-
#1220 by Rodal on 05 Mar, 2015 00:39
-
The flight thruster is shown as what appears to be an oscillator configuration, so it is still possible that the loaded Q is higher than the unloaded Q. But that is speculation. I do worry about his concept of "no static thrust" as well.
Shawyer's EM Drive was never tested in a vacuum as far as I know. One wonders whether thermal effects (convection current generation) are very significant for Shawyer's and the Chinese tests due to the very large power input, in other words: is a big portion of Shawyer's (and the Chinese) thrust measurement due to thermal effects ?
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)
Question: your expression has mode-dependence (unlike Shawyer's and McCulloch's expressions which have no mode dependence). Why is there a mode dependence, and why does it depend only on the circular cross-section quantum numbers "m" and "n" but it does not depend on the axial quantum number "p"? -
#1221 by Rodal on 05 Mar, 2015 00:49
-
...
I understand that your expression is based on the exact solution for a cylindrical cavity. Is there a reason why a cylindrical cavity thrust should depend on the cross-sectional quantum numbers "m" and "n" but not depend on the axial quantum number "p"?
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)
The exact solution for the cylindrical cavity has a frequency dependent on "p". Why isn't thrust dependent on "p" ? -
#1222 by Notsosureofit on 05 Mar, 2015 00:51
-
The mode dependence is already there in the waveguide dispersion relation. "p" drops out when you evaluate it at the k values of each end. That's the simple case. Higher terms in a more exact solution (or a dielectric solution) need not do that.
-
#1223 by Notsosureofit on 05 Mar, 2015 00:54
-
...
I understand that your expression is based on the exact solution for a cylindrical cavity. Is there a reason why a cylindrical cavity thrust should depend on the cross-sectional quantum numbers "m" and "n" but not depend on the axial quantum number "p"?
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)
The exact solution for the cylindrical cavity has a frequency dependent on "p". Why isn't thrust dependent on "p" ?
Thrust for a completely cylindrical cavity is zero, all the terms cancel out. "p" cancels out in the simple tapered cavity, but that is not an exact solution.
[now if you were to calculate the thrust of a cylindrical cavity in an accelerated frame of reference you would have an exact solution but the "thrust" would oppose the acceleration, ie. Nhfg/c^2]
-
#1224 by Rodal on 05 Mar, 2015 00:59
-
...
The problem is that the exact solution is not closed-form: it requires the solution of two eigenvalue problems which have no closed-form solution.
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)
Without a closed-form solution for the frequency and mode shape for a truncated cone there is no hope of a closed-form solution for the thrust.
Maybe when I have the time I can come up with a closed-from piecewise approximation (comprising several simple piecewise functions in subregions that can be spliced together) that can closely approximate the exact solution. -
#1225 by ThinkerX on 05 Mar, 2015 01:07
-
So, then in order to produce 'thrust,' this device MUST be a truncated cone - preferably with a rounded base?
Also a wild and stupid thought about Sawyer's commentary: maybe he envisions this device as some sort of 'turbo-charger?' That is something that adds to an already existing velocity, but won't function well, if at all when 'at rest.' Say you have a spacecraft in motion at speed X. Flip this drive on, and the speed becomes X+1. -
#1226 by Notsosureofit on 05 Mar, 2015 01:10
-
So, then in order to produce 'thrust,' this device MUST be a truncated cone - preferably with a rounded base?
Also a wild and stupid thought about Sawyer's commentary: maybe he envisions this device as some sort of 'turbo-charger?' That is something that adds to an already existing velocity, but won't function well, if at all when 'at rest.' Say you have a spacecraft in motion at speed X. Flip this drive on, and the speed becomes X+1.
The device must have an asymmetric dispersion relation in some axis. A truncated cone is one example.
I havn't been able to get my head around the velocity based explanations. -
#1227 by Rodal on 05 Mar, 2015 01:23
-
...maybe he envisions this device as some sort of 'turbo-charger?' That is something that adds to an already existing velocity, but won't function well, if at all when 'at rest.' ...
A turbocharger, is a device that increases an engine's efficiency and power by forcing extra air into the combustion chamber.
What is the "engine" that the EM Drive is turbocharging?
The bizarre nature of something needing to be free to accelerate for it to produce a force doesn't apply to the turbocharger or to the engine: the engine that is being turbocharged does not need to be accelerating or even be in rigid body motion. Its center of mass can be completely stationary, and the turbocharged engine can then be used for electric power generation, for example, instead of for transporting people as in an automobile.
-
#1228 by Rodal on 05 Mar, 2015 01:43
-
...
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)...
I think I might be able to obtain a closed-form expression for the frequency of a cylindrical cavity with a dielectric at one end. (Or really a cylindrical cavity having two dielectrics in contact with each other: one dielectric at one end and another dielectric with different permitivity and permeability at the other end).
The device must have an asymmetric dispersion relation in some axis.
.....
Would that be of any use to you to improve your equation? -
#1229 by zen-in on 05 Mar, 2015 03:21
-
@ Rodal :
See attached picture to share my mental image. Tilt over-exaggerated for illustration.
Grey : solid rotating assembly (no deformation implied)
Orange/brown : fixed assembly (no deformation implied)
Blue : the ground slab of the vacuum chamber (no deformation implied)
For now, assume a perfect axis of rotation around Z : only one degree of freedom of Grey relative to Orange, the "official" rotation around Z, no compliance implied, Grey kept in the XY plane, plane has same tilt as Orange (XY parallel to Orange platform).
@Star-Drive
Can you confirm this is a correct way to understand that there is a tilt in the axis of rotation ?
One aspect of this tilt in the axis of rotation that hasn't been discussed is the requirement for imbalance. The difference in moments on either side of the beam only has to be very slight for it to always come to rest at the same location +/- a micron or two.
A small CW tilt along the X axis of the balance arm would explain the apparent drift in the baseline seen in some of the thrust waveforms. When the cavity has the orientation shown below its CM shifts to the left. This would reduce the tilt, resulting in an increase in brightness of the reflected light the LDS measures; due to the mirror position being closer to an optimal perpendicular position wrt the light beam. The increase in brightness corresponds to a decrease in distance; hence the negative slope. With the device mounted the other way the shift in CM increases the tilt. This reduces the reflected light and is registered as an increase in distance. No actual motion of the beam occurs. This apparent motion is an optical artifact. This assumes the Philtec distance sensor is used on the far side. If it is used on the near side a small CCW tilt along the X axis of the beam would produce the same effect; except requiring much less rotation from the change in CM.
-
#1230 by Star-Drive on 05 Mar, 2015 03:56
-
@ Rodal :
See attached picture to share my mental image. Tilt over-exaggerated for illustration.
Grey : solid rotating assembly (no deformation implied)
Orange/brown : fixed assembly (no deformation implied)
Blue : the ground slab of the vacuum chamber (no deformation implied)
For now, assume a perfect axis of rotation around Z : only one degree of freedom of Grey relative to Orange, the "official" rotation around Z, no compliance implied, Grey kept in the XY plane, plane has same tilt as Orange (XY parallel to Orange platform).
@Star-Drive
Can you confirm this is a correct way to understand that there is a tilt in the axis of rotation ?
Yes there is a tilt in the torque pendulum's (TP) axis of rotation with the RF power supply lower than the test article as you have it shown in your schematic diagram. Using my 24" long level I use about a 1/4 bubble of tilt. From that point on its a matter of fine tuning the TP response by adding or removing small weights around the test article and/or adjusting the tilt angle with a micrometer adjusting the length of the TP support under the test article.
Best, Paul M. -
#1231 by ThinkerX on 05 Mar, 2015 06:37
-
Ok, its late...but...Quote
What is the "engine" that the EM Drive is turbocharging?
