Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1805275 times)

Offline txdrive

  • Member
  • Posts: 48
  • Liked: 47
  • Likes Given: 14
...
The equation should be    Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.
I'm using D'Alembert's principle looking at Shawyer's diagram.  His force convention does not follow any of the books I have in Mechanics (the fact that he has these two equal an opposite forces which should result in a body in equilibrium, hence having no acceleration).


Let's say that we instead interpret Shawyer as you suggest.

Then work out the bullet/gun split: one comes up with the accelerations having different signs which I agree is a more conventional view.  If one consistently follows the same convention all the way through, for the bullet and the gun to both have real positive masses, then one ends up with the same result I have above that the mass of the Big End is the negative of the mass of the Small End and that the Total Mass of the EM Drive must be zero, according to Shawyer.

I fear that the available documents from Mr. Shawyer are unusable for any reasonable discussion. Judging by the available reports of a number of groups telling that something seems to or is going on, I feel that Mr. Shawyer might have found something by sheer coincidence. It reminds me of the logical implication that tells us that starting from a wrong premise, any conclusion is possible - even the right one.
;)
Well, what he stumbled across is neither new nor profound. It is the fact that accurate measurement of forces of roughly the magnitude of the radiation pressure upon a surface, in the presence of said radiation heating the surface being irradiated, is tricky.

edit: I think people here really under-estimate conventional physics, and the ease with which one can stumble upon some perfectly normal forces (that either do not work for accelerating a spaceship, or are already used to that end).
« Last Edit: 05/23/2015 11:31 PM by txdrive »

Offline LasJayhawk

It seems like we are trying to apply Newtonian physics to something that may be a quantum level effect.

But the thought had occurred to me that when Iulian got less thrust in the down direction, something else might come into play besides hot air. If the force is between his D.U.T and the floor, the difference might be caused by the inverse square law. By eyeball, the big end to floor distance increased by about 125-150%

Offline Blaine

  • Member
  • Posts: 58
  • Spring Hill, KS
  • Liked: 45
  • Likes Given: 122
It seems like we are trying to apply Newtonian physics to something that may be a quantum level effect.

But the thought had occurred to me that when Iulian got less thrust in the down direction, something else might come into play besides hot air. If the force is between his D.U.T and the floor, the difference might be caused by the inverse square law. By eyeball, the big end to floor distance increased by about 125-150%

Yes, but by intensity of what? What exactly are we dealing with? The inverse square law would only make sense if we had some sort of system going.
« Last Edit: 05/23/2015 11:55 PM by Blaine »
Weird Science!

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5927
  • Likes Given: 5270
Impedance?

locus of 50+j0
I would have to dedicate some time to write some more code to calculate the Impedance and I lost that motivation when I calculated the average Poynting vector due to the dielectric Tan Delta loss, some time ago.

Perhaps you can elaborate on the reasons why to calculate the Impedance.  ;)

Meanwhile I'm looking at the thermodynamics of this thing.

Waiting to see further data from Iulian, Shawyer's superconducting EM  Drive, Cannae's latest, Prof. Yang, and Paul March's.
« Last Edit: 05/24/2015 12:51 AM by Rodal »

Offline Dortex

  • Member
  • Posts: 31
  • United States
  • Liked: 21
  • Likes Given: 12
But the thought had occurred to me that when Iulian got less thrust in the down direction, something else might come into play besides hot air.

Frankly, it's probably mostly hot air in this case. The bigger end is facing up this time, and unless I'm mistaken, the holes are closer the the small end/middle. It can't vent out quite as well as it could last time. I noticed the rsults became unusable after it was turned on a few times.

Offline zaphod_vi

  • Member
  • Posts: 9
  • UK
  • Liked: 9
  • Likes Given: 0
This is a supposed quote from Roger Shawyer that I have copy/pasted.

Quote
What the EmDrive thruster does is to produce a force, which we call the thrust, in one direction. This is a force that you can measure. If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction. So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.

It's a little confusing, but i hope i am not stating the obvious in saying that the interesting part is where he says that if you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. That is, your hand is pushed away from the plate. In which case its behaving similar to a common or garden rocket. i.e. mass is thrown out the back end of the frustrum, which you feel bouncing off your hand. Momentum is conserved, and the frustrum goes in the opposite direction.

As LasJayhawk suggested, it might be an idea if Iulian varied the distance between the thrust plate and the floor if more tests are done.

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5927
  • Likes Given: 5270
Interesting. Maybe he should have a chat with Professor Woodward  8)
We know Prof. Woodward's opinion from the previous threads, reportedly (as conveyed by others):

1) He thinks that the only reliable tests are the ones performed by NASA with the dielectric, and the reason for the measured thrust was due to Woodward/Mach Effect from the HDPE dielectric.  Ditto for Cannae's drive with the PTFE dielectric.

2) He thinks that an EM Drive without a dielectric insert should not be able to produce thrust, because Prof. Woodward thinks that the law of conservation of momentum negates any such thrust in an empty cavity.

3) He does not think that the Quantum Vacuum plasma hypothesis from Dr. White is viable because the QV is immutable and not degradable, and because one cannot push against the QV.
« Last Edit: 05/24/2015 01:10 AM by Rodal »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Is there anyone here who can speak to the care and feeding of tapered fibre lasers?
Thank you.

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5927
  • Likes Given: 5270
This is a supposed quote from Roger Shawyer that I have copy/pasted.

Quote
What the EmDrive thruster does is to produce a force, which we call the thrust, in one direction. This is a force that you can measure. If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction. So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.

It's a little confusing, but i hope i am not stating the obvious in saying that the interesting part is where he says that if you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. That is, your hand is pushed away from the plate. In which case its behaving similar to a common or garden rocket. i.e. mass is thrown out the back end of the frustrum, which you feel bouncing off your hand. Momentum is conserved, and the frustrum goes in the opposite direction.

As LasJayhawk suggested, it might be an idea if Iulian varied the distance between the thrust plate and the floor if more tests are done.
OK, let's think about this. As I understand it, he says that he can feel the force pushing when the BigEnd is not moving.  Let's assume that he very gently places his hand on the BigEnd (because if he pushes it, what he will feel is the inertia of the EM Drive resisting motion).

If the end plate is rigid, then he is feeling some particles. Is he invoking quantum tunneling of photons through the copper? Can he really feel the pressure of photons? I doubt it. Can he feel evanescent waves? I doubt it.

If the end plate is very compliant (a thin copper copper membrane) then he is feeling either vibrations or static bending of the membrane.  This could be produced by classical forces like internal pressure from heated moist air (PV=nRT), or it could be produced by thermal buckling (see my paper), or it could be produced by thermal expansion of a pre-buckled membrane.  Many classical explanations for what he may feel...
« Last Edit: 05/24/2015 01:19 AM by Rodal »

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364
(I step out of the playroom for one minute and ....)

Quotes:


The topology is different.  The wave packet in the EM case has the shape and phase distribution set by the cavity.  If the cavity walls disappeared the trajectory of the wave packet would curve in 4-space (accelerate).  It can't do that because the cavity is still there and has much more mass-equivalent than the wavepacket, so all you see is the reaction force.
Last Edit: 05/23/2015 02:42 PM by Notsosureofit

OK, I would have to work out the math to convince myself that the "walls dissapeared".  If they dissapeared we are in agreement. But to get there I need a proof, as you said  :)
 05/23/2015 02:43 PM by Rodal

You are still too quick for me !
4-D "curve" is acceleration.  The "holographic" representation is 3-D in the EM cavity.  The fixed plane is time.

It should be reducible to a x,y version w/ z,t in the propagation direction (?) but again the walls must disappear for it to propagate.  ??  does the Poynting vector satisfy that condition if the walls are removed ?  Probably not when I try to visualize it. ? although the standing waves are then propagating waves.  Still sounds like you need to integrate all point spherical waves over the cavity volume using their instantaneous amplitude and phase when the walls disappear.

(Sorry about thinking out loud)
Last Edit: 05/23/2015 03:14 PM by Notsosureofit

The only way I can see having a non-zero period-time-averaged Poynting vector in a cavity is either through a nonlinearity (example: Marco Frasca's second order nonlinearity due to GR, or van Tiggelen's 4th order nonlinearity due to magneto-chiral effect), or through an energy gradient (radiative heat transfer, etc.).

The example you gave with the "backbone curve" (as it is known in the literature, where one has a nonlinear spring) is a nonlinearity.
Last Edit: 05/23/2015 03:33 PM by Rodal

Of course, if the Poynting vector stays zero then momentum is conserved.  Is that the case in a self-accelerating wavefunction ?  I havn't seen it explicitly mentioned but they do claim CoM.
05/23/2015 03:19 PM by Notsosureofit

To be specific, let's point out that we are talking about the time-average (over an integer number of periods) of the Poynting vector being zero, as the Poynting vector itself is a non-zero harmonic function of time even as a solution of Maxwell's equations (the Poynting vector in that case having twice the frequency of the electromagnetic field frequency).
Last Edit: 05/23/2015 03:26 PM by Rodal

Yes, only the "disappearance" of the wall for mathematical reasons would be instantaneous.
Last Edit: 05/23/2015 03:31 PM by Notsosureofit

Quote from: Notsosureofit on 05/23/2015 03:57 PM


FYI

Here we go:

http://physics.technion.ac.il/~msegev/publications/Maxwell_accelerating_beams.pdf

"For both TE and TM polarizations, the beams exhibit shape-preserving bending which can
have subwavelength features, and the Poynting vector of the main lobe displays a turn of more than 90"

"of the main lobe"

In our case the cavity keeps the shape from changing, so we see the force necessary to maintain the Poynting vector.


Added:  in the conclusions...

". To complete
the picture, future work should study the possibility of 3D
accelerating beams, including those with trajectories that
do not lie in a single plane. In practical terms, this work
brings accelerating beam optics into the subwavelength
regime, through the less-than-wavelength features of our
solutions, facilitating higher resolution for particle
manipulation."


Last Edit: 05/23/2015 04:32 PM by Notsosureofit

Unquotes:

...


At least on a cylindrical (ie symetrical) cavity, dropping the wall still integrates to a zero Poynting vector over the far-field sphere.  No interesting ramifications yet.

« Last Edit: 05/24/2015 01:37 AM by Notsosureofit »

Offline aero

  • Senior Member
  • *****
  • Posts: 2744
  • 92129
  • Liked: 705
  • Likes Given: 240
Totally off topic, but on a lighter note, if they can get the beams to turn ~90 degrees, does that mean that your laser rifle can shoot around corners?
Retired, working interesting problems

Offline zaphod_vi

  • Member
  • Posts: 9
  • UK
  • Liked: 9
  • Likes Given: 0
So, in a static test, with the unit powered up, he feels the end plate pushing against his hand (the thrust coming out of it and hitting his hand). Does this mean he doesn't feel the end plate pushing against his hand in a moving test, or that he hasn't tried it, or that his hand isn't sensitive enough in this circumstance, or that it is a bit dangerous to try.
« Last Edit: 05/24/2015 01:56 AM by zaphod_vi »

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364
Totally off topic, but on a lighter note, if they can get the beams to turn ~90 degrees, does that mean that your laser rifle can shoot around corners?

Maybe.  They claim to be able to scan the beam of a laser welder (I should look that up)

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5927
  • Likes Given: 5270
(I step out of the playroom for one minute and ....)

....

At least on a cylindrical (ie symetrical) cavity, dropping the wall still integrates to a zero Poynting vector over the far-field sphere.  No interesting ramifications yet.

OK, they get some very interesting, non-intuitive solutions to Maxwell's equations: nondiffracting spatially accelerating solutions, with a Poynting vector that displays a turn of more than 90 degrees. 

Let's say that this would involve an acceleration of the EM Drive when the Poynting vector >0

But then when the Poynting vector < 0 shouldn't the acceleration be in the opposite direction?

And aren't we back to the same situation we are with standing waves in a cavity? : even with standing waves we have a non-zero Poynting vector, the problem is that it keeps switching direction back and forth at a frequency twice as high as the electromagnetic field frequency.

It seems to me like we need a nonlinearity (at least 2nd order) in order for the Poynting vector average to be different from zero.

Or we need a thermodynamic loss that will produce an energy flux preferentially in one direction
« Last Edit: 05/24/2015 01:49 AM by Rodal »

Offline kdhilliard

  • Full Member
  • *
  • Posts: 171
  • Kirk
  • Tanstaa, FL
  • Liked: 162
  • Likes Given: 523
This is a supposed quote from Roger Shawyer that I have copy/pasted.

Quote
What the EmDrive thruster does is to produce a force, which we call the thrust, in one direction. This is a force that you can measure. If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction. So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.

It's a little confusing, ...

The quote is from 0:33 of , and it's not a little confusing; it's a lot confusing.

I'm pretty sure he is not saying that he has actually felt the thrust with his hand from any of his exiting drives, but instead that were you to build a model with higher thrust you could feel it.  In other words, he is arguing that it is real, not imaginary.

Beyond that, confusion lies.  His published theory only deals with radiation pressure imbalance between the inner surfaces of the two end plates, and the resulting net force on the cavity toward the large end plate.  But instead of allowing it to accelerate big end first, he invokes the powers of the phrase "Conservation of Momentum" to declare that this force, which he has already described as a net force on the cavity, is now somehow a thrust which invokes a reaction force on the cavity causing it to accelerate in the opposite direction.

I'm very reluctant to believe that he would mangle the simple mechanics problem this way without a deeper theory, but I've not found any other explanation.

~Kirk

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364


OK, they get some very interesting, non-intuitive solutions to Maxwell's equations: nondiffracting spatially accelerating solutions, with a Poynting vector that displays a turn of more than 90 degrees. 

Let's say that this would involve an acceleration of the EM Drive when the Poynting vector >0

But then when the Poynting vector < 0 shouldn't the acceleration be in the opposite direction?

And aren't we back to the same situation we are with standing waves in a cavity? : even with standing waves we have a non-zero Poynting vector, the problem is that it keeps switching direction back and forth at a frequency twice as high as the electromagnetic field frequency.

It seems to me like we need a nonlinearity (at least 2nd order) in order for the Poynting vector average to be different from zero.

Or we need a thermodynamic loss that will produce an energy flux preferentially in one direction

Dropping the wall (ie letting it propagate) turns out to be the same as replacing w/ a dielectric resonator.  That, of course, radiates.  It "looks-like" the frustum resonator might radiate in a preferred and mode-dependent direction w/ or w/o the Poynting vector integrating to zero over the complete sphere (after all, antennas do it)

The thermodynamic argument also seems to have asymmetrical promise, what with the up-conversion to ambient and the hot end of the cavity (wish we had that for no dielectric)

I was just thinking that the dielectric resonator could not beat the photon rocket, but that is not true, at that instant the PQ is that of the metallic resonator.  The PQ of the dielectric resonator is much lower due to radiation, but that radiation is frustrated in the metallic resonator.  Remember we are talking static force in this example, no work required.
« Last Edit: 05/24/2015 02:10 AM by Notsosureofit »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5927
  • Likes Given: 5270
...
Dropping the wall (ie letting it propagate) turns out to be the same as replacing w/ a dielectric resonator.  That, of course, radiates.  It "looks-like" the frustum resonator might radiate in a preferred and mode-dependent direction w/ or w/o the Poynting vector integrating to zero over the complete sphere (after all, antennas do it)
...
would the preferred direction of radiation occur from the Small End directed towards the Big End ?

and if so, why would this be more effective than an open waveguide?

is it because of the Q factor ?
« Last Edit: 05/24/2015 02:10 AM by Rodal »

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1364
See Edit above

Got to work out the preferred direction

OK, yes the preferred direction is from the Small End directed towards the Big End at least from impedance considerations.  Still don't know if the integrated vector is zero as a dielectric cavity radiates in all directions  (some modes have no radiation exactly on the axis, etc) but it looks to be non-zero so far.

Still doesn't show the momentum balance, we are looking at a differential radiation as it goes to zero.
« Last Edit: 05/24/2015 02:55 AM by Notsosureofit »

Offline aero

  • Senior Member
  • *****
  • Posts: 2744
  • 92129
  • Liked: 705
  • Likes Given: 240
Ok - I posted this image back on page 42 of thread one. That's about 400 pages back.
http://forum.nasaspaceflight.com/index.php?topic=29276.msg1259471#msg1259471
But now we know that "something" carrying energy and momentum can change direction without an externally applied force. That something being the light beams discussed in the paper referenced above. Doesn't that mean that by replacing the electrons in my attached drawing with photons, and the driving field with resonate
RF, that we have a mechanism to cause unbalanced forces on the ends of the cavity?

If someone can figure out how these RF beams could maintain resonance then we're home free.

When the beam bounces off the wall, or glances off the base, if the force (Poynting vector) reverses direction then wouldn't the RF beam retrace it's path in reverse? That allows resonance and we are all very familiar with the relationship of thrust to Q, but wouldn't this mechanism give a maximum upper limit of 1Q as apposed to 2Q as the force equation multiplier factor? Unless it turns through a full 90 degrees +, but to much of that would have the radiation going in circles and never hitting anything.

Added: As for the needed nonlinearity, it is the curving path of the radiation.
« Last Edit: 05/24/2015 02:55 AM by aero »
Retired, working interesting problems

Offline zaphod_vi

  • Member
  • Posts: 9
  • UK
  • Liked: 9
  • Likes Given: 0
Ah, kdhilliard, i think i might have stolen the quote you typed out. However, this is speculation.

Quote
I'm pretty sure he is not saying that he has actually felt the thrust with his hand from any of his exiting drives, but instead that were you to build a model with higher thrust you could feel it

From the video you could take it either way. Either he has, or you would, feel a force. What is odd though is how do you get from a reaction-less drive that operates by bouncing microwaves around in a cavity, to something that behaves more like a traditional rocket. How do you make that leap. Perhaps Shawyer found the rocket like behaviour when he built his prototype, has mashed the two together because he is not quite sure what is going on, and has been hand-waving ever since.

Tags: