Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2101123 times)

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?

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.

I assume this is the final product that SPR shipped to Boeing. Has been plated and I'm sure other issues addressed and fixed. Also suggest the photo was distorted to make it harder to work out the dimensions.

On a NASA EW slide, this unit was labeled a "High Fidelity Test Article".
« Last Edit: 05/17/2015 05:09 PM by TheTraveller »
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
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Offline txdrive

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With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.
« Last Edit: 05/17/2015 05:21 PM by txdrive »

Offline phaseshift

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Yes, that version has a bunch of improvements over the copper version.

Building it in SketchUp really highlights issues that would come upduring construction. I have had this photo and really should have been going off of it. I am not trying to guess dimensions from the photos but instead are using the equations and dimensions in this forum. The script currently takes 3 parameters and generates the model from that.
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline dustinthewind

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I have been meaning to ask.  If anyone gets the chance to test this out for me I would be fascinated to hear the results.  It's a dual cavity experiment.  I think it would require a klystron or twt with a continuous phase.  Cylindrical cavities should work.  They would need to have a gap between the two of them of 1/4 lambda wavelength with their phase out of alignment by 90 degrees or pi/2 radians.   

What should happen is the plate of cavity 1 should have circulating current stimulated by the radiation inside that builds up depending on the Q of the cavity.  Information is limited by the speed of light so that appearance of current should be on its way to the next cavity which has its current out of phase by 90 degrees.  When the information reaches cavity 2 the current in cavity 2 should appear to be moving with cavity 1 so it is attracted.  Cavity 1 on the other hand when the signal reaches it appears to have its current moving against cavity 2 and should be repulsed (similar to current in wires).  I would guess this effect should also scale with the Q of the cavity. 

It may be possible for a dielectric to slow the information transfer between the cavities and move them closer together and increase the effect.  I would guess some radiation may tunnel through the cavities into the other one and attenuation of one cavity may be required to keep them equal in amplitude and out of phase by 90 degrees. 

My guess is it should be similar to attraction between magnets but the force is in the same direction for both cavities.  Images attached for ease of understanding. 

Fig1 Simple.png key:
1. Forces
2. Current
3. Current
4. Appearance of current limited by the speed of light (bottom wire)
5. Appearance of current limited by the speed of light (top wire)

8, 9, 10, 11 are frames of current over time.  Frame 12 is a repeat of frame 8 and frame 13 is a repeat of 9.  (Only 4 frames needed).  The illustration of force between wires can parallel to the cavity plate I think. 

Thanks!

I would also be curious to hear if we think it won't work for some specific reason as a counter argument.  I'm open minded to counter arguments. 

Offline WarpTech

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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.

I think the difference is, that it's not the radiation pressure of a transverse wave anymore. It's Lorentz force due to EM induction. When the waves become evanescent, the interaction becomes a near-field induction of eddy currents and a rapid loss of momentum. Where, in the other direction, waves traveling toward the large end do not see this evanescent wave effect. I'm concentrated on these evanescent waves because this is where the asymmetry is, in the attenuation.

Offline TheTraveller

With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.

Last line say it all:

Quote
If the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.

The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.
« Last Edit: 05/17/2015 05:50 PM by TheTraveller »
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
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Offline txdrive

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With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.

Last line say it all:

Quote
If the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.

The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.
You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.

edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.
« Last Edit: 05/17/2015 05:56 PM by txdrive »

Offline SeeShells

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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. 

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.
Let's say your hitting the steering wheel very fast again and again and let's leave out it was someone who cut you off in traffic. Lets say each time you brought your hand back it gained energy in the action of the backward movement , it was from let's say a quantum/force which has no relation or attachment to the car but into the underlying quantum world (weird and neat stuff), then each time you would strike the wheel you would impart force and momentum to the car.  Off you go. Not so much with a flashlight on the windshield as light is a little different in it's construction and it would piss off the car in front of you but that's another story.
Shell

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. 

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.
Let's say your hitting the steering wheel very fast again and again and let's leave out it was someone who cut you off in traffic. Lets say each time you brought your hand back it gained energy in the action of the backward movement , it was from let's say a quantum/force which has no relation or attachment to the car but into the underlying quantum world (weird and neat stuff), then each time you would strike the wheel you would impart force and momentum to the car.  Off you go. Not so much with a flashlight on the windshield as light is a little different in it's construction and it would piss off the car in front of you but that's another story.
Shell
Then, unless this force and momentum is very small, the mechanics of motion of a hand through space would have to differ from what was observed in high precision experiments that did not involve cars.

edit: Suppose you computed the average force f that you'd obtain if you ignored the pressure of your back against the car seat - just the hand banging on the steering wheel.

If you had previously done a wide variety of highly precise parts-per-million experiments on the motion of the hand through the weird quantum world (never noticing any weird changes in energy or momentum, to a high precision), then you can very confidently exclude any forces much greater than a millionth of f.
« Last Edit: 05/17/2015 06:27 PM by txdrive »

Offline LasJayhawk

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.

I assume this is the final product that SPR shipped to Boeing. Has been plated and I'm sure other issues addressed and fixed. Also suggest the photo was distorted to make it harder to work out the dimensions.

On a NASA EW slide, this unit was labeled a "High Fidelity Test Article".


If I want to maximize the Q of a cavity I would silver plate the copper, and if I wanted maximum repeatablity I would gold flash the silver.

The cinder blocks in the wall should be 8" high. The rf connector with the cap on it looks to be a type SMA and not N, iirc the SMA flange connectors are about .5" square.

Offline TheTraveller

With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.

Last line say it all:

Quote
If the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.

The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.
You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.

edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.

We were discussing radiation pressure generated from EM waves according to Maxwell's equations (attached). This pressure is apparently subject to cosine angle loss and thus when the EM wave moves along the frustum side walls and is not bouncing off the frustum spherical end plates, there is no Maxwell radiation pressure generated on the frustum side walls.

Or did I not understand what the cosine loss factor is for in the lower of the 2 equations?
« Last Edit: 05/17/2015 06:25 PM by TheTraveller »
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
Herman Melville, Moby Dick

Offline Flyby

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Yes, that version has a bunch of improvements over the copper version.

Building it in SketchUp really highlights issues that would come upduring construction. I have had this photo and really should have been going off of it. I am not trying to guess dimensions from the photos but instead are using the equations and dimensions in this forum. The script currently takes 3 parameters and generates the model from that.
As I showed about a hundred pages ago, building a 3D model based upon photographs requires adjustments to compensate the camera lens distortions. Unless you have the specific data of the camera and its lens, it will be impossible to accurately build a 3D model. At best you'll have an approximation.

Do not underestimate the distortion caused by targeting under the horizon line. They are important and increase when using a wide angle lens. When targeting under the horizon, the lens distortions  cause vertical lines to focus on a point below, making the top of the frustum appear larger to the bottom plate.

I did stop a further image analysis on the alu frustum because somebody found the exact dimensions of the frustum, as it appears to be an industrial manufactured piping part. It is readily available to everybody...There was no need anymore for an estimate on that frustum...

I'll try to dig it up... it's somewhere inhere, i suspect in the first 50 pages...


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?

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.

I assume this is the final product that SPR shipped to Boeing. Has been plated and I'm sure other issues addressed and fixed. Also suggest the photo was distorted to make it harder to work out the dimensions.

On a NASA EW slide, this unit was labeled a "High Fidelity Test Article".


If I want to maximize the Q of a cavity I would silver plate the copper, and if I wanted maximum repeatablity I would gold flash the silver.

The cinder blocks in the wall should be 8" high. The rf connector with the cap on it looks to be a type SMA and not N, iirc the SMA flange connectors are about .5" square.

That small connector near the small end plate of the Flight Thruster may be for an e field sense feedback. Here is a view from the other side.

Note what looks like a cavity resonance tuner to the left of the red Rf input fed.
"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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With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.

Last line say it all:

Quote
If the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.

The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.
You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.

edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.

We were discussing radiation pressure generated from EM waves according to Maxwell's equations (attached). This pressure is apparently subject to cosine angle loss and thus when the EM wave moves along the frustum side walls and is not bouncing off the frustum spherical end plates, there is no Maxwell radiation pressure generated on the frustum side walls.

Or did I not understand what the cosine loss factor is for in the lower of the 2 equations?
It's a geometric optics approximation. Doesn't hold precisely.

For example suppose you got a beam of coherent light, 5mm across, 500nm wavelength, in space. It's not going all parallel to it's original direction, it's spreading, to about 1m across at 10 000m from the source (Makes a fuzzy "Airy disk" pattern).

edit: The Poynting vector on the sides of a perfect cylinder over this beam, is not parallel to the axis. It's this Poynting vector that matters.
« Last Edit: 05/17/2015 06:47 PM by txdrive »

Offline phaseshift

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Yes, that version has a bunch of improvements over the copper version.

Building it in SketchUp really highlights issues that would come upduring construction. I have had this photo and really should have been going off of it. I am not trying to guess dimensions from the photos but instead are using the equations and dimensions in this forum. The script currently takes 3 parameters and generates the model from that.
As I showed about a hundred pages ago, building a 3D model based upon photographs requires adjustments to compensate the camera lens distortions. Unless you have the specific data of the camera and its lens, it will be impossible to accurately build a 3D model. At best you'll have an approximation.

Do not underestimate the distortion caused by targeting under the horizon line. They are important and increase when using a wide angle lens. When targeting under the horizon, the lens distortions  cause vertical lines to focus on a point below, making the top of the frustum appear larger to the bottom plate.

I did stop a further image analysis on the alu frustum because somebody found the exact dimensions of the frustum, as it appears to be an industrial manufactured piping part. It is readily available to everybody...There was no need anymore for an estimate on that frustum...

I'll try to dig it up... it's somewhere inhere, i suspect in the first 50 pages...

I've mentioned a few times - I'm not getting the dimensions from the photos. :)
"It doesn't have to be a brain storm, a drizzle will often do" - phaseshift

Offline Rodal

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Dear Jose,

Thanks a lot for spending some time on my calculations. You tried with b but did you check for a?...

http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=830137

Here are the expressions including both a and b, to give a value to L

Assuming r1 =<r2

as per the geometry in the following image




Taking (from Eq. 39 in your paper)

b = (r1^2 - r2^2)/(4 (lo^2) Log[r2/r1])

Define

r1bar = r1 / lo
r2bar = r2 / lo

Then

b = (r1bar^2 - r2bar^2)/(4 Log[r2bar/r1bar])

Define:

r1bar = r2bar/c  (where c >= 1 since r1bar =< r2bar )

bb = b / (r2bar^2)

then

bb = - cc/4

where

cc = (c^2 - 1)/( (c^2) Log[c])

and

b = - ( (r2bar^2) /4 ) *cc

Taking (from Eq. 39 in your paper)

a = (1/Log[r2/r1]) (1/(4*(lo^2)))*((r2^2)*Log[r1/lo] - (r1^2)*Log[r2/lo])

then

a = (1/Log[r2bar / r1bar]) (1/4)*((r2bar^2)*Log[r1bar] - (r1bar ^2)*Log[r2bar])

a = ( (r2bar^2) /4 ) * (Log[r2bar] *cc - 1 )

and we obtain

L = a + b Log[rbar] + (rbar^2) / 4

substituting the expressions for a and b

L = ( (r2bar^2)/4 ) * (Log[r2bar] *cc - 1 ) - ( (r2bar^2)/4) *cc * Log[rbar] + (rbar^2)/4

L = (r2bar^2)/4)*cc*( Log[r2bar] - Log[rbar]) - (r2bar^2 - rbar^2)/4

L = (r2bar^2)/4)*cc* Log[r2/r] - (r2bar^2 - rbar^2)/4

( since ( Log[r2bar] - Log[rbar]) = Log[r2bar/rbar] )



Recall that

rbar =< r2bar   (all material points must be at a radius vector r smaller than or equal to r2)

and that we want to maximize the following quantity:

L = (r2bar^2)/4)*cc* Log[r2/r] - (r2bar^2 - rbar^2)/4

Now, for r=r2,

L(r->r2) = (r2bar^2)/4)*cc* Log[r2/r2]  - (r2bar^2 - r2bar^2)/4
              =0

which is a minimum. While for r=r1,


L(r=r1) = (r2bar^2)/4)*cc* Log[r2/r1]  - (r2bar^2 - r1bar^2)/4
             =(r2bar^2)/4)*cc* Log[r2bar/r1bar]  - (r2bar^2 - r1bar^2)/4
             =(r2bar^2)/4)*cc* Log[c] - (r2bar^2 - r1bar^2)/4
             =(r2bar^2)/4)* (c^2 - 1)/(c^2) - (r2bar^2 - r1bar^2)/4                 

Now, we have the following limits:

For Limit[(c^2 - 1)/(c^2), c -> 1] = 0  (this corresponds to the maximum possible value of r1, r1 ~ r2 )

giving

L(r=r1, r1->r2)  = - (r2bar^2 - r1bar^2)/4 

While Limit[(c^2 - 1)/(c^2), c -> Infinity] = 1  (this corresponds to the minimum possible value of r1, r1 ~ 0 )

gives

L(r=r1, r1->0) = (r2bar^2)/4) (Log[r2/r1] - 1)  where Log[r2/r1] -> Infinity for r1 approaching zero

It is evident we also want to also maximize r2

Therefore in order to maximize L in Equation 39, we want to have:

1) r1 as close as possible to 0
2) r2 as large as possible



This is a long pointy cone. All the interesting action takes place near r=r1 for r1 -> 0

"The geometry comes to the rescue" as you said.