The bizarre nature of something needing to be free to accelerate for it to produce a force doesn't apply to the turbocharger or to the engine: the engine that is being turbocharged does not need to be accelerating or even be in rigid body motion. Its center of mass can be completely stationary, and the turbocharged engine can then be used for electric power generation, for example, instead of for transporting people as in an automobile.
Maybe what initially sets the spacecraft in motion doesn't matter. What matters is that the craft is in motion (inertial radiation/force as per Doctor McCulloch's theory, maybe). Maybe this device acts as a sort of amplifier/concentrator for that force when in operation. But if inertial force/radiation is absent, the device either doesn't work, or works poorly because there is nothing to amplify/concentrate.
Like I said, a wild and stupid idea. -
#1232 by Notsosureofit on 05 Mar, 2015 11:13
-
...
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)...
I think I might be able to obtain a closed-form expression for the frequency of a cylindrical cavity with a dielectric at one end. (Or really a cylindrical cavity having two dielectrics in contact with each other: one dielectric at one end and another dielectric with different permitivity and permeability at the other end).
The device must have an asymmetric dispersion relation in some axis.
.....
Would that be of any use to you to improve your equation?
Absolutely ! That should be a very interesting case as it is 2 coupled cavities. Most of these kind of problems use iterative solutions but I think that case may have some unique analytic solutions !
-
#1233 by Rodal on 05 Mar, 2015 13:35
-
Yes, it is amenable to a closed-form solution since it involves a quartic: 4th order polynomial. OK, exact, closed-form solution for a cylindrical cavity containing two coupled dielectrics coming your way soon, courtesy of Rodal-exact-solutions. I have the solution now for the 4 roots in general. Just have to figure out which root is the real root for which cases....
(and yes the simple dispersion relation drops dependence on "p". Maybe we can get that in there w/ your exact solution ??)...
I think I might be able to obtain a closed-form expression for the frequency of a cylindrical cavity with a dielectric at one end. (Or really a cylindrical cavity having two dielectrics in contact with each other: one dielectric at one end and another dielectric with different permitivity and permeability at the other end).
The device must have an asymmetric dispersion relation in some axis.
.....
Would that be of any use to you to improve your equation?
Absolutely ! That should be a very interesting case as it is 2 coupled cavities. Most of these kind of problems use iterative solutions but I think that case may have some unique analytic solutions !
Solution will contain SquareRoots and our friends Xmn and X'mn the Bessel and Bessel Prime zeros.
-
#1234 by frobnicat on 05 Mar, 2015 17:54
-
@ Rodal :
See attached picture to share my mental image. Tilt over-exaggerated for illustration.
Grey : solid rotating assembly (no deformation implied)
Orange/brown : fixed assembly (no deformation implied)
Blue : the ground slab of the vacuum chamber (no deformation implied)
For now, assume a perfect axis of rotation around Z : only one degree of freedom of Grey relative to Orange, the "official" rotation around Z, no compliance implied, Grey kept in the XY plane, plane has same tilt as Orange (XY parallel to Orange platform).
@Star-Drive
Can you confirm this is a correct way to understand that there is a tilt in the axis of rotation ?
One aspect of this tilt in the axis of rotation that hasn't been discussed is the requirement for imbalance. The difference in moments on either side of the beam only has to be very slight for it to always come to rest at the same location +/- a micron or two.
Yes, a difference in centre of mass can make long lasting difference in position. The following statement tries to summarize rigorously :
Statement A : Any shift in centre of mass of a part (relative to fixation to the balance arm) along a direction orthogonal to a compliant (not infinite stiffness) axis that is not strictly vertical can induce a change in angular rest position around said compliant axis.
This can be quite important as a (thermal) longitudinal (Y+ or Y-) shift in CoM of the frustum assembly would be orthogonal to axis Z, but so long as this most compliant by far axis Z (the natural axis of the pendulum) was believed to be strictly vertical there would be no change in angular rest position of the arm (around Z). Only transient angular positions shifts could be induced by such thermal CoM shifts, the arm position (around Z) would be quickly enough restored to unmodified rest position (return to baseline) by the spring restoring torque of flexure bearings, and then time of ~45s during which a sustained displacement where recorded could be used as an effective argument that this couldn't be due to thermal expansions alone.
Note that now that the Z axis is no longer vertical the argument is no longer valid.QuoteA small CW tilt along the X axis of the balance arm would explain the apparent drift in the baseline seen in some of the thrust waveforms. When the cavity has the orientation shown below its CM shifts to the left. This would reduce the tilt, resulting in an increase in brightness of the reflected light the LDS measures; due to the mirror position being closer to an optimal perpendicular position wrt the light beam. The increase in brightness corresponds to a decrease in distance; hence the negative slope. With the device mounted the other way the shift in CM increases the tilt. This reduces the reflected light and is registered as an increase in distance.
Is this what you are saying ?
Indeed the X axis is not vertical (it is horizontal, or almost), so the above statement A does apply if we consider rotations around X are not against infinite stiffness. But preliminary rough estimations gave me relative displacements of rest optical length of at least one order of magnitude below signal. Even if not infinitely stiff, the added torsion compliance of faztek beam around X and compliance of tandem flexure bearings around X is not enough. For reasonable thermal CoM's shifts, this is stiff enough , and the optical lever is small enough (optical length d measured at small distance below X) that it would make very little contribution to the signal.QuoteNo actual motion of the beam occurs. This apparent motion is an optical artifact. This assumes the Philtec distance sensor is used on the far side. If it is used on the near side a small CCW tilt along the X axis of the beam would produce the same effect; except requiring much less rotation from the change in CM.
For this twist around X effect, no actual rotation of the beam around Z would occur, but the beam is deformed. I wouldn't qualify that as an "optical artifact". Sorry I'm becoming quite finicky on wordings lately That would be a real mechanical motion, only not a motion around the "official" Z axis.
We have all reasons to believe the Philtec linear sensor is used in the far range, see attached chart's horizontal units. The initial report (anomalous...) clearly states that it is used around 500µm, the vertical readings of the charts are consistent, and Paul March confirms :...
...
- That the vertical scale in the charts (indicated in µm, around 500) are relevant or not relevant.
...
The Philtec D63 fiber-optic displacement sensor measures distance from its target mirror in microns, so the numbers on the left hand side of the force plots measure the distance from the end of the fiber-optic laser head to its mirror target mounted on the torque pendulum arm. The data sheet for same is attached.
...
Still, there is a factor 10 disparity between the apparent stiffness readings (LDS readings against calibrations pulses alone) and the stiffness needed to explain the 4.5s pseudo-period of oscillations, so this can put a doubt on the readings of vertical scale, a factor 10 here could explain a lot of things.
Regardless of this nagging problem of late, now that we know that Z deviates from strict verticality by a quarter bubble, from statement A above there is now a real possibility that a thermal CoM shift along Y changes angular rest position around the official Z, of relative magnitude compatible with signals, as rotation around Z has a very low stiffness (the restoring torque of flexure bearing around their natural axis of rotation). And such change in angular rest position around Z would record as false "sustained thrusts" in the charts.
Statement B : The validity of real sustained thrust signals now rests on the amount of deviation of Z from vertical, resolution of the contradictions between apparent stiffness around Z, and careful assessment of thermal expansions in vacuum. The sustained duration alone no longer suffice.
-
#1235 by Rodal on 05 Mar, 2015 18:52
-
EXACT, CLOSED-FORM SOLUTION FOR A CYLINDRICAL CAVITY ELECTROMAGNETIC RESONATOR CONTAINING TWO COUPLED ADJOINING DIELECTRICS FILLING THE CAVITY by Jose' J. Rodal, Ph.D.
There are two positive roots for the frequency fm,n,p to consider, if root1 is real take frequency = root1, otherwise take root2 (if it is real). If both roots are imaginary, there is no resonance. Remember to use the cutt-off frequency rule to cut-off frequencies whose wavelength do not fit.
fm,n,p = root1 if root1 and root2 are both real numbers,
fm,n,p = root2 if root1 is not a real number, and root 2 is a real number
fm,n,p = 0 if neither root1 and root2 are real numbers
root1= Sqrt[((-4*b^2*(dDielectric1 - dDielectric2)*(dDielectric1 + dDielectric2)*(dielectricLength1/cMedium1 - dielectricLength2/cMedium2)*(dielectricLength1/cMedium1 + dielectricLength2/cMedium2))/
(dDielectric1^2*dDielectric2^2) + (dielectricLength1^2/cMedium1^2 + dielectricLength2^2/cMedium2^2)*p^2 -
2*Sqrt[(-4*b^2*(dielectricLength1/cMedium1 - dielectricLength2/cMedium2)*(dielectricLength1/cMedium1 + dielectricLength2/cMedium2)*((dDielectric1*dielectricLength1)/cMedium1 - (dDielectric2*dielectricLength2)/cMedium2)*
((dDielectric1*dielectricLength1)/cMedium1 + (dDielectric2*dielectricLength2)/cMedium2)*p^2)/(dDielectric1^2*dDielectric2^2) + (dielectricLength1^2*dielectricLength2^2*p^4)/(cMedium1^2*cMedium2^2)])/
(dielectricLength1^2/cMedium1^2 - dielectricLength2^2/cMedium2^2)^2]/2
root2= Sqrt[((-4*b^2*(dDielectric1 - dDielectric2)*(dDielectric1 + dDielectric2)*(dielectricLength1/cMedium1 - dielectricLength2/cMedium2)*(dielectricLength1/cMedium1 + dielectricLength2/cMedium2))/
(dDielectric1^2*dDielectric2^2) + (dielectricLength1^2/cMedium1^2 + dielectricLength2^2/cMedium2^2)*p^2 +
2*Sqrt[(-4*b^2*(dielectricLength1/cMedium1 - dielectricLength2/cMedium2)*(dielectricLength1/cMedium1 + dielectricLength2/cMedium2)*((dDielectric1*dielectricLength1)/cMedium1 - (dDielectric2*dielectricLength2)/cMedium2)*
((dDielectric1*dielectricLength1)/cMedium1 + (dDielectric2*dielectricLength2)/cMedium2)*p^2)/(dDielectric1^2*dDielectric2^2) + (dielectricLength1^2*dielectricLength2^2*p^4)/(cMedium1^2*cMedium2^2)])/
(dielectricLength1^2/cMedium1^2 - dielectricLength2^2/cMedium2^2)^2]/2
where we define the following dimensionless ratios as follows:
dDielectric1 = diameter/dielectricLength1;
dDielectric2 = diameter/dielectricLength2;
and where
dielectricLength1 = length - dielectricLength2;
where length is the total internal length of the cylindrical cavity.
The quantity "b" is defined as follows:
b := If[modetype == "TM", xbesselzeros[[m + 1, n]]/Pi, If[modetype == "TE", xprimebesselzeros[[m + 1, n]]/Pi]]
Where Xmn=xbesselzeros[[m+1,n]] (the zeros of the Bessel function)
xbesselzeros = {{2.40483, 5.52008, 8.65373, 11.7915, 14.9309}, {3.83171, 7.01559,
10.1735, 13.3237, 16.4706}, {5.13562, 8.41724, 11.6198, 14.796,
17.9598}, {6.38016, 9.76102, 13.0152, 16.2235, 19.4094}, {7.58834,
11.0647, 14.3725, 17.616, 20.8269}, {8.77148, 12.3386, 15.7002,
18.9801, 22.2178}, {9.93611, 13.5893, 17.0038, 20.3208,
23.5861}, {11.0864, 14.8213, 18.2876, 21.6415, 24.9349}, {12.2251,
16.0378, 19.5545, 22.9452, 26.2668}, {13.3543, 17.2412, 20.807,
24.2339, 27.5837}, {14.4755, 18.4335, 22.047, 25.5095, 28.8874}}
and where X'mn=xprimebesselzeros [[m+1,n]] (the zeros of the derivative of the Bessel function)
xprimebesselzeros = {{3.83171, 7.01559, 10.1735, 13.3237, 16.4706}, {1.84118, 5.33144,
8.53632, 11.706, 14.8636}, {3.05424, 6.70613, 9.96947, 13.1704,
16.3475}, {4.20119, 8.01524, 11.3459, 14.5858, 17.7887}, {5.31755,
9.2824, 12.6819, 15.9641, 19.196}, {6.41562, 10.5199, 13.9872,
17.3128, 20.5755}, {7.50127, 11.7349, 15.2682, 18.6374,
21.9317}, {8.57784, 12.9324, 16.5294, 19.9419, 23.2681}, {9.64742,
14.1155, 17.774, 21.2291, 24.5872}, {10.7114, 15.2867, 19.0046,
22.5014, 25.8913}, {11.7709, 16.4479, 20.223, 23.7607, 27.182}}
For example, for mode TE01, b= xprimebesselzeros[[1,1]]/Pi= 3.83171 / Pi = 3.83171 / 3.14159
NUMERICAL EXAMPLE: we take the case used by @aero, containing a dielectric2 of HD PE material with dielectric constant (relative electric permitivity) = 2.3, and a dielectric1 being the empty portion of the cavity, under vacuum:
diameter = 0.08278945 meter;
length = 0.1224489 meter;
dielectricLength2 = 0.027282494103102 meter;
cMedium1 = cVacuum; cVacuum = 299792458 meter/second;
cMedium2 = cVacuum/Sqrt[2.3] ; (relative electric permittivity=2.3;relative magnetic permeability=1)
Results:
modetype = "TE"; m = 1; n = 1; p = 1; root1 = 2.26774 GHz
modetype = "TE"; m = 1; n = 1; p = 2; root1 = 2.93557 GHz
modetype = "TM"; m = 0; n = 1; p = 2; root1= 3.37114 GHz
For this case, root1 is real, there is no need to consider root2. A number of modes are cut-off, for example modes TE011 and TM011 are cut-off
-
#1236 by Notsosureofit on 05 Mar, 2015 20:52
-
@ RODAL
Can you check this ? (working it down)
root1= Sqrt[(
(-4*b^2*(dD1^2-dD2^2)*((L1/c1)^2-(L2/c2)^2))
/ (dD1^2*dD2^2)
+ ((L1/c1)^2+(L2/c2)^2)*p^2
- 2*Sqrt[
(-4*b^2*((L1/c1)^2-(L2/c2)^2)*((dD1*L1/c1)^2-(dD2*L2/c2)^2)*p^2)
/ (dD1^2*dD2^2)
+ (L1^2*L2^2*p^4)
/ (c1^2*c2^2)]) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> )] ??
/ (L1^2/c1^2-L2^2/c2^2)^2]/2
root2= Sqrt[(
(-4*b^2*(dD1^2-dD2^2)*((L1/c1)^2-(L2/c2)^2))
/ (dD1^2*dD2^2)
+ ((L1/c1)^2+(L2/c2)^2)*p^2
+ 2*Sqrt[
(-4*b^2*((L1/c1)^2 - (L2/c2)^2)*((dD1*L1)/c1)^2-(dD2*L2/c2)^2)*p^2)
/ (dD1^2*dD2^2)
+ (L1^2*L2^2*p^4)
/ (c1^2*c2^2)]) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> )] ??
/ (L1^2/c1^2-L2^2/c2^2)^2]/2
where we define the following dimensionless ratios as follows:
dD1 = diameter/L1;
dD2 = diameter/L2;
and where
L1 = length - L2;
where length is the total internal length of the cylindrical cavity.
