Quote from: WarpTech on 05/12/2015 08:43 pmQuote from: Rodal on 05/12/2015 07:59 pmQuote from: rfmwguy on 05/12/2015 07:48 pmQuote from: Rodal on 05/12/2015 07:25 pmQuote from: WarpTech on 05/12/2015 02:16 pmFrom the thermal and A/m plots from EW, most of the resonance is happening at the big end. I would not put the magnetron in that space because the input there will probably perturb the waves. Shawyer put the input near the small end. I would put it "at" the small end, depending on wave polarization. The walls should do most of the reflecting, not the small end.Todd D.Please take a gander at this Demo Drive by Shawyer: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=829850It seems to have the feed near the big diameter end.Good call doctor. I find the stub coupler/feed at the larger diameter of the cavity interesting...thinking this was the end with the highest return loss/standing wave at 2.4 GHz...perhaps not. The stepper motor appears to be adjusting the length of a tuning stub centered in the cavity on the narrow end, probably something like this: http://i.stack.imgur.com/Vfdxq.png Its and old tried and true methodology.Regardless, this tuning stub is simply a matching element which can be fixed (non-adjustable) once a center frequency is set and a tuning stub length measurement can be made. Normally, this tuning stub is adjusted for best S parameter match/bandwidth: http://www.antenna-theory.com/definitions/sparameters.phpThank you. I agree with you If that stub extends inside the frustum, that would explain why he put the input at the big end. I still say, it should be at the small end to avoid perturbing the harmonics.ToddSomebody else pointed out that Shawyer's Demonstrator was feed via a waveguide instead of via coax. Does that make any difference to your point that feeding should preferentially occur (if possible) at the small end to avoid perturbing harmonics? In other words do you think that feeding with a waveguide avoids perturbing harmonics and therefore if one feeds with a waveguide (fed upstream from a magnetron) you could just as well feed the waveguide at the big end ?
Quote from: Rodal on 05/12/2015 07:59 pmQuote from: rfmwguy on 05/12/2015 07:48 pmQuote from: Rodal on 05/12/2015 07:25 pmQuote from: WarpTech on 05/12/2015 02:16 pmFrom the thermal and A/m plots from EW, most of the resonance is happening at the big end. I would not put the magnetron in that space because the input there will probably perturb the waves. Shawyer put the input near the small end. I would put it "at" the small end, depending on wave polarization. The walls should do most of the reflecting, not the small end.Todd D.Please take a gander at this Demo Drive by Shawyer: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=829850It seems to have the feed near the big diameter end.Good call doctor. I find the stub coupler/feed at the larger diameter of the cavity interesting...thinking this was the end with the highest return loss/standing wave at 2.4 GHz...perhaps not. The stepper motor appears to be adjusting the length of a tuning stub centered in the cavity on the narrow end, probably something like this: http://i.stack.imgur.com/Vfdxq.png Its and old tried and true methodology.Regardless, this tuning stub is simply a matching element which can be fixed (non-adjustable) once a center frequency is set and a tuning stub length measurement can be made. Normally, this tuning stub is adjusted for best S parameter match/bandwidth: http://www.antenna-theory.com/definitions/sparameters.phpThank you. I agree with you If that stub extends inside the frustum, that would explain why he put the input at the big end. I still say, it should be at the small end to avoid perturbing the harmonics.Todd
Quote from: rfmwguy on 05/12/2015 07:48 pmQuote from: Rodal on 05/12/2015 07:25 pmQuote from: WarpTech on 05/12/2015 02:16 pmFrom the thermal and A/m plots from EW, most of the resonance is happening at the big end. I would not put the magnetron in that space because the input there will probably perturb the waves. Shawyer put the input near the small end. I would put it "at" the small end, depending on wave polarization. The walls should do most of the reflecting, not the small end.Todd D.Please take a gander at this Demo Drive by Shawyer: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=829850It seems to have the feed near the big diameter end.Good call doctor. I find the stub coupler/feed at the larger diameter of the cavity interesting...thinking this was the end with the highest return loss/standing wave at 2.4 GHz...perhaps not. The stepper motor appears to be adjusting the length of a tuning stub centered in the cavity on the narrow end, probably something like this: http://i.stack.imgur.com/Vfdxq.png Its and old tried and true methodology.Regardless, this tuning stub is simply a matching element which can be fixed (non-adjustable) once a center frequency is set and a tuning stub length measurement can be made. Normally, this tuning stub is adjusted for best S parameter match/bandwidth: http://www.antenna-theory.com/definitions/sparameters.phpThank you. I agree with you
Quote from: Rodal on 05/12/2015 07:25 pmQuote from: WarpTech on 05/12/2015 02:16 pmFrom the thermal and A/m plots from EW, most of the resonance is happening at the big end. I would not put the magnetron in that space because the input there will probably perturb the waves. Shawyer put the input near the small end. I would put it "at" the small end, depending on wave polarization. The walls should do most of the reflecting, not the small end.Todd D.Please take a gander at this Demo Drive by Shawyer: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=829850It seems to have the feed near the big diameter end.Good call doctor. I find the stub coupler/feed at the larger diameter of the cavity interesting...thinking this was the end with the highest return loss/standing wave at 2.4 GHz...perhaps not. The stepper motor appears to be adjusting the length of a tuning stub centered in the cavity on the narrow end, probably something like this: http://i.stack.imgur.com/Vfdxq.png Its and old tried and true methodology.Regardless, this tuning stub is simply a matching element which can be fixed (non-adjustable) once a center frequency is set and a tuning stub length measurement can be made. Normally, this tuning stub is adjusted for best S parameter match/bandwidth: http://www.antenna-theory.com/definitions/sparameters.php
Quote from: WarpTech on 05/12/2015 02:16 pmFrom the thermal and A/m plots from EW, most of the resonance is happening at the big end. I would not put the magnetron in that space because the input there will probably perturb the waves. Shawyer put the input near the small end. I would put it "at" the small end, depending on wave polarization. The walls should do most of the reflecting, not the small end.Todd D.Please take a gander at this Demo Drive by Shawyer: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=829850It seems to have the feed near the big diameter end.
From the thermal and A/m plots from EW, most of the resonance is happening at the big end. I would not put the magnetron in that space because the input there will probably perturb the waves. Shawyer put the input near the small end. I would put it "at" the small end, depending on wave polarization. The walls should do most of the reflecting, not the small end.Todd D.
Quote from: WarpTech on 05/12/2015 09:39 pmQuote from: deltaMass on 05/12/2015 09:04 pmThere's still the issue of drifting off tune with temperature... and acceleration. That is why I believe now that it should be pulsed, not steady state operation.ToddThat also makes sense to me. Separately, but interestingly the Serrano Field Effect Boeing Darpa device tested by Dr. White displayed the highest thrust/InputPower of any device, yet it only showed very short time impulses (like Dirac Delta Functions) instead of steady state operation (although to me its principle of operation is very different from the EM Drive, Dr. White classified this device also as a Q-thuster).This is the text for Boeing/DARPA in slide 40 of Dr. White's presentation (http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140000851.pdf ):<<SFE Test Article at JSCIn 2013, Boeing/DARPA sent Eagleworks Lab an SFE test article for testing and evaluationEvaluation of the test article in and out of a Faraday Shield performed from Feb through June 2013.• There is a consistent transient thrust at device turn-on and turn-off that is consistent with Qthruster physics• The magnitude of the thrust scaled approximately with the cube of the input voltage (20-110uN).• The magnitude of the thrust is dependent on the AC content of the turn-on and turn-off pulse• Specific force of transient thrust was in the ~1- 20 N/kW range.~20-110 uN Thrust PulsesSpecific Force ~1-20N/kW>>NASA Eagleworks also provided this information in a 2013 Newsletter, which is available in the Internet from this link: https://xa.yimg.com/kq/groups/86787010/513081407/name/Eagleworks+Newsletter+2013.pdfthat reads:<<NASA/Boeing/SFE Campaign: Boeing/DARPA sent Eagleworks Lab an SFE test article for testing andevaluation. The guest thruster was evaluated in numerous test configurations using varying degrees ofFaraday shielding and vacuum conditions. Observations show that there is a consistent transient thrustat device turn-on and turn-off that is consistent with Q-thruster physics. The magnitude of the thrustscaled approximately with the cube of the input voltage (20-110uN). The magnitude of the thrust isdependent on the AC content of the turn-on and turn-off pulse. Thrust to power of transient thrust wasin the ~1-20 N/kW rangeYes, that's twenty Newtons per kiloWatt on the upper range
Quote from: deltaMass on 05/12/2015 09:04 pmThere's still the issue of drifting off tune with temperature... and acceleration. That is why I believe now that it should be pulsed, not steady state operation.Todd
There's still the issue of drifting off tune with temperature
Quote from: TheTraveller on 05/12/2015 02:52 pm...I know infinite values are impossible. Likewise negative Dfs. That was not the point.The point is the Df equation, applied to a variable frequency, shows there is an ideal frequency that will generate the best cavity Df. Driving the cavity with some chosen frequency, like 2.45GHZ will probably NOT make anything happen. Like trying to drive a tuned LC circuit with a frequency far away from it's resonate frequency. Waste of time. Likewise driving the cavity with the calculated best Df frequency will probably do the same thing. No thrust in the real world. What the exercise shows is that the Rf generator driving the cavity should be operating 2x or 3x the best Df cavity frequency and that Rf frequency generating system must be able to vary the driving frequency so to search for the best frequency in the real world. The spreadsheet give me a starting place and an understand the cavity best Df frequency should be 1/2 or 1/3 the applied Rf frequency.To me as an engineer starting a replication of the Flight Thruster and associated variable Rf generation system, it is very new and valuable information. This is all related to real world engineering (building actual hardware) to give the best chance of generating thrust.To assisting theory development, well it may not be of much value.As to why Shawyer is using a particular excitation frequency I suggest that you use your spreadsheet to check the above vs. the calculated natural frequencies and mode shapes here:http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=634741(see in the slide the drawing insert detail containing the two dimensions (the diameter of the big base and the length of the truncated cone) given by Shawyer while the third dimension (the diameter of the small base) of the truncated cone is parametrized on the horizontal axis (ranging from a pointy cone to a cylinder) to ascertain its correct value)Shawyer (to my knowledge) has never provided all three geometrical dimensions of the truncated cone. Hence one of the dimensions (the diameter of the small base) has to be estimated (obtained from the inverse expression for the Design Factor -when Shawyer has provided the Design Factor, parametrized as per the attached file, or estimated from images, as done by @aero and others). Since Shaywer has not provided all three dimensions, there is uncertainty as to what he actually did and why@aero had correspondence with Shawyer asking for the dimension of the small diameter, to my recollection Shawyer cryptically answered "small base diameter chosen on the basis of the cut-off frequency" (hence still unknown how close to the cutoff wavelength was Shawyer's chosen small base diameter).
...I know infinite values are impossible. Likewise negative Dfs. That was not the point.The point is the Df equation, applied to a variable frequency, shows there is an ideal frequency that will generate the best cavity Df. Driving the cavity with some chosen frequency, like 2.45GHZ will probably NOT make anything happen. Like trying to drive a tuned LC circuit with a frequency far away from it's resonate frequency. Waste of time. Likewise driving the cavity with the calculated best Df frequency will probably do the same thing. No thrust in the real world. What the exercise shows is that the Rf generator driving the cavity should be operating 2x or 3x the best Df cavity frequency and that Rf frequency generating system must be able to vary the driving frequency so to search for the best frequency in the real world. The spreadsheet give me a starting place and an understand the cavity best Df frequency should be 1/2 or 1/3 the applied Rf frequency.To me as an engineer starting a replication of the Flight Thruster and associated variable Rf generation system, it is very new and valuable information. This is all related to real world engineering (building actual hardware) to give the best chance of generating thrust.To assisting theory development, well it may not be of much value.
How can the principle of operation of the ring laser gyroscope be compared to the principle of operation of EM Drive thrust, escapes me...
Serrano's 0.02 N/W means breakeven speed is only 100 m/s = 2/k
Quote from: Rodal on 05/12/2015 09:51 pmQuote from: WarpTech on 05/12/2015 09:39 pmQuote from: deltaMass on 05/12/2015 09:04 pmThere's still the issue of drifting off tune with temperature... and acceleration. That is why I believe now that it should be pulsed, not steady state operation.ToddThat also makes sense to me. Separately, but interestingly the Serrano field effect device tested by Dr. White displayed the highest thrust/InputPower of any device, yet it only showed very short time impulses (like Dirac Delta Functions) instead of steady state operation (although to me its principle of operation is very different from the EM Drive, Dr. White classified this device also as a Q-thuster).Forgive me for interrupting, but with pulsing, do you mean Pulse Width Modulation or PWM?
Quote from: WarpTech on 05/12/2015 09:39 pmQuote from: deltaMass on 05/12/2015 09:04 pmThere's still the issue of drifting off tune with temperature... and acceleration. That is why I believe now that it should be pulsed, not steady state operation.ToddThat also makes sense to me. Separately, but interestingly the Serrano field effect device tested by Dr. White displayed the highest thrust/InputPower of any device, yet it only showed very short time impulses (like Dirac Delta Functions) instead of steady state operation (although to me its principle of operation is very different from the EM Drive, Dr. White classified this device also as a Q-thuster).
Quote from: PaulF on 05/12/2015 09:55 pmQuote from: Rodal on 05/12/2015 09:51 pmQuote from: WarpTech on 05/12/2015 09:39 pmQuote from: deltaMass on 05/12/2015 09:04 pmThere's still the issue of drifting off tune with temperature... and acceleration. That is why I believe now that it should be pulsed, not steady state operation.ToddThat also makes sense to me. Separately, but interestingly the Serrano field effect device tested by Dr. White displayed the highest thrust/InputPower of any device, yet it only showed very short time impulses (like Dirac Delta Functions) instead of steady state operation (although to me its principle of operation is very different from the EM Drive, Dr. White classified this device also as a Q-thuster).Forgive me for interrupting, but with pulsing, do you mean Pulse Width Modulation or PWM?Yes. Ramp it up to peak thrust and let it exponentially decay back to zero... and Repeat. Since it can't spend more than the input power supplies and remain in steady state operation, and we already know that the input power will only provide uN of thrust. Higher thrust cannot be sustained at steady state operation. Higher thrust levels can only be momentary, and will quickly decay to nothing when output exceeds input.Todd
...The larger the Q, the larger the number of photons that can be red-shifted into attenuation in response to acceleration (and presumably providing “thrust” in an attempt to counter said acceleration when in "generator" mode). A similar thought experiment applies to the opposite end plate moving "towards" the rest frame, which should introduce a blue shift. How does blue or red shifting correlate to "thrust"? Or is such an observation merely another Red Herring? I have no idea......
