Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1796342 times)

Offline rfmwguy

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Video alert: A couple of weeks ago, NY hosted the world science festival. One of the panel discussions was about our friend, dark energy and the expanding universe. It remains a little understood force, but progress has been made since 1998. Video is long, but panel commentary is very interesting imo. http://pdvod.new.livestream.com/events/00000000003e0055/a7e1cbca-faa0-4b68-8689-6096c858a39e_678.mp4?start=1298&end=6900&__gda__=1434172981_c0fafb8efe0110980ff2e55b4730aa11

Offline Rodal

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@frobnicat, thank you for taking the time to write up a very nice reply and you have some great ideas as well as some nasty problems. You're right on with the thinking that I could increase sensitivity by submerging the whole drive and the problems that would occur because of it. Heating as in a hot tub environment isn't going to happen, sorry that's just for me and... ;)

This was the thing we saw in ASW when we did a frustum shape cavity and submerged it with a little antenna popping out of the water. Power it on and the length on the antenna would change. Totally baffled everyone. 

I'm going to think some more on this and I might have an idea but let me bounce it around between my ears for a bit.

And thanks again,
Shell
@frobnicat's idea has merit: it increases sensitivity. 

Keeping it afloat like a submarine midwater is an issue.

The length on the antenna changing is an issue.

Stronger convection currents are an issue.

Stirring the whole bath, with hard to predict delayed currents bouncing back from the container walls, is an issue.

I don't see inertial effects as an issue, given what we know about the extremely small acceleration (if any) expected from the EM Drive. (And my understanding is that @frobnicat expects the EM Drive acceleration to be zero, as an artifact  :) )

I looked at the inertia effects and considered the added mass as well.  If (and only if) we consider Shawyer's experiment on an air bearing to be representative of the expected inertial effects, we know that the Delta V was a maximum of 2 centimeters/second, and that the RPM of his rig was as slow as watching paint dry: one revolution every 6 minutes

As such, the inertial effects are very small (due to the very low acceleration) -if any- imparted by the EM Drive.  The Reynolds number is very small, so that most likely one will have Stokes flow (flow dominated by viscosity and not by the inertial effects).

Once you know the mass of the drive set-up in your experiment, then the maximum acceleration expected would be

acceleration = 2* F/mass

Where F is the expected force from the EM Drive.  The factor of "2" is the inertial overshoot from steady state.
The forces measured in experiments range from 0.0099 milliNewtons (NASA Eagleworks vacuum, with drive turned around 180 degrees)  to 0.27 Newtons (Prof. Yang's maximum).
« Last Edit: 06/13/2015 01:47 AM by Rodal »

Online WarpTech

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...




(*)  This is also obvious in Todd's picture:



where it is obvious that as θ -> Pi/2, or 2θ -> Pi, the cone angle becomes 180 degrees, the walls of the truncated cone  (denoted as "tapered waveguide" in Todd's drawing) become a line, and the "truncated cone" height goes to zero, such that the truncated cone becomes the base in Todd's picture.  The cylinder drawn by Todd to the right of the cone (denoted as "straight waveguide" in Todd's drawing) is a completely different structure that is not present in Zeng and Fan's discussion.  Zeng and Fan only discuss the cone (there is no cylinder next to it). Keep your focus on the cone in Todd's picture and you will see the cone flatten out to a line as θ -> Pi/2, or 2θ -> Pi.

Jose,

We are talking around each other. Thank you for going through all that. I understand what you are saying about spherical coordinates and r => infinity as theta => 0, gives a cylinder now. However, my drawing represents a traveling wave propagating down a cylindrical hollow waveguide, that is terminated by either a plate, cone, frustum or coupler. This interpretation comes from what Zeng and Fan wrote;

"In recent years, hollow metallic waveguides with a conical taper have been investigated in view of their applicability as couplers."

They are talking about traveling waves in a waveguide, so I drew the diagram as such... In this application, the cylinder is terminated with a conical section called an "attenuator". The half-angle theta of that section goes from a cylinder at theta=0, just as you said above, to a flat plate at theta=pi/2. At any angle in between, you can truncate the cone with another cylindrical wave guide, plate, sphere, cone or whatever, which is what I attempted to show, that the end section could be "anything".

I hope we are on the same page now. :)

The smaller the angle, the greater the phase shift of the reflected wave relative to the standing waves. If the angle is large, the phase difference between a flat-ended cylinder and a truncated cone will be negligible, where a small angle will have a large phase difference between a flat-ended cylinder and a longer truncated cone.
Todd

Offline Rodal

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(*)  This is also obvious in Todd's picture:



where it is obvious that as θ -> Pi/2, or 2θ -> Pi, the cone angle becomes 180 degrees, the walls of the truncated cone  (denoted as "tapered waveguide" in Todd's drawing) become a line, and the "truncated cone" height goes to zero, such that the truncated cone becomes the base in Todd's picture.  The cylinder drawn by Todd to the right of the cone (denoted as "straight waveguide" in Todd's drawing) is a completely different structure that is not present in Zeng and Fan's discussion.  Zeng and Fan only discuss the cone (there is no cylinder next to it). Keep your focus on the cone in Todd's picture and you will see the cone flatten out to a line as θ -> Pi/2, or 2θ -> Pi.

Jose,

We are talking around each other. Thank you for going through all that. I understand what you are saying about spherical coordinates and r => infinity as theta => 0, gives a cylinder now. However, my drawing represents a traveling wave propagating down a cylindrical hollow waveguide, that is terminated by either a plate, cone, frustum or coupler. This interpretation comes from what Zeng and Fan wrote;

"In recent years, hollow metallic waveguides with a conical taper have been investigated in view of their applicability as couplers."

They are talking about traveling waves in a waveguide, so I drew the diagram as such... In this application, the cylinder is terminated with a conical section called an "attenuator". The half-angle theta of that section goes from a cylinder at theta=0, just as you said above, to a flat plate at theta=pi/2. At any angle in between, you can truncate the cone with another cylindrical wave guide, plate, sphere, cone or whatever, which is what I attempted to show, that the end section could be "anything".

I hope we are on the same page now. :)

The smaller the angle, the greater the phase shift of the reflected wave relative to the standing waves. If the angle is large, the phase difference between a flat-ended cylinder and a truncated cone will be negligible, where a small angle will have a large phase difference between a flat-ended cylinder and a longer truncated cone.
Todd

OK, these conclusions apply  :) :

1) A cylinder is the limit for both cone angle going to zero and the spherical radii r1 and r2 going to Infinity.

2) The attenuation (at a location specified by a given cone angle θ and a given spherical radius r) is a function of both the cone angle and the spherical radius r as shown by Zeng and Fan.

3) For spherical radius going to Infinity (which means a cylinder) the attenuation goes to zero.

4) Maximum attenuation occurs for small cone angles and small r1

 :)

Practical example:

Description           r1(m)    r2(m)       Cone Half Angle (degrees)
Shawyer Demo     0.2260   0.4241   19.28
Yang                     0.6953   0.9367     6.159

Observe how Yang's geometry (closer to a cylinder) has the half angle 3 times smaller, and the spherical radius r1 about 3 times longer

Attachment: Zeng and Fan's figures showing that attenuation goes to zero for radius r (horizontal axis) going to Infinity (notice that the vertical and horizontal scales are both non-dimensionalized by the wavenumber k )

Vertical axis: attenuation divided by k
Horizontal axis:  spherical radius r multiplied by k

k = 2 Pi /wavelength = 2 Pi f / c

f= frequency (Hz)
c = speed of light

For large attenuation we want both small k * r and small cone half-angle θ 



« Last Edit: 06/13/2015 02:04 PM by Rodal »

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My interpretation of kr on these diagrams is, 2pi x number of wavelengths. So these diagrams show 5, 10, 15 wavelengths. You can understand that a frustum that is 1/2 wavelength long, does not have enough length to absorb much energy, and it should be several wavelengths long I think. It makes a good resonator, but a very poor thruster.

Todd

Offline aero

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I don't mean to stick my nose into your discussions, but doesn't the half angle also go to zero when height goes to infinity?
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Offline Rodal

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My interpretation of kr on these diagrams is, 2pi x number of wavelengths. So these diagrams show 5, 10, 15 wavelengths. You can understand that a frustum that is 1/2 wavelength long, does not have enough length to absorb much energy, and it should be several wavelengths long I think. It makes a good resonator, but a very poor thruster.

Todd

r is the spherical radial distance from the apex of the cone as defined here:



Don't forget the factor of 2 Pi in the definition of k

k r =  2 Pi r /wavelength

in the horizontal axis is a dimensionless expression of spherical radial distance, so that it can be applied for any size fustrum.

For example,  k r = 5 means

2 Pi r /wavelength = 5

so that it means a spherical radial distance of

r = 5 wavelength / ( 2 Pi )
  = 0.80 wavelength

so:

kr                                   5               10           15
r (wavelengths)            0.80          1.59         2.39


You have to get very close to the apex of the cone to get geometrical attenuation (unless θ is small)

« Last Edit: 06/13/2015 05:26 PM by Rodal »

Offline Rodal

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I don't mean to stick my nose into your discussions, but doesn't the half angle also go to zero when height goes to infinity?
Yes, for non-zero difference between the diameters.  Take a gander at my proof in this message:
http://forum.nasaspaceflight.com/index.php?topic=37642.msg1388641#msg1388641

No for the case approaching a cylinder, where one imposes the conditions of the diameters being equal.

The Limit of a mathematical expression depends on what assumptions are made on the behavior of the parameters in the expression in the limit.

For example, the case of 0/0 (zero over zero)  is Indeterminate in general.  To find out one has to specify the behavior of the numerator and the denominator in the Limit.

From geometry it follows:

Tan[θ] =(Db - Ds)/ (2 h)

where

Db= big diameter of the truncated cone
Ds= small diameter of the truncated cone
h = height (length distance between the flat bases of the truncated cone

For the cone angle going to zero, :

θ -> 0  we have 

Tan[θ] -> 0,

(the tangent of zero degrees is zero) so that the following expression approaches zero:

(Db - Ds)/ (2 h)  -> 0

Now, for the limit case approaching a cylinder, we specify the diameters approaching each other at constant longitudinal distance h between them,: Db = Ds, approaching the diameter of the cylinder, so that the difference is zero (Db - Ds) =0, therefore

0/ (2 h)  -> 0

which is satisfied for any h greater than zero.   

In other words, θ -> 0 will satisfy Db = Ds, the diameter of the cylinder for any h greater than zero.

We are perfectly free to contemplate that Limit behavior. 

What comes out of the analysis is that a cylinder is the Limit not just for the cone angle going to zero, but also for the radii going to infinity keeping a distance h between them.
« Last Edit: 06/13/2015 02:31 AM by Rodal »

Offline aero

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...I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.

Since I went to a bit of trouble installing it, I'll run through the tutorial and the try loading the file you posted. Hopefully I can figure it out.

A question (perhaps the first of many); I assume Meep is solving Maxwell's 4 equations, two static equations (Gauss' law for E and B) and two for electrodynamics.

Do the near-field/evanescent waves result from the contributions of the two static equations?

Does the fine-structure constant have anything to do with the near-field/evanescent waves and Maxwell's equations?

I read some paper posted here about (4-wave mixing?) and photon-photon scattering/acceleration.

Obviously, I don't even understand enough to ask the right question the right way.

I'm sorry, but I'm no expert on the internal operation of Meep. I've found a few background papers but not one that describes FDTD and Meep in the same context.

Here is one paper that may interest you but doesn't answer your specific questions.

http://web.ics.purdue.edu/~pbermel/pdf/Farjadpour09.pdf

And if you have problems operating meep, ask and I'll try to answer .
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Offline Rodal

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...I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.

Since I went to a bit of trouble installing it, I'll run through the tutorial and the try loading the file you posted. Hopefully I can figure it out.

A question (perhaps the first of many); I assume Meep is solving Maxwell's 4 equations, two static equations (Gauss' law for E and B) and two for electrodynamics.

Do the near-field/evanescent waves result from the contributions of the two static equations?

Does the fine-structure constant have anything to do with the near-field/evanescent waves and Maxwell's equations?

I read some paper posted here about (4-wave mixing?) and photon-photon scattering/acceleration.

Obviously, I don't even understand enough to ask the right question the right way.

I'm sorry, but I'm no expert on the internal operation of Meep. I've found a few background papers but not one that describes FDTD and Meep in the same context.

Here is one paper that may interest you but doesn't answer your specific questions.

http://web.ics.purdue.edu/~pbermel/pdf/Farjadpour09.pdf


And if you have problems operating meep, ask and I'll try to answer .


It is a very flexible code, one can do a lot of things with it.

1) In Meep one can write your own constitutive equations  (  http://ab-initio.mit.edu/~meep/meep.pdf  )

2) Meep uses the Finite Difference method in the time domain in order to allow the solution of nonlinear constitutive equations, which is the particular attraction for using Meep in optics, particularly.


I have not used it up to now because:

A) the FD scheme is extremely demanding of computer resources.  It can be mainly justified for nonlinear constitutive equations, that need small time incrementation in the time domain.

B) I haven't seen enough information yet on what one would be modelling with Meep for the EM Drive.  What constitutive equations?  where do we get the constitutive parameters from ? imaginary values of susceptibility?  coupling coefficients? nonlinear constitutive parameters? (not available in the Internet)

@aero discussed the issues associated with the Drude model, etc.

We are still debating how we can have any thrust.

I can move a spacecraft by hitting it with tennis balls from the outside.  Or by using a magnet on it from the outside.

I cannot move a spacecraft by using a magnet on the inside or hitting its internal walls with tennis balls.

I can move something by using evanescent waves from the outside.

I cannot see how to move something by using internal  evanescent waves.

Unless you emit something to the outside:  one needs to either emit mass or energy to the outside to have propulsion.



« Last Edit: 06/13/2015 03:03 AM by Rodal »

Offline Dortex

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2 Pi r /wavelength = 5


I feel compelled to plug Tau in here: http://netalyzr.icsi.berkeley.edu/

Offline aero

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I'm still fooling with meep, trying to get some 3D images and yesterday I had an interesting accident with the code. Using what I thought was a sealed "Bradycone" cavity in 3D, I continued to see extensive RF energy outside the cone. Of course that can not be as we understand things. I made a Force/Power run and meep measured 10+ times ideal photon rocket.

On further investigation i discovered that I had modeled the dielectric shifted in the Y direction - sideways - instead of axially. (Meep uses different coordinates in 3D that in 2D, it seems) So with the dielectric penetrating the center of one of the sidewalls of the cone (still oriented with dielectric and cavity ends parallel) F/P was 10 times higher than typical. OH, and this model won't resonate, that's where my investigation started.

I don't know if the above is meaningful but I found it interesting. I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.

I have Meep up and running on Ubuntu 15.04 and have been posting on thread 2 an image and video of a 2d simulation with a dielectric.  If you point to the latest version of your model I can try to run it.

I centered the dielectric and got meep to calculate zero forces, then increased the dielectric radius to equal the big end radius, so the dielectric penetrated the cone sides all around, (see the attached image). Also attached is an image of the center slice of the 3D cavity showing field patterns after 16 cycles. I ran resonances and did reliably get a low value of Q ~= 100. I then ran forces and meep calculated Force/Flux ~=5 times a photon rocket. I haven't experimented further but my control file is attached (as .txt because NSF doesn't allow .ctl extensions). If you find things you wonder about, ask or PM me and ask, I'll try to explain. And of course, tell me when you find errors.

Oh, I am also running under Ubuntu 15.04 so I expect the attached control file will run on your system.

And would you post a link to "thread 2" as I have no idea where to view your meep posts.
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Offline mwvp

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And if you have problems operating meep, ask and I'll try to answer .

Thanks. I ran a couple examples & looked up your Bradycone.txt file from your recent 5/31 post. Any reason you're using the Harminv rather than the frequency domain solver? It would seem to be a better choice for speed, if all that's desired is the steady-state mode pattern.

Offline apoc2021

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Just a note about DIY professional experiments..

There are several theories of operation and a dearth of experimental data. EW is constrained by a shoestring budget, but collectively our community is not so constrained. If we want to test a new theory or a higher power implementation, we can! One credible DIYer carefully does the build, experiments and data, and the rest of us simply pool our financial resources via Patreon or GoFundMe (or bitcoin for that matter).

I don't know about you guys, but given the importance of this work, I think a few thousand dollars should be seen as a drop in the bucket. I would be eager to directly contribute resources to an experimental setup that would otherwise not be possible, given the chance and a given a credible community member who was willing and able to carry it out.

WarpTech, Rodal and DeltaMass dream team??  ;)
« Last Edit: 06/13/2015 05:11 AM by apoc2021 »

Offline mwvp

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...
A) the FD scheme is extremely demanding of computer resources.  It can be mainly justified for nonlinear constitutive equations, that need small time incrementation in the time domain.

I've noticed the difference between time and frequency domain in Spice. I went to a Comsol seminar last November. Very nice software and user friendly, but was disappointed with the few modes it could handle FDTD, IIRC. But that, AFAIK is true of any code running on a PC with normal resources.

...
B) I haven't seen enough information yet on what one would be modelling with Meep for the EM Drive.  What constitutive equations?  where do we get the constitutive parameters from ? imaginary values of susceptibility?  coupling coefficients? nonlinear constitutive parameters? (not available in the Internet)

Yes, you can hardly simulate a design based on unknown operating principles.

I would find a simulation useful to determine feed placement, type and loading. To predict Q. To map fields and phase, for comparison with a prototype, to know if there is something unexpected going on.

...Unless you emit something to the outside:  one needs to either emit mass or energy to the outside to have propulsion.

That's the question. Is the system open or closed, and how. It open to infrared heat, x-rays and up, gravity.

I had a nasty thought few days back; what if its heating up, detuning, automatic frequency control chirps at it until it limits, and then repeats in a sawtooth cycle, resulting in acoustic vibrations that interact with the scale bearings and environment. This could be one of those flaky phenomena, like LENR, that you could make a career out of getting marginal and intermittent results.

Offline ElizabethGreene

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I now suspect there are two discrete phenomenon at work inside the emDrive.

The first effect is very close to that described in Mr. Sawyer's theory paper.  By ray tracing the paths of photons in a continuously tapered asymmetric waveguide you can see a small force is generated.

The second effect is that discovered by the Nasa team, that a dielectric placed in an otherwise non-thrusting cavity will cause it to generate a force.  I no longer believe this is due to the Sawyer effect.  I believe this second effect is due to a very non-intuitive behavior of radiation pressure.

My physics textbook (Serway) says that Radiation pressure P is equal to the Plank constant times the frequency of light divided by C. (1)

The speed of light in a medium is equal to the speed of light in a vacuum divided by the index of refraction (n) (2)

Substituting 2 into 1 yields:

P= hf/(c/n)
or, to make it more obvious...
(3)  P = hfn/c

Radiation Pressure is increased by the index of refraction of the material surrounding the bounce.

This effect has been confirmed experimentally (Jones, 1978) Radiation pressure is greater if the target mirror is immersed in a material with a higher index of refraction.

(citation)
The Measurement of Optical Radiation Pressure in Dispersive Media
R. V. Jones and B. Leslie
Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences
Vol. 360, No. 1702 (Apr. 4, 1978) , pp. 347-363
Published by: The Royal Society
http://www.jstor.org/stable/79586
(/citation)

Serway also says that radiation pressure P is valid for emission or absorption, but is actually 2P for reflection.  The cause for this is obvious.  A reflection is actually an absorption followed by an emission. (4)

The combined effect of (3) and (4) is, I suspect, the cause for the Nasa/dielectric effect.

Consider a one-dimensional system where a photon is trapped in the vacuum space between two loss-less front-surface mirrors.  it will bounce back and forth practically forever, generating no net thrust.

At the left mirror impact the mirror receives a left push of -1 on absorption, and a left push of -1 on emission.  The acceleration on the left mirror is -2.  When the photons strike the right mirror, +1 for absorption, +1 for emission.  The net thrust is zero.

Now consider the case where the left mirror is immersed in another lossy material with the index of refraction of 1.5.

ActionThrust from ActionNet thrust
(start)0
Transition v-dAbsorption at vacuum-fluid boundary.-1-1
Transition v-dEmission at vac/fluid boundary.1.50.5
Dielectric ReflectionAbsorption at left mirror-1.5-1
Dielectric ReflectionEmission at left mirror.-1.5-2.5
Transition d-vAbsorption at the fluid-vacuum boundary1.5-1
Transition d-vEmission at the fluid-vacuum boundary-1-2
Vacuum ReflectionAbsorption at the right Mirror1-1
Vacuum ReflectionEmission at the right Mirror10
(return to start condition)

The radiation pressure is greater during the phase where the photons are in the lossy dielectric.  Loss of one of these these photons during that phase results in a asymmetric force.  This asymmetry is, I propose, the cause of the Nasa effect.  The index of refraction of PTFE at microwave frequencies is complex, so this absorption and loss can occur.

I propose to confirm the second effect experimentally by creating a traditional round symmetrical microwave resonator and operating it with and without a PTFE endplate inside the existing metal plate.  If I am correct, this circular-non-tapered resonator will generate thrust when the PTFE endplate is in place.

Do I fundamentally misunderstand any concepts here?

Elizabeth Greene
elizabeth.a.greene@gmail.com

Edited: copy and pasted the wrong theory, added table.
« Last Edit: 06/13/2015 07:03 AM by ElizabethGreene »

Offline deltaMass

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Video alert: A couple of weeks ago, NY hosted the world science festival. One of the panel discussions was about our friend, dark energy and the expanding universe. It remains a little understood force, but progress has been made since 1998. Video is long, but panel commentary is very interesting imo. http://pdvod.new.livestream.com/events/00000000003e0055/a7e1cbca-faa0-4b68-8689-6096c858a39e_678.mp4?start=1298&end=6900&__gda__=1434172981_c0fafb8efe0110980ff2e55b4730aa11
Great link - but it's gone bad for reasons unknown. Do you have a replacement?

Offline deuteragenie

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Quote from: aero
And would you post a link to "thread 2" as I have no idea where to view your meep posts.

Here is the video: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=830280

Offline deltaMass

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Re. the Baby EmDrive data: if friction be modelled going as w2, then in the absence of a driving force
w(t) -> e-t
Unfortunately, the shape of that does not fit the shape of the undriven data, which is roughly a straight line of negative slope.
Therefore the friction model isn't right.

Neither does friction going as w fit: it yields
w(t) -> 1/(1 + t)

The friction model which fits the data is constant frictional force, independent of the rotational velocity.
I've already analysed that and shown that the driving force cannot be determined without more data about the experiment. Specifically we need to know either
a) the frictional torque, or
b) the moment of inertia of the cavity platform (I) AND the lever arm (R) of the cavity

In any case, I am abandoning the w2 friction model.
« Last Edit: 06/13/2015 09:02 AM by deltaMass »

Offline Prunesquallor

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Re. the Baby EmDrive data: if friction be modelled going as w2, then in the absence of a driving force
w(t) -> e-t
Unfortunately, the shape of that does not fit the shape of the undriven data, which is roughly a straight line of negative slope.
Therefore the friction model isn't right.

Neither does friction going as w fit: it yields
w(t) -> 1/(1 + t)

The friction model which fits the data is constant frictional force, independent of the rotational velocity.
I've already analysed that and shown that the driving force cannot be determined without more data about the experiment. Specifically we need to know either
a) the frictional torque, or
b) the moment of inertia of the cavity platform (I) AND the lever arm (R) of the cavity

In any case, I am abandoning the w2 friction model.

Too bad it's a dead end, but I'm not surprised. Like I speculated on an earlier post, after the rig has been spinning for a time, the air in that bell jar is probably in some kind of "steady state" vortex losing energy through boundary layer interactions with the jar walls.  Maybe an analogy is a ducted fan.
Retired, yet... not

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