Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2166642 times)

Offline TheTraveller

...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan. 

It is not his theory.  Greg Egan is showing again a well-known result.

Instead, Greg Egan showed again, an already known-proof that the stresses due to standing waves on the inner walls of a resonating cavity of any arbitrary shape whatsoever perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

Egan exploited the computations of what the community only stated: Momentum is not conserved and so this does not work. This is the reason why we are looking elsewhere than only Maxwell equations.

What is your understanding of how an EM wave, acting at right angles to the cavity wall, can exert any force on it?

I assume the result would be zero, except for eddy current losses and as the cavity has a Q of 50,000, those losses are very small and are not involved in thrust generation.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline txdrive

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Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan.  Instead, Greg Egan showed again, a known-proof that the stresses on the inner walls of a resonating cavity of any arbitrary shape whatsoever (as long as it is closed) perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

If you can, please answer my simple question:

How can a curved EM wave, as in the attachment, touching the cavity wall at right angles to the wall, cause any force to be generated on the wall?
It's being reflected off the wall, why do you think it wouldn't generate a force orthogonally to the wall? I know it doesn't look like it's being reflected off the wall, but it has to be otherwise it would simply go through the wall.
« Last Edit: 05/17/2015 03:44 PM by txdrive »

Offline TheTraveller

...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan.  Instead, Greg Egan showed again, a known-proof that the stresses on the inner walls of a resonating cavity of any arbitrary shape whatsoever (as long as it is closed) perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

If you can, please answer my simple question:

How can a curved EM wave, as in the attachment, touching the cavity wall at right angles to the wall, cause any force to be generated on the wall?

You cannot stare at a pictorial representation claiming something is there. Physics means mathematics and mathematics, through Maxwell equations, says no net momentum.

Are you saying the picture is wrong?

How? Where?

If you had a EM antenna at the vertex of the cone, EM waves of the same wavelength would radiate outward. When constrained in a frustum, the frustum guide wavelength will alter from the free wavelength as the diameter of the frustum alters and the wave edges will be at right angles to the cavity walls. Other than that the EM waves will radiate from small end to big end as if the cavity did not exist. That is until they hit the spherical big end plate, do an inphase bounce, head back toward the vertex, hit the small end plate, bounce and repeat the process until cavity eddy current losses eat up all the energy in the wave.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline SeeShells

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I just want to throw this up here:
http://www.nature.com/ncomms/2014/140306/ncomms4300/full/ncomms4300.html
http://arxiv.org/ftp/arxiv/papers/1308/1308.0547.pdf

I think if you're looking in classical physics for answers, you won't find them.

Yes! Running down that same path myself!

Offline StrongGR

...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan. 

It is not his theory.  Greg Egan is showing again a well-known result.

Instead, Greg Egan showed again, an already known-proof that the stresses due to standing waves on the inner walls of a resonating cavity of any arbitrary shape whatsoever perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

Egan exploited the computations of what the community only stated: Momentum is not conserved and so this does not work. This is the reason why we are looking elsewhere than only Maxwell equations.

What is your understanding of how an EM wave, acting at right angles to the cavity wall, can exert any force on it?

I assume the result would be zero, except for eddy current losses and as the cavity has a Q of 50,000, those losses are very small and are not involved in thrust generation.

By the third principle, there is a movement of the cavity compensating the hit of the wave. It is the same that happens when you push on the steering wheel to move your car from inside. Your car stands still.
« Last Edit: 05/17/2015 03:46 PM by StrongGR »

Offline phaseshift

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I'm working on a Ruby script that builds a Sketchup model of Sawyer's drive according to the equations and concepts being generated by this group. So as work proceeds a few questions have come up.

Most of these questions refer to the recently posted photo of Sawyer's drive.

1. What possible material is the large end plate made of?

It appears to be aluminum. One purpose of the o-ring is therefore to prevent galvanic corrosion. This also means the bolts are made of the same metal and they do appear to be the same color/luster and are seated in rubber grommets.

2. Why a different metal?

A possible answer (see question 4) is that it is far easier to put a curve in an aluminum blank (at least for me it is) than in copper (cheaper and lighter in weight as well).

3. What possible material is the small end plate made of?

It appears to be copper - yet has aluminum looking bolts also seated in grommets!

4. Do we know with fair certainty that the end plates have a spherical curve?

This might explain why the large end plate is so thick - it is much like a telescope mirror. It needs thickness to accommodate the curvature and provides rigidity.  The outside of the large plate appears to be flat to me and I see no reason for a curve there.

5. What are people's thoughts on the plates being silvered on the curved surfaces?
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline TheTraveller

...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan.  Instead, Greg Egan showed again, a known-proof that the stresses on the inner walls of a resonating cavity of any arbitrary shape whatsoever (as long as it is closed) perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

If you can, please answer my simple question:

How can a curved EM wave, as in the attachment, touching the cavity wall at right angles to the wall, cause any force to be generated on the wall?
It's being reflected off the wall, why do you think it wouldn't generate a force orthogonally to the wall? I know it doesn't look like it's being reflected off the wall, but it has to be otherwise it would simply go through the wall.

It is not reflection off the wall any more than it would in free space if the walls did not exist. All the altering frustum diameter does is to alter the guide wavelength and the group velocity based on the cutoff frequency. There is no side wall bounce involved here.

BTW this is not a laser with a very short wavelength compared to the frustum dimensions and guide wavelength. The wave front does not bounce from side wall to side wall as the guide wavelength is way too long to allow that to happen. This is a resonate cavity, with the Em wave bouncing back and forth between the end plates and altering guide wavelength, frequency and group velocity as the wave slides by each diameter change.

Additionally if the EM wave behaved like a very much shorted laser EM wave, bouncing from side wall to side wall, there would be no need for end plates curved as Shawyer shows nor any change in guide wavelength, not guide frequency nor group velocity as the frustum diameter varies. So the laser bouncing off the side walls image is wrong.

You can see the effective end plate wavelengths, frequencies and group velocities here
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline deuteragenie

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Quote
By the third principle, there is a movement of the cavity compensating the hit of the wave. It is the same that happens when you push on the steering wheel to move your car from inside. Your car stands still.

Enlighten me: what happens if I point a flashlight to the car windshield?

Offline TheTraveller

...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan. 

It is not his theory.  Greg Egan is showing again a well-known result.

Instead, Greg Egan showed again, an already known-proof that the stresses due to standing waves on the inner walls of a resonating cavity of any arbitrary shape whatsoever perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

Egan exploited the computations of what the community only stated: Momentum is not conserved and so this does not work. This is the reason why we are looking elsewhere than only Maxwell equations.

What is your understanding of how an EM wave, acting at right angles to the cavity wall, can exert any force on it?

I assume the result would be zero, except for eddy current losses and as the cavity has a Q of 50,000, those losses are very small and are not involved in thrust generation.

By the third principle, there is a movement of the cavity compensating the hit of the wave. It is the same that happens when you push on the steering wheel to move your car from inside. Your car stands still.

http://en.wikipedia.org/wiki/Radiation_pressure and as attached:

Radiant pressure has a cosine angle adjustment.

As the EM wave, at the wall boundary, is moving at 0 deg relative to the wall, the cosine loss angle is at max and no pressure is delivered to the side wall by the passing by EM wave. At the spherical end plate, the EM wave is moving at 90 deg relative to all the surface area of the end plate and delivers max pressure. Pressure delivered to the spherical end plates depends on the EM waves group velocity at impact / bounce.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline deltaMass

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For fear of stating the obvious:
With appropriately curved internal endplates, as TheTraveller has drawn, the Poynting vector is precisely parallel to the side walls at all locations along the side walls, at all times. Ergo zero thrust on the side walls anywhere, at any time. Another way this is expressed is in the maths of light sail thrust (see any of Geoff Landis's papers for example). The thrust goes as the cosine of the angle made by the Poynting vector with the surface normal. Thus in this case that angle is a right angle so the cosine is zero so the thrust is zero.

Offline phaseshift

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Scratch question #4 - doh - yes its certain. :)
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline TheTraveller

I'm working on a Ruby script that builds a Sketchup model of Sawyer's drive according to the equations and concepts being generated by this group. So as work proceeds a few questions have come up.

Most of these questions refer to the recently posted photo of Sawyer's drive.

1. What possible material is the large end plate made of?

It appears to be aluminum. One purpose of the o-ring is therefore to prevent galvanic corrosion. This also means the bolts are made of the same metal and they do appear to be the same color/luster and are seated in rubber grommets.

2. Why a different metal?

A possible answer (see question 4) is that it is far easier to put a curve in an aluminum blank (at least for me it is) than in copper (cheaper and lighter in weight as well).

3. What possible material is the small end plate made of?

It appears to be copper - yet has aluminum looking bolts also seated in grommets!

4. Do we know with fair certainty that the end plates have a spherical curve?

This might explain why the large end plate is so thick - it is much like a telescope mirror. It needs thickness to accommodate the curvature and provides rigidity.  The outside of the large plate appears to be flat to me and I see no reason for a curve there.

5. What are people's thoughts on the plates being silvered on the curved surfaces?

Assume copper spherical end caps.

Hard to see a Q of 50,000 to 60,000 with flat end plates.

End caps are not necessarily the frustum EM wave bounce end plates.

End caps may just be to provide pressure seal and sandwich end plates between cavity flange and the external end plates.

Also believe the bottom end cap is so think as it has to allow the big end curve to exist. Also note the big end cap is not sitting flush on the alum beam. It may have a convex curve or maybe other big end sense ports are down there as shown in the 2nd attachment.

Bolts are probably space grade stainless. Should be no H20 in a satellite so no galvanic action. I'm not a sat guy, so just my guestimation.

Once you know the frustum vertex you can work out the end plate radii from this:
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline WarpTech

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...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan.  Instead, Greg Egan showed again, a known-proof that the stresses on the inner walls of a resonating cavity of any arbitrary shape whatsoever (as long as it is closed) perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

If you can, please answer my simple question:

How can a curved EM wave, as in the attachment, touching the cavity wall at right angles to the wall, cause any force to be generated on the wall?

If the walls are perfectly conducting, the NET will be zero, but they are not perfectly conducting and these are transverse waves. The electric field parallel to the wall may be zero but the field perpendicular is not. It attracts/repels electrons in the metal, driving currents. Also, magnetic permeability of copper is very low and conductivity is very high, so the rapidly changing magnetic field also drive currents. This is "work" being done on the walls of the cone.

Likewise, as work is done by the waves, the waves lose energy and decay to longer wavelengths. Giving up momentum to the Frustum, asymmetrically. As the waves are heading toward the small end, the guide wavelength eventually becomes imaginary and so the waves decay much faster. What Maxwell's equations and Greg Egan show is that the reflected waves that generate standing waves, play no part in the thrust, and they're right.

Todd

Offline phaseshift

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I'm working on a Ruby script that builds a Sketchup model of Sawyer's drive according to the equations and concepts being generated by this group. So as work proceeds a few questions have come up.

Most of these questions refer to the recently posted photo of Sawyer's drive.

1. What possible material is the large end plate made of?

It appears to be aluminum. One purpose of the o-ring is therefore to prevent galvanic corrosion. This also means the bolts are made of the same metal and they do appear to be the same color/luster and are seated in rubber grommets.

2. Why a different metal?

A possible answer (see question 4) is that it is far easier to put a curve in an aluminum blank (at least for me it is) than in copper (cheaper and lighter in weight as well).

3. What possible material is the small end plate made of?

It appears to be copper - yet has aluminum looking bolts also seated in grommets!

4. Do we know with fair certainty that the end plates have a spherical curve?

This might explain why the large end plate is so thick - it is much like a telescope mirror. It needs thickness to accommodate the curvature and provides rigidity.  The outside of the large plate appears to be flat to me and I see no reason for a curve there.

5. What are people's thoughts on the plates being silvered on the curved surfaces?

Assume copper spherical end caps.

Hard to see a Q of 50,000 to 60,000 with flat end plates.

End caps are not necessarily the frustum EM wave bounce end plates.

End caps may just be to provide pressure seal and sandwich end plates between cavity flange and the external end plates.

Also believe the bottom end cap is so think as it has to allow the big end curve to exist. Also note the big end cap is not sitting flush on the alum beam. It may have a convex curve or maybe other big end sense ports are down there as shown in the 2nd attachment.

Bolts are probably space grade stainless. Should be no H20 in a satellite so no galvanic action. I'm not a sat guy, so just my guestimation.

Once you know the frustum vertex you can work out the end plate radii from this:

Yep on the galvanic corrosion. 
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline TheTraveller

...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan.  Instead, Greg Egan showed again, a known-proof that the stresses on the inner walls of a resonating cavity of any arbitrary shape whatsoever (as long as it is closed) perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

If you can, please answer my simple question:

How can a curved EM wave, as in the attachment, touching the cavity wall at right angles to the wall, cause any force to be generated on the wall?

If the walls are perfectly conducting, the NET will be zero, but they are not perfectly conducting and these are transverse waves. The electric field parallel to the wall may be zero but the field perpendicular is not. It attracts/repels electrons in the metal, driving currents. Also, magnetic permeability of copper is very low and conductivity is very high, so the rapidly changing magnetic field also drive currents. This is "work" being done on the walls of the cone.

Likewise, as work is done by the waves, the waves lose energy and decay to longer wavelengths. Giving up momentum to the Frustum, asymmetrically. As the waves are heading toward the small end, the guide wavelength eventually becomes imaginary and so the waves decay much faster. What Maxwell's equations and Greg Egan show is that the reflected waves that generate standing waves, play no part in the thrust, and they're right.

Todd

Yes. Correct. Understand eddy current losses.

A Q of 50,000 say it takes 50,000 up and back cycles to drain off all the energy in the EM wave into wall eddy current losses and increased cavity wall temperature. The Chinese did measure and report on this as attached.

Due to the side wall cosine angle to the EM wave, there is no resultant radiation pressure on the side walls.
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline TheTraveller

Yep on the galvanic corrosion.

Galvanic corrosion occurs in space? Doesn't there needs to be an electrolyte between the different metals for it to happen? Thought spacecraft were ultra clean?
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline deltaMass

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Yes. Correct. Understand eddy current losses.

A Q of 50,000 say it takes 50,000 up and back cycles to drain off all the energy in the EM wave into wall eddy current losses and increased cavity wall temperature. The Chinese did measure and report on this as attached.

Due to the side wall cosine angle to the EM wave, there is no resultant radiation pressure on the side walls.
Yes, that's a more complete description. 100% concur.

ETA Almost concur :)  Most losses occur at the end plates, I would imagine. Due to copper's high thermal conductivity, this heat gets conducted to the side walls too.
« Last Edit: 05/17/2015 05:00 PM by deltaMass »

Offline phaseshift

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Yep on the galvanic corrosion.

Galvanic corrosion occurs in space? Doesn't there needs to be an electrolyte between the different metals for it to happen? Thought spacecraft were ultra clean?

Yep, as in I agree about there being none.
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline txdrive

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...
Shawyer published this measured thrust as did the Chinese and why they think it does what it does. Someone who publishes a paper claiming it can't work because of their theory, just might be politely ignored as they get on with business.
Greg Egan did not claim  that the EM Drive cannot work because of a theory invented by Greg Egan.  Instead, Greg Egan showed again, a known-proof that the stresses on the inner walls of a resonating cavity of any arbitrary shape whatsoever (as long as it is closed) perfectly balance, and therefore there is zero net force in any direction, according to Maxwell's equations.

If you can, please answer my simple question:

How can a curved EM wave, as in the attachment, touching the cavity wall at right angles to the wall, cause any force to be generated on the wall?
It's being reflected off the wall, why do you think it wouldn't generate a force orthogonally to the wall? I know it doesn't look like it's being reflected off the wall, but it has to be otherwise it would simply go through the wall.

It is not reflection off the wall any more than it would in free space if the walls did not exist. All the altering frustum diameter does is to alter the guide wavelength and the group velocity based on the cutoff frequency. There is no side wall bounce involved here.
Your wave inside and outside the waveguide is merely a sum of what you have if the wall did not exist and what is radiated by the wall.

Quote
For fear of stating the obvious:
With appropriately curved internal endplates, as TheTraveller has drawn, the Poynting vector is precisely parallel to the side walls at all locations along the side walls, at all times. Ergo zero thrust on the side walls anywhere, at any time. Another way this is expressed is in the maths of light sail thrust (see any of Geoff Landis's papers for example). The thrust goes as the cosine of the angle made by the Poynting vector with the surface normal. Thus in this case that angle is a right angle so the cosine is zero so the thrust is zero.
At the surface of a perfectly reflective light sail, E is orthogonal to the sail and E x H is parallel to the sail, which merely means that there's nothing being absorbed, not that there's no force.

What works for the sail calculation is the Poynting vector without the sail reflecting the light back.

Without his walls, the wave will diffract out.
« Last Edit: 05/17/2015 05:00 PM by txdrive »

Offline phaseshift

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Yep on the galvanic corrosion.

Galvanic corrosion occurs in space? Doesn't there needs to be an electrolyte between the different metals for it to happen? Thought spacecraft were ultra clean?

Yep, as in I agree about there being none.

Though that might be the original reason. I think the drive is several years from heading to space and he would want to be able to disassemble/reassemble in the near term. 
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

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