Why has no one noticed superconducting resonant cavities floating or ripping themselves apart due to supposed EM drive thrust effects? Answer: no EM drive thrust possible within a symmetrical resonant cavity
Quote from: frobnicat on 05/10/2015 11:55 pmQuote from: WarpTech on 05/10/2015 08:27 pmQuote from: Notsosureofit on 03/10/2015 11:20 amQuote from: Rodal on 03/10/2015 01:42 amQuote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.The previous expression is only valid approximation for a "uniformly varying dielectric". There is no L1 and L2 in that case.What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )I was discussing this last night and we made some interesting observations. In a variable dielectric, like in the frustum, when the waves are accelerating to a higher group velocity, they are losing momentum. This momentum is lost to the material "in the direction of the wave". It is similar to frame dragging. The wave is losing energy trying to drag the waveguide or the dielectric with it.After the wave is reflected, it again tries to drag the dielectric or frustum with it, and this time it meets more resistance. It becomes an evanescent wave and decays faster. I do not believe a small end cap is needed and the frustum should taper all the way down to the wave guide feeding it. The reflected waves cannot reach the small plate. That's what the thermal images show as well. Most of the energy I think should be trapped at the big end.Todd D.Looking for a mechanical analogy :Let's play with a bended pipe and a ball rolling in it. The pipe can constrain the ball to various path, it can rise or fall, at various steepness. Height of the pipe at a given location defines gravitational potential energy of the ball there. The ball is launched with a given velocity, and then turns around the pipe if it is a closed circuit, or goes back and forth if the two ends of the pipe are high enough, should make no difference.Assuming no friction, the ball goes-on forever. When rising the ball loses kinetic energy, slows, and imparts momentum to the pipe. When on the return path (different part of pipe if circuit path or same part of pipe if going back and forth), the same delta height will make ball regain same kinetic energy as lost when rising, accelerate, and imparts momentum again. When taking curves, ball also imparts momentum on pipe. Integrating all those momentum exchanges on a cycle yields 0 net momentum. Not depending on path details.Assuming a closed circuit path and friction (dry, viscous, magnetic... whatever dissipative interaction), including parts with low friction (forth) and parts with high friction (back) and arbitrary height profile (potential well whatever). After a number of cycles the ball will come to rest. Integrating all the momentum exchanges of ball on pipe (changes of height, curves, friction) will yield a total momentum equal to the initial momentum of the ball when launched. Not depending on path details and what parts are more or less dissipative.I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?The ball (photon) doesn't fall back down the well. There is nothing to give it back enough energy to do so. It dissipates in multiple reflections between the walls and the big end. They are not getting more out than they put in, so it does not violate conservation of energy. They are simply getting more NET momentum on one direction than in the other direction because there is more dissipation and attenuation in one direction than there is in the other. Dissipative systems are typically "not" conservative, loses prevent a true equal measure from occuring in both directions. Todd D.
Quote from: WarpTech on 05/10/2015 08:27 pmQuote from: Notsosureofit on 03/10/2015 11:20 amQuote from: Rodal on 03/10/2015 01:42 amQuote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.The previous expression is only valid approximation for a "uniformly varying dielectric". There is no L1 and L2 in that case.What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )I was discussing this last night and we made some interesting observations. In a variable dielectric, like in the frustum, when the waves are accelerating to a higher group velocity, they are losing momentum. This momentum is lost to the material "in the direction of the wave". It is similar to frame dragging. The wave is losing energy trying to drag the waveguide or the dielectric with it.After the wave is reflected, it again tries to drag the dielectric or frustum with it, and this time it meets more resistance. It becomes an evanescent wave and decays faster. I do not believe a small end cap is needed and the frustum should taper all the way down to the wave guide feeding it. The reflected waves cannot reach the small plate. That's what the thermal images show as well. Most of the energy I think should be trapped at the big end.Todd D.Looking for a mechanical analogy :Let's play with a bended pipe and a ball rolling in it. The pipe can constrain the ball to various path, it can rise or fall, at various steepness. Height of the pipe at a given location defines gravitational potential energy of the ball there. The ball is launched with a given velocity, and then turns around the pipe if it is a closed circuit, or goes back and forth if the two ends of the pipe are high enough, should make no difference.Assuming no friction, the ball goes-on forever. When rising the ball loses kinetic energy, slows, and imparts momentum to the pipe. When on the return path (different part of pipe if circuit path or same part of pipe if going back and forth), the same delta height will make ball regain same kinetic energy as lost when rising, accelerate, and imparts momentum again. When taking curves, ball also imparts momentum on pipe. Integrating all those momentum exchanges on a cycle yields 0 net momentum. Not depending on path details.Assuming a closed circuit path and friction (dry, viscous, magnetic... whatever dissipative interaction), including parts with low friction (forth) and parts with high friction (back) and arbitrary height profile (potential well whatever). After a number of cycles the ball will come to rest. Integrating all the momentum exchanges of ball on pipe (changes of height, curves, friction) will yield a total momentum equal to the initial momentum of the ball when launched. Not depending on path details and what parts are more or less dissipative.I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?
Quote from: Notsosureofit on 03/10/2015 11:20 amQuote from: Rodal on 03/10/2015 01:42 amQuote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.The previous expression is only valid approximation for a "uniformly varying dielectric". There is no L1 and L2 in that case.What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )I was discussing this last night and we made some interesting observations. In a variable dielectric, like in the frustum, when the waves are accelerating to a higher group velocity, they are losing momentum. This momentum is lost to the material "in the direction of the wave". It is similar to frame dragging. The wave is losing energy trying to drag the waveguide or the dielectric with it.After the wave is reflected, it again tries to drag the dielectric or frustum with it, and this time it meets more resistance. It becomes an evanescent wave and decays faster. I do not believe a small end cap is needed and the frustum should taper all the way down to the wave guide feeding it. The reflected waves cannot reach the small plate. That's what the thermal images show as well. Most of the energy I think should be trapped at the big end.Todd D.
Quote from: Rodal on 03/10/2015 01:42 amQuote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.The previous expression is only valid approximation for a "uniformly varying dielectric". There is no L1 and L2 in that case.What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )
Quote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.
@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !
...@WarpTech or other proponents of "dependence of the effect on acceleration", please run the numbers. What would be the acceleration of frustum needed to really drift out of bandwidth for Q around 10000 or otherwise reach a magnitude significant to behaviour of waves inside ?
Quote from: WarpTech on 05/11/2015 01:26 amQuote from: frobnicat on 05/10/2015 11:55 pmQuote from: WarpTech on 05/10/2015 08:27 pmQuote from: Notsosureofit on 03/10/2015 11:20 amQuote from: Rodal on 03/10/2015 01:42 amQuote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.The previous expression is only valid approximation for a "uniformly varying dielectric". There is no L1 and L2 in that case.What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )I was discussing this last night and we made some interesting observations. In a variable dielectric, like in the frustum, when the waves are accelerating to a higher group velocity, they are losing momentum. This momentum is lost to the material "in the direction of the wave". It is similar to frame dragging. The wave is losing energy trying to drag the waveguide or the dielectric with it.After the wave is reflected, it again tries to drag the dielectric or frustum with it, and this time it meets more resistance. It becomes an evanescent wave and decays faster. I do not believe a small end cap is needed and the frustum should taper all the way down to the wave guide feeding it. The reflected waves cannot reach the small plate. That's what the thermal images show as well. Most of the energy I think should be trapped at the big end.Todd D.Looking for a mechanical analogy :Let's play with a bended pipe and a ball rolling in it. The pipe can constrain the ball to various path, it can rise or fall, at various steepness. Height of the pipe at a given location defines gravitational potential energy of the ball there. The ball is launched with a given velocity, and then turns around the pipe if it is a closed circuit, or goes back and forth if the two ends of the pipe are high enough, should make no difference.Assuming no friction, the ball goes-on forever. When rising the ball loses kinetic energy, slows, and imparts momentum to the pipe. When on the return path (different part of pipe if circuit path or same part of pipe if going back and forth), the same delta height will make ball regain same kinetic energy as lost when rising, accelerate, and imparts momentum again. When taking curves, ball also imparts momentum on pipe. Integrating all those momentum exchanges on a cycle yields 0 net momentum. Not depending on path details.Assuming a closed circuit path and friction (dry, viscous, magnetic... whatever dissipative interaction), including parts with low friction (forth) and parts with high friction (back) and arbitrary height profile (potential well whatever). After a number of cycles the ball will come to rest. Integrating all the momentum exchanges of ball on pipe (changes of height, curves, friction) will yield a total momentum equal to the initial momentum of the ball when launched. Not depending on path details and what parts are more or less dissipative.I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?The ball (photon) doesn't fall back down the well. There is nothing to give it back enough energy to do so. It dissipates in multiple reflections between the walls and the big end. They are not getting more out than they put in, so it does not violate conservation of energy. They are simply getting more NET momentum on one direction than in the other direction because there is more dissipation and attenuation in one direction than there is in the other. Dissipative systems are typically "not" conservative, loses prevent a true equal measure from occuring in both directions. Todd D.Come to think of it, it is not particularly surprising that a gradient is more easily interpreted in the comoving frame than in covariant form: dissipative phenomena are by nature alien to covariance. They are associated with the production of entropy, they have a thermodynamical arrow of time.
Quote from: frobnicat on 05/13/2015 12:14 am...@WarpTech or other proponents of "dependence of the effect on acceleration", please run the numbers. What would be the acceleration of frustum needed to really drift out of bandwidth for Q around 10000 or otherwise reach a magnitude significant to behaviour of waves inside ?I read it in Shawyer's recent paper, he simulated it on a computer. I don't have that type of software. I've been wishing I did for decades. Since the system moves by attenuation and dissipation, the photons lose momentum and are red-shifted, while the frustum either gains momentum or gains heat from them. You're correct, that the momentum it gains from the photons is in the direction of the photons, but results from the difference in the attenuation in each direction. The photons lose more momentum moving inward than moving outward, because they become evanescent waves. They do not increase their energy, except what they can take away from the frustum. So I see it like ringing a bell. It is the exponential decay from a higher energy state that is giving the thrust. Attempting to make the Q very high to sustain resonance requires reducing the losses, but it is the losses that give it thrust. So... Shawyer increases the angle to make it more like a pill box. Anything over pi/6 is very close to a pill box. Then it should have a higher Q, but it should also have less efficient use of it.If a photon rocket is: F/P = 1/cand the Frustum is: F/P ~ Q/c x pulse widthDesign efficiency should then target: (F*c)/(P*Q) = 1 but in practice < 1This would imply maximizing thrust with a lower value of Q, i.e., we do not want to maximize Q, we want to maximize asymmetry in the attenuation, which is what I'm working on at the moment.Best Regards,Todd
Quote from: WarpTech on 05/13/2015 01:05 amQuote from: frobnicat on 05/13/2015 12:14 am...Best Regards,ToddBarely able to keep pace Todd...its a good thing. maximizing asymmetry in attenuation different from absorption like the stuff I used to work with?http://www.westernrubber.com/products/himag-microwave-absorbers/himag-cavity-resonance-absorbers/It is getting pretty hot and heavy in here and I'm not sure I am keeping up either. Love it though.I've been mulling around the ideas of harmonics and wondered if anyone has considered injecting 2 RF sources into the cavity, One set and the other variable in frequency? I've been slowly working my way through this but like I said it's been slow. I welcome and inputs and thoughts.
Quote from: frobnicat on 05/13/2015 12:14 am...Best Regards,ToddBarely able to keep pace Todd...its a good thing. maximizing asymmetry in attenuation different from absorption like the stuff I used to work with?http://www.westernrubber.com/products/himag-microwave-absorbers/himag-cavity-resonance-absorbers/
...Best Regards,Todd
Quote from: rfmwguy on 05/13/2015 02:13 amQuote from: WarpTech on 05/13/2015 01:05 amQuote from: frobnicat on 05/13/2015 12:14 am...Best Regards,ToddBarely able to keep pace Todd...its a good thing. maximizing asymmetry in attenuation different from absorption like the stuff I used to work with?http://www.westernrubber.com/products/himag-microwave-absorbers/himag-cavity-resonance-absorbers/It is getting pretty hot and heavy in here and I'm not sure I am keeping up either. Love it though.I've been mulling around the ideas of harmonics and wondered if anyone has considered injecting 2 RF sources into the cavity, One set and the other variable in frequency? I've been slowly working my way through this but like I said it's been slow. I welcome and inputs and thoughts.From what I gather a magnetron source is full of harmonics and subharmonics...one called it "dirty" which is a good visual... spectrum-wise.