Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1797160 times)

Online Rodal

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http://arxiv.org/pdf/hep-ph/0409292

Hydrodynamics of the Vacuum
P. M. Stevenson
T. W. Bonner Laboratory, Department of Physics and Astronomy
Rice University

“anti-Galilean” invariance

Although 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.pdf
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 it
as an approximation to the full Lorentz transformations, valid in the stated context. One
is used to dealing with small objects that move slowly, so that their density distributions
vary rapidly in space, but slowly in time. In the present case one is dealing with large
objects, slowly varying in space, but relatively rapidly varying in time. This is related to
the fact that the Higgs vacuum, as a spontaneous Bose-Einstein condensate, has almost
all its particles in the same quantum state. Small disturbances of this state involve vast
numbers of particles, spread over long distances, all moving nearly in lockstep, so that
the disturbance varies only slowly with position while the whole collective has the same,
relatively rapid time dependence.
« Last Edit: 06/09/2015 06:08 PM by Rodal »

Offline SeeShells

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Once 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.

Offline vulture4

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Let us assume the incident radiation is perpendicular and is perfectly reflected from the large end of the frustrum, and that there is no pressure at all on the small end.

Measured scale force = 5 gm = .05N  (from the Shawyer Demo file above)

Radiation pressure for perfect normal reflection:
P = 2E/c
where:
P=radiation pressure (N/m^2)
E=energy flux (W/m^2)
c= light speed in vacuum = 3E8 m/s
(see https://en.wikipedia.org/wiki/Radiation_pressure#Radiation_pressure_by_particle_model:_photons )

Multiplying by area,

F=2P/c
where
F=force (N)
P=power (W)

P=Fc/2
 =.05N(3*10^8m/s)/2
=1.5*10^7 W
=15,000,000 watts incident power is required on the frustum endplate to produce a 5gm scale reading
« Last Edit: 06/09/2015 06:33 PM by vulture4 »

Offline deltaMass

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Once 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.
Just highlighting Shawyer's lack of experimental professionalism. YMMV

Offline ort

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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
This 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.

Offline Prunesquallor

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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.

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 W
http://www.diyspaceexploration.com/power-system-budget-analysis/

Typical CubeSat mass: 1.3 kg
http://en.wikipedia.org/wiki/CubeSat

Now, 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.
« Last Edit: 06/09/2015 07:02 PM by Prunesquallor »
Retired, yet... not

Offline deltaMass

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...
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
This 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.
Yes, of course. You get deltaMass for free. The whole idea is to show that if you can do that - even briefly, but repeatedly - then you get free energy and perpetual motion. Naturally nobody knows how to actually do this.

Now, if you think that you can do this, we should start a company :)
« Last Edit: 06/09/2015 06:54 PM by deltaMass »

Offline deltaMass

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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.

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 W
http://www.diyspaceexploration.com/power-system-budget-analysis/

Typical CubeSat mass: 1.3 kg
http://en.wikipedia.org/wiki/CubeSat

Now, 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.

Online Rodal

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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 W
http://www.diyspaceexploration.com/power-system-budget-analysis/

Typical CubeSat mass: 1.3 kg
http://en.wikipedia.org/wiki/CubeSat

Now, 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-gs

The lowest reported value is NASA Eagleworks, in 5*10^(-4) Torr partial vacuum turned around 180 degrees, 0.000283 N/kW giving 0.01 micro-gs, which is lower than the lowest value reported in the chart (0.1 micro g)

« Last Edit: 06/09/2015 07:16 PM by Rodal »

Offline Prunesquallor

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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.

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 W
http://www.diyspaceexploration.com/power-system-budget-analysis/

Typical CubeSat mass: 1.3 kg
http://en.wikipedia.org/wiki/CubeSat

Now, 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-gs

The lowest reported value is NASA Eagleworks, in 5*10^(-4) Torr turned around 180 degrees, 0.000283 N/kW giving 0.01 micro-gs

Hence the log-log plot  ;)
Retired, yet... not

Offline Prunesquallor

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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.

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 W
http://www.diyspaceexploration.com/power-system-budget-analysis/

Typical CubeSat mass: 1.3 kg
http://en.wikipedia.org/wiki/CubeSat

Now, 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 missing piece is the orbital decay.  You would want have to (IMHO) stay at least an order of magnitude above the predicted drag deceleration, since that thing is quite variable and hard to predict.  I'll try to dig up some info.
Retired, yet... not

Offline vulture4

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The winner is Prof. Yang, reporting 1 N/kW (for ambient air)which gives 40 micro-gs
The lowest reported value is NASA Eagleworks, in 5*10^(-4) Torr turned around 180 degrees, 0.000283 N/kW giving 0.01 micro-gs
So thrust varies directly with atmospheric pressure, eh? Wait! I've got a theory...

Seriously, theory and verification must come first. All spacecraft are subject to perturbations, even the Pioneers in interstellar space. If the theory we are considering is the one proposed by Shawyer, that reactionless thrust is produced by unequal radiation pressures, then the measured thrusts would require unattainable energy densities in the resonator. If some other theory is to be considered and tested, it must first be defined.
« Last Edit: 06/09/2015 07:45 PM by vulture4 »

Offline WarpTech

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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

"...is alterable by means unspecified"

In the paper you wrote; "...by whatever means."

The "means" you are referring to, I refer to as a self-charging capacitor. If you have a device that can spontaneous gain mass, i.e., a self-charging capacitor, you don't need the cart or the wheel to extract it.

Good luck with that! I wouldn't buy Woodward's device either.





Offline SeeShells

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http://arxiv.org/pdf/hep-ph/0409292

Hydrodynamics of the Vacuum
P. M. Stevenson
T. W. Bonner Laboratory, Department of Physics and Astronomy
Rice University

“anti-Galilean” invariance

Although 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.pdf
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 it
as an approximation to the full Lorentz transformations, valid in the stated context. One
is used to dealing with small objects that move slowly, so that their density distributions
vary rapidly in space, but slowly in time. In the present case one is dealing with large
objects, slowly varying in space, but relatively rapidly varying in time. This is related to
the fact that the Higgs vacuum, as a spontaneous Bose-Einstein condensate, has almost
all its particles in the same quantum state. Small disturbances of this state involve vast
numbers of particles, spread over long distances, all moving nearly in lockstep, so that
the disturbance varies only slowly with position while the whole collective has the same,
relatively rapid time dependence.
I LOVE this! Great find Dr. Rodal!
I've always wondered before the inflationary period where there seemed to be no controls on the expansion of spacetime that the Higgs manifested itself into a state that was quantum entangled and then the Higgs interaction with matter and mass slowed the expansion to present day values we see. It would make sense that to be felt throughout the universe in a slow down the Higgs would need to be in a quantum entanglement lockstep.

I was just reading this before reading your post.
http://www.nature.com/news/higgs-data-could-spell-trouble-for-leading-big-bang-theory-1.12804
<edit>
More thoughts before I go out and clean the hot tub area. From superluminal thoughts to more mundane...such is life.

You know a particle in lockstep ie:entanglement means there is no spoon. Ok, it means that the walls of the EM Cavity are not there and any effect it has on the particles within the cavity effects those outside that are entangled. Seeing this correctly?
 
« Last Edit: 06/09/2015 08:34 PM by SeeShells »

Offline SeeShells

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https://hackaday.io/project/5596-em-drive/log/19253-emdrive-tests

Still not sure if they have thrust yet...looking at data.
« Last Edit: 06/09/2015 08:27 PM by SeeShells »

Online Rodal

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https://hackaday.io/project/5596-em-drive/log/19253-emdrive-tests

Still 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!

« Last Edit: 06/09/2015 08:34 PM by Rodal »

Offline SeeShells

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https://hackaday.io/project/5596-em-drive/log/19253-emdrive-tests

Still 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.

Online Rodal

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https://hackaday.io/project/5596-em-drive/log/19253-emdrive-tests

Still 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)?

Offline Prunesquallor

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https://hackaday.io/project/5596-em-drive/log/19253-emdrive-tests

Still 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)?

This is their magnetically suspended rig?
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Offline sghill

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At that power level and those small sizes, they could conceivably sell this thing as a kit to hobbyists who want to test  their own EMDrives!


Kickstarter powers, Activate!!!

(P.S.  we could all chip in and mail one to Paul March for Christmas).
« Last Edit: 06/09/2015 08:53 PM by sghill »
Bring the thunder Elon!

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