Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1872340 times)

Offline dustinthewind

  • Full Member
  • ****
  • Posts: 611
  • U.S. of A.
  • Liked: 246
  • Likes Given: 267
Quote
Note: Shawyer's analogy to a rocket is non-viable because a rocket has variable mass , it is the propellant exiting the rocket (like a bullet exiting a gun results in the gun's recoil force), the variable mass of the rocket, that is responsible for a rocket's acceleration.  The EM Drive is a closed cavity and is described by Shawyer as propellant-less with nothing exiting the EM Drive.

Alternatively, a rocket throws momentum out of its back end. Perhaps this is what Shawyer means when he talks about thrust. If the net force from the microwaves on the cavity is towards the small end, then the cavity must accelerate towards the small end. However being in violation of CoM there must be momentum ejected in the opposite direction, ergo Shawyers thrust. This would then act as a pushing force in the opposite direction. Quite what the ejected momentum consists of is perhaps another matter.

Since the year 1900 by Lebedev (see:  http://web.ihep.su/dbserv/compas/src/lebedev01/eng.pdf ), experiments have been conducted, confirming that Maxwell was correct that the radiation pressure of photons against a surface push the surface in the same direction as the force , such that positive work is performed. 

This (Maxwell's theory and the experiments that have verified it)  is in sharp contrast with Shawyer's theory claiming that the Thrust pressure of photons towards the Big Diameter results in motion of the EM Drive in the opposite direction, towards the Small Diameter .  (This results, as I have shown, in the curious behavior that the Work being done is negative, according to Shawyer's theory)

What about running the current in the aluminum ring in the picture in reverse of the direction the electric field of light would drive the current.  That would take work to do but wouldn't that in effect be reversing the direction of push to pull? 

optical tweezers also strike me as changing the pressure of light from positive to negative.
« Last Edit: 05/24/2015 08:49 PM by dustinthewind »

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1301
  • Do it!
  • Vista, CA
  • Liked: 1350
  • Likes Given: 1813
...
When you say "toward the small end plate", do you mean from the outside or the inside? That's the ambiguity with using that sort of nomenclature.

Once again, I recommend using "small end forward" etc. as the least ambiguous designator of the direction of the resultant thrust vector - the one which produces acceleration.
what matters is what she means:  her words: "net EM ... directs towards the minor end plate". The microwave EM is inside not outside, hence no ambiguity

What she means by "Thrust" is very clear. It is the sum off all the electric and magnetic forces acting the surfaces of the frustum. What Shawyer refers to as thrust is ambiguous, since he says the side walls have no contribution to the forces.

Her use of complex fields incorporates the attenuation factors. With her equations, 12 thru 14, and the graphs in Zeng and Fan, predictions can be extrapolated based on theta. I said I had a lot of reading to do. I wish I had read this one a week ago.

On another note: You might be able to excite a TM01 mode from the side of the frustum using a loop rather than stub. Where the plane of the loop is parallel to the axis of the frustum.

Thanks!
Todd

Offline wallofwolfstreet

  • Full Member
  • *
  • Posts: 165
  • Liked: 169
  • Likes Given: 436
This discussion is clearly interminable, in the literal sense of that word. Whatever experimental results accrue, in past, present or future, there will always be doubt. This is why I am so strongly in favour of a space-based test. I've already laid out my reasons. Without that, I'd lay odds that one could return to this forum in years to come and people would still be arguing the toss.

Let's cut the Gordian Knot!

The test program I  plan to run will remove ALL doubt that the EMDrive generates real propellantless thrust without needing a space test.

One of my goals is to be able to hold the EMDrive & while switching it on and off, to be able to FEEL the thrust.

You may have posted it somewhere before Traveller, but I haven't had the chance to read the entirety of threads 1 and 2.  Do you have a general idea of what kind of timeline you are going to be working off of with your test program?  Do you expect to have a build ready in the next two months, with another month or two for testing, or do you expect it take longer?

Just trying to get a sense of experimental development on this device given we can't expect EW results until at least late July.

Offline Flyby

  • Full Member
  • ***
  • Posts: 385
  • Belgium
  • Liked: 447
  • Likes Given: 48
This discussion is clearly interminable, in the literal sense of that word. Whatever experimental results accrue, in past, present or future, there will always be doubt. This is why I am so strongly in favour of a space-based test. I've already laid out my reasons. Without that, I'd lay odds that one could return to this forum in years to come and people would still be arguing the toss.

