Quote from: Mulletron on 10/12/2014 05:27 PMWe must be sure to understand that emdrive doesn't need MiHsC to work if dielectric thrust holds true. MiHsC is an optimization factor. Here on earth in strong gravity MiHsC's effect is essentially zero. Once in microgravity is it helpful. The effect is pretty much nothing otherwise unless you have fancy meta materials.OK, let's analyze the meaning of "dielectric thrust"To analyze this concept we need to specify what it entails. Are you referring to the paper you quoted on chirality of the molecule and the quantum vacuum? or are you referring to something else?Thanks

We must be sure to understand that emdrive doesn't need MiHsC to work if dielectric thrust holds true. MiHsC is an optimization factor. Here on earth in strong gravity MiHsC's effect is essentially zero. Once in microgravity is it helpful. The effect is pretty much nothing otherwise unless you have fancy meta materials.

I've attached a sketch of the Eagleworks cavity with the area ratios. Note that the base is 2.25 times larger than the throat. Does this mean that a particle oscillating between the base the throat at constant speed will impart 2.25 times the momentum on the throat as it does on the base?Well and good, but if that particle has inertial mass of 1 unit as measured in the laboratory frame, what is it's inertial mass at the base? Of course it is 2.25 times that amount at the throat but what is the reference frame?It is an important consideration because the answer will tell us how many oscillating particles are needed to produce the measured thrust.

Quote from: Rodal on 10/12/2014 05:29 PMQuote from: Mulletron on 10/12/2014 05:27 PMWe must be sure to understand that emdrive doesn't need MiHsC to work if dielectric thrust holds true. MiHsC is an optimization factor. Here on earth in strong gravity MiHsC's effect is essentially zero. Once in microgravity is it helpful. The effect is pretty much nothing otherwise unless you have fancy meta materials.OK, let's analyze the meaning of "dielectric thrust"To analyze this concept we need to specify what it entails. Are you referring to the paper you quoted on chirality of the molecule and the quantum vacuum? or are you referring to something else?ThanksThe perceived importance of the dielectric media to the operation of the device. And yes. Also Dr. White's research on QVT.

Quote from: Mulletron on 10/12/2014 05:42 PMQuote from: Rodal on 10/12/2014 05:29 PMQuote from: Mulletron on 10/12/2014 05:27 PMWe must be sure to understand that emdrive doesn't need MiHsC to work if dielectric thrust holds true. MiHsC is an optimization factor. Here on earth in strong gravity MiHsC's effect is essentially zero. Once in microgravity is it helpful. The effect is pretty much nothing otherwise unless you have fancy meta materials.OK, let's analyze the meaning of "dielectric thrust"To analyze this concept we need to specify what it entails. Are you referring to the paper you quoted on chirality of the molecule and the quantum vacuum? or are you referring to something else?ThanksThe perceived importance of the dielectric media to the operation of the device. And yes. Also Dr. White's research on QVT.The chirality theory has been already dealt with. It is not-explanatory due to the extremely small size of the effect predicted (nanometer per second velocity due to 10 Tesla magnetic field) and due to the fact that a 9 inch solid rod injection molded has no directionality.

Quote from: Rodal on 10/12/2014 05:52 PMQuote from: Mulletron on 10/12/2014 05:42 PMQuote from: Rodal on 10/12/2014 05:29 PMQuote from: Mulletron on 10/12/2014 05:27 PMWe must be sure to understand that emdrive doesn't need MiHsC to work if dielectric thrust holds true. MiHsC is an optimization factor. Here on earth in strong gravity MiHsC's effect is essentially zero. Once in microgravity is it helpful. The effect is pretty much nothing otherwise unless you have fancy meta materials.OK, let's analyze the meaning of "dielectric thrust"To analyze this concept we need to specify what it entails. Are you referring to the paper you quoted on chirality of the molecule and the quantum vacuum? or are you referring to something else?ThanksThe perceived importance of the dielectric media to the operation of the device. And yes. Also Dr. White's research on QVT.The chirality theory has been already dealt with. It is not-explanatory due to the extremely small size of the effect predicted (nanometer per second velocity due to 10 Tesla magnetic field) and due to the fact that a 9 inch solid rod injection molded has no directionality.That's fine and all. The key point to get is that modified inertia doesn't make things move. It makes things easier to move. You still have to thrust against something in order to move.

Just a note: MiHsC does predict movement. If you assume conservatn of m'tum, and change m, v changes. See my papers on the flyby anomaly (in MNRAS, 2008) or the Tajmar effect (in EPL, 2011).

From my knowledge, the formula from Prof. McCulloch's and the formula from Shawyer remain the only formulas that come close to predicting the experimental thrusts in the USA, UK and China experiments. Everything else is orders of magnitude off.

Quote from: Rodal on 10/12/2014 06:14 PMFrom my knowledge, the formula from Prof. McCulloch's and the formula from Shawyer remain the only formulas that come close to predicting the experimental thrusts in the USA, UK and China experiments. Everything else is orders of magnitude off.http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdfThe paper says clearly and provides data that the dielectric is key. The question is how. Now on the flipside, with no dielectric present, you would get uneven heat of the cavity from fore to aft and some movement of the device. How else could it move? Internal stresses would convert to wall strain. Where is the reaction mass?

Quote from: aero on 10/12/2014 05:31 PMI've attached a sketch of the Eagleworks cavity with the area ratios. Note that the base is 2.25 times larger than the throat. Does this mean that a particle oscillating between the base the throat at constant speed will impart 2.25 times the momentum on the throat as it does on the base?Well and good, but if that particle has inertial mass of 1 unit as measured in the laboratory frame, what is it's inertial mass at the base? Of course it is 2.25 times that amount at the throat but what is the reference frame?It is an important consideration because the answer will tell us how many oscillating particles are needed to produce the measured thrust.For McCulloch's formula and for Shawyer's formula it is not only the ratio of the large and small flat areas that matter, but instead it is the dimensional difference of their reciprocals:F = (PowerInput* Q / frequency) *(1/SmallRadius- 1/LargeRadius)So you see, the dimensions of both surfaces enter the equation and not just their ratioOf course, as it is obvious from the equation, they are inapplicable as the limit of SmallRadius --> 0 is approached, since the force goes to Infinity as that limit (a pointy cone) is approached.EDIT: Also observe that it is not the surface area and it is not therefore the square of the radius of the flat surfaces that enters the formula but instead is the radius itself.

What if the resonant cavity walls acted like a Hubble horizon, especially for Unruh waves of a similar length (as they are in this case)? Then the inertial mass of the photons would increase towards the cavity's wide end, since more Unruh waves would fit there, since mi=m(1-L/2w), where w is the cavity width.

Quote from: Rodal on 10/12/2014 05:46 PMQuote from: aero on 10/12/2014 05:31 PMI've attached a sketch of the Eagleworks cavity with the area ratios. Note that the base is 2.25 times larger than the throat. Does this mean that a particle oscillating between the base the throat at constant speed will impart 2.25 times the momentum on the throat as it does on the base?Well and good, but if that particle has inertial mass of 1 unit as measured in the laboratory frame, what is it's inertial mass at the base? Of course it is 2.25 times that amount at the throat but what is the reference frame?It is an important consideration because the answer will tell us how many oscillating particles are needed to produce the measured thrust.For McCulloch's formula and for Shawyer's formula it is not only the ratio of the large and small flat areas that matter, but instead it is the dimensional difference of their reciprocals:F = (PowerInput* Q / frequency) *(1/SmallDiameter- 1/LargeDiameter)So you see, the dimensions of both surfaces enter the equation and not just their ratioOf course, as it is obvious from the equation, they are inapplicable as the limit of SmallRadius --> 0 is approached, since the force goes to Infinity as that limit (a pointy cone) is approached.EDIT: Also observe that it is not the surface area and it is not therefore the square of the radius of the flat surfaces that enters the formula but instead is the radius itself.QuoteWhat if the resonant cavity walls acted like a Hubble horizon, especially for Unruh waves of a similar length (as they are in this case)? Then the inertial mass of the photons would increase towards the cavity's wide end, since more Unruh waves would fit there, since mi=m(1-L/2w), where w is the cavity width.This looks to me like a 1-D derivation. More Unruh waves would also fit is a larger area, not simply on a 1-D line. The equation would look the same in 2-D, with area replacing width. With a little algebra the equation becomes F = PowerInput*Q * (Large area-Small area)/(frequency*Large area*Small area)Not quite what I had before but the area difference is explicit.

Quote from: aero on 10/12/2014 05:31 PMI've attached a sketch of the Eagleworks cavity with the area ratios. Note that the base is 2.25 times larger than the throat. Does this mean that a particle oscillating between the base the throat at constant speed will impart 2.25 times the momentum on the throat as it does on the base?Well and good, but if that particle has inertial mass of 1 unit as measured in the laboratory frame, what is it's inertial mass at the base? Of course it is 2.25 times that amount at the throat but what is the reference frame?It is an important consideration because the answer will tell us how many oscillating particles are needed to produce the measured thrust.For McCulloch's formula and for Shawyer's formula it is not only the ratio of the large and small flat areas that matter, but instead it is the dimensional difference of their reciprocals:F = (PowerInput* Q / frequency) *(1/SmallDiameter- 1/LargeDiameter)So you see, the dimensions of both surfaces enter the equation and not just their ratioOf course, as it is obvious from the equation, they are inapplicable as the limit of SmallRadius --> 0 is approached, since the force goes to Infinity as that limit (a pointy cone) is approached.EDIT: Also observe that it is not the surface area and it is not therefore the square of the radius of the flat surfaces that enters the formula but instead is the radius itself.

Apparantly the dielectric block in Figure 15. was just the starting point for the COMSOLanalysis iterationprocess used prior to assembly to determine the optimal thickness and diameter of the dielectric RF resonator disc located at the small end of the thruster.From Figure 18:"Cu Frustrum Test Configuration:2, 6.25" x 1.06" PE Discs at small End"

I wonder if it will make a difference in the acceleration required to establish the Unruh waves within the cavity?

Quote from: Notsosureofit on 10/12/2014 09:34 PMApparantly the dielectric block in Figure 15. was just the starting point for the COMSOLanalysis iterationprocess used prior to assembly to determine the optimal thickness and diameter of the dielectric RF resonator disc located at the small end of the thruster.From Figure 18:"Cu Frustrum Test Configuration:2, 6.25" x 1.06" PE Discs at small End"Fantastic information @notsosureofit !!!!!You determined the material: Polyethylene (PE)and the dimensions 6.25" x 1.06" apparently what they used was 2 (two?) solid disks 6.25 inches in diameter by 1.06 inches thickThere was apparently no inner hole, or they did not record the inner diameter in the report?

Quote from: Rodal on 10/12/2014 05:46 PM.../...For McCulloch's formula and for Shawyer's formula it is not only the ratio of the large and small flat areas that matter, but instead it is the dimensional difference of their reciprocals:(1) F = (PowerInput* Q / frequency) *(1/SmallRadius- 1/LargeRadius)So you see, the dimensions of both surfaces enter the equation and not just their ratioOf course, as it is obvious from the equation, they are inapplicable as the limit of SmallRadius --> 0 is approached, since the force goes to Infinity as that limit (a pointy cone) is approached..../...This looks to me like a 1-D derivation. More Unruh waves would also fit is a larger area, not simply on a 1-D line. The equation would look the same in 2-D, with area replacing width. With a little algebra the equation becomes (2)F = PowerInput*Q * (Large area-Small area)/(frequency*Large area*Small area)Not quite what I had before but the area difference is explicit.

.../...For McCulloch's formula and for Shawyer's formula it is not only the ratio of the large and small flat areas that matter, but instead it is the dimensional difference of their reciprocals:(1) F = (PowerInput* Q / frequency) *(1/SmallRadius- 1/LargeRadius)So you see, the dimensions of both surfaces enter the equation and not just their ratioOf course, as it is obvious from the equation, they are inapplicable as the limit of SmallRadius --> 0 is approached, since the force goes to Infinity as that limit (a pointy cone) is approached..../...