L(r=r2 ) = 0

L(r=r1  r1->0) =(r2bar^2)/4) (Log[r2/r1] -1)


As you said, the "lo" constant is a really huge number, unless the EM Drive happens to be near a magnestar.

We conclude that as Marco, said, the geometry comes to the rescue for L(r=r1  r1->0), a pointy cone
We have at r=r2 L=0, but at r=r1 for r1 approaching zero we have a small number (r2bar^2)/4) multiplying a large number (Log[r2/r1] -> Infinity  if r ~ 0 (a pointy cone))


We need to run some numerical results because Log[r2/r1] goes to Infinity very slowly for r1 approaching zero, so the size of L is very dependent on the magnitude of lo and the magnitude of r1.


« Last Edit: 05/18/2015 03:06 AM by Rodal »

Offline TheTraveller

With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.

Last line say it all:

Quote
If the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.

The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.
You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.

edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.

We were discussing radiation pressure generated from EM waves according to Maxwell's equations (attached). This pressure is apparently subject to cosine angle loss and thus when the EM wave moves along the frustum side walls and is not bouncing off the frustum spherical end plates, there is no Maxwell radiation pressure generated on the frustum side walls.

Or did I not understand what the cosine loss factor is for in the lower of the 2 equations?
It's a geometric optics approximation. Doesn't hold precisely.

For example suppose you got a beam of coherent light, 5mm across, 500nm wavelength, in space. It's not going all parallel to it's original direction, it's spreading, to about 1m across at 10 000m from the source (Makes a fuzzy "Airy disk" pattern).

edit: The Poynting vector on the sides of a perfect cylinder over this beam, is not parallel to the axis. It's this Poynting vector that matters.

Built and used many optical telescopes, including a 12 inch, Schmidt Cassegrain. Diffraction and I got to be good enemies.

There is no natural spread for the beam inside the cavity. It is controlled by the guide and cutoff wavelengths determined by the cavity diameter it travels through and by bouncing off the spherical end plates, which as I see it orient the EM waves so they travel / slide along the side walls at a 0 bounce radiant cosine angle.

I do see that using spherical end plates and getting them highly aligned, pointing at the same vertex, is very necessary to getting an EM Drive to work correctly. If they are out of alignment even say 1 deg, the reflected wave will be not moving along the side wall properly, will partly bounce off it and hit the other end plate as a phase distorted mess.

Is like early days playing with Helium-Neon lasers with adjustable end mirrors. Bloody hard to get the mirrors properly aligned, even with confocal mirrors.
« Last Edit: 05/17/2015 07:28 PM by TheTraveller »
"As for me, I am tormented with an everlasting itch for things remote. I love to sail forbidden seas.
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Offline Notsosureofit

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@RODAL

It seems (to me) that that is quite correct, but that does not mean that that is for max force since it neglects frequency.  I can see how that might be for max laser sideband generation w/o constraining the frequency. (ie very high radial modes)

I know I said I didn't think the cylindrical cavity should show sidebands outside of the cavity but I am rethinking that as the laser propagates in one direction and I'm used to thinking of symmetrical systems.

« Last Edit: 05/17/2015 07:59 PM by Notsosureofit »

Offline StrongGR

...

Dear Jose,

Thanks a lot for spending some time on my calculations. You tried with b but did you check for a?...

http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=830137

Here are the expressions including both a and b, to give a value to L

Assuming r1 =<r2

as per the geometry in the following image




Taking (from Eq. 39 in your paper)

b = (r1^2 - r2^2)/(4 (lo^2) Log[r2/r1])

Define

r1bar = r1 / lo
r2bar = r2 / lo

Then

b = (r1bar^2 - r2bar^2)/(4 Log[r2bar/r1bar])

Define:

r1bar = r2bar/c  (where c >= 1 since r1bar =< r2bar )

bb = b / (r2bar^2)

then

bb = - cc/4

where

cc = (c^2 - 1)/( (c^2) Log[c])

and

b = - ( (r2bar^2) /4 ) *cc

Taking (from Eq. 39 in your paper)

a = (1/Log[r2/r1]) (1/(4*(lo^2)))*((r2^2)*Log[r1/lo] - (r1^2)*Log[r2/lo])

then

a = (1/Log[r2bar / r1bar]) (1/4)*((r2bar^2)*Log[r1bar] - (r1bar ^2)*Log[r2bar])

a = ( (r2bar^2) /4 ) * (Log[r2bar] *cc - 1 )

and we obtain

L = a + b Log[rbar] + (rbar^2) / 4

substituting the expressions for a and b

L = ( (r2bar^2)/4 ) * (Log[r2bar] *cc - 1 ) - ( (r2bar^2)/4) *cc * Log[rbar] + (rbar^2)/4

L = (r2bar^2)/4)*cc*( Log[r2bar] - Log[rbar]) - (r2bar^2 - rbar^2)/4

Limit[cc, c -> 1] = 2  (this corresponds to the maximum possible value of r1, r1 ~ r2 )

Limit[cc, c -> Infinity] = 0  (this corresponds to the minimum possible value of r1, r1 -> 0 )


so for c -> 1,  r1 ~ r2

L = (r2bar^2)/2)( Log[r2bar] - Log[rbar]) - (r2bar^2 - rbar^2)/4

and for c -> Infinity, r1 -> 0

L = - (r2bar^2 - rbar^2)/4



Since

rbar =< r2bar   (all material points must be at a radius vector r smaller than or equal to r2)

We want to maximize the following quantity:

 ( (r2bar^2) /4 ) * (Log[r2bar] *cc

which means that we must have maximum cc

Maximum cc occurs for

Limit[cc, c -> 1] = 2  (this corresponds to the maximum possible value of r1, r1 ~ r2 )

Therefore, for a maximum L we want to have:

r1 as close as possible to r2

This says that the axial length of the truncated cone should be close to zero.In other words, this optimized geometry, according to Eq. 39 in your paper seems to be much closer to Cannae's device.
And it is actually not far from the geometry presently used by Dr. White. It is certainly not the geometry of a pointy cone



As you said, the "lo" constant is a really huge number, unless the EM Drive happens to be near a magnestar.

This means that the variable  ((r2/lo)^2) is very small.

---->However, it get multiplied by Log[ r2/lo] which can be an even larger negative number<----

To be more specific, we need a number for lo, so that we can calculate the term  (r2bar^2)/2)( Log[r2bar] - Log[rbar]) in

L = (r2bar^2)/2)( Log[r2bar] - Log[rbar]) - (r2bar^2 - rbar^2)/4


(The above expression is for r1~r2, such that c~1 and cc~2)

 :)

NOTE: while the condition r1~r2 (r1 as close as possible to r2) still holds (since the smaller r1 with respect to r2 the smaller the term cc), notice that I removed a condition on r2 that I had previously stated.  This is because while to maximize the term (r2bar^2)/2) we also want r2 as large as possible, to maximize (-Log[r2bar]) one wants r2 as small as possible. So to be specific one needs a number for lo.



Dear Jose,

Thank you a lot for your effort to this question. Also thanks to @notsureofit for his help. You are right. I have checked this with Maple and optimized the function L(r) as a function of r1, r2 and U0 also. And yes, r1=r2 is the solution! I have other checks to do and my notebook just keeps on freezing for a computation of these but I hope to complete all in a few days.

This means that Harold White experiments could represent a great leap beyond in experimental general relativity. This was my initial hope. As a theoretical physicist please don't ask me about applications!

Give me a few days for further checks and I will update my draft, with due acknowledgements.

Regards.
« Last Edit: 05/17/2015 07:55 PM by StrongGR »

Offline StrongGR

@RODAL

It seems (to me) that that is quite correct, but that does not mean that that is for max force since it neglects frequency.  I can see how that might be for max laser sideband generation w/o constraining the frequency. (ie very high radial modes)
Thanks.  But where does the frequency enter into the equation for L? all I see is a, b, r2bar, r1bar, lo geometrical parameters.  It seems that the electromagnetic field Power only enters through the parameter (Uo)2 which is built inside the length "lo", so as we said, we need a number for "lo" to make any  more progress.

See Marco's Eq. 19, 20 and 28 in http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=830137

It is the choice of the resonant mode that fixes l0 and this depends on frequency (indirectly). You should check Egan link for some numbers to put in.

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