Although the flow velocity is nonrelativistic (v ≪ 1), disturbances tend to “propagate” superluminally, at 1/v. Hence, the NFA here is not a normal nonrelativistic reduction. The resulting equations are “anti-Galilean” invariant...This is certainly strange, and takes some getting used to, but one should simply view itas an approximation to the full Lorentz transformations, valid in the stated context. Oneis used to dealing with small objects that move slowly, so that their density distributionsvary rapidly in space, but slowly in time. In the present case one is dealing with largeobjects, slowly varying in space, but relatively rapidly varying in time. This is related tothe fact that the Higgs vacuum, as a spontaneous Bose-Einstein condensate, has almostall its particles in the same quantum state. Small disturbances of this state involve vastnumbers of particles, spread over long distances, all moving nearly in lockstep, so thatthe disturbance varies only slowly with position while the whole collective has the same,relatively rapid time dependence.
Once again, I must mention that there are spinning fans on the test rig. A HD drive also.
Quote from: deltaMass on 06/09/2015 05:26 pmOnce again, I must mention that there are spinning fans on the test rig. A HD drive also.Yep there are but we also see the abnormality in tests without fans and spinning drives.
Quote from: WarpTech on 06/09/2015 05:30 pmQuote from: deltaMass on 06/09/2015 03:53 pmQuote from: A_M_Swallow on 06/09/2015 01:45 pm...You would have to exceed the top speed to extract excess energy. As energy is removed the mass would go down.You do not actually get to the top speed because that would produce infinite mass and zero acceleration. The system will have rest mass.I can show you a very simple machine utilising the hypothetical principle of variable mass that readily produces free energy forever. A version can be built for either linear motion in free space or, using a different approach, rotary motion in a gravitational field. The bottom line is that if you have variable mass then you have perpetual motion and free energy.Really? How? Where does the extra free energy come from?The rotary device is the simplest. A balanced wheel with two equal masses A,B diametrically placed, one of which (A) is alterable by means unspecified. When A is descending it is made heavier. Thus each half cycle the wheel undergoes acceleration. The linear version requires no gravity and can operate in free space. It consists of a variable mass "puck" losslesssly bouncing between two walls of a container. When it strikes the "front" wall it is made heavier. The container experiences a steady acceleration in the forward direction.The full descriptions are attached
Quote from: deltaMass on 06/09/2015 03:53 pmQuote from: A_M_Swallow on 06/09/2015 01:45 pm...You would have to exceed the top speed to extract excess energy. As energy is removed the mass would go down.You do not actually get to the top speed because that would produce infinite mass and zero acceleration. The system will have rest mass.I can show you a very simple machine utilising the hypothetical principle of variable mass that readily produces free energy forever. A version can be built for either linear motion in free space or, using a different approach, rotary motion in a gravitational field. The bottom line is that if you have variable mass then you have perpetual motion and free energy.Really? How? Where does the extra free energy come from?
Quote from: A_M_Swallow on 06/09/2015 01:45 pm...You would have to exceed the top speed to extract excess energy. As energy is removed the mass would go down.You do not actually get to the top speed because that would produce infinite mass and zero acceleration. The system will have rest mass.I can show you a very simple machine utilising the hypothetical principle of variable mass that readily produces free energy forever. A version can be built for either linear motion in free space or, using a different approach, rotary motion in a gravitational field. The bottom line is that if you have variable mass then you have perpetual motion and free energy.
...You would have to exceed the top speed to extract excess energy. As energy is removed the mass would go down.You do not actually get to the top speed because that would produce infinite mass and zero acceleration. The system will have rest mass.
Quote from: deltaMass on 06/08/2015 05:44 pmRe. the recent flyby reference to the Aachen group's Baby EmDrive and CubeSats, I'm reminded that their team leader has already flown a couple of amateur space missions with an outfit called PoqetQub (from memory). This is a NASA forum, so presumably packed to the brim with orbital mechanics specialists!! So... what value of k (N/W) is needed to get EmDrive up from LEO, O Experts?ETA: On reflection that's a dumb question Any positive k value will do.You would have to determine what constitutes an orbit change that is outside the natural decay forces. Cubesats don't have much power, so they may not get much thrust. Would a retardation of orbital decay convince anyone? That is a tricky deal, because orbit decay is sensitive to upper atmosphere expansion/contraction, which is affected by solar activity, etc.If the thrust was significantly greater than the decay forces, you can use something like the Edelbaum approximation to determine the altitude change you should see with constant, tangential acceleration. If there is interest, I'll run some quick parametrics to see what that might be.
Re. the recent flyby reference to the Aachen group's Baby EmDrive and CubeSats, I'm reminded that their team leader has already flown a couple of amateur space missions with an outfit called PoqetQub (from memory). This is a NASA forum, so presumably packed to the brim with orbital mechanics specialists!! So... what value of k (N/W) is needed to get EmDrive up from LEO, O Experts?ETA: On reflection that's a dumb question Any positive k value will do.
Quote from: deltaMass on 06/09/2015 05:59 pmQuote from: WarpTech on 06/09/2015 05:30 pmQuote from: deltaMass on 06/09/2015 03:53 pmQuote from: A_M_Swallow on 06/09/2015 01:45 pm...You would have to exceed the top speed to extract excess energy. As energy is removed the mass would go down.You do not actually get to the top speed because that would produce infinite mass and zero acceleration. The system will have rest mass.I can show you a very simple machine utilising the hypothetical principle of variable mass that readily produces free energy forever. A version can be built for either linear motion in free space or, using a different approach, rotary motion in a gravitational field. The bottom line is that if you have variable mass then you have perpetual motion and free energy.Really? How? Where does the extra free energy come from?The rotary device is the simplest. A balanced wheel with two equal masses A,B diametrically placed, one of which (A) is alterable by means unspecified. When A is descending it is made heavier. Thus each half cycle the wheel undergoes acceleration. The linear version requires no gravity and can operate in free space. It consists of a variable mass "puck" losslesssly bouncing between two walls of a container. When it strikes the "front" wall it is made heavier. The container experiences a steady acceleration in the forward direction.The full descriptions are attachedThis is my first time posting but it seems to me that this "free energy" machine only works if you assume that the device that alters the mass doesn't require energy input to work. (Or atleast less energy then whatever the break even point would work out to) Shutting up now.
Quote from: Prunesquallor on 06/08/2015 07:58 pmQuote from: deltaMass on 06/08/2015 05:44 pmRe. the recent flyby reference to the Aachen group's Baby EmDrive and CubeSats, I'm reminded that their team leader has already flown a couple of amateur space missions with an outfit called PoqetQub (from memory). This is a NASA forum, so presumably packed to the brim with orbital mechanics specialists!! So... what value of k (N/W) is needed to get EmDrive up from LEO, O Experts?ETA: On reflection that's a dumb question Any positive k value will do.You would have to determine what constitutes an orbit change that is outside the natural decay forces. Cubesats don't have much power, so they may not get much thrust. Would a retardation of orbital decay convince anyone? That is a tricky deal, because orbit decay is sensitive to upper atmosphere expansion/contraction, which is affected by solar activity, etc.If the thrust was significantly greater than the decay forces, you can use something like the Edelbaum approximation to determine the altitude change you should see with constant, tangential acceleration. If there is interest, I'll run some quick parametrics to see what that might be.Here's some analysis of what kind of orbital raising one could expect given constant, tangential, in-plane orbit thrusting starting from a 600 km circular orbit (an average CubeSat altitude). It is not dependent on the type of thruster.Now if one wanted to apply this to a CubeSat with a little bitty EM Drive, here is how the numbers might stack up:Typical CubeSat available power: 0.5 Whttp://www.diyspaceexploration.com/power-system-budget-analysis/Typical CubeSat mass: 1.3 kghttp://en.wikipedia.org/wiki/CubeSatNow, choose your assumed EM Drive efficiency and compute acceleration. For example if you want to assume 0.1 N/kW, your acceleration would be (0.1 N/kW)*(0.5 W)*(0.001 kW/W)/(1.3 kg)/(9.81 m/s2/g) = around 4 micro-gs.You can then look at the chart, find the 4 micro-g line and see the altitude gain as a function of thruster on-time. You can decide for yourself if you want it to have constant thrust at constant power or if you want compute the time you think the universe will let the thruster operate and see how high it will get.
Here's some analysis of what kind of orbital raising one could expect given constant, tangential, in-plane orbit thrust acceleration starting from a 600 km circular orbit (an average CubeSat altitude). It is not dependent on the type of thruster.Now if one wanted to apply this to a CubeSat with a little bitty EM Drive, here is how the numbers might stack up:Typical CubeSat available power: 0.5 Whttp://www.diyspaceexploration.com/power-system-budget-analysis/Typical CubeSat mass: 1.3 kghttp://en.wikipedia.org/wiki/CubeSatNow, choose your assumed EM Drive efficiency and compute acceleration. For example if you want to assume 0.1 N/kW, your acceleration would be (0.1 N/kW)*(0.5 W)*(0.001 kW/W)/(1.3 kg)/(9.81 m/s2/g) = around 4 micro-gs.You can then look at the chart, find the 4 micro-g line and see the altitude gain as a function of thruster on-time. You can decide for yourself if you want it to have constant thrust at constant power or if you want compute the time you think the universe will let the thruster operate and see how high it will get.
Quote from: Prunesquallor on 06/09/2015 06:51 pmQuote from: Prunesquallor on 06/08/2015 07:58 pmQuote from: deltaMass on 06/08/2015 05:44 pmRe. the recent flyby reference to the Aachen group's Baby EmDrive and CubeSats, I'm reminded that their team leader has already flown a couple of amateur space missions with an outfit called PoqetQub (from memory). This is a NASA forum, so presumably packed to the brim with orbital mechanics specialists!! So... what value of k (N/W) is needed to get EmDrive up from LEO, O Experts?ETA: On reflection that's a dumb question Any positive k value will do.You would have to determine what constitutes an orbit change that is outside the natural decay forces. Cubesats don't have much power, so they may not get much thrust. Would a retardation of orbital decay convince anyone? That is a tricky deal, because orbit decay is sensitive to upper atmosphere expansion/contraction, which is affected by solar activity, etc.If the thrust was significantly greater than the decay forces, you can use something like the Edelbaum approximation to determine the altitude change you should see with constant, tangential acceleration. If there is interest, I'll run some quick parametrics to see what that might be.Here's some analysis of what kind of orbital raising one could expect given constant, tangential, in-plane orbit thrust acceleration starting from a 600 km circular orbit (an average CubeSat altitude). It is not dependent on the type of thruster.Now if one wanted to apply this to a CubeSat with a little bitty EM Drive, here is how the numbers might stack up:Typical CubeSat available power: 0.5 Whttp://www.diyspaceexploration.com/power-system-budget-analysis/Typical CubeSat mass: 1.3 kghttp://en.wikipedia.org/wiki/CubeSatNow, choose your assumed EM Drive efficiency and compute acceleration. For example if you want to assume 0.1 N/kW, your acceleration would be (0.1 N/kW)*(0.5 W)*(0.001 kW/W)/(1.3 kg)/(9.81 m/s2/g) = around 4 micro-gs.You can then look at the chart, find the 4 micro-g line and see the altitude gain as a function of thruster on-time. You can decide for yourself if you want it to have constant thrust at constant power or if you want compute the time you think the universe will let the thruster operate and see how high it will get.The winner is Prof. Yang, reporting 1 N/kW (for ambient air)which gives 40 micro-gsThe lowest reported value is NASA Eagleworks, in 5*10^(-4) Torr turned around 180 degrees, 0.000283 N/kW giving 0.01 micro-gs
Quote from: Prunesquallor on 06/08/2015 07:58 pmQuote from: deltaMass on 06/08/2015 05:44 pmRe. the recent flyby reference to the Aachen group's Baby EmDrive and CubeSats, I'm reminded that their team leader has already flown a couple of amateur space missions with an outfit called PoqetQub (from memory). This is a NASA forum, so presumably packed to the brim with orbital mechanics specialists!! So... what value of k (N/W) is needed to get EmDrive up from LEO, O Experts?ETA: On reflection that's a dumb question Any positive k value will do.You would have to determine what constitutes an orbit change that is outside the natural decay forces. Cubesats don't have much power, so they may not get much thrust. Would a retardation of orbital decay convince anyone? That is a tricky deal, because orbit decay is sensitive to upper atmosphere expansion/contraction, which is affected by solar activity, etc.If the thrust was significantly greater than the decay forces, you can use something like the Edelbaum approximation to determine the altitude change you should see with constant, tangential acceleration. If there is interest, I'll run some quick parametrics to see what that might be.Here's some analysis of what kind of orbital raising one could expect given constant, tangential, in-plane orbit thrust acceleration starting from a 600 km circular orbit (an average CubeSat altitude). It is not dependent on the type of thruster.Now if one wanted to apply this to a CubeSat with a little bitty EM Drive, here is how the numbers might stack up:Typical CubeSat available power: 0.5 Whttp://www.diyspaceexploration.com/power-system-budget-analysis/Typical CubeSat mass: 1.3 kghttp://en.wikipedia.org/wiki/CubeSatNow, choose your assumed EM Drive efficiency and compute acceleration. For example if you want to assume 0.1 N/kW, your acceleration would be (0.1 N/kW)*(0.5 W)*(0.001 kW/W)/(1.3 kg)/(9.81 m/s2/g) = around 4 micro-gs.You can then look at the chart, find the 4 micro-g line and see the altitude gain as a function of thruster on-time. You can decide for yourself if you want it to have constant thrust at constant power or if you want compute the time you think the universe will let the thruster operate and see how high it will get.
Quote from: Prunesquallor on 06/09/2015 06:51 pmQuote from: Prunesquallor on 06/08/2015 07:58 pmQuote from: deltaMass on 06/08/2015 05:44 pmRe. the recent flyby reference to the Aachen group's Baby EmDrive and CubeSats, I'm reminded that their team leader has already flown a couple of amateur space missions with an outfit called PoqetQub (from memory). This is a NASA forum, so presumably packed to the brim with orbital mechanics specialists!! So... what value of k (N/W) is needed to get EmDrive up from LEO, O Experts?ETA: On reflection that's a dumb question Any positive k value will do.You would have to determine what constitutes an orbit change that is outside the natural decay forces. Cubesats don't have much power, so they may not get much thrust. Would a retardation of orbital decay convince anyone? That is a tricky deal, because orbit decay is sensitive to upper atmosphere expansion/contraction, which is affected by solar activity, etc.If the thrust was significantly greater than the decay forces, you can use something like the Edelbaum approximation to determine the altitude change you should see with constant, tangential acceleration. If there is interest, I'll run some quick parametrics to see what that might be.Here's some analysis of what kind of orbital raising one could expect given constant, tangential, in-plane orbit thrusting starting from a 600 km circular orbit (an average CubeSat altitude). It is not dependent on the type of thruster.Now if one wanted to apply this to a CubeSat with a little bitty EM Drive, here is how the numbers might stack up:Typical CubeSat available power: 0.5 Whttp://www.diyspaceexploration.com/power-system-budget-analysis/Typical CubeSat mass: 1.3 kghttp://en.wikipedia.org/wiki/CubeSatNow, choose your assumed EM Drive efficiency and compute acceleration. For example if you want to assume 0.1 N/kW, your acceleration would be (0.1 N/kW)*(0.5 W)*(0.001 kW/W)/(1.3 kg)/(9.81 m/s2/g) = around 4 micro-gs.You can then look at the chart, find the 4 micro-g line and see the altitude gain as a function of thruster on-time. You can decide for yourself if you want it to have constant thrust at constant power or if you want compute the time you think the universe will let the thruster operate and see how high it will get.Thanks for the analysis! Well, the performance is parlous, as expected.The idea here is to ask whether we can actually tell that the drive is working. It looks like this is not too hard.
The winner is Prof. Yang, reporting 1 N/kW (for ambient air)which gives 40 micro-gsThe lowest reported value is NASA Eagleworks, in 5*10^(-4) Torr turned around 180 degrees, 0.000283 N/kW giving 0.01 micro-gs
Quote from: WarpTech on 06/09/2015 05:30 pm...Really? How? Where does the extra free energy come from?The rotary device is the simplest. A balanced wheel with two equal masses A,B diametrically placed, one of which (A) is alterable by means unspecified. When A is descending it is made heavier. Thus each half cycle the wheel undergoes acceleration. The linear version requires no gravity and can operate in free space. It consists of a variable mass "puck" losslesssly bouncing between two walls of a container. When it strikes the "front" wall it is made heavier. The container experiences a steady acceleration in the forward direction.The full descriptions are attached
...Really? How? Where does the extra free energy come from?
http://arxiv.org/pdf/hep-ph/0409292Hydrodynamics of the VacuumP. M. StevensonT. W. Bonner Laboratory, Department of Physics and AstronomyRice University“anti-Galilean” invarianceAlthough the flow velocity is nonrelativistic (v ≪ 1), disturbances tend to “propagate” superluminally, at 1/v. page 9:Quote from: page 9 of Hydrodynamics of the Vacuum_0409292v2.pdfAlthough the flow velocity is nonrelativistic (v ≪ 1), disturbances tend to “propagate” superluminally, at 1/v. Hence, the NFA here is not a normal nonrelativistic reduction. The resulting equations are “anti-Galilean” invariant...This is certainly strange, and takes some getting used to, but one should simply view itas an approximation to the full Lorentz transformations, valid in the stated context. Oneis used to dealing with small objects that move slowly, so that their density distributionsvary rapidly in space, but slowly in time. In the present case one is dealing with largeobjects, slowly varying in space, but relatively rapidly varying in time. This is related tothe fact that the Higgs vacuum, as a spontaneous Bose-Einstein condensate, has almostall its particles in the same quantum state. Small disturbances of this state involve vastnumbers of particles, spread over long distances, all moving nearly in lockstep, so thatthe disturbance varies only slowly with position while the whole collective has the same,relatively rapid time dependence.
https://hackaday.io/project/5596-em-drive/log/19253-emdrive-testsStill not sure if they have thrust yet...looking at data.
Quote from: SeeShells on 06/09/2015 08:21 pmhttps://hackaday.io/project/5596-em-drive/log/19253-emdrive-testsStill not sure if they have thrust yet...looking at data.What torque makes the device spin around then ?It's a beautiful set-up, this baby EM Drive.Neat!
Quote from: Rodal on 06/09/2015 08:33 pmQuote from: SeeShells on 06/09/2015 08:21 pmhttps://hackaday.io/project/5596-em-drive/log/19253-emdrive-testsStill not sure if they have thrust yet...looking at data.What torque makes the device spin around then ?It's a beautiful set-up, this baby EM Drive.Neat!I think they had to impart some spin before they switched it on. Just early in the testing and I'm sure they will refine it more.
Quote from: SeeShells on 06/09/2015 08:37 pmQuote from: Rodal on 06/09/2015 08:33 pmQuote from: SeeShells on 06/09/2015 08:21 pmhttps://hackaday.io/project/5596-em-drive/log/19253-emdrive-testsStill not sure if they have thrust yet...looking at data.What torque makes the device spin around then ?It's a beautiful set-up, this baby EM Drive.Neat!I think they had to impart some spin before they switched it on. Just early in the testing and I'm sure they will refine it more.They had to impart the spin in order to stabilize the set-up so that it would stay centered? (using the gyroscopic effect, conservation of rotational momentum)?