Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1804471 times)

Online Rodal

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Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?
OK, I need your help here Shell.
I understand Zeng and Fan's paper analytically, that the smaller the half-angle the greater the attenuation, but I don't have a physical understanding of this to explain it in simple terms.  Which means that when it comes down to it, I don't understand really understand it   :) .

This is my problem in understanding this:

1) A perfect cylinder has no geometrical attenuation: it has no evanescent waves, no progressive cut-off of modes.  The modes that are cut-off are the ones that have a bigger diameter than the cylinder, that's it.

2) So, how can it be that the biggest geometrical attenuation occurs for the smallest half-angle ? in other words for the geometry that is closest to a cylinder?

3) What happens in the limit with the geometrical attenuation as the cone becomes a cylinder (as the half angle becomes zero)? How does one go from the largest geometrical attenuation for a cone to NO geometrical attenuation for a cylinder?

In Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;

alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.

If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.

Todd

So, your answer boils down to that the explanation is that maximum attenuation occurs when "  the cut-off and the wavelength are close to the same value."

But what's the problem with a larger cone angle?  as I see it the difference is that for a larger cone angle there are extra modes that get cut-off.




EXAMPLE:

Imagine two different truncated cones, both having the same length  and the same small diameter.

The only difference is that truncated cone A has a much larger big diameter than truncated cone B. Therefore cone A has a significantly larger cone angle than cone B.

In cone A, you have other modes that get cut off.   All the modes that get cut off in cone B also get cut of in cone A.

So, why is the geometrical attenuation of cone A worse than the geometrical attenuation of cone B?
« Last Edit: 06/12/2015 07:18 PM by Rodal »

Offline mwvp

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One possible source of drag is eddy currents.   As the Aluminum cavity sweeps through the magnetic field created by the levitator any change in the field strength will induce currents in the Aluminum.   This can happen because of the wobbles seen in the rotation vs time curves.   The Aluminum block moves out of center slightly and the magnetic field strength around it is slightly weaker.   That will introduce Ohmic losses and the rotation rate will decrease faster than it would if there was no metal.

It's an interesting apparatus; very crafty construction.   I hope they collect lots of data from it and see what averages out.

I thought of eddy currents too, though not in the Aluminum block far away from the magnet in a weak field, but in the electromagnet core itself.

There was chat about variable mass and CoE, which brings to mind parametric resonance and pumping. I suspect it would be possible for an unscrupulous hacker to oscillate the magnetic field, such that a ball bearing in a viscous pool in the axis of the device could induce rotation.

Offline mwvp

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I'm currently in bed, dealing with a few long term health issues. Not much EMDrive build work will happen by me for 4 weeks or so.

Very sorry to hear that, I'm especially looking forward to your results, since you've got such (IMHO) a proficient grasp of the hardware and knowledge. Get well soon!

Offline WarpTech

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In Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;

alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.

If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.

Todd

So, your answer boils down to that the explanation is that maximum attenuation occurs when "  the cut-off and the wavelength are close to the same value."

But what's the problem with a larger cone angle?  as I see it the difference is that for a larger cone angle there are extra modes that get cut-off.




EXAMPLE:

Imagine two different truncated cones, both having the same length  and the same small diameter.

The only difference is that truncated cone A has a much larger big diameter than truncated cone B. Therefore cone A has a significantly larger cone angle than cone B.

In cone A, you have other modes that get cut off.   All the modes that get cut off in cone B also get cut of in cone A.

So, why is the geometrical attenuation of cone A worse than the geometrical attenuation of cone B?

I think you are misinterpreting Zeng & Fan, when you think the half-angle of a cylinder is zero. My interpretation would be that the half-angle of a cylinder = pi/2. It has a "Flat" end. As you reduce the cone angle, the "end" of the cylinder becomes pointed, and the permitter of the conical end shifts back toward the large end to become a cone. Then the longer the the cone, the smaller the half-angle. A half-angle of zero is not a cylinder, it's a solid rod.

Now, for a cylinder of length L, with resonant modes, of N*lambda/2 = L, the allowed wavelengths will be;

lambda = 2L/N

Reflections in this case will have phase Beta = k, and will form standing waves.
Attenuation = j(k - Beta) = 0.

In the case of a cone at the end of a cylinder, there will be partial reflections from the side walls that are shorter than L. So the allowed wavelengths will be;

lamda2 = 2(L-dL)/N,

Where, dL is a variable depending on where the wave was reflected along the length of the conical section.

Beta =/= k anymore, so now there will be attenuation. Do you see how that causes a phase shift in Beta, of the reflected wave?

A larger dL value, coming from a smaller half-angle, will generate larger phase shifts on the returning wave, causing destructive interference moving backwards toward the large end. This relieves the pressure on the big end, allowing the small end to feel a force forward.

I'm still working on the hard core mathematics of all this, but you should have a better understanding now, I hope.

Todd







Online Rodal

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In Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;

alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.

If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.

Todd

So, your answer boils down to that the explanation is that maximum attenuation occurs when "  the cut-off and the wavelength are close to the same value."

But what's the problem with a larger cone angle?  as I see it the difference is that for a larger cone angle there are extra modes that get cut-off.




EXAMPLE:

Imagine two different truncated cones, both having the same length  and the same small diameter.

The only difference is that truncated cone A has a much larger big diameter than truncated cone B. Therefore cone A has a significantly larger cone angle than cone B.

In cone A, you have other modes that get cut off.   All the modes that get cut off in cone B also get cut of in cone A.

So, why is the geometrical attenuation of cone A worse than the geometrical attenuation of cone B?

I think you are misinterpreting Zeng & Fan, when you think the half-angle of a cylinder is zero. My interpretation would be that the half-angle of a cylinder = pi/2. It has a "Flat" end. As you reduce the cone angle, the "end" of the cylinder becomes pointed, and the permitter of the conical end shifts back toward the large end to become a cone. Then the longer the the cone, the smaller the half-angle. A half-angle of zero is not a cylinder, it's a solid rod.

Now, for a cylinder of length L, with resonant modes, of N*lambda/2 = L, the allowed wavelengths will be;

lambda = 2L/N

Reflections in this case will have phase Beta = k, and will form standing waves.
Attenuation = j(k - Beta) = 0.

In the case of a cone at the end of a cylinder, there will be partial reflections from the side walls that are shorter than L. So the allowed wavelengths will be;

lamda2 = 2(L-dL)/N,

Where, dL is a variable depending on where the wave was reflected along the length of the conical section.

Beta =/= k anymore, so now there will be attenuation. Do you see how that causes a phase shift in Beta, of the reflected wave?

A larger dL value, coming from a smaller half-angle, will generate larger phase shifts on the returning wave, causing destructive interference moving backwards toward the large end. This relieves the pressure on the big end, allowing the small end to feel a force forward.

I'm still working on the hard core mathematics of all this, but you should have a better understanding now, I hope.

Todd

This is the geometry defined by Zeng and Fan:



here the angle θo is the cone half-angle

it is exactly the same definition as in Egan for θw :



and elsewhere in the literature.

A cylinder is the limit for radius r1 approaching radius r2, both going to infinity

A cylinder has cone half-angle θo =0  and r1-> r2 ->Infinity


A cylinder is not at all the limit for θo  =90 degrees . That limit gives you a line.  It only defines one line, instead of the two walls of a cylinder.




Instead think of what the geometry looks like just before the limit:

1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines.  However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero.

2) Therefore you need to increase both r1 and r2 as the cone angle becomes smaller, in order to get closer to a cylinder.  You need r1 and r2 both going to infinity as the cone angle goes to zero.

As the cone angle goes to zero, the lines become closer to parallel (like in a cylinder).
As r1 and r2 go to Infinity, the end caps become flat, as the end caps of a cylinder are flat.

As the cone angle goes to zero, you need r1 and r2 go to Infinity, so that the base diameters remain finite.  As they go to Infinity, they become the same diameter. And the figure becomes a cylinder.

I think that the resolution of the problem is that one has to consider the fact that r1 and r2 go to Infinity for a cylinder, one cannot ignore r1 and r2 in the solution.  The description of geometrical attenuation solely on the basis of the cone half angle, only works for moderate size r1 and r2.  A cylinder has infinite spherical radius r1 and r2, and this has to be taken into account.

Take a gander at what happens for large r (in the horizontal axis) in the pictures below:  the geometrical attenuation goes to zero.

« Last Edit: 06/13/2015 01:18 AM by Rodal »

Offline deuteragenie

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It is amazing that Yang achieves record thrust force and record thrust force/powerInput by doing the complete opposite of common wisdom:

Maybe this should read "by doing the complete opposite of what I expect" or ".. of what would be expected". 
Wisdom, on the other hand, would commend doing exactly the opposite of what you would expect... once in a while.
This is what Yang did: applying wisdom by doing the complete opposite of what yould be expected.  This is how discoveries are made.

Offline mwvp

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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.

Offline deuteragenie

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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.

Online SeeShells

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As a thought experiment. I've thought of making a Frustum cone collapsing to a point without the small end end plate, but in place of the plate inserting a wire grid where endcap would be.  I'm sure it would work just like a normal Frustum until the waveform collapses into the wire grid and effervescent waves. That's where my thought experiment falls apart.

On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.

Online Rodal

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...On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.
Yes, displacing water acts like a spring that is much stiffer than the torsion of a thin wire for example.  Effectively, you would need a liquid with orders of magnitude smaller density than water to have a spring constant small enough such that the displacement for these small forces would be significant.

Still, water will work if you do get the magnitude of forces obtained by Prof. Yang (0.3 Newtons).
It won't work if you get the forces measured by NASA.

Offline rfmwguy

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As a thought experiment. I've thought of making a Frustum cone collapsing to a point without the small end end plate, but in place of the plate inserting a wire grid where endcap would be.  I'm sure it would work just like a normal Frustum until the waveform collapses into the wire grid and effervescent waves. That's where my thought experiment falls apart.

On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.

I'll let you know how it turns out, both my endcaps are mesh, swappable with solid copper plated boards. Expect first test after July 4th holiday weekend.

Note: spelling police will get you "Evanescent Waves"  ;)

Online SeeShells

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As a thought experiment. I've thought of making a Frustum cone collapsing to a point without the small end end plate, but in place of the plate inserting a wire grid where endcap would be.  I'm sure it would work just like a normal Frustum until the waveform collapses into the wire grid and effervescent waves. That's where my thought experiment falls apart.

On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.

I'll let you know how it turns out, both my endcaps are mesh, swappable with solid copper plated boards. Expect first test after July 4th holiday weekend.

Note: spelling police will get you "Evanescent Waves"  ;)
Silly spelling program, flew right by be, so sorry. Dr. Rodal sent me an email telling me the same. I had to laugh at my mistake and tell him it did sound like Mr. Bubble in a hot tub.

I am very interested in just what you'll find! I expect sometime in July I should have everything together to do my test but no set date yet.

Online SeeShells

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...On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.
Yes, displacing water acts like a spring that is much stiffer than the torsion of a thin wire for example.  Effectively, you would need a liquid with orders of magnitude smaller density than water to have a spring constant small enough such that the displacement for these small forces would be significant.

Still, water will work if you do get the magnitude of forces obtained by Prof. Yang (0.3 Newtons).
It won't work if you get the forces measured by NASA.

True but I need to get some measurement first and if it's within Prof. Yang's scale then I'll do the water.

Shell

Offline WarpTech

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A cylinder is not at all the limit for θo  =90 degrees . That limit gives you a line.  It only defines one line, instead of the two walls of a cylinder.




Instead think of what the geometry looks like just before the limit:

1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines.  However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero.
...

No, you are completely misinterpreting the angle theta. Theta is measured at the "Vertex" of the cone, not the difference between the cone wall and a cylinder wall.

See attached....
Todd

Online Rodal

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A cylinder is not at all the limit for θo  =90 degrees . That limit gives you a line.  It only defines one line, instead of the two walls of a cylinder.




Instead think of what the geometry looks like just before the limit:

1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines.  However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero.
...

No, you are completely misinterpreting the angle theta. Theta is measured at the "Vertex" of the cone, not the difference between the cone wall and a cylinder wall.

See attached....
Todd

Gee, we are not in a good mood today are we :)  (I am completely misrepresenting ?)

I never defined the angle  θ as "the difference between the cone wall and a cylinder wall. "  :)

I even included the figures.

This is their figure for  θ



As θ -> 0  the lines of the cone become parallel, that's what I said.  Look at the figure to see what happens as θ -> 0,

To describe what happens in the limit you need to consider the spherical radius r.

Look at the fact that attenuation is a function of both θ and r in Zeng and Fan's paper, and that for r -> Infinity, the attenuation goes to zero in their figures.

Also look at their equations, and what happens for r -> Infinity.

You cannot ignore the fact that attenuation is a function of r for large r going to Infinity.




« Last Edit: 06/12/2015 08:53 PM by Rodal »

Offline WarpTech

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A cylinder is not at all the limit for θo  =90 degrees . That limit gives you a line.  It only defines one line, instead of the two walls of a cylinder.




Instead think of what the geometry looks like just before the limit:

1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines.  However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero.
...

No, you are completely misinterpreting the angle theta. Theta is measured at the "Vertex" of the cone, not the difference between the cone wall and a cylinder wall.

See attached....
Todd

Gee, we are not in a good mood today are we :)  (I am completely misrepresenting ?)

I never defined the angle  θ as "the difference between the cone wall and a cylinder wall. "  :)

I even included the figures.

This is their figure for  θ



As θ -> 0  the lines of the cone become parallel, that's what I said.  Look at the figure to see what happens as θ -> 0,

To describe what happens in the limit you need to consider the spherical radius r.

Look at the fact that attenuation is a function of both θ and r in Zeng and Fan's paper, and that for r -> Infinity, the attenuation goes to zero in their figures.

Also look at their equations, and what happens for r -> Infinity.

You cannot ignore the fact that attenuation is a function of r for large r going to Infinity.

I apologize if I appear to be rude, it wasn't intended that way. Just trying to squeeze this in "fast" while I'm working.

When I look at the diagram above, I see the following:
As theta => 0, R => 0, there is no waveguide anymore, the opening closes and it becomes a solid bar.

What you are saying is:
As theta => 0, R(zin) = R(z0)

To me, that's not what the diagram shows. To me, as R(z0) => R(zin), theta => pi/2. The vertex of the cone doesn't just "disappear" it gets pulled inward toward the center of the end plate of the cylinder and the angle 2*theta => pi.

Large values of r, are located at the big end. The bigger it is, the less effect the waveguide has on the free-space wavelength.
Todd


EDIT: The difference is, in your interpretation, the end result of theta => 0 is an "infinitely long" cylinder. Where in my interpretation, we have a cylinder of finite length L at theta = pi/2.
« Last Edit: 06/12/2015 09:55 PM by WarpTech »

Offline madsci

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I once during an experiment encountered some non-symmetry that also baffled me.  We had a high resistance volt meter connected to a capacitor that was outside and concentric around a large solenoid.  It had, I think around 180 picofarads and was aluminum.  I could apply DC current through the solenoid in one direction and the voltage would rise on the capacitor and stay that way but decay as charge flowed off slowly.  If I discharged it then reversed the voltage wires so current flowed the other way through the Edit:(solenoid) then give it current the magnitude of increase in the voltage on the capacitor was about an order less.

  That's interesting.
  Do you think that you could reproduce that experiment today ?
  If you can, that might open another promising line of (experimental) research.

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...I apologize if I appear to be rude, it wasn't intended that way. Just trying to squeeze this in "fast" while I'm working.

When I look at the diagram above, I see the following:
As theta => 0, R => 0, there is no waveguide anymore, the opening closes and it becomes a solid bar.

What you are saying is:
As theta => 0, R(zin) = R(z0)

To me, that's not what the diagram shows. To me, as R(z0) => R(zin), theta => pi/2. The vertex of the cone doesn't just "disappear" it gets pulled inward toward the center of the end plate of the cylinder and the angle 2It *theta => pi.

Large values of r, are located at the big end. The bigger it is, the less effect the waveguide has on the free-space wavelength.
Todd

PROOF dedicated to Todd.




SPHERICAL GEOMETRY DEFINITIONS

In spherical coordinates, such as used by Zeng and Fan, and by Egan, defined as follows in the following figures:





In the figures above, the following quantities are defined:

θ = cone's half angle
r1 = spherical radius of small base of truncated cone
r2 = spherical radius of large base of truncated cone

It is trivial to show that:

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

In the following, " ->" is standard notation for "approaching in the Limit"




PROOF THAT IF THE CONE'S HALF ANGLE APPROACHES 90 DEGREES, THE LENGTH OF THE TRUNCATED CONE APPROACHES ZERO


Then, for:

θ -> Pi/2 (the cone half-angle approaching a right angle, as proposed by Todd) 

Tan[θ] -> Infinity,

(tangent of 90 degrees is Infinite) so that the following expression approaches Infinity

(Db - Ds)/ (2 h)  -> Infinity

For finite Db  and Ds this is only possible for h -> 0

So that θ -> Pi/2 (as proposed by Todd)  means a truncated cone with zero height, in other words a line (which is obvious from construction or inspection).  Obviously θ -> Pi/2 does not represent the limit of the truncated cone going to a cylinder  (*)





TO APPROACH A LIMIT CYLINDER, THE SPHERICAL RADIUS MUST APPROACH INFINITY AS THE CONE ANGLE APPROACHES ZERO

On the other hand, 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 a cylinder, Db = Ds, the diameters of both ends of a cylinder are equal: 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




It is also trivial to show that

2 Sin[θ] r1 = Ds  in general

which for  θ -> 0, and therefore  Sin[θ] ->θ  becomes

2 θ r1 = Ds   (which is a trivial re-statement that the arc length becomes equal to the subtended secant)

or

θ r1 = Ds/2

So, in order to converge to a cylinder of diameter   Ds all we need is to keep this ratio constant,

therefore

θ -> 0 means

Ds/(2θ) -> Infinity  or r1 -> Infinity (for finite diameter Ds )

To approach a cylinder, having the cone half angle approach zero is tantamount to the spherical radius of the truncated cone approach Infinity



NECESSARY LIMIT CONSTRAINT FOR THE LIMIT CYLINDER TO HAVE A SPECIFIED LENGTH  (or "height") "h"

It is also trivial to show that

r2 - r1 = (Db - Ds) /(2Sin[θ])

Now, recall that for θ -> 0, it follows that Sin[θ] ~ Tan[θ] ~ θ, therefore for  θ -> 0

r2 - r1 = h




Now, we put the above results together, to gather that indeed,  for

θ -> 0

in order to converge to a cylinder of diameter Db ~ Ds

what is needed is to also simultaneously have r1 approach Infinity such that

θ r1 = Ds/2

and such that

r2 = r1 + h

QED

In words: a cylinder is the limit of a truncated cone, in spherical coordinates such that the cone half angle approaches zero, and simultaneously that the spherical radii r1 and r2 both go to Infinity such that

the product of the cone half-angle and the spherical small radius

θ r1 = Ds/2

is kept constant, equal to half the diameter of the cylinder, and such that r2 is kept larger than r1 by a distance h as both r1 and r2 go to Infinity as follows:

 r2 = r1 + h


This is a general result, for any truncated cone defined in spherical coordinates, r1, r2 and θ.





(*)  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.
« Last Edit: 06/13/2015 12:59 AM by Rodal »

Offline frobnicat

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...On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.
Yes, displacing water acts like a spring that is much stiffer than the torsion of a thin wire for example.  Effectively, you would need a liquid with orders of magnitude smaller density than water to have a spring constant small enough such that the displacement for these small forces would be significant.

Still, water will work if you do get the magnitude of forces obtained by Prof. Yang (0.3 Newtons).
It won't work if you get the forces measured by NASA.

It is easy to lower the apparent "stiffness" of a floating object by immersing the main body and letting emerge just a thin cylinder, as in an alcoholmeter :





Compared to your previous apparent stiffness calculations, there is a gain in sensitivity as the ratio of section of the main cylinder by the section of the small emerging cylinder.

Obviously, the more sensitive it gets (the thinner the emerging cylinder) the more stringent gets the requirement to balance the overall weight so that it floats midwater (or mid alcohol). Needs tuning with leads or other ballast. The sensitivity could in principle be made arbitrarily high (within the limits of surface tension effects), but at some point it will require precise temperature control of water bath to avoid change of density and, ahem, thermal drift of baseline (again) ...

Also the move of a large volume underwater will not only add apparent inertia to the mass of the tested rig, but it will stir the whole bath, with hard to predict delayed currents bouncing back from the container walls. High thermal inertia of the bath is a good point to mitigate high temperature at cavities wall (good dissipation), but higher density and viscosity of water means also stronger convection currents. Maybe using a thermal blanket around ? Could be wet, like open cell foam (neoprene for diving...) so that it's heavy enough not to float like cork.

So I would say, this could be made sensitive enough as far as static displacement/force at equilibrium goes, but I'm afraid this is going to have poor dynamics, high mass (thing has to have same density as water) so high inertia => slow to reach equilibrium, which for an EM drive means running the power for long time, so lots of heat, with a lot of potential spurious delayed effects hard to foretell. Unless explicitly dealing with liquids from the start (characterisation of surface tension...), precision force measurement are usually done in dry set-ups, vacuum ideally, don't they ? Let's stay sober.

Online SeeShells

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...On another note I think my water tank idea for the frustum is trashed. I would need to  increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.
Yes, displacing water acts like a spring that is much stiffer than the torsion of a thin wire for example.  Effectively, you would need a liquid with orders of magnitude smaller density than water to have a spring constant small enough such that the displacement for these small forces would be significant.

Still, water will work if you do get the magnitude of forces obtained by Prof. Yang (0.3 Newtons).
It won't work if you get the forces measured by NASA.

It is easy to lower the apparent "stiffness" of a floating object by immersing the main body and letting emerge just a thin cylinder, as in an alcoholmeter :





Compared to your previous apparent stiffness calculations, there is a gain in sensitivity as the ratio of section of the main cylinder by the section of the small emerging cylinder.

Obviously, the more sensitive it gets (the thinner the emerging cylinder) the more stringent gets the requirement to balance the overall weight so that it floats midwater (or mid alcohol). Needs tuning with leads or other ballast. The sensitivity could in principle be made arbitrarily high (within the limits of surface tension effects), but at some point it will require precise temperature control of water bath to avoid change of density and, ahem, thermal drift of baseline (again) ...

Also the move of a large volume underwater will not only add apparent inertia to the mass of the tested rig, but it will stir the whole bath, with hard to predict delayed currents bouncing back from the container walls. High thermal inertia of the bath is a good point to mitigate high temperature at cavities wall (good dissipation), but higher density and viscosity of water means also stronger convection currents. Maybe using a thermal blanket around ? Could be wet, like open cell foam (neoprene for diving...) so that it's heavy enough not to float like cork.

So I would say, this could be made sensitive enough as far as static displacement/force at equilibrium goes, but I'm afraid this is going to have poor dynamics, high mass (thing has to have same density as water) so high inertia => slow to reach equilibrium, which for an EM drive means running the power for long time, so lots of heat, with a lot of potential spurious delayed effects hard to foretell. Unless explicitly dealing with liquids from the start (characterisation of surface tension...), precision force measurement are usually done in dry set-ups, vacuum ideally, don't they ? Let's stay sober.
@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

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