Before spending the effort and money to send something into space, we should first try to build something here on earth that works beyond reasonable doubt. This means a thrust greater then what the sum of all possible secondary effects might cause. Personally, I think if a device could generate 50 to 100 gf in a prolonged manner, without noticing any jet/exhausts, I'd consider the EMdrive thrust as very, very likely...
Only then additional research, up there (pointing up), in the hard vacuum would make sense...

Until we first get good solid results here on earth, there is really no reason to send it up... You don't want to spend millions to find out you made an error, or that it simply doesn't work...you most definitely want a real space test, for sure, but not yet at this stage...

Let's wait a bit, till TheTravler get's his stuff ready and the guys at Eagleworks get green light to bring us the (good ?) news.  Let's hope P.March has not been permanently muted... :'(

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1301
  • Do it!
  • Vista, CA
  • Liked: 1350
  • Likes Given: 1813
If Yang "nailed it" with "no new physics", which of Noether or Einstein did she decide to trash? Because you have to choose, and either way, it's new physics. To recap: for the free motion:

1. If P = F v, then magically v is known, which implies a preferred rest frame, and Einstein shrieks
2. If P = F = constant, then free energy is available in profusion, and Noether shrieks.

No new physics?

I choose 1, and Einstein would not shriek. For example. In a gravitational field, if I have 2 identical clocks, held stationary at different altitudes in the field, they do not run at the same speed. How do they know? Because there is a difference in potential energy, i.e. the integrated accelerations along each clocks worldliness are not the same. The resulting "velocity", or in this case, gravitational potential is not the same. When a force F is exerted, and the integral over time results in a velocity v, the refractive index at that potential is not the same as it was in the frame it started from at rest. Regardless if it is an inertial frame when the engine is turned off. You can have 2 inertial frames, no acceleration, that do not have the same relative refractive index. It is referred to as a conformal transformation I believe.

Todd






Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
...
What she means by "Thrust" is very clear. It is the sum off all the electric and magnetic forces acting the surfaces of the frustum. What Shawyer refers to as thrust is ambiguous, since he says the side walls have no contribution to the forces.

Her use of complex fields incorporates the attenuation factors. With her equations, 12 thru 14, and the graphs in Zeng and Fan, predictions can be extrapolated based on theta. I said I had a lot of reading to do. I wish I had read this one a week ago.

On another note: You might be able to excite a TM01 mode from the side of the frustum using a loop rather than stub. Where the plane of the loop is parallel to the axis of the frustum.

Thanks!
Todd

Notice that Prof. Yang writes Section 3.2 page 9 of the translated paper http://www.emdrive.com/NWPU2010translation.pdf :

Quote from: Yang
Mode TM011 thrusters has the worst performance 

this is diametrically opposite to Shawyer's reported aim at TM01

(Recall that TM010 is an impossible mode for truncated cones because p=0 stands for constant electromagnetic field in the longitudinal direction which is impossible for a truncated cone.  Hence it is not immediately clear to me what degenerate mode of TM011 she is referring to as there are two degenerate TM011 modes in a truncated cone near each other)
« Last Edit: 05/24/2015 08:58 PM by Rodal »

Offline flux_capacitor

  • Full Member
  • ****
  • Posts: 563
  • France
  • Liked: 679
  • Likes Given: 927
...That is not new news.  Shawyer has stated the Chinese developed another approach to his many times. End result of both approaches is the same level of measured versus predicted thrust.
Well,let's forget that a short time ago you were stating that Prof Yang was using the same theory as Shawyer, that she was using waveguide modes like TM01, cut off length and other Shawyer stuff, and the fact that she predicts the force in the complete opposite direction as to Shawyer.  But let's forget about that.  How do you know that "both approaches is the same level of measured versus predicted thrust." ?

Do you have dimensions for Prof Yang's EM Drive so that we can assess such a prediction?

Did you already ask her directly? yangjuan@nwpu.edu.cn

I found her email in her published papers and on her profile page on the Northwestern Polytechnical University web site.

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
If Yang "nailed it" with "no new physics", which of Noether or Einstein did she decide to trash? Because you have to choose, and either way, it's new physics. To recap: for the free motion:

1. If P = F v, then magically v is known, which implies a preferred rest frame, and Einstein shrieks
2. If P = F = constant, then free energy is available in profusion, and Noether shrieks.

No new physics?

I choose 1, and Einstein would not shriek. For example. In a gravitational field, if I have 2 identical clocks, held stationary at different altitudes in the field, they do not run at the same speed. How do they know? Because there is a difference in potential energy, i.e. the integrated accelerations along each clocks worldliness are not the same. The resulting "velocity", or in this case, gravitational potential is not the same. When a force F is exerted, and the integral over time results in a velocity v, the refractive index at that potential is not the same as it was in the frame it started from at rest. Regardless if it is an inertial frame when the engine is turned off. You can have 2 inertial frames, no acceleration, that do not have the same relative refractive index. It is referred to as a conformal transformation I believe.

Todd
If you'd be good enough to confine the discussion to a field-free (at least to zeroth and first order) region of space, you will find that you are paying a price to ascribe an absolute value to velocity. As I wrote before, consider this thought experiment:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1370943#msg1370943

EDIT and here are my two "position posts" rolled into one
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1369875#msg1369875
 

« Last Edit: 05/24/2015 09:09 PM by deltaMass »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
...That is not new news.  Shawyer has stated the Chinese developed another approach to his many times. End result of both approaches is the same level of measured versus predicted thrust.
Well,let's forget that a short time ago you were stating that Prof Yang was using the same theory as Shawyer, that she was using waveguide modes like TM01, cut off length and other Shawyer stuff, and the fact that she predicts the force in the complete opposite direction as to Shawyer.  But let's forget about that.  How do you know that "both approaches is the same level of measured versus predicted thrust." ?

Do you have dimensions for Prof Yang's EM Drive so that we can assess such a prediction?

Did you already ask her directly? yangjuan@nwpu.edu.cn

I found her email in her published papers and on her profile page on the Northwestern Polytechnical University web site.

No.  I (at least) never contacted her.  Thanks.

First I have to take a very close look at her paper because I just now realized (hat tip to Todd) that she does take into account the pressure in all faces (upon further reading now I realize that it was only early in the paper when she is going over Shawyer's theory that she refers to neglecting the pressure on the side walls.  She does not neglect those pressures), to understand how she approaches the conservation of momentum issues.

It seems to me at first glance that she addresses it by the heat dissipation which gives direction to the Poynting vector.

EDIT: At this point I don't understand how this can result a closed cavity having a performance thousands of time better than a perfect photon rocket.   ???
« Last Edit: 05/24/2015 09:22 PM by Rodal »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
If Yang "nailed it" with "no new physics", which of Noether or Einstein did she decide to trash? Because you have to choose, and either way, it's new physics. To recap: for the free motion:

1. If P = F v, then magically v is known, which implies a preferred rest frame, and Einstein shrieks
2. If P = F = constant, then free energy is available in profusion, and Noether shrieks.

No new physics?

I choose 1, and Einstein would not shriek. For example. In a gravitational field, if I have 2 identical clocks, held stationary at different altitudes in the field, they do not run at the same speed. How do they know? Because there is a difference in potential energy, i.e. the integrated accelerations along each clocks worldliness are not the same. The resulting "velocity", or in this case, gravitational potential is not the same. When a force F is exerted, and the integral over time results in a velocity v, the refractive index at that potential is not the same as it was in the frame it started from at rest. Regardless if it is an inertial frame when the engine is turned off. You can have 2 inertial frames, no acceleration, that do not have the same relative refractive index. It is referred to as a conformal transformation I believe.

Todd

It also appears that Prof. Yang solves the CoM puzzle by taking into account dissipation losses, which involves the 2nd law of thermodynamics and hence entropy.

Entropy gives a preferred direction (causality).   
« Last Edit: 05/24/2015 09:40 PM by Rodal »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Yang doesn't take the free dynamics bull by the horns.

Also, if she's using heat dissipation as an explanation, how does this system manage  not to be a photon rocket?

Offline Star One

  • Senior Member
  • *****
  • Posts: 8251
  • UK
  • Liked: 1335
  • Likes Given: 168

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1301
  • Do it!
  • Vista, CA
  • Liked: 1350
  • Likes Given: 1813
...
On another note: You might be able to excite a TM01 mode from the side of the frustum using a loop rather than stub. Where the plane of the loop is parallel to the axis of the frustum.


Notice that Prof. Yang writes Section 3.2 page 9 of the translated paper http://www.emdrive.com/NWPU2010translation.pdf :

Quote from: Yang
Mode TM011 thrusters has the worst performance 

this is diametrically opposite to Shawyer's reported aim at TM01

(Recall that TM010 is an impossible mode for truncated cones because p=0 stands for constant electromagnetic field in the longitudinal direction which is impossible for a truncated cone.  Hence it is not immediately clear to me what degenerate mode of TM011 she is referring to as there are two degenerate TM011 modes in a truncated cone near each other)

She said; "(2) The calculation of the different modes and different cavity structure, the mode TM012 which has smallest cavity Large-End has the largest thrust, so has the highest quality factor and thrust. Mode TM011 thrusters has the worst performance."

So a narrower cone angle, and a TM012 mode. What's wrong with that? The TM011 mode is not bad, but requires a larger cone, so of course it will be weaker. I actually think this is theoretically correct.

Note, TM010 mode I think would be the case of a DC current carrying wire running down the axis of the frustum. And TE011 would be a solenoid. (Like the primary winding of a transformer perhaps?) 

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5895
  • USA
  • Liked: 6045
  • Likes Given: 5325
Yang doesn't take the free dynamics bull by the horns.

Also, if she's using heat dissipation as an explanation, how does this system manage  not to be a photon rocket?

Yes, thousands of times better than a perfect photon rocket.  The easy answer is "Q", but Q is the opposite of dissipation.  I think we have to take a look at this in detail.  Already some very interesting things:

1) Yang has the thrust force (from the EM field) directed towards the Small End. This is completely opposite to Shawyer's unorthodox "thrust force".

2) Yang experimentally measured the highest temperature (by far) at the Small End.  (Opposite to NASA, but NASA had a dielectric HDPE at the small end isolating the small end).

3) Effective Q used by Yang in her experiments (calculated the same way as in the West) is significantly lower than Shawyer's Q.

4) Yang writes that Shawyer's preferred mode shape (TM011, notice that TM010 is impossible for a truncated cone) transverse magnetic with m=0, n=1,"has the worst performance" for production of thrust (the complete opposite has been reported for Shawyer in this thread).
(EDIT: but see Todd's note just above this one concerning geometry used in that assessment)

No idea how she can get Force/InputPower thousands of times better than perfect photon rocket.


QUESTION: Has anyone gone over Yang's calculation of the Input Power to make sure she calculates it the same way as in the West ? (I never did)
« Last Edit: 05/24/2015 11:05 PM by Rodal »

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1301
  • Do it!
  • Vista, CA
  • Liked: 1350
  • Likes Given: 1813
If Yang "nailed it" with "no new physics", which of Noether or Einstein did she decide to trash? Because you have to choose, and either way, it's new physics. To recap: for the free motion:

1. If P = F v, then magically v is known, which implies a preferred rest frame, and Einstein shrieks
2. If P = F = constant, then free energy is available in profusion, and Noether shrieks.

No new physics?

I choose 1, and Einstein would not shriek. For example. In a gravitational field, if I have 2 identical clocks, held stationary at different altitudes in the field, they do not run at the same speed. How do they know? Because there is a difference in potential energy, i.e. the integrated accelerations along each clocks worldliness are not the same. The resulting "velocity", or in this case, gravitational potential is not the same. When a force F is exerted, and the integral over time results in a velocity v, the refractive index at that potential is not the same as it was in the frame it started from at rest. Regardless if it is an inertial frame when the engine is turned off. You can have 2 inertial frames, no acceleration, that do not have the same relative refractive index. It is referred to as a conformal transformation I believe.

Todd

It also appears that Prof. Yang solves the CoM puzzle by taking into account dissipation losses, which involves the 2nd law of thermodynamics and hence entropy.

Entropy gives a preferred direction (causality).

Exactly right! I was going to mention where she said;

"|Pe|=|Ph|= Qcavity*Pr= Qcavity*Pinput"

Note that if Pr = 0, . i.e, no dissipation losses, then the integrated forces will be zero, because all energy AND EM momentum will be stored in the capacitance and inductance, like a battery. If you could switch ON, Pr at time t=0, then that battery would discharge asymmetrically, into thrust.

It appears however that she is making the same mistake Shawyer makes, in neglecting the duty cycle. The loaded Q must decrease quickly, or else there will be no thrust.
 

Offline Raj2014

  • Full Member
  • **
  • Posts: 233
  • United Kingdom
  • Liked: 8
  • Likes Given: 20
They have tested it in a vacuum. What will they do next with the EM drive? Will they possibly build a spacecraft with the EM drive to be tested in orbit?

Offline flux_capacitor

  • Full Member
  • ****
  • Posts: 563
  • France
  • Liked: 679
  • Likes Given: 927
1) Yang has the thrust force (from the EM field) directed towards the Small End. This is completely opposite to Shawyer's unorthodox "thrust force".

I'm starting to wonder if Shawyer doesn't use the term "thrust" as something similar to the "thrust the exhaust gas of a rocket would produce if the cavity was, er, a rocket" (i.e. opposite to the direction of movement), and the term "reaction" as the meaning of "direction of the movement of the cavity" relatively to the "virtual" thrust. I'm not sure, since EmDrive does not expel anything. We don't understand what Shawyer tries to explain with his scheme "thrust vs reaction" (CoM evidently, but it is evident only to him).

Everyone else (Eagleworks, NWPU, Cannae LLC, you, me) uses the word "thrust" as the force in the direction of movement, i.e. small end forward.

So if Shawyer use the word "reaction" instead of our thrust, how could we understand each other? And I insist, I'm not even sure what term he chose for the direction of movement. That's so trivially weird :(
« Last Edit: 05/24/2015 10:07 PM by flux_capacitor »

Offline Notsosureofit

  • Full Member
  • ****
  • Posts: 656
  • Liked: 704
  • Likes Given: 1367
If Yang "nailed it" with "no new physics", which of Noether or Einstein did she decide to trash? Because you have to choose, and either way, it's new physics. To recap: for the free motion:

1. If P = F v, then magically v is known, which implies a preferred rest frame, and Einstein shrieks
2. If P = F = constant, then free energy is available in profusion, and Noether shrieks.

No new physics?

No. 2 is not quite correct since all of the "free energy" schemes have circular terms in the Hamiltonian and only the acceleration to spacial seperation looks that way.  Noether snoozes.

Offline aero

  • Senior Member
  • *****
  • Posts: 2784
  • 92129
  • Liked: 724
  • Likes Given: 249
A few pages back, pg. 4, I commented on the possibility that curving RF beams (see http://physics.technion.ac.il/~msegev/publications/Maxwell_accelerating_beams.pdf)
could be responsible for the thrust. I have calculated how much turn the beams would need in order to provide the measured thrust. These are for older experiments, Eagleworks 50 W experiments reported this year are missing, as are all the Canne thrusters including the superconducting thruster.
                                                  Big end incident angle
    Test Data used                          degrees
Shawyer Experimental                   31.4508375
Shawyer Demonstrator                  16.6880148
Yaun a -                                        -0.139288
Yaun b -                                         3.1903738
 Brady - a                                     62.8096712
 Brady - b                                     77.2015432
 Brady - c                                       58.7399415 This is the 2.6 Watt test.
 Shawyer Flight                             -0.0278019

Here, the incident angle of the RF on the small end is assumed to be 90 degrees giving small end force
Fs = 2*P*Q/c * sin(90)  in accordance with Maxwell's equations for a plane wave. The RF radiation within the cavity is assumed to turn as allowed by the mathematics in the referenced paper:
http://physics.technion.ac.il/~msegev/publications/Maxwell_accelerating_beams.pdf

The fly in the ointment is the thermal camera images taken at Eagleworks and confirmed mathematically by Dr. Rodal. The images and theory would seem to allow only small turn angles while the test data, applying my theory gives quite large turn angles. Not impossibly large but large enough that the offset of the heating of the big end would likely have been noticed.

So another hopeful theory damaged by real data. Can't I just pick and choose the data I want to use  :'(         


Edit add: I don't know the uncertainty of the measurement data, force, Q and P but it is quite likely large enough to justify reducing the turn angles to less than 90 degrees for the Yaun a -, and Shawyer Flight cases.
« Last Edit: 05/24/2015 10:24 PM by aero »
Retired, working interesting problems

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1301
  • Do it!
  • Vista, CA
  • Liked: 1350
  • Likes Given: 1813
Yang doesn't take the free dynamics bull by the horns.

Also, if she's using heat dissipation as an explanation, how does this system manage  not to be a photon rocket?

(...)
No idea how she can get Force/InputPower thousands of times better than perfect photon rocket.

(...)

Make Q a cyclic, time dependent, PWM duty cycle controlled function, not a constant!

It charges until it has enough thrust to overcome the resistance of the mass, but once it moves, that energy is dissipated. P drops back to Pin and Q has to build up again. I don't see any other way...




Tags: