton's equations. A correct statement you could make is that you, personally, don't see a way that enough thrust/InputPower can result from an asymmetric microwave cavity that is in excess of a perfect photon rocket thrust/inputPower by several orders of magnitude, but that is not the statement you made. Quote from: Anson MountI'm an enemy of exposition. I feel there's no need to overstate.
Quote from: WarpTech on 07/04/2015 02:28 pmQuote from: deltaMass on 07/04/2015 06:46 amYou guys tickle me pink. It doesn't matter where you put the antenna if you are using Maxwell/Poynting to figure that out. That theory says that you get zero thrust. You're better off with a Hail Mary. It's logical.You need to look at the equation I posted yesterday. I shows there is a huge thrust, many orders larger than a photon rocket, going like 1/r^2 from the apex of the cone.http://forum.nasaspaceflight.com/index.php?topic=37642.msg1399882#msg1399882ToddNeed further exposition to understand your message. For example, when you state:Quote from: WarpTechfor a tapered rectangular waveguide of width and height;x*A*tanθ and x*B*tanθx is previously defined as the longitudinal coordinate. Is x a dimensional coordinate or a dimensionless coordinate?What is θ ? Is that the cone half-angle measured from the axisymmetry longitudinal axis x?
Do A and B have dimensions?
In that case, and if x has dimensions, did you meana tapered rectangular waveguide of width and height;A*tanθ and B*tanθas xA would otherwise have the dimensions of an area.
_______________When you state Quotethrust... , going like 1/r^2 from the apex of the cone.does that mean that the truncated cones should terminate near the apex of the cone, and that therefore they are presently grossly underdesigned, since they terminate way before the apex of the cone?
Quote from: WarpTech on 07/03/2015 11:30 pmQuote from: Rodal on 07/03/2015 07:58 pmQuote from: aero on 07/03/2015 04:52 pmDr. Rodal- Time slices 3 thru 7 have been added to the csv folder on Google Drive. Same place as the other ones, I just changed the name to reflect 3 thru 13.@aero:two important questions to investigate this further:1) How do you impose boundary conditions? What are your boundary conditions and how do you actually implement them in Meep (I read that boundary conditions can be imposed such that the problem becomes nonlinear, which would also partly explain the results)2) TS013 : does this mean that you only marched the FD solution through 13 time steps total? If so, this is way insufficient to make sure that this is not just a transient, if so we would need to investigate marching forwards many more time steps to investigate the time evolutionIt appears from what we're seeing in the gif movies and what I've calculated for a tapered waveguide, that the two match. For a very long frustum waveguide, starting at the small end with a k vector perpendicular to the x-axis, the axis of the frustum;k2 = 0 + ky2 + kz2, initially.This is the rocket equation (per photon);dpx = (E(w)/(c2*px))*[dE(w) + (c/p)*hbar2*(Xmn2/x^3)*dx]Xmn2 = [(m*pi/A*tanθ)2 + (n*pi/B*tanθ)2], for a tapered rectangular waveguide of width and height;x*A*tanθ and x*B*tanθMultiple reflections from the side wall will rotate k into kx, thereby transferring all of the momentum that was perpendicular to the axis, to be parallel to the axis. That rotation occurs a little bit, each time it reflects from the side walls. You can see this clearly in the Hx-z images from @aero's work. (Thank you @aero and @VAXheadroom) This is WHY a smaller cone angle and a longer frustum is better. "If it were long enough", there would be so many bounces that no energy would reach the big end, it would all be converted into thrust.I retract my previous statement that I've given up on the microwaves, thanks to all the work done here. Theory and simulation seem to be merging. ToddIf I understand your explanation correctly, there should not be a need for the big end to be closed. Furthermore, linear sides may also not be optimal. An (inverse) hyperbolic shape would give better results.Would something like this be optimal according to your theory? http://www.mathworks.com/matlabcentral/fileexchange/screenshots/4304/original.jpgOr is it so that the number of bounces need to be maximized and two almost parallel lines going to infinity would be optimal? I can hardly imagine this would be the case, but just to understand...
...The k vectors look a lot like this...Todd
Quote from: SeeShells on 07/04/2015 02:24 pm...I've just fried my brain on reading a great paper from MIT even considering my math skills felt like pidgin math. after. Good paper and I highly recommend it. http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDF...Those are the class notes chapter 12 of MIT's Course in the Nuclear Engineering Department (Dept. 22), Principles of Nuclear Radiation Measurement and Protection, Undergraduate Course 22.09The following chapters (not as easy to find) may also be of interest:Chapter 13 Phase Dynamicshttps://stuff.mit.edu/afs/athena/course/22/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP13.PDFand Chapter 14 Radio-Frequency Linear Acceleratorshttps://stuff.mit.edu/afs/athena/course/22/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP14.PDF
Quote from: WarpTech on 07/04/2015 04:13 pm...The k vectors look a lot like this...ToddThanks for taking the time to explain this. With the ends closed (as claimed by Shawyer) how can anything get out to result in acceleration of the closed truncated cone without violating conservation of momentum? (if so what gets out and how does it get out?). If nothing gets out, it appears that acceleration of the copper cone would imply a violation of CoM.Or are you considering that the use of choke joints ( https://en.wikipedia.org/wiki/Waveguide_flange#Choke_connection ) (hat tip R. L. for the references) may allow emission out of the frustum thus preserving conservation of momentum?