Author Topic: Resonant Cavity Space-Propulsion: institutional experiments and theory  (Read 59555 times)

Offline X_RaY

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The quote attached is a copy of my post in the EMDrive main thread #9.

The simulations are related to the discussion of the {Q}uality -factor when the small end of a truncated conical cayity resonator is below, at or above the cutoff diameter of a cylindrically waveguide.
To use the related value is frequently stated by TheTraveller(TT) (he says he quotes Shawyer in this regard) and have suggested to take it as a rule, i.e. to make the small diameter equal to the cutoff diameter.
Several explanations where given by TT but nothing conclusive.
For example, that there is no reflection at all if the small end plate diameter is below this cutoff rule. This was debunked as nonsense by Dr.Rodal and others.
As one of lastest "explanations" TT stated that the Q for such a cavity is much smaller  ( or even: "...below anything usefull..") than for a cavity that fits the so called cutoff rule.
The results shown here debunk this to be nonsense also.

Please note that this tells nothing about differences related to (possible) thrust generation.

As you can see in the first two results the situation differs from calculation to calculate. The reason is slightly different mesh size and coupling factors.

Don't forget to include the spherical end-plate frustum with Q of 111,454. That is a pretty significant increase.
Again, at the moment I don't believe in this Q values based on calculations with HOBF. I get freaky inconclusive results when using it.

EDIT
This is the same frustum as used for the Q compare but using HOBF and fine mesh. Instead of natural possible QL~36000, i get 207000 loaded Q! ???
Can EmPro calculate Q?
If yes can you check the results with EmPro?
https://forum.nasaspaceflight.com/index.php?topic=41732.msg1626572#msg1626572

FEKO SE calculation is above, EMPro (FEM Eigenresonance solver) below.
Dimensions are the same, material is copper in both cases. EMPro results should show Q0.

Can you lower the frequency of the one on the right to 2.45 GHz, by expanding the Big diameter? I need to know what those dimensions would be. 2.6 GHz is too far outside the amplifier's range.
Todd I am not sure this is what you like to see. However, the field pattern looks very interesting.
Q0=74382
« Last Edit: 01/22/2017 11:31 AM by X_RaY »

Offline WarpTech

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Can you lower the frequency of the one on the right to 2.45 GHz, by expanding the Big diameter? I need to know what those dimensions would be. 2.6 GHz is too far outside the amplifier's range.
Todd I am not sure this is what you like to see. However, the field pattern looks very interesting.
Q0=74382

Very interesting indeed! I like it. Don't know what will happen, but the fact that the Q when up and the surface current went up is promising. My model and @notsosureofit's both predict that the larger angle should help. It definitely concentrated sidewall losses to one end. Thank you!

Edit: is there a particular location for the antenna, that gives this nice 50 Ohm impedance?
« Last Edit: 01/15/2017 05:37 PM by WarpTech »

Offline X_RaY

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Can you lower the frequency of the one on the right to 2.45 GHz, by expanding the Big diameter? I need to know what those dimensions would be. 2.6 GHz is too far outside the amplifier's range.
Todd I am not sure this is what you like to see. However, the field pattern looks very interesting.
Q0=74382

Very interesting indeed! I like it. Don't know what will happen, but the fact that the Q when up and the surface current went up is promising. My model and @notsosureofit's both predict that the larger angle should help. It definitely concentrated sidewall losses to one end. Thank you!

Edit: is there a particular location for the antenna, that gives this nice 50 Ohm impedance?
I used a loop antenna driven by a voltage source (as Monomorphic did within several models).
Location:
x=0; y=0; z=6.3mm*
loop radius=10mm
wire radius=1mm

*the exact value depends on mesh conditions. I used the finest conditions I can get with FEKO SE.

While thinking on this, I could try to use a finer mesh near the antenna region, but then I must use a little coarser mesh for the rest of the model because of the restrictions of the software I have to deal with. I put this on the list of what to try next days to see what happens. :) 

Online Monomorphic

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I've set up all the formulas into FEKO so I only have to adjust the three dimensions BIG_D, SMALL_D, and FRUST_H - and FEKO generates the spherical end-plates. It is vastly easier to check different dims now. The axially mounted loop antenna is also controlled via the three FEED variables.

A bug that was fixed: FEKO's parabolic function uses the focal length, which is 1/2 radius. This is not correct for spherical end-plates since the curve depth is apex radius length. I had to use the curve depth and diameter in the sagitta focal length equation to get the quasi focal length to input into FEKO so it generates the proper curve with proper depth: Focal length = (D^2) / (16 * depth)   ***credit goes to TT for providing that***


Online Monomorphic

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X_Ray noted earlier a sphere at Vertex didn't align correctly. That was fixed. Once meshed, a sphere with radius Big_R aligns perfectly now. Thanks X_Ray!

Offline WarpTech

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The quote attached is a copy of my post in the EMDrive main thread #9.

The simulations are related to the discussion of the {Q}uality -factor when the small end of a truncated conical cayity resonator is below, at or above the cutoff diameter of a cylindrically waveguide.
To use the related value is frequently stated by TheTraveller(TT) (he says he quotes Shawyer in this regard) and have suggested to take it as a rule, i.e. to make the small diameter equal to the cutoff diameter.
Several explanations where given by TT but nothing conclusive.
For example, that there is no reflection at all if the small end plate diameter is below this cutoff rule. This was debunked as nonsense by Dr.Rodal and others.
As one of lastest "explanations" TT stated that the Q for such a cavity is much smaller  ( or even: "...below anything usefull..") than for a cavity that fits the so called cutoff rule.
The results shown here debunk this to be nonsense also.

Please note that this tells nothing about differences related to (possible) thrust generation.

As you can see in the first two results the situation differs from calculation to calculate. The reason is slightly different mesh size and coupling factors.

Don't forget to include the spherical end-plate frustum with Q of 111,454. That is a pretty significant increase.
Again, at the moment I don't believe in this Q values based on calculations with HOBF. I get freaky inconclusive results when using it.

EDIT
This is the same frustum as used for the Q compare but using HOBF and fine mesh. Instead of natural possible QL~36000, i get 207000 loaded Q! ???
Can EmPro calculate Q?
If yes can you check the results with EmPro?
https://forum.nasaspaceflight.com/index.php?topic=41732.msg1626572#msg1626572

FEKO SE calculation is above, EMPro (FEM Eigenresonance solver) below.
Dimensions are the same, material is copper in both cases. EMPro results should show Q0.

Can you lower the frequency of the one on the right to 2.45 GHz, by expanding the Big diameter? I need to know what those dimensions would be. 2.6 GHz is too far outside the amplifier's range.
Todd I am not sure this is what you like to see. However, the field pattern looks very interesting.
Q0=74382

Can you simply scale the frustum on the right, of "your" original image, until the frequency is 2.45GHz please? Just to compare to my wide base idea. It's hard to choose which one I want to spend money on.

Thanks.

Offline X_RaY

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The quote attached is a copy of my post in the EMDrive main thread #9.

The simulations are related to the discussion of the {Q}uality -factor when the small end of a truncated conical cayity resonator is below, at or above the cutoff diameter of a cylindrically waveguide.
To use the related value is frequently stated by TheTraveller(TT) (he says he quotes Shawyer in this regard) and have suggested to take it as a rule, i.e. to make the small diameter equal to the cutoff diameter.
Several explanations where given by TT but nothing conclusive.
For example, that there is no reflection at all if the small end plate diameter is below this cutoff rule. This was debunked as nonsense by Dr.Rodal and others.
As one of lastest "explanations" TT stated that the Q for such a cavity is much smaller  ( or even: "...below anything usefull..") than for a cavity that fits the so called cutoff rule.
The results shown here debunk this to be nonsense also.

Please note that this tells nothing about differences related to (possible) thrust generation.

As you can see in the first two results the situation differs from calculation to calculate. The reason is slightly different mesh size and coupling factors.

Don't forget to include the spherical end-plate frustum with Q of 111,454. That is a pretty significant increase.
Again, at the moment I don't believe in this Q values based on calculations with HOBF. I get freaky inconclusive results when using it.

EDIT
This is the same frustum as used for the Q compare but using HOBF and fine mesh. Instead of natural possible QL~36000, i get 207000 loaded Q! ???
Can EmPro calculate Q?
If yes can you check the results with EmPro?
https://forum.nasaspaceflight.com/index.php?topic=41732.msg1626572#msg1626572

FEKO SE calculation is above, EMPro (FEM Eigenresonance solver) below.
Dimensions are the same, material is copper in both cases. EMPro results should show Q0.

Can you lower the frequency of the one on the right to 2.45 GHz, by expanding the Big diameter? I need to know what those dimensions would be. 2.6 GHz is too far outside the amplifier's range.
Todd I am not sure this is what you like to see. However, the field pattern looks very interesting.
Q0=74382

Can you simply scale the frustum on the right, of "your" original image, until the frequency is 2.45GHz please? Just to compare to my wide base idea. It's hard to choose which one I want to spend money on.

Thanks.
I have parameterized all dimensions now, therefore I can scale the whole thing by changing one number only*. After increasing the frustum step by step, it matched 2.45GHz almost exactly.

*This means all dimensions(!) including the antenna radius, antenna wire radius and its position, as well as the frustum size. All of this depend on this simple scaling factor.

Regards
« Last Edit: 01/20/2017 03:43 PM by X_RaY »

Offline WarpTech

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I have parameterized all dimensions now, therefore I can scale the whole thing by changing one number only*. After increasing the frustum step by step, it matched 2.45GHz almost exactly.

*This means all dimensions(!) including the antenna radius, antenna wire radius and its position, as well as the frustum size. All of this depend on this simple scaling factor.

Regards

That's cool! I would be curious then, how the frequency transforms with general scaling. Such as;

1:1
1:2
1:3
1:4

Does doubling the size give 1/2 the frequency, or 1/4th the frequency?


Offline X_RaY

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I have parameterized all dimensions now, therefore I can scale the whole thing by changing one number only*. After increasing the frustum step by step, it matched 2.45GHz almost exactly.

*This means all dimensions(!) including the antenna radius, antenna wire radius and its position, as well as the frustum size. All of this depend on this simple scaling factor.

Regards

That's cool! I would be curious then, how the frequency transforms with general scaling. Such as;

1:1
1:2
1:3
1:4

Does doubling the size give 1/2 the frequency, or 1/4th the frequency?

Spreadsheet data:

Scale factor      f(GHz)
4:1                  9,7764943955
3:1                  7,3323707966
2:1                  4,8882471977
1:1                  2,4441235989
 1:2                1,2220617994
1:3                  0,8147078663
1:4                  0,6110308997

As you can see your first idea is true  "doubling the size give 1/2 the frequency" :)
Also please look at this post of Dr. Rodal https://forum.nasaspaceflight.com/index.php?topic=39214.msg1474351#msg1474351
« Last Edit: 01/20/2017 09:24 PM by X_RaY »

Offline X_RaY

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The quote attached is a copy of my post in the EMDrive main thread #9.

The simulations are related to the discussion of the {Q}uality -factor when the small end of a truncated conical cayity resonator is below, at or above the cutoff diameter of a cylindrically waveguide.
To use the related value is frequently stated by TheTraveller(TT) (he says he quotes Shawyer in this regard) and have suggested to take it as a rule, i.e. to make the small diameter equal to the cutoff diameter.
Several explanations where given by TT but nothing conclusive.
For example, that there is no reflection at all if the small end plate diameter is below this cutoff rule. This was debunked as nonsense by Dr.Rodal and others.
As one of lastest "explanations" TT stated that the Q for such a cavity is much smaller  ( or even: "...below anything usefull..") than for a cavity that fits the so called cutoff rule.
The results shown here debunk this to be nonsense also.

Please note that this tells nothing about differences related to (possible) thrust generation.

As you can see in the first two results the situation differs from calculation to calculate. The reason is slightly different mesh size and coupling factors.

Don't forget to include the spherical end-plate frustum with Q of 111,454. That is a pretty significant increase.
Again, at the moment I don't believe in this Q values based on calculations with HOBF. I get freaky inconclusive results when using it.

EDIT
This is the same frustum as used for the Q compare but using HOBF and fine mesh. Instead of natural possible QL~36000, i get 207000 loaded Q! ???
Can EmPro calculate Q?
If yes can you check the results with EmPro?
https://forum.nasaspaceflight.com/index.php?topic=41732.msg1626572#msg1626572

FEKO SE calculation is above, EMPro (FEM Eigenresonance solver) below.
Dimensions are the same, material is copper in both cases. EMPro results should show Q0.

Can you lower the frequency of the one on the right to 2.45 GHz, by expanding the Big diameter? I need to know what those dimensions would be. 2.6 GHz is too far outside the amplifier's range.
Todd I am not sure this is what you like to see. However, the field pattern looks very interesting.
Q0=74382

Can you simply scale the frustum on the right, of "your" original image, until the frequency is 2.45GHz please? Just to compare to my wide base idea. It's hard to choose which one I want to spend money on.

Thanks.
I have parameterized all dimensions now, therefore I can scale the whole thing by changing one number only*. After increasing the frustum step by step, it matched 2.45GHz almost exactly.

*This means all dimensions(!) including the antenna radius, antenna wire radius and its position, as well as the frustum size. All of this depend on this simple scaling factor.

Regards
The geometrie as shown here should be fine to go with, other modes are ~50MHz away.
« Last Edit: 01/21/2017 06:49 PM by X_RaY »

Offline Rodal

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This is a formal paper that summarizes and updates the previous posts:

https://forum.nasaspaceflight.com/index.php?topic=39214.msg1474347#msg1474347
and
https://forum.nasaspaceflight.com/index.php?topic=39214.msg1474351#msg1474351

Here I formally prove that the thrust force per input power (for all three EM-Drive theories) scales like the square root of any geometrical dimension, for constant resistivity and magnetic permeability of the interior wall of the cavity and for constant geometrical ratios, constant medium properties and for the same mode shape. To maximize the thrust per input power, according to all three theories the most efficient EM-Drive would be as large as possible, this being due to the fact that the quality of factor of resonance Q (all else being equal) scales like the square root of the geometrical dimensions. Small cavity EM-Drives (all else being equal) are predicted to have smaller quality of resonance Q and therefore smaller thrust force/input power.

Paper with proof can be downloaded from here:

https://www.researchgate.net/project/Assessing-the-EM-Drive-claims
« Last Edit: 02/04/2017 06:43 PM by Rodal »

Offline rfmwguy

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« Last Edit: 02/06/2017 08:13 PM by rfmwguy »

Offline Rodal

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EM DRIVE TESTING in CAVENDISH PENDULUM:
PROOF that CHANGE in WIRE's CROSS-SECTIONAL AREA is NEGLIGIBLE for CAVENDISH TORSIONAL PENDULUM'S MUSIC WIRE FOR LOADS that RESULT in STRESS EQUAL TO OR BELOW YIELD STRESS


I had promised to post this proof that the concern expressed by some that the change in area was relevant is not borne by the facts.

Of course, for those knowledgeable about mechanics of materials, this is obvious, since it is well-known to Engineers that design mechanical and aerospace structures with metals, that elastic strains in metals are infinitesimal and hence such changes in diameter due to longitudinal strain are negligible.  I proceed to prove this for this case, and provide specific numerical examples for Monomorphic's rig.

Since some people expressed a concern with change in area, this is a concern about the strains not being infinitesimal, hence we should use finite strain and finite stress measures to explore the validity of this concern.  Therefore I use the "true strain" (Hencky or Logarithmic measure) and the "true stress" (Cauchy measure: force divided by actual present cross sectional area).

Transverse strain

Since the torsional wire is made of steel, the material is isotropic.  Using Cartesian coordinates, with z being the longitudinal direction of the wire, and x and y are the transverse directions, the transverse (Hencky or logarithmic measure of) strains εx  in the x direction and εy in the y direction are:

εx = εy  = ln (D/Do)

where Do is the undeformed diameter of the wire and D is the deformed diameter of the wire

therefore the ratio of the deformed to the undeformed diameter of the wire

D/Do = eεx =  eεy

Area ratio


The ratio of the present cross-sectional area "A" of the wire to the original cross-sectional area "Ao" of the wire is:

A/Ao = (D/Do)2

substituting the expression for the diameter ratio in terms of the transverse strains:

A/Ao = ex =  ey

Longitudinal strain

From the Poisson's ratio "ν" (Greek letter nu) definition, the longitudinal strain is:

εx = εy= - ν εz

substituting this equation into the expression for the area ratio, one gets an expression for the ratio A/Ao of the deformed to undeformed cross-sectional area as a function of the longitudinal strain  εz:

A/Ao =  e- 2 ν εz

Yield stress

Using the "true stress" (Cauchy stress: ratio of force to present cross-sectional area) value of the yield stress:

σz = E  εz

where E=modulus of elasticity. Therefore the yield stress is related to yield strain as follows:

σz YIELD = E  εz YIELD

and therefore the ratio of the cross-sectional area deformed at yield, to the undeformed cross-sectional area is:

A/Ao =  e- 2 ν σz YIELD/E




Torsional stiffness

The torsional stiffness "k" of the wire is defined as follows, as the constant relating the torsional moment to the angle of rotation of the torsional pendulum:

T = k θ

T=Torsional moment (Newton-meter or pound force-inch)

θ= angle of rotation (radians)

k = torsional spring stiffness (Newton-meter per radian or inch-pound force per radian)

where

k = (Pi * D4/(32 L))  E/(2(1+ν))

L= length of torsional wire between support and hanging mass

Therefore, expressing this in terms of the stress-free diameter Do of the wire:

k = (Pi * Do4/(32 L))  E/(2(1+ν)) (D/Do)4

and since it was previously shown that:

A/Ao =  e- 2 ν σz YIELD/E
        = (D/Do)2

it follows that

k = (Pi * Do4/(32 L))  E/(2(1+ν)) (D/Do)4
   = ko  e- 4 ν σz YIELD/E

where "ko" is the torsional stiffness based on the original stress-free diameter: the torsional stiffness for infinitesimally small load

ko= (Pi * Do4/(32 L))  E/(2(1+ν))

and the ratio of the actual torsional stiffness at yield to the torsional stiffness under no load is simply:

k/ ko = e- 4 ν σz YIELD/E

FREQUENCY

The angular frequency is

ω = √(k/I)

and therefore the natural frequency of the torsional pendulum (analyzed as a single degree of freedom) is:

f =(1/(2 π)) √(k/I)

where

I = moment of inertia of suspended mass

Expressing this in terms of the torsional stiffness ko for the undeformed wire (under no load):

f =(1/(2 π)) (√(ko/I)) ( √(k/ko) )

since the natural frequency of the torsional pendulum, calculated on the basis of the undeformed wire is:

fo = (1/(2 π)) (√(ko/I) )

using this, and expressing the ratio of the torsional stiffness k under load to the  torsional stiffness ko for the undeformed wire (under no load), the following follows:

f  =fo  e- 2 ν σz YIELD/E

where it is assumed that the moment of inertia of the wire is negligibly small compared with the moment of inertia of the suspended mass (assumed to remain constant during the test)



Numerical Example

Numerical values are shown first in US units instead of SI units because Monomorphic is in the USA and has expressed the weight of his rig in pounds force. Values are converted to SI units, for international purposes.

Using the actual value for music wire chosen by Monomorphic's torsional pendulum:

Music wire ASTM A228
σz YIELD =  420 MPa = 60,916 psi  (Yield stress)
E = 207 GPa = 30*106 psi  (Young's modulus of elasticity)
εz YIELDz YIELD / E = 0.00203 = 0.203%  (Yield strain)
ν = 0.3  (Poisson's ratio)

Substituting these values into the expression for the area ratio at yield:

A/Ao =   e- 2 * 0.3 *  0.00203

        =   0.99878242 The change in cross-sectional area at yield being only - 0.12%

natural frequency at yield

Using the previous expression, it is obvious that the frequency decreases in direct proportion to the decrease in cross-sectional area

f  =fo  (A/Ao)

   =fo  e- 2 ν σz YIELD/E

and therefore, at yield:

f  =fo   e- 2 * 0.3 *  0.00203

   = fo   0.99878242 The natural frequency at yield is lower by - 0.12%

where it is assumed that the moment of inertia of the wire is negligibly small compared with the moment of inertia of the suspended mass (assumed to remain constant during the test)

torsional stiffness ratio at yield

k/ ko = e- 4 ν σz YIELD/E
         = e - 4 * 0.3 *  0.00203
         = 0.99756696    The torsional stiffness at yield is lower by - 0.24%

change in diameter at yield:

D/Do = √ (A/Ao) = 0.99939119

εx = εy  = ln (D/Do) = - 0.00060900 = -0.06%

(D - Do)/Do = - 0.00060897 = - 0.06 %

Notice: since the transverse strain measure on the diameter (using the Engineering Strain measure) is only 0.06%, this strain is infinitesimal, and hence it was not necessary to use finite strain or finite stress measures, as the strain on the cross-section is so small that at these values the stress measures are equivalent (True Stress ~ Engineering Stress and True Strain ~ Engineering Strain).

Actual Diameter of music wire used by Monomorphic

Do = 0.033 inches = 0.838 mm

Change in diameter:
D - Do =  - 0.00060897 *  0.033 inches = - 0.00002010 inches
                                                             = - 0.5105 μm


Tolerance on diameter of music wire used by Monomorphic


Music Wire ASTM A228
Minimum tolerance on diameter = 8 μm = 0.0003 inches (another supplier had a greater tolerance of 0.0004 inches)

Therefore, the ratio of the change in diameter to the wire's tolerance in diameter, for a load high enough to result in yield stress (permanent deformation threshold) is:

(Change in diameter)/(Tolerance on diameter) = - (0.00002010 inches)/(0.0003 inches) = 6.7%

Therefore, the change in diameter of the wire, even at a stress high enough to result in incipient yield of the material, is only 6.7% of the tolerance in diameter.  Hence the change in diameter, and the change in cross-sectional area, for stresses below the yield limit is negligible, and concerns about change in diameter are ill-founded,  since the dimensional tolerance changes of the wire itself are 15 times greater.

Calculation of Maximum Load for Monomorphic's wire

σz YIELD =  420 MPa = 60,916 psi

Longitudinal Force = Fz

σz YIELD = 60,916 psi  = Fz / A = ( Fz / Ao ) *(Ao/A)

 60,916 psi  =   ( Fz / (Pi*(Do/2)2) ) *(Ao/A)

                   =   ( Fz / (Pi*(0.033 inches/2)2) ) /0.99878242

 Fz  =  60,916 psi *0.99878242 * (Pi*(0.033 inches/2)2)
       = 52.038 lbf
       = 231.476 Newtons
       = 23.6040 kg force

The maximum load, the load that will result in permanent yield, is 52.038 pound force (23.604 kgf)

Due to the fact that the change in diameter, and hence the change in cross-sectional area is negligible, this load could also have been calculated for engineering purposes, based on the initial, unloaded, cross-sectional area:

 Fz  =  60,916 psi * (Pi*(0.033 inches/2)2)
       = 52.101 lbf
       =231.758 Newtons
       =23.6328 kg force

The difference due to the change in cross-sectional area, being only 0.12%

Since Monomorphic's rig is only 25 lbf (111 Newtons or 11.3 kg force), it is less than 1/2 of the weight that would result in permanent deformation of the torsional wire

The change in diameter of the music wire is completely negligible, particularly in comparison with the manufacturing tolerance of the music wire itself and with respect to the variation in yield stress of the wire itself.
« Last Edit: 02/17/2017 05:00 PM by Rodal »

Offline X_RaY

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Short FEKO visualization/animation (v.avi) for the folks interested in the EM- fields inside a truncated conical cavity resonator with conductive walls.
It shows how H-field E-field and surface currents are arranged over a full cycle.
It clearly shows the direction of the vectors of the surface current and its dependency from the H-Field.
The mode shape is TE012. Left side shows the H-component of the field inside the cavity across a plane and the right side shows the E-component respectively.
The arrows represents the field vectors which is alternating over the full cycle for each component. Please note that the direction of the fieldsvectors are complementary for neighboring lobes of the field pattern.
« Last Edit: 02/23/2017 08:21 PM by X_RaY »

Offline X_RaY

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QU dependency for different resistivity (or equivalent conductivity) related to temperature differences

Monomorphic's frustum dimensions
LD=0.299m SD=0.178m H=0.24m
mode=TE013
frequency=2.451GHz
material=Copper101

calculated QU@293K gives ~101631

https://www.copper.org/resources/properties/cryogenic/images/Electrical-Resistivity-of-Alloys.gif
resistivity of
Copper101 at 293K = 1.71E-8 Ωm
Copper101 at 77K   = 0.19E-8 Ωm



QU@77K = 101631 * sqrt(1.71E-8 / 0.19E-8) =  304893


Please note:
This approximation does not involve the change in frequency because of the thermal expansion coefficient or other secound order effects like the change in Q for smaller (cavity-) volume at lower temperature !!
« Last Edit: 03/04/2017 07:35 PM by X_RaY »

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