Looking over the last several posts, I see absolutely nothing related to space flight applications. All I see is more of exactly what got the previous version of this thread to disappear for a while.

Quote from: Rodal on 12/30/2014 03:21 pmNone of these experiments have demonstrated a linear acceleration: all of them have measured rotational accelerations. None of the EM Drives have been tested in a vacuum. None of the measured forces are high enough to levitate the drive. Forget about levitation. I havent read much or anything about it, but are the forces even in the same ballpark as existing magnetic torquer rods for cubesats ? If yes, in theory this could assist with attitude control in deep space, at least for desaturation.

None of these experiments have demonstrated a linear acceleration: all of them have measured rotational accelerations. None of the EM Drives have been tested in a vacuum. None of the measured forces are high enough to levitate the drive.

....Beyond unorthodox pseudo-forces orientation conventions, this "dynamic environment" condition seems quite problematic and ill defined. Ignoring theoretical musings, Shawyer makes 3 phenomenological predictions : ....- That a horizontal EM thruster restrained from accelerating horizontally (through an opposing spring) will record no thrust, page 3 figure 3 "Because the thruster is at rest, no force will be measured on the load cell"....

Quote from: savuporo on 12/31/2014 05:15 pmQuote from: Rodal on 12/30/2014 03:21 pmNone of these experiments have demonstrated a linear acceleration: all of them have measured rotational accelerations. None of the EM Drives have been tested in a vacuum. None of the measured forces are high enough to levitate the drive. Forget about levitation. I havent read much or anything about it, but are the forces even in the same ballpark as existing magnetic torquer rods for cubesats ? If yes, in theory this could assist with attitude control in deep space, at least for desaturation.Anyone ? Anyone ? I mean, actual spaceflight application. Desaturation spends fuel. Can we get a propellantless desaturation device, with main attitude control provided by reaction wheels ? ...

I thought I would make up a summary of the dispersion relation approach, as I keep doing this in bits and pieces.....

The difference from other calculations is that there is a term dependent on the particular mode of the cavity, (X[subm,n])^2, not just the area of the end plates.

I got some help making those equations look pretty with LaTeX, but I'm not sure it's rendered correctly. Are these accurate?Thrust per photon:\mathrm T = \left( \mathrm X _{m,n} \right) ^2 \left( \frac{\hslash}{4} \pi ^2 \right) \lambda \left( \frac{1}{a^2} - \frac{1}{b^2} \right)Force on the photons:\mathrm{T = P} \mathbb Q \left( \mathrm X_{m,n} \right) ^2 \left( \frac{1}{\mathrm{c}} 4 \pi ^2 \right) \lambda^2 \left( \frac{1}{a^2} - \frac{1}{b^2} \right)

Thrust per photon, with Planck's constant instead of the reduced constant:

I think that the factors of (Pi^2) in the first Latex equation and 4 Pi^2 in the 2nd Latex equation are in an incorrect position (they should be in the denominator instead of the numerator): the factors should be (h/(4 Pi^2)) in the first equation and (1/(c 4 Pi^2)) in the second equation.

Quote from: Rodal on 01/06/2015 01:56 amI think that the factors of (Pi^2) in the first Latex equation and 4 Pi^2 in the 2nd Latex equation are in an incorrect position (they should be in the denominator instead of the numerator): the factors should be (h/(4 Pi^2)) in the first equation and (1/(c 4 Pi^2)) in the second equation.Hmm. Are you sure? Oh well, here they are just the same:

PS: I agree with Notsosureofit, the second equation would better read NT, where "N" stands for the thrust of all the photons, instead of the thrust of a single photon "T".

Guys, Guys......I stopped reading some time ago.....now we are into the 2nd thread.Time to "let it go"........prove it one way or another. The energy spent back and forth could have been put into a cad file, exported into STEP or IGES format by now....(someone please do it)Then maybe if time permits i'll print out a test model.Someone then talk maybe to Nanoracks, and lets get it tested.

Quote from: Prober on 01/06/2015 08:25 amGuys, Guys......I stopped reading some time ago.....now we are into the 2nd thread.Time to "let it go"........prove it one way or another. The energy spent back and forth could have been put into a cad file, exported into STEP or IGES format by now....(someone please do it)Then maybe if time permits i'll print out a test model.Someone then talk maybe to Nanoracks, and lets get it tested. What are STEP and IGES ?

what is the correct force expression for the force on the solenoid andsecond, the assumption that Newton’s third law holds in the sense that thechange of the solenoid’s momentum is compensated by the change of theelectron’s momentum. The discussion of “Feynman’s paradox” shows thatthe latter is not always the case. It is possible that a change in field momentumis an essential part of the Aharonov-Bohm discussion, which is exactlywhat Aharonov and Casher claim in 1984 [45]. Many theoretical papers havediscussed this issue [16, 17, 36, 37]. These discussions involve imbalancedforces, field momentum and relativistic terms, all of which are present in ourabove discussion. However, none of the discussions gives an explicit and exactderivation of the delicate balance of all the momentum terms, but oftenresort to a treatment of simplified systems. For example, Aharonov and D.Rohrlich [16] discuss a flux tube with a radially moving charge, instead of acharge passing by the flux tube. While the issue of whether the charge distributionof the solenoid is perturbed has been addressed [17, 36, 46], none ofthe discussions mention the relativistic electric field imbalance.As it is possible to describe a solenoid as a collection of moving chargedparticles, the above treatment of the Feynman paradox provides hope to settlethe theoretical discussion on forces. Integration over a solenoidal currentdistribution would provide an exact derivation of momentum conservationfor the Aharonov-Bohm case.

Guys, Guys......I stopped reading some time ago.....now we are into the 2nd thread.