Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1799593 times)

Offline Flyby

  • Full Member
  • ***
  • Posts: 382
  • Belgium
  • Liked: 447
  • Likes Given: 46
Didn't someone bring up a technical datasheet from the company that sells these premade aluminium cones?
That's the very reason I stopped trying to find the right dimensions based upon the photograph, because it was all there...
it was about.. 200 - 250 pages ago?

I also recall we established that the flanges were so thick because there is a high possibility Shawyer started to experiment with convex and concave surfaces.

Also, note that in that later photo, where he's standing next to a later (dummy?) setup with the Flight Thruster, the thruster itself has a different color then the bottom/top plates.
Either the body has undergone some color changes due to heat, with the top/bottom plates being replaced by different "models" or it has been copper plated.
« Last Edit: 06/11/2015 03:58 PM by Flyby »

Offline aceshigh

  • Full Member
  • ****
  • Posts: 606
  • Liked: 171
  • Likes Given: 16
That 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.

It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...



Well, someone might built a one meter frustum and connect it to a couple megawatts (continuous power) klystron at 300 Mhz but no one will spend such amount of money without clear evidence.

what would be the theorized (so far) thrust of such experimented device?

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1227
  • Do it!
  • Vista, CA
  • Liked: 1293
  • Likes Given: 1743
That 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.

It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...



Well, someone might built a one meter frustum and connect it to a couple megawatts (continuous power) klystron at 300 Mhz but no one will spend such amount of money without clear evidence.

what would be the theorized (so far) thrust of such experimented device?

If my Theory of Operation is correct, damping factor = attenuation/frequency. So a lower frequency should have a lower Q but a higher thrust output, due to higher damping on one side. The greater the differential of phase between the forward and backward standing waves, the higher the power factor, the more work that can be done as thrust.

Todd

Offline RotoSequence

  • Full Member
  • ****
  • Posts: 751
  • Liked: 554
  • Likes Given: 762
Disturbing to se the EM Drive's strange preference of one direction vs. another (reminds me of the issue with NASA turning it around by 180 degrees).

Has anyone checked for a North/South/East/West orientation dependency in the measured thrust? If the quantum vacuum conjectures are correct (or at least on the right track), could it be that the EM drive's measured thrust is affected by frame dragging from the rotation of the Earth?

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
That 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.

It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...



Well, someone might built a one meter frustum and connect it to a couple megawatts (continuous power) klystron at 300 Mhz but no one will spend such amount of money without clear evidence.

what would be the theorized (so far) thrust of such experimented device?

If my Theory of Operation is correct, damping factor = attenuation/frequency. So a lower frequency should have a lower Q but a higher thrust output, due to higher damping on one side. The greater the differential of phase between the forward and backward standing waves, the higher the power factor, the more work that can be done as thrust.

Todd

It has been characterized that Prof. Yang was advised by Shawyer, but it looks that she greatly improved on his results:

                                      Force/InputPower (N/kW)     Q                 ModeShape     Frequency (GHz)
Shawyer Demonstrator    0.080 to 0.243                   45,000         TE012            2.45
Prof.  Yang                      0.9 to 1.07                            1,531         TE012            2.45

Same frequency, same mode shape, but Yang achieved 5 to 10 times greater Force/PowerInput using a Q that was 29 times smaller than Shawyer's



Yang's Q calculated from Fig.5 "frustum microwave cavity actual resonance curve"of "Net thrust measurement of propellantless microwave thrusters", 2011, where Frequency Bandwidth=0.0016GHz, Frequency=2.45 GHz, hence Q=2.45/0.0016=1531. This is the value of Q calculated according to the convention in the West. See definition of quality factor Q . Notice that Prof. Yang reports different values in her tables because of her different convention.
« Last Edit: 06/11/2015 05:23 PM by Rodal »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
Disturbing to se the EM Drive's strange preference of one direction vs. another (reminds me of the issue with NASA turning it around by 180 degrees).

Has anyone checked for a North/South/East/West orientation dependency in the measured thrust? If the quantum vacuum conjectures are correct (or at least on the right track), could it be that the EM drive's measured thrust is affected by frame dragging from the rotation of the Earth?
Nobody has checked this, to my knowledge.

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1227
  • Do it!
  • Vista, CA
  • Liked: 1293
  • Likes Given: 1743
...
We are "guessing" what Yang did with her design and @Rodal suggested she followed Shawyer's DF making the small end equal to (half?) the input wavelength. I'd be surprised if it wasn't 50% smaller than that, given her thrust measurements and low Q values.

What Shawyer suggest for the small end is it should operate just ABOVE cutoff, defined as Guide Wavelength at small end still above zero. That way there is an EM wave to bounce back off the small end plate

That gives max Df as either small end diameter, excitation mode or input frequency will need to be adjusted / chosen to obtain that result.

Next he suggests getting big end as big as possible while obtaining desired internal 1/2 wave multiples of the effective overall guide wavelength to fit inside end plate spacing.

I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations. The difference is, Yang is getting orders of magnitude greater thrust, but she did not disclose the details of her designs. I'm suggesting it is because she made the small end smaller to increase the attenuation due to phase shift. Shawyer's design is not much better than a straight cylinder. It's intended to maximize Q, based on his incorrect theory. Yang's equation "proves", just as Egan's does, that standing waves alone will not produce thrust. I agree with this 100% since they have 0 Power Factor, they can't do "work" to create thrust. All the power is stored energy, like an LC circuit. Yang's equation has only 1 obvious way to create thrust, and that is through evanescent waves that are phase shifted relative to the standing waves. The stored energy can either generate heat or thrust. Asymmetrical attenuation will cause thrust. Copper losses will cause heat. Reduce the losses, but increase the attenuation by making the small end smaller and the frustum longer, "should" result in much higher thrust.

I'll publish this when it's ready, but for now I'm just offering this info to help where I can.

Todd


Offline rfmwguy

  • EmDrive Builder (retired)
  • Senior Member
  • *****
  • Posts: 2165
  • Liked: 2681
  • Likes Given: 1124
Updated Preliminary project page with bill of materials.

Offline Flyby

  • Full Member
  • ***
  • Posts: 382
  • Belgium
  • Liked: 447
  • Likes Given: 46
...... Asymmetrical attenuation will cause thrust. Copper losses will cause heat. Reduce the losses, but increase the attenuation by making the small end smaller and the frustum longer, "should" result in much higher thrust.
Todd
If you continue on that mental path - by making the small end smaller and smaller - you end up with a simple cone shape, no?
Or do the front and back plates have a role to play?

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
History repeats itself. Which reminds me of Leo Szilard in the 1930'es. Skeptics need a proof first.  :(
As it looks now, it could easily happen that the Chinese will provide it...

Shawyer has stated he will release a new peer reviewed Superconducting EMDrive in 2015, created with the assistance of other companies SPR works with. He normally releases at IAC meetings. Next is mid Oct 2015.

Perhaps yes, perhaps not ?

Presentations at conferences like the IAC meetings are not what is usually meant by "peer reviewed publications". [  http://en.wikipedia.org/wiki/Peer_review  ]
« Last Edit: 06/11/2015 05:02 PM by Rodal »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
...... Asymmetrical attenuation will cause thrust. Copper losses will cause heat. Reduce the losses, but increase the attenuation by making the small end smaller and the frustum longer, "should" result in much higher thrust.
Todd
If you continue on that mental path - by making the small end smaller and smaller - you end up with a simple cone shape, no?
Or do the front and back plates have a role to play?
All frequencies (up to infinity) are cut-off in a perfect cone ending at a point.

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations.
I dispute that Yang's maths is correct.
Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.
« Last Edit: 06/11/2015 05:15 PM by deltaMass »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations.
I dispute that Yang's maths is correct.
Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.
Prof. Yang did not publish the actual equations she (may have?) used to model thrust.  How she can get thrust from the equations she shows in her paper is a mystery...

Therefore: no basis to tell whether her equations for thrust are correct or incorrect ...

She invites the reader to think about ions inside the cavity and she displays equations including J terms.  Then she switches the story saying that there are no ions inside the cavity but that the conclusions still apply...

I would have expected the journal referees in her  "peer reviewed paper" to further inquire as to where the thrust comes from.  The word "evanescent waves" or geometrical attenuation are not found in her paper, to my recollection.
« Last Edit: 06/11/2015 05:44 PM by Rodal »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
I'm trying to deduce the thrust from the Baby Emdrive data, and failing.

Let q represent dw/dt or the angular acceleration (always < 0), where
q0 is the acceleration with the drive off = -29.408
q+ is the acceleration with the drive prograde (impeding decay) = -24.987
q- is the acceleration with the drive retrograde (assisting decay) = -35.007
where the numbers have been read off from the interpolated graphical slopes.

Let b = the constant frictional force acting against the motion (Newtons)
Let F = the drive force (Newtons)
Let a = I/R, so that a*q is also in Newtons (torque = moment of inertia * dw/dt)

Then we have 3 equations in 3 unknowns:
1. a q0 = -b
2. a q+ = -b + F
3. a q- = -b - F

For consistency we are forced to have
2 q0 = q+ + q-
and from the interpolated graphical slopes, this is quite nicely the case (58.82 vs. 59.99).

However, we cannot solve for F independently of a or b.
In order to get the value of F, either the friction needs be known (which is the value 'b'), or the moment of inertia of the cavity and its lever arm needs be known (which yields the value 'a').
Note that the numerical values used for q are not yet in the correct (SI) units at this point.

Its difficult for me to wrap my head around what the natural anguiar deceleration should be, which is why I would like to see more of the non-thrusting behavior. On the one hand, if it were due primarily to "smooth" aerodynamics, the decelerating torque should be proportional to the square of the angular rate. On the other hand, the airflow in the jar may be completely chaotic or vortical making prediction impossible. On the third hand, is there a magnetic dampening torque?
We seem to be the only two so far discussing the Baby EmDrive data - thanks for the valuable feedback.

If I take your suggestion that "the decelerating torque should be proportional to the square of the angular rate", then we solve by integration to end up with
w(t) = w0 / (b w0 t + 1)
where b is a frictional constant of the motion. Two comments:
1. The test was done in vacuum, so I don't think this applies. That's why I adopted a friction model that was independent of w.
2. We see to 1st order a constant negative slope of w(t) in all 3 cases (drive off, drive on prograde, drive on retrograde), which is the observation that leads to my equations.

Still, I have provided an w(t) based on your hypothesis. Do you think you can fit it to the data?

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
...
We seem to be the only two so far discussing the Baby EmDrive data - thanks for the valuable feedback.

If I take your suggestion that "the decelerating torque should be proportional to the square of the angular rate", then we solve by integration to end up with
w(t) = w0 / (b w0 t + 1)
where b is a frictional constant of the motion. Two comments:
1. The test was done in vacuum, so I don't think this applies. That's why I adopted a friction model that was independent of w.
2. We see to 1st order a constant negative slope of w(t) in all 3 cases (drive off, drive on prograde, drive on retrograde), which is the observation that leads to my equations.

Still, I have provided an w(t) based on your hypothesis. Do you think you can fit it to the data?
I was not participating since so little is known at this point.

In this message you state "The test was done in vacuum" yet in prior messages others state that it was not done in vacuum.  How do you know it was done in vacuum ?  What was the partial vacuum pressure?

You state you use a friction model.  I thought they had the Baby EM Drive levitated with magnets, so if there is no air, where is the friction coming from in your model ?
« Last Edit: 06/11/2015 05:34 PM by Rodal »

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
...
We seem to be the only two so far discussing the Baby EmDrive data - thanks for the valuable feedback.

If I take your suggestion that "the decelerating torque should be proportional to the square of the angular rate", then we solve by integration to end up with
w(t) = w0 / (b w0 t + 1)
where b is a frictional constant of the motion. Two comments:
1. The test was done in vacuum, so I don't think this applies. That's why I adopted a friction model that was independent of w.
2. We see to 1st order a constant negative slope of w(t) in all 3 cases (drive off, drive on prograde, drive on retrograde), which is the observation that leads to my equations.

Still, I have provided an w(t) based on your hypothesis. Do you think you can fit it to the data?
I was not participating since so little is known at this point.

In this message you state "The test was done in vacuum" yet in prior messages others state that it was not done in vacuum.  How do you know it was done in vacuum ?  What was the partial vacuum pressure?

You state you use a friction model.  I thought they had the Baby EM Drive levitated with magnets, so if there is no air, where is the friction coming from in your model ?
That's not a vacuum chamber? Are you saying that it was not evacuated?

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
...
That's not a vacuum chamber? Are you saying that it was not evacuated?

I'm saying that it appears that there is no basis to assume that there was a partial vacuum inside the transparent container.  Others wrote that it was an enclosure to prevent drafts affecting the experiment.  Ambient pressure.


Notice that the transparent jar is sitting on top of tapes, and there are big air gaps between its bottom and the table' surface:



Quote from: K wrote 9 hours ago
Would be nice to create enclosure from which you'd be able to remove most of the air. Next, initiate spin using external electromagnet pulses. Record the curve of the decay (chart #1). Next, Restart the spinning, this time in the middle of recording (chart #2), turn on the EMD via laser beam (you probably would have to install sensor and indicator for that). Comparison of curve #1 and #2 would be great!

Great work!

The Aachen fellow answered:

Quote from: movax wrote 9 hours ago
Thanx. We got a vacuum chamber, but the levitator is too high, so we need to modify it.
The thruster can be turned on by radio.
« Last Edit: 06/11/2015 05:48 PM by Rodal »

Offline MyronQG

  • Member
  • Posts: 14
  • Liked: 9
  • Likes Given: 5
I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations.
I dispute that Yang's maths is correct.
Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.
Prof. Yang did not publish the actual equations she (may have?) used to model thrust.  How she can get thrust from the equations she shows in her paper is a mystery...

Therefore: no basis to tell whether her equations for thrust are correct or incorrect ...




She invites the reader to think about ions inside the cavity and she displays equations including J terms.  Then she switches the story saying that there are no ions inside the cavity but that the conclusions still apply...

I would have expected the journal referees in her  "peer reviewed paper" to further inquire as to where the thrust comes from.  The word "evanescent waves" or geometrical attenuation are not found in her paper, to my recollection.

In the attached paper Prof. Yang get thrust from Eq. 13, i.e., by integrating the Maxwell's tensor over the thruster's boundary.

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5835
  • USA
  • Liked: 5915
  • Likes Given: 5253
I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations.
I dispute that Yang's maths is correct.
Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.
Prof. Yang did not publish the actual equations she (may have?) used to model thrust.  How she can get thrust from the equations she shows in her paper is a mystery...

Therefore: no basis to tell whether her equations for thrust are correct or incorrect ...




She invites the reader to think about ions inside the cavity and she displays equations including J terms.  Then she switches the story saying that there are no ions inside the cavity but that the conclusions still apply...

I would have expected the journal referees in her  "peer reviewed paper" to further inquire as to where the thrust comes from.  The word "evanescent waves" or geometrical attenuation are not found in her paper, to my recollection.

In the attached paper Prof. Yang get thrust from Eq. 13, i.e., by integrating the Maxwell's tensor over the thruster's boundary.
Eq. 13 is just an integration (over the surfaces) of the energy densities (for the electric and magnetic fields). 

How does she calculate the energy density? If she calculates the energy densities based on the standing waves solution, then the net force results should be zero.

So either:

1. She made an elementary mistake, which would be hard to understand that the referees would let go by

or

2. She calculates the energy density including terms that are not based  on standing waves harmonic solutions). If that is so, she does not disclose what those terms could be.
« Last Edit: 06/11/2015 07:53 PM by Rodal »

Offline WarpTech

  • Full Member
  • ****
  • Posts: 1227
  • Do it!
  • Vista, CA
  • Liked: 1293
  • Likes Given: 1743
I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations.
I dispute that Yang's maths is correct.
Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.

Maxwell's equations predict zero thrust with "standing waves". If you integrate evanescent waves instead of standing waves, Maxwell's equations absolutely do predict thrust. Yang just didn't show what she was plugging in for E and H fields. She is concealing her design details.

Didn't you read that paper I wrote?

Todd

Tags: