Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1797345 times)

Offline WarpTech

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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?

It doesn't look like a vacuum chamber. It looks like a bell jar resting on two rolls of electrical tape.

Offline Prunesquallor

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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?

As others have noted, the test was in a bell jar to (I assume) isolate it from room currents, but I don't think they have pumped it down yet.

My comment about deceleration was just based on that fact that in free stream, drag is proportional to velocity squared, so the drag of each of the components should have caused a decelerating torque (drag x radius) proportional to the square of the angular rate (angular rate2 = v2/r2).  But I doubt the air flow in the jar is anywhere close to free stream, it has probably been propelled into a vortex that is transmitting losses to the boundary layer with the sides of the jar.  It has been too long since aerodynamics courses for me to figure out how to compute the drag in that situation.


So, I can't figure out a way to apply any analytics to the deceleration.  What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing.  THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope.  Then copilot everything normalizing start time to the thruster-on angular velocity.  THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.
« Last Edit: 06/11/2015 06:41 PM by Prunesquallor »
Retired, yet... not

Offline Rodal

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So, I can't figure out a way to apply any analytics to the deceleration.  What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing.  THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope.  Then copilot everything normalizing start time to the thruster-on angular velocity.  THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.

Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube):  http://arxiv.org/pdf/1201.6402

Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see  Eq. 3.4

http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf

or Eq. 13 here http://www.dtic.mil/dtic/tr/fulltext/u2/a365512.pdf

« Last Edit: 06/11/2015 06:55 PM by Rodal »

Offline deltaMass

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So, I can't figure out a way to apply any analytics to the deceleration.  What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing.  THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope.  Then copilot everything normalizing start time to the thruster-on angular velocity.  THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.

Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube):  http://arxiv.org/pdf/1201.6402

Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see  Eq. 3.4

http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf
Power  = Torque * w
So if P = k1 w3, then

F = k2 w2

using constants k.
This is what Prunesquallor also suggests as a model.
But I agree with him that much more experimental data is even better.

« Last Edit: 06/11/2015 07:02 PM by deltaMass »

Offline Rodal

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So, I can't figure out a way to apply any analytics to the deceleration.  What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing.  THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope.  Then copilot everything normalizing start time to the thruster-on angular velocity.  THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.

Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube):  http://arxiv.org/pdf/1201.6402

Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see  Eq. 3.4

http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf
Power  = Torque * w
So if P = k1 w3, then

F = k2 w2

using constants k.

Yes for a the blades in a helicopter.  A little less for the empirical relationship for a disk (since it is 2.8 power instead of 3).

F = k2 w1.8  (for a disk, empirically)

So at this point, to do rough calculations one might as well assume square of velocity (as originally assumed by @prunesquallor).  It will be affected by the walls of the jar, but it is better than assuming a linear relationship with velocity
« Last Edit: 06/11/2015 07:12 PM by Rodal »

Offline Rodal

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Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube):  http://arxiv.org/pdf/1201.6402

Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see  Eq. 3.4

http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf

or Eq. 13 here http://www.dtic.mil/dtic/tr/fulltext/u2/a365512.pdf

The "drag power" is typically D x V IIRC.  Since D is proportional to V2, that makes sense.  I still think the change in angular acceleration is driven by drag (at least, dimensional analysis would say so).
I agree.  Perhaps it should be power of 1.8 (2.8 for the power, then 1.8=2.8-1 for the drag force dependence on velocity) but 2 is a better model than assuming a linear coefficient of friction (power of 1). 
« Last Edit: 06/11/2015 07:11 PM by Rodal »

Offline Fugudaddy

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...... 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?

It matters *where* the wave is cut off, if I am understanding Todd's conjecture correctly.
Shawyer cut his waves off too early resulting in higher 'Q', but lower thrust.
Yang cut her waves off a bit later, resulting in lower 'Q', but (reportedly) higher thrust.
If the shape goes all the way to a cone, all the waves just get chopped up and there's none of that transfer of energy when the wave goes from one form to another.
The idea then of tuning is to find that 'sweet spot' exactly where the waves are cut in just the right way to produce the 'thrust' effect.

Todd- please correct me if I'm wrong, but the idea in even more basic terms is that a wave that moves down this fustrum gains (mass/energy). When the wave is cut off, it 'changes form' (standing vs. evenescent). Normally, the (mass/energy) that the wave gains is then converted right back in whatever scattered direction when a wave is cut or bounced or changed. But if it's done in the right place/time of wave cycle, where the two wave-forms sort of 'meet' and change over, the mass/energy that's given to the new wave does so in a specific direction. So CoM and CoE are conserved, it's just that the EM Drive is 'taking advantage' of that conversion at a point where that energy exchange 'pushes' in a single direction.

Am I even in the ballpark? :D
« Last Edit: 06/11/2015 07:21 PM by Fugudaddy »

Offline Rodal

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Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube):  http://arxiv.org/pdf/1201.6402

Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see  Eq. 3.4

http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf

or Eq. 13 here http://www.dtic.mil/dtic/tr/fulltext/u2/a365512.pdf

The "drag power" is typically D x V IIRC.  Since D is proportional to V2, that makes sense.  I still think the change in angular acceleration is driven by drag (at least, dimensional analysis would say so).
I agree.  Perhaps it should be power of 1.8 (2.8 for the power, then 1.8=2.8-1 for the drag force dependence on velocity) but 2 is a better model than assuming a linear coefficient of friction (power of 1).

This paper: http://physics.wooster.edu/JrIS/Files/grugel.pdf  has data going over the whole range:

from a power of 1 for Stokes flow (very low Reynolds number) F= k omega.
So deltaMass relationship applies for Stokes flow (very low Reynolds number, very slow velocity).

it goes to a power of 2 for Newtonian  F= k omega2

The empirical relationship for the disk falls in between

F= k omega1.8

closer to the Newtonian

This is still missing the effects of the transparent jar inner walls (no slip boundary condition)
« Last Edit: 06/11/2015 08:05 PM by Rodal »

Offline deltaMass

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How does she calculate the energy density? If she calculates the energy densities based on the standing waves solution, then the results (over an integer number of periods) should be zero.
How can Integral[E2] = 0 under any circumstances? (for nonzero E)
« Last Edit: 06/11/2015 07:39 PM by deltaMass »

Offline Rodal

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How does she calculate the energy density? If she calculates the energy densities based on the standing waves solution, then the results (over an integer number of periods) should be zero.
How can Integral[E2] = 0 under any circumstances? (for nonzero E)
Look at the section Energy, pressure and forces

http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html

The reason is because the Maxwell stress tensor is a 2nd order rank tensor, when you  integrate the stress components over the surface you get the vector forces  The net sum of the vector forces are zero because the cavity is a closed surface.

If you think of the energy density as a scalar (which it is, since the components of tensors are scalars) then think of the inner surfaces as having a unit vector normal to the surface, this unit vector appears in the differential of the Integral (it is a surface integral) which gives the directionality of the vector resulting from the integration.

Yang's cavity has a big hole for the waveguide feed, but it leads to another closed chamber, so if the system is not open to the outside, it should form a closed system.



I should have written "then the net force results should be zero" in my message (what I wrote before was referring to the Poynting vector instead). 
« Last Edit: 06/11/2015 08:06 PM by Rodal »

Offline deltaMass

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This means Egan, who showed his work, integrated the Maxwell stress-energy tensor correctly, and it therefore means that Yang, also using  the Maxwell stress-energy tensor, but not showing her work, did it wrong.

Which is kinda what I said to WarpTech.

Offline Rodal

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This means Egan, who showed his work, integrated the Maxwell stress-energy tensor correctly, and it therefore means that Yang, also using  the Maxwell stress-energy tensor, but not showing her work, did it wrong.

Which is kinda what I said to WarpTech.
And if she would have modeled the cavity as having an open hole to the outside (where the waveguide feeds the RF to the cavity), and she ends up with a non-zero net force (let's say because she neglected to model the waveguide), shouldn't the net force be directed opposite to the hole ?  (it would be like a balloon with an open hole).  But this is not a symmetrical balloon, it is a tapered balloon...
It is Greek to me (as to how she ends with a non-zero net force).
« Last Edit: 06/11/2015 08:20 PM by Rodal »

Offline JasonAW3

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Crazy thought here;

     Is it not possible that some sort of inverted gravity force, similar to that that is causing expansion in the universe, may have been accidentally produced in these experiments?  I would imagine that the shape of the EM device itself could act as a sort of focus, projecting the majority of the force in one direction rather than a spherical direction.

     Such a force has been theorized in current physics and could produce such results, pretty much violating Newton's Laws, by simply expanding space in one direction, or conversely, compressing it in the opposite direction.  We pretty much think that that is what happened shortly after the initial Event that produced the universe, so why couldn't this be an extension, by other means, of that same phenomena?
My God!  It's full of universes!

Offline deltaMass

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Von Daeniken, author of more crazy thoughts than you can shake a stick at, was fond of starting his sentences like that too.

"Could it be that <insert preposterous hypothesis here>... ?"

Offline WarpTech

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This means Egan, who showed his work, integrated the Maxwell stress-energy tensor correctly, and it therefore means that Yang, also using  the Maxwell stress-energy tensor, but not showing her work, did it wrong.

Which is kinda what I said to WarpTech.

There is nothing "wrong" about integrating over evanescent waves instead of standing waves. With standing waves, the time-average energy density is a constant, that results in NET-zero force. Where, evanescent waves are exponentially decaying, Zeng & Fan have shown the attenuation is asymmetrical. Therefore, the NET force when integrated over the surface area will be asymmetrical and not NET-zero. It is the same Force equation used by Egan and Yang, but the functions input for E and H are not standing wave sin(wt), cos(wt), the are e-t/T. Where T ~ lambda0/c. If the frustum does not provide at least 1 wavelength past the cut-off diameter, then the wave will be reflected rather than attenuated. That's why Shawyer's design has such a low thrust. His theory is wrong and he's optimizing it for the wrong variable, Q rather than F.

Todd

Offline deltaMass

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Assuming you're right, can you write down your equation for thrust as a function of parameters that can be engineered?

Offline WarpTech

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Assuming you're right, can you write down your equation for thrust as a function of parameters that can be engineered?

The equation is already written by Egan and by Yang. What I don't have is the expressions for E and H, that are dependent on the design of the frustum. I can only say, the function will be of the form E0*e-t/T, where T will be a function dependent on the direction of the unit Normal vectors and the shape of the frustum. If you have software that can crunch Hankel functions, we could determine what they are.

Todd

Offline Rodal

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Assuming you're right, can you write down your equation for thrust as a function of parameters that can be engineered?

The equation is already written by Egan and by Yang. What I don't have is the expressions for E and H, that are dependent on the design of the frustum. I can only say, the function will be of the form E0*e-t/T, where T will be a function dependent on the direction of the unit Normal vectors and the shape of the frustum. If you have software that can crunch Hankel functions, we could determine what they are.

Todd

Well I can crunch Hankel functions all day long.  What I don't get is how one is going to satisfy the boundary conditions for the cavity.

The geometrical attenuation equation derived by the Chinese authors is for an open waveguide, which has no standing waves.  To get standing waves one has to impose Boundary Conditions, as done for example by Egan.

What is the solution (if there is one) that has simultaneous standing waves and evanescent waves and yet it respects the boundary conditions of the problem?

Zeng and Fan obtain the attenuation γ as Hankel functions from the derivative of the Log of the Electric field for a travelling wave

If we take the derivative of the Log of the Electric field standing waves we will not get a correct expression for an evanescent wave.

Are we supposed to assume that some of the travelling waves (with a particular frequency) traveling towards the apex become evanescent waves? and other ones (at another frequency) become standing waves?

I can do that, but then how do you impose a condition that restricts what travelling waves become evanescent waves and which ones form standing waves?

I have not seen an exact solution for this problem, I'm not sure there is one. 
I have only seen solutions for whispering gallery modes coupling with evanescent waves, and some of those solutions show chaos (which is interesting because the Finite Difference numerical solutions obtained by aero did look like chaotic fractal patterns)




It may have to be solved numerically
« Last Edit: 06/11/2015 10:11 PM by Rodal »

Offline deltaMass

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Evanescent waves are a wholly natural phenomenon - for example, the skin depth calculation for e/m waves incident upon a conductor shows a negative exponential decay of field strength inside the conductor. So as such, they cannot violate CoM.  How then are they supposed to explain the EmDrive's violation of CoM?

Offline Rodal

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Evanescent waves are a wholly natural phenomenon - for example, the skin depth calculation for e/m waves incident upon a conductor shows a negative exponential decay of field strength inside the conductor. So as such, they cannot violate CoM.  How then are they supposed to explain the EmDrive's violation of CoM?
Todd is not saying that the evanescent waves violate conservation of momentum (CoM) on the contrary he says that to satisfy CoM the cavity has to move.

Evanescent waves carry momentum.  Evanescent waves outside the cavity, emitted from another object, directed at the cavity, can make the cavity move.

But the problem here is that the evanescent waves are produced inside the cavity: how can internal evanescent waves make the cavity move?

I can make a space object move by throwing balls at the object (with whatever coefficient of restitution, from 0 to 1) from the outside.

I cannot make a space vehicle get a deltaV by throwing balls inside the vehicle at its walls  (with whatever coefficient of restitution, from 0 to 1)
« Last Edit: 06/11/2015 10:19 PM by Rodal »

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