Various explanations, both of spacecraft behavior and of gravitation itself, were proposed to explain the anomaly. Over the period 1998–2012, one particular explanation became accepted. The spacecraft, which are surrounded by an ultra-high vacuum and are each powered by a radioisotope thermoelectric generator (RTG), can shed heat only via thermal radiation. If, due to the design of the spacecraft, more heat is emitted in a particular direction—what is known as a radiative anisotropy—then the spacecraft would exhibit a small acceleration in the direction opposite that of the excess emitted radiation due to radiation pressure. Because this force is due to the recoil of thermal photons, it is also called the thermal recoil force. If the excess radiation and attendant radiation pressure were pointed in a general direction opposite the Sun, the spacecrafts’ velocity away from the Sun would be decelerating at a greater rate than could be explained by previously recognized forces, such as gravity and trace friction, due to the interplanetary medium (imperfect vacuum).By 2012 several papers by different groups, all reanalyzing the thermal radiation pressure forces inherent in the spacecraft, showed that a careful accounting of this could explain the entire anomaly, and thus the cause was mundane and did not point to any new phenomena or need for a different physical paradigm.[2][3] The most detailed analysis to date, by some of the original investigators, explicitly looks at two methods of estimating thermal forces, then states "We find no statistically significant difference between the two estimates and conclude that once the thermal recoil force is properly accounted for, no anomalous acceleration remains

{snip}Wild speculation 2:If this is some sort of thermal drive, then perhaps...thrust remains constant only for so long as the temperature increases? So, if the temperature plateau's, then thrust declines. This might provide a solution of sorts to the Conservation of Energy issue.If this is the case, then the EM drive might make for a useful orbital thruster, and perhaps power interplanetary probes...but would it be adequate for interstellar propulsion?

Something I have thought about before and am thinking about once again is thermal effects. Virtually every engineer or theorist to jump into these EM Drive threads has cited thermal artefacts as a possible solution for EM Drive thrust. Doctor Rodal's first posts in the original thread dealt with this. Back then, the reasoning was:1) thermal effects could mimic significant thrust in ambient air conditions;2) those same thermal effects would still produce thrust, albeit far smaller thrust, in a vacuum on earth;

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 !

@zellerium:70^{o}C temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.

Continuing with the "Test In Space" theme.We lack cheap space access for unmanned cargo. We have no railgun running from the coast of Ecuador up into the Andes to the east, and we have no Skylon/SABRE SSTO yet. So we must pay many thousands of dollars per launched kilogram rather than what could be only tens of dollars. What we do have is Cubesat and SpaceX. The problem is that the devices under consideration here won't fit even into the largest Cubesat. So let's talk miniaturisation.What we have is photons in an asymmetric cavity. So let's use light instead of microwaves. I'll stop there for now.

I know Shawyer and EW have tried a dielectric in the frustum. Are there any specifications for that dielectric? Material properties? Absorption properties at microwave frequencies?I was looking at Pyramid Absorbers for microwaves, they can attenuate up to -55dB. A high power microwave source, pumped through a diode into such an absorber, seems to me should have a higher probability of thrust than the EM Drive and relatively simple to construct.Todd D.

Quote from: deltaMass on 05/11/2015 12:05 am@zellerium:70^{o}C temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.The base is made of glass-fiber reinforced epoxy with an extremely thin layer (deposited) of copperon the outside: 0.063 inch thick FR4 printed circuit board and ~35 microns thick of Cu on the insideThermal conductivity, through-plane 0.29 W/(m·K)

Quote from: Rodal on 05/11/2015 12:16 amQuote from: deltaMass on 05/11/2015 12:05 am@zellerium:70^{o}C temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.The base is made of glass-fiber reinforced epoxy with an extremely thin layer (deposited) of copperon the outside: 0.063 inch thick FR4 printed circuit board and ~35 microns thick of Cu on the insideThermal conductivity, through-plane 0.29 W/(m·K)A rough calc says that the Cu thickness has to be less than 44 microns. Looks like it's possible. But I calculated heat flow around the copper, not through-plane. Someone should recheck both.P = 50 W, dT = 70^{o}K

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: frobnicat on 05/10/2015 11:55 pm...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.

...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 ?

Do higher em frequencies deliver increased net effect?

Quote from: frobnicat on 05/10/2015 11:55 pm...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.

...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 ?