...Restating my earlier conjecture:EM Momentum is asymmetric in the frustrum due to copper loss, and selectively thermally dissipated.The loss is greater in the wide back, more than the narrow frontThe wide back is inductive, the narrow front capacitive; the vacuum is a loss-less dielectric.
The wide back gets more low-sideband current and heat, the narrow front more E-field and momentum.The frustrum selectively dissipates energy supplied at a point on its tuning-slope.If its accelerated, sidebands appear, one of which may be selectively dissipated, resulting in the acceleration being enhanced or retarded depending on the side of the center-frequency energy is supplied at....
Quote from: WarpTech on 05/29/2015 06:22 pmWhat I'm saying is, standing waves are not evanescent waves. The standing waves in TE013 mode have frequency well above the cut-off of the cavity. I do not know precisely "how" standing waves can be shifted to a longer wavelength where they can become evanescent waves, without a mechanism for those waves to lose energy, i.e, dissipation before attenuation. I'll have to go re-read Zeng and Fan again, but their derivation is for waves propagating down a tapered waveguide, not standing waves in a cavity. I do not assume that the standing waves will be attenuated, based on what I remember of their results. I am assuming that dissipation will cause the standing wave to decay to longer wavelength, where attenuation can take place. Only the evanescent waves, not the standing waves can produce thrust.How would you consider converting the standing waves to evanescents? What mechanism would you choose?ToddIn this picture the TEM-waves/photons are red shifted cause of the loss factor of the cavity, lower energy leads to bigger wavelength --> larger bandwidth and therefore lower Q - what is well known. If a given evanescent wave reaches cutoff frequency at the smaller diameter but not at the bigger the boundary conditions to resonate are not given. What happens to the energy of these photons/field?? Would the energy of the field lose by thermal heating or trust producing ?Is the latter option what you think is the right?
What I'm saying is, standing waves are not evanescent waves. The standing waves in TE013 mode have frequency well above the cut-off of the cavity. I do not know precisely "how" standing waves can be shifted to a longer wavelength where they can become evanescent waves, without a mechanism for those waves to lose energy, i.e, dissipation before attenuation. I'll have to go re-read Zeng and Fan again, but their derivation is for waves propagating down a tapered waveguide, not standing waves in a cavity. I do not assume that the standing waves will be attenuated, based on what I remember of their results. I am assuming that dissipation will cause the standing wave to decay to longer wavelength, where attenuation can take place. Only the evanescent waves, not the standing waves can produce thrust.How would you consider converting the standing waves to evanescents? What mechanism would you choose?Todd
A thin copper fustrum would deform if you tried to induce a vacuum inside it (and not outside as well). Plus it would be really difficult to hermetically seal, especially around the RF input. A full vacuum enclosure is the best way to go. NASA did the one test that way, and @zellerium and his colleagues at Cal Poly are planning vacuum tests this summer. The guys at hackaday.io are also thinking about a vacuum test of their 25ghz thruster. That one is machined out of an aluminum block and small enough to fit the entire test in a vacuum bell jar.
Quote from: X_RaY on 05/29/2015 07:11 pmQuote from: WarpTech on 05/29/2015 06:22 pmWhat I'm saying is, standing waves are not evanescent waves. The standing waves in TE013 mode have frequency well above the cut-off of the cavity. I do not know precisely "how" standing waves can be shifted to a longer wavelength where they can become evanescent waves, without a mechanism for those waves to lose energy, i.e, dissipation before attenuation. I'll have to go re-read Zeng and Fan again, but their derivation is for waves propagating down a tapered waveguide, not standing waves in a cavity. I do not assume that the standing waves will be attenuated, based on what I remember of their results. I am assuming that dissipation will cause the standing wave to decay to longer wavelength, where attenuation can take place. Only the evanescent waves, not the standing waves can produce thrust.How would you consider converting the standing waves to evanescents? What mechanism would you choose?ToddIn this picture the TEM-waves/photons are red shifted cause of the loss factor of the cavity, lower energy leads to bigger wavelength --> larger bandwidth and therefore lower Q - what is well known. If a given evanescent wave reaches cutoff frequency at the smaller diameter but not at the bigger the boundary conditions to resonate are not given. What happens to the energy of these photons/field?? Would the energy of the field lose by thermal heating or trust producing ?Is the latter option what you think is the right?Correct! I believe it is a probability distribution of several options; Thrust, scattering, rotation, vibration (heat), emission, torque, (more?). It depends on the collision of the photons with the material at the quantum scale. At the engineering scale, that factor is the unknown variable to be determined by experiment or simulation.The only OTHER way I can imagine it to work with standing waves is if the Forward wave has a different attenuation factor from the Backward wave, that create the standing wave. AND, that somehow their Power Factor is no longer 0. Standing waves have zero PF because E & B are 90deg out of phase. Evanescent waves have unity PF, because E & B are in phase, like a resistor. There may be some mechanism that allows standing waves to decay to evanescent waves without the need of losses to heat, etc. But as of now, I am still uncertain how that happens. So can we agree that it is the evanescent waves that do work to exert thrust, and what we really need to understand is how standing waves are coupled and/or converted to evanescent waves?Todd
Quote from: deltaMass on 05/29/2015 06:46 pmQuote from: RodalObviously, in a cylinder dE/dr (where r is the spherical radial coordinate) is zeroNot obvious to me I'm afraid.I re-worded the statement this better way:"Conical waveguides with a small cone angle and low value of kr appear to have the most attenuation, according to Zeng and Fan."Zeng and Fan place a lot of the emphasis (in their statements) on the cone angle, but kr is equally important to understand this, as the cone with a small cone angle is different from a cylinder only at small values of r. As r1 goes to infinity, the cone becomes a cylinder.
Quote from: RodalObviously, in a cylinder dE/dr (where r is the spherical radial coordinate) is zeroNot obvious to me I'm afraid.
Obviously, in a cylinder dE/dr (where r is the spherical radial coordinate) is zero
Quote from: Rodal on 05/29/2015 07:37 pm..I re-worded the statement this better way:"Conical waveguides with a small cone angle and low value of kr appear to have the most attenuation, according to Zeng and Fan."Zeng and Fan place a lot of the emphasis (in their statements) on the cone angle, but kr is equally important to understand this, as the cone with a small cone angle is different from a cylinder only at small values of r. As r1 goes to infinity, the cone becomes a cylinder.Nope. Still not getting it. Is r the coordinate along the central axis?
..I re-worded the statement this better way:"Conical waveguides with a small cone angle and low value of kr appear to have the most attenuation, according to Zeng and Fan."Zeng and Fan place a lot of the emphasis (in their statements) on the cone angle, but kr is equally important to understand this, as the cone with a small cone angle is different from a cylinder only at small values of r. As r1 goes to infinity, the cone becomes a cylinder.
We want for geometrical attenuation according to Zeng & Fan:1) MOST IMPORTANT: Low wavenumber ( https://en.wikipedia.org/wiki/Wavenumber ) in the longitudinal direction. Low number of half-waves in the longitudinal direction. You want to get the wavelength cut-off as it approaches the small end. If the wavenumber is high enough (small wavelength), the wavelength doesn't get cut-off, and hence it is not going to become an evanescent wave. You want evanescent waves at the small end to get geometrical attenuation.2) LESS IMPORTANT: Small cone-angleIs geometrical attenuation what we want ? I'm not sure.If a travelling wave coming from the feed section gets cut-off as it approaches the small end, can that lead to acceleration of the cone ?I know that losses (resistive and dielectric) lead to a non-zero Poynting vector, but that leads to heat. It does not look like an efficient means of propulsion. Not clear what happens with geometrical attenuation and evanescent waves yet.
Quote from: frobnicat on 05/29/2015 05:14 pmQuote from: dustinthewind on 05/29/2015 07:25 am...Quote from: Paul Novy on 05/28/2015 07:49 am..."Macroscopic and Direct Light Propulsion of Bulk Graphene Material"http://arxiv.org/ftp/arxiv/papers/1505/1505.04254.pdfMy question here is what do the CoM (conservation of mass) and CoE (conservation of energy) people think about this force on the graphene? Obviously it is much larger than that of just light. Do we still have a violation here? What is interesting is it is being observed in matter. Whats to stop them from using a mirror behind a ship and reflecting the laser to propel the ship? It might not be the right thing to do but we could even shove a graphene sponge in the narrow end of the cavity and let that magnetron go. Well I diverge but my question is regarding the CoM and CoE argument. I read this, nice research, nice find, still struggling to understand how the sample, when levitated, can neutralize the charge build-up that would occur though. There is no CoM or CoE argument against this find, this is no fundamentally different from a beamed ablative propulsion scheme. The injected energy is used to expel particles of mass>0, therefore it is to be expected that the thrust/power ratio is 1/ Bigger than that of a photon rocket if power counts only the imparted energy2/ Lower than that of a photon rocket if power counts imparted energy + mass equivalent energy of expelled particlesThis is not propellantless, the floated device loses mass. The exotic aspect is that the expelled mass is electrons, but this is sooner or later (rather sooner) to be paid back as the devices builds up charge and it gets more and more difficult to release electrons against this growing voltage. For continuous operation it has to be replenished in electrons. Recycling the ejected electrons would cancel any net thrust, like a traditional rocket that would try to swallow back its exhaust. So this needs a source of "free" electrons, if possible harvested at low velocity relative to system. This is not unlike a jet engine that needs to swallow ambient mass and uses power to accelerate it and expel it faster backward, everything is ok as far as CoM and CoE is concerned, that can't be used as an apparent free energy generator, unlike an EM drive thrusting continuously (constant thrust for constant power) at constant velocity (no acceleration) above a certain velocity.I am embarrassed as I meant to type CoM for conservation of momentum and not mass.
Quote from: dustinthewind on 05/29/2015 07:25 am...Quote from: Paul Novy on 05/28/2015 07:49 am..."Macroscopic and Direct Light Propulsion of Bulk Graphene Material"http://arxiv.org/ftp/arxiv/papers/1505/1505.04254.pdfMy question here is what do the CoM (conservation of mass) and CoE (conservation of energy) people think about this force on the graphene? Obviously it is much larger than that of just light. Do we still have a violation here? What is interesting is it is being observed in matter. Whats to stop them from using a mirror behind a ship and reflecting the laser to propel the ship? It might not be the right thing to do but we could even shove a graphene sponge in the narrow end of the cavity and let that magnetron go. Well I diverge but my question is regarding the CoM and CoE argument. I read this, nice research, nice find, still struggling to understand how the sample, when levitated, can neutralize the charge build-up that would occur though. There is no CoM or CoE argument against this find, this is no fundamentally different from a beamed ablative propulsion scheme. The injected energy is used to expel particles of mass>0, therefore it is to be expected that the thrust/power ratio is 1/ Bigger than that of a photon rocket if power counts only the imparted energy2/ Lower than that of a photon rocket if power counts imparted energy + mass equivalent energy of expelled particlesThis is not propellantless, the floated device loses mass. The exotic aspect is that the expelled mass is electrons, but this is sooner or later (rather sooner) to be paid back as the devices builds up charge and it gets more and more difficult to release electrons against this growing voltage. For continuous operation it has to be replenished in electrons. Recycling the ejected electrons would cancel any net thrust, like a traditional rocket that would try to swallow back its exhaust. So this needs a source of "free" electrons, if possible harvested at low velocity relative to system. This is not unlike a jet engine that needs to swallow ambient mass and uses power to accelerate it and expel it faster backward, everything is ok as far as CoM and CoE is concerned, that can't be used as an apparent free energy generator, unlike an EM drive thrusting continuously (constant thrust for constant power) at constant velocity (no acceleration) above a certain velocity.
...Quote from: Paul Novy on 05/28/2015 07:49 am..."Macroscopic and Direct Light Propulsion of Bulk Graphene Material"http://arxiv.org/ftp/arxiv/papers/1505/1505.04254.pdfMy question here is what do the CoM (conservation of mass) and CoE (conservation of energy) people think about this force on the graphene? Obviously it is much larger than that of just light. Do we still have a violation here? What is interesting is it is being observed in matter. Whats to stop them from using a mirror behind a ship and reflecting the laser to propel the ship? It might not be the right thing to do but we could even shove a graphene sponge in the narrow end of the cavity and let that magnetron go. Well I diverge but my question is regarding the CoM and CoE argument.
..."Macroscopic and Direct Light Propulsion of Bulk Graphene Material"http://arxiv.org/ftp/arxiv/papers/1505/1505.04254.pdf
Also the way I understood it, there is no loss of electrons. You kick off an electron and the graphene sponge immediately becomes charged and is immediately attracted to the electron.
I didn't see anything about supplying current to the graphene in the vacuum tube.
This attraction of the electron to the graphene should give more force than just the light striking a surface. I saw some arguments earlier way back that suggested if we got over a certain force per power input that maybe we would have a perpetual motion device on our hands. The idea is taking a really light ship with a laser and mounting it pointing in the direction of the ship such that the light strikes in the direction the ship wants to go. photons coming out of the laser don't provide much force but the light striking the gaphene provide more force than simply light so we have again propellantless propulsion but with force much greater than that of photons.
Third attempt (please be direct this time)1. Is r the coordinate along the central axis? (this is generally designated z)2. Why do you say that dE/dr = 0, for any and all modes that can exist?
Quote from: dustinthewind on 05/29/2015 05:46 pmBack to the graphene article : the illuminated graphene sponge is, basically, a ion thruster..if it ain't leavin' it ain't thrustin'QuoteI didn't see anything about supplying current to the graphene in the vacuum tube. Unfortunately this is a point that is not very clear in this otherwise good article. It does appear clearly from fig. 4a that for a continuous or long term operation of the effect one needs to close the current loop (replenish the lost electrons)...QuoteThis attraction of the electron to the graphene should give more force than just the light striking a surface....Do same musings with a ion thruster : this illuminated graphene sponge is just a fancy solar/beam powered ion thruster system. Does it make sense ?In a perfect vacuum (with no exterior resources) this can't be used as a perpetuum mobile of the second kind (providing energy from apparently nothing) unlike EM drives with thrust/power>1/c (averaged).I went back and looking at the article and indeed they draw the electrons leaving in the direction of the incoming laser. This is problematic as in order for them to get thrust positive ions would also need to be ejected. I guess I thought the electrons were being ejected in the opposite direction of the incoming laser and then the graphene was attracted in the direction of the kicked electron not requiring any current but what they are suggesting in order for it to be an ion drive requires ejection of positive charge, else them electrons are coming back home, and as you said if it isn't leaving it isn't thrusting. There does appear to be some mystery here. I guess the idea was the electron kicked in the opposite direction of the incoming photon made capture of the incoming energy more efficient some how. It appears they were thinking about it differently in the article.
Back to the graphene article : the illuminated graphene sponge is, basically, a ion thruster..if it ain't leavin' it ain't thrustin'QuoteI didn't see anything about supplying current to the graphene in the vacuum tube. Unfortunately this is a point that is not very clear in this otherwise good article. It does appear clearly from fig. 4a that for a continuous or long term operation of the effect one needs to close the current loop (replenish the lost electrons)...QuoteThis attraction of the electron to the graphene should give more force than just the light striking a surface....Do same musings with a ion thruster : this illuminated graphene sponge is just a fancy solar/beam powered ion thruster system. Does it make sense ?In a perfect vacuum (with no exterior resources) this can't be used as a perpetuum mobile of the second kind (providing energy from apparently nothing) unlike EM drives with thrust/power>1/c (averaged).
This attraction of the electron to the graphene should give more force than just the light striking a surface....
That's pretty crappy nomenclature, especially whena) the axis you reference is labelled 'z'b) there's a dimension orthogonal to it labelled 'R'....
Quote from: dustinthewind on 05/29/2015 05:46 pmQuote from: frobnicat on 05/29/2015 05:14 pmQuote from: dustinthewind on 05/29/2015 07:25 am...Quote from: Paul Novy on 05/28/2015 07:49 am..."Macroscopic and Direct Light Propulsion of Bulk Graphene Material"http://arxiv.org/ftp/arxiv/papers/1505/1505.04254.pdfMy question here is what do the CoM (conservation of mass) and CoE (conservation of energy) people think about this force on the graphene? Obviously it is much larger than that of just light. Do we still have a violation here? What is interesting is it is being observed in matter. Whats to stop them from using a mirror behind a ship and reflecting the laser to propel the ship? It might not be the right thing to do but we could even shove a graphene sponge in the narrow end of the cavity and let that magnetron go. Well I diverge but my question is regarding the CoM and CoE argument. I read this, nice research, nice find, still struggling to understand how the sample, when levitated, can neutralize the charge build-up that would occur though. There is no CoM or CoE argument against this find, this is no fundamentally different from a beamed ablative propulsion scheme. The injected energy is used to expel particles of mass>0, therefore it is to be expected that the thrust/power ratio is 1/ Bigger than that of a photon rocket if power counts only the imparted energy2/ Lower than that of a photon rocket if power counts imparted energy + mass equivalent energy of expelled particlesThis is not propellantless, the floated device loses mass. The exotic aspect is that the expelled mass is electrons, but this is sooner or later (rather sooner) to be paid back as the devices builds up charge and it gets more and more difficult to release electrons against this growing voltage. For continuous operation it has to be replenished in electrons. Recycling the ejected electrons would cancel any net thrust, like a traditional rocket that would try to swallow back its exhaust. So this needs a source of "free" electrons, if possible harvested at low velocity relative to system. This is not unlike a jet engine that needs to swallow ambient mass and uses power to accelerate it and expel it faster backward, everything is ok as far as CoM and CoE is concerned, that can't be used as an apparent free energy generator, unlike an EM drive thrusting continuously (constant thrust for constant power) at constant velocity (no acceleration) above a certain velocity.I am embarrassed as I meant to type CoM for conservation of momentum and not mass. So far on this thread the usage for CoM is to stand for conservation of momentum. This is how I understood your remarks : is it (the illuminated graphene sponge) apparently breaking conservation of momentum (centre of mass to be more precise) and conservation of energy, the same way that EM drive is apparently doing, as this is very apparent to many observers (in spite of proponents saying it is not) ? My answer is, from the point of view of someone who thinks that violation of CoM and CoE is very manifest in EM drive claimed phenomenology : no, the illuminated graphene sponge shows no such apparent CoE or CoM breaking.QuoteAlso the way I understood it, there is no loss of electrons. You kick off an electron and the graphene sponge immediately becomes charged and is immediately attracted to the electron. Then you are saying that a classical rocket can swallow back its exhaust to recycle its ejected mass and use it again and again ? Make it a ion thruster to avoid concerns about chemical potential energy : you have in one hand a flow of reaction mass (ejected) and in the other hand a flow of energy (power) that's used to give a kick to the ejected mass (at some velocity relative to spacecraft). Please read carefully the points 1/ and 2/ above, this is key to proper understanding of comparison between propellant reaction propulsion (rockets, ion thrusters) and propellantless propulsion (the only admitted case being the photon rocket, I mean if we consider a spacecraft in perfect vacuum, in perfect deep space, in perfect darkness, i.e. with no external resources).Actually a photon rocket is not really "propellantless" as it throws away photons. We could say that its exhaust is "pure energy" (in the sense that the exhaust particles have no rest mass), while with a ion thruster for instance the exhaust is composed of particles with a rest mass. A ship with ion thruster loses mass (ejected propellant) but so does a photon rocket also because a mass energy equivalent must be converted into photons. This would be clearly visible with a nuclear reactor : while only photons left the spacecraft (assuming perfect shielding of other radiations, neutrinos... other story) the spacecraft would weigh less after some years of operation (isotopes decaying to lower total mass). With a (perfect) photon rocket there is no separation between a mass flow and an energy flow : the later being the total spent power and this power is such that thrust/power=1/c at best.With a ion thruster we have two distinct flows, flow of energy (onboard power generator), and flow of mass (expelled propellant, say Xenon). We can take the two flows separately to discuss some trade off between power starved designs (higher velocity ejection) and mass starved designs (lower velocity ejection). Using the power of onboard generator, thrust/power can be much higher than 1/c, obviously. But if we take together all that is spent forever, that is both spent energy and spent mass, and count spent mass on its energy equivalent content mc², then we arrive at colossal spent powers such that thrust/power will always be much lower than 1/c. This is only on this equal footing ground that a ion thruster can be compared to a photon rocket, if we want to discuss CoE aspects, and perfect photon rocket thrust/power=1/c is an absolute optimum when counting all that is spent in the power term. Only system that could do better, following those terms, is hypothetical tachyon rocket.EM drive is supposed to have thrust/power>1/c with all taken into account in power term since by definition it's supposed to lose no mass. Even with ion thruster that apparently does much better than 1/c, this is no longer the case when all that is lost (from the point of view of an isolated spacecraft with no external resources) is integrated in power term. This is so blatantly obvious that the EM drive has this "little" CoE problem, and the answers about that so far by proponents are so weak, that it is no wonder that science pundits come down on that like a metric ton of bricks.Back to the graphene article : the illuminated graphene sponge is, basically, a ion thruster, just consider the emitted electrons as very light ions, and the illumination process as a smart but not exotic equivalent way to get power, as could be done through solar panels to drive a ion thruster, or with beamed power.Electrons can't be attracted back to the sponge and yet impart the high thrust. Throw a ball attached to a spring. At throw launcher gets p kicks opposite direction, when ball slows and come back spring imparts -2p kick, and when ball strikes back launcher hand it again kicks p : total=p-2p+p=0. Forget the spring, say farewell to the ball, then launcher gets a net kick total=p, but has no longer the ball.if it ain't leavin' it ain't thrustin'QuoteI didn't see anything about supplying current to the graphene in the vacuum tube. Unfortunately this is a point that is not very clear in this otherwise good article. It does appear clearly from fig. 4a that for a continuous or long term operation of the effect one needs to close the current loop (replenish the lost electrons). For the levitated sample, well, maybe they have enough free electrons floating around in their vacuum to neutralize the lost ones ? Short of a closed circuit (fig 4a) the effect is either short lived or requires free electrons around like an air breathing jet engine needs free air around. Not working in a perfect vacuum (i.e. no external resources).QuoteThis attraction of the electron to the graphene should give more force than just the light striking a surface. I saw some arguments earlier way back that suggested if we got over a certain force per power input that maybe we would have a perpetual motion device on our hands. The idea is taking a really light ship with a laser and mounting it pointing in the direction of the ship such that the light strikes in the direction the ship wants to go. photons coming out of the laser don't provide much force but the light striking the gaphene provide more force than simply light so we have again propellantless propulsion but with force much greater than that of photons. Do same musings with a ion thruster : this illuminated graphene sponge is just a fancy solar/beam powered ion thruster system. Does it make sense ?In a perfect vacuum (with no exterior resources) this can't be used as a perpetuum mobile of the second kind (providing energy from apparently nothing) unlike EM drives with thrust/power>1/c (averaged).
I reproduce the charts from Zeng and Fan belowVertical axis: (attenuation coefficient) * (c /(2 Pi frequency))Horizontal axis wavenumber in longitudinal direction (an irrational number governing # of half-waves in the longitudinal direction)What we see:1) High attenuation for small wavenumber kr2) Higher attenuation for smaller cone half-angle θo
Quote from: mwvp on 05/29/2015 04:58 pm...The loss is greater in the wide back, more than the narrow frontThe wide back is inductive, the narrow front capacitive; the vacuum is a loss-less dielectric.I thought so too, but then I looked at where the "B" field is strongest, and it is stronger near the small end, making it higher inductance than the large end. I "think" the reason why is that inductance is proportional to Area*Turns^2. The large end has more area, and therefore you would assume there is more inductance, but one circuit around the circumference at the large end has about twice the resistance as the small end. So at the small end, there is half the area but the current can loop around twice for the same voltage drop. I think it is correct to say, the big end has more resistance and is lossier, but the energy is stored at the small end, as @Rodal's images show.QuoteThe wide back gets more low-sideband current and heat, the narrow front more E-field and momentum.The frustrum selectively dissipates energy supplied at a point on its tuning-slope.If its accelerated, sidebands appear, one of which may be selectively dissipated, resulting in the acceleration being enhanced or retarded depending on the side of the center-frequency energy is supplied at....Can you elaborate on that statement please? Some equations or references?
...The loss is greater in the wide back, more than the narrow frontThe wide back is inductive, the narrow front capacitive; the vacuum is a loss-less dielectric.
AFAIK, I haven't seen any "high-Q" (low-loss) ferrites used above ~1 GHz. They're used in broadband hybrid transformers for lower HF-VHF frequencies, but mu-r disappears in the UHF. They're used circulators, but are low Q. I would think metglass would be very lossy, but I haven't looked it up. The Q of the cavity is only going to be as good as the dielectric or magnetic material loss will allow...