The flight thruster is shown as what appears to be an oscillator configuration, so it is still possible that the loaded Q is higher than the unloaded Q. But that is speculation. I do worry about his concept of "no static thrust" as well.(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)

...(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)

Quote from: Notsosureofit on 03/05/2015 12:35 am...(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)I understand that your expression is based on the exact solution for a cylindrical cavity. Is there a reason why a cylindrical cavity thrust should depend on the cross-sectional quantum numbers "m" and "n" but not depend on the axial quantum number "p"?The exact solution for the cylindrical cavity has a frequency dependent on "p". Why isn't thrust dependent on "p" ?

So, then in order to produce 'thrust,' this device MUST be a truncated cone - preferably with a rounded base?Also a wild and stupid thought about Sawyer's commentary: maybe he envisions this device as some sort of 'turbo-charger?' That is something that adds to an already existing velocity, but won't function well, if at all when 'at rest.' Say you have a spacecraft in motion at speed X. Flip this drive on, and the speed becomes X+1.

...maybe he envisions this device as some sort of 'turbo-charger?' That is something that adds to an already existing velocity, but won't function well, if at all when 'at rest.' ...

...The device must have an asymmetric dispersion relation in some axis. .....

@ Rodal :See attached picture to share my mental image. Tilt over-exaggerated for illustration. Grey : solid rotating assembly (no deformation implied) Orange/brown : fixed assembly (no deformation implied) Blue : the ground slab of the vacuum chamber (no deformation implied) For now, assume a perfect axis of rotation around Z : only one degree of freedom of Grey relative to Orange, the "official" rotation around Z, no compliance implied, Grey kept in the XY plane, plane has same tilt as Orange (XY parallel to Orange platform).@Star-Drive Can you confirm this is a correct way to understand that there is a tilt in the axis of rotation ?

What is the "engine" that the EM Drive is turbocharging?The bizarre nature of something needing to be free to accelerate for it to produce a force doesn't apply to the turbocharger or to the engine: the engine that is being turbocharged does not need to be accelerating or even be in rigid body motion. Its center of mass can be completely stationary, and the turbocharged engine can then be used for electric power generation, for example, instead of for transporting people as in an automobile.

Quote from: Notsosureofit on 03/05/2015 12:35 am...(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)Quote from: Notsosureofit on 03/05/2015 01:10 am...The device must have an asymmetric dispersion relation in some axis. .....I think I might be able to obtain a closed-form expression for the frequency of a cylindrical cavity with a dielectric at one end. (Or really a cylindrical cavity having two dielectrics in contact with each other: one dielectric at one end and another dielectric with different permitivity and permeability at the other end).Would that be of any use to you to improve your equation?

Quote from: Rodal on 03/05/2015 01:43 amQuote from: Notsosureofit on 03/05/2015 12:35 am...(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)Quote from: Notsosureofit on 03/05/2015 01:10 am...The device must have an asymmetric dispersion relation in some axis. .....I think I might be able to obtain a closed-form expression for the frequency of a cylindrical cavity with a dielectric at one end. (Or really a cylindrical cavity having two dielectrics in contact with each other: one dielectric at one end and another dielectric with different permitivity and permeability at the other end).Would that be of any use to you to improve your equation?Absolutely ! That should be a very interesting case as it is 2 coupled cavities. Most of these kind of problems use iterative solutions but I think that case may have some unique analytic solutions !

Quote from: frobnicat on 03/04/2015 08:46 pm@ Rodal :See attached picture to share my mental image. Tilt over-exaggerated for illustration. Grey : solid rotating assembly (no deformation implied) Orange/brown : fixed assembly (no deformation implied) Blue : the ground slab of the vacuum chamber (no deformation implied) For now, assume a perfect axis of rotation around Z : only one degree of freedom of Grey relative to Orange, the "official" rotation around Z, no compliance implied, Grey kept in the XY plane, plane has same tilt as Orange (XY parallel to Orange platform).@Star-Drive Can you confirm this is a correct way to understand that there is a tilt in the axis of rotation ?One aspect of this tilt in the axis of rotation that hasn't been discussed is the requirement for imbalance. The difference in moments on either side of the beam only has to be very slight for it to always come to rest at the same location +/- a micron or two.

A small CW tilt along the X axis of the balance arm would explain the apparent drift in the baseline seen in some of the thrust waveforms. When the cavity has the orientation shown below its CM shifts to the left. This would reduce the tilt, resulting in an increase in brightness of the reflected light the LDS measures; due to the mirror position being closer to an optimal perpendicular position wrt the light beam. The increase in brightness corresponds to a decrease in distance; hence the negative slope. With the device mounted the other way the shift in CM increases the tilt. This reduces the reflected light and is registered as an increase in distance.

No actual motion of the beam occurs. This apparent motion is an optical artifact. This assumes the Philtec distance sensor is used on the far side. If it is used on the near side a small CCW tilt along the X axis of the beam would produce the same effect; except requiring much less rotation from the change in CM.

Quote from: frobnicat on 03/01/2015 09:12 pm...- That the vertical scale in the charts (indicated in µm, around 500) are relevant or not relevant.......The Philtec D63 fiber-optic displacement sensor measures distance from its target mirror in microns, so the numbers on the left hand side of the force plots measure the distance from the end of the fiber-optic laser head to its mirror target mounted on the torque pendulum arm. The data sheet for same is attached....

...- That the vertical scale in the charts (indicated in µm, around 500) are relevant or not relevant....

@ RODALCan you check this ? (working it down)root1= Sqrt[( (-4*b^2*(dD1^2-dD2^2)*((L1/c1)^2-(L2/c2)^2)) / (dD1^2*dD2^2) + ((L1/c1)^2+(L2/c2)^2)*p^2 - 2*Sqrt[ (-4*b^2*((L1/c1)^2-(L2/c2)^2)*((dD1*L1/c1)^2-(dD2*L2/c2)^2)*p^2) / (dD1^2*dD2^2) + (L1^2*L2^2*p^4) / (c1^2*c2^2)]) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> )] ?? / (L1^2/c1^2-L2^2/c2^2)^2]/2root2= Sqrt[( (-4*b^2*(dD1^2-dD2^2)*((L1/c1)^2-(L2/c2)^2)) / (dD1^2*dD2^2) + ((L1/c1)^2+(L2/c2)^2)*p^2 + 2*Sqrt[ (-4*b^2*((L1/c1)^2 - (L2/c2)^2)*((dD1*L1)/c1)^2-(dD2*L2/c2)^2)*p^2) / (dD1^2*dD2^2) + (L1^2*L2^2*p^4) / (c1^2*c2^2)]) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> )] ?? / (L1^2/c1^2-L2^2/c2^2)^2]/2where we define the following dimensionless ratios as follows:dD1 = diameter/L1;dD2 = diameter/L2;and whereL1 = length - L2;where length is the total internal length of the cylindrical cavity.The quantity "b" is defined as follows:b := If[modetype == "TM", xbesselzeros[[i + 1, j]]/Pi, If[modetype == "TE", xprimebesselzeros[[i + 1, j]]/Pi]]Thanks