The quantity "b" is defined as follows:
b := If[modetype == "TM", xbesselzeros[[i + 1, j]]/Pi, If[modetype == "TE", xprimebesselzeros[[i + 1, j]]/Pi]]
Thanks -
#1237 by Rodal on 05 Mar, 2015 20:58
-
@ RODAL
Can you check this ? (working it down)
root1= Sqrt[(
(-4*b^2*(dD1^2-dD2^2)*((L1/c1)^2-(L2/c2)^2))
/ (dD1^2*dD2^2)
+ ((L1/c1)^2+(L2/c2)^2)*p^2
- 2*Sqrt[
(-4*b^2*((L1/c1)^2-(L2/c2)^2)*((dD1*L1/c1)^2-(dD2*L2/c2)^2)*p^2)
/ (dD1^2*dD2^2)
+ (L1^2*L2^2*p^4)
/ (c1^2*c2^2)]) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> )] ??
/ (L1^2/c1^2-L2^2/c2^2)^2]/2
root2= Sqrt[(
(-4*b^2*(dD1^2-dD2^2)*((L1/c1)^2-(L2/c2)^2))
/ (dD1^2*dD2^2)
+ ((L1/c1)^2+(L2/c2)^2)*p^2
+ 2*Sqrt[
(-4*b^2*((L1/c1)^2 - (L2/c2)^2)*((dD1*L1)/c1)^2-(dD2*L2/c2)^2)*p^2)
/ (dD1^2*dD2^2)
+ (L1^2*L2^2*p^4)
/ (c1^2*c2^2)]) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> )] ??
/ (L1^2/c1^2-L2^2/c2^2)^2]/2
where we define the following dimensionless ratios as follows:
dD1 = diameter/L1;
dD2 = diameter/L2;
and where
L1 = length - L2;
where length is the total internal length of the cylindrical cavity.
The quantity "b" is defined as follows:
b := If[modetype == "TM", xbesselzeros[[i + 1, j]]/Pi, If[modetype == "TE", xprimebesselzeros[[i + 1, j]]/Pi]]
Thanks
root1 = Sqrt[((-4*b^2*(dD1 - dD2)*(dD1 + dD2)*(L1/c1 - L2/c2)*(L1/c1 + L2/c2))/(dD1^2*dD2^2) + (L1^2/c1^2 + L2^2/c2^2)*p^2 -
2*Sqrt[(-4*b^2*(L1/c1 - L2/c2)*(L1/c1 + L2/c2)*((dD1*L1)/c1 - (dD2*L2)/c2)*((dD1*L1)/c1 + (dD2*L2)/c2)*p^2)/(dD1^2*dD2^2) + (L1^2*L2^2*p^4)/(c1^2*c2^2)])/(L1^2/c1^2 - L2^2/c2^2)^2]/2
root2 = Sqrt[((-4*b^2*(dD1 - dD2)*(dD1 + dD2)*(L1/c1 - L2/c2)*(L1/c1 + L2/c2))/(dD1^2*dD2^2) + (L1^2/c1^2 + L2^2/c2^2)*p^2 +
2*Sqrt[(-4*b^2*(L1/c1 - L2/c2)*(L1/c1 + L2/c2)*((dD1*L1)/c1 - (dD2*L2)/c2)*((dD1*L1)/c1 + (dD2*L2)/c2)*p^2)/(dD1^2*dD2^2) + (L1^2*L2^2*p^4)/(c1^2*c2^2)])/(L1^2/c1^2 - L2^2/c2^2)^2]/2
and yes, you can always substitute, if you prefer
(A-B)(A+B)=A^2 - B^2 -
#1238 by Rodal on 05 Mar, 2015 21:13
-
Since the denominator is the SquareRoot of a square, you can take the denominator out of the Square Root
Take a gander
at the image below:
(it reads ROOT2 and ROOT4 because there are 4 roots and root1 above is the 2nd root of the 4 roots). I only showed two of the roots, since the other two roots give negative frequencies.
-
#1239 by Notsosureofit on 05 Mar, 2015 23:01
-
Great !
Next need to solve for the expression (f0^2 - f2^2) where (f0 => L1+L2=0 and f2 =>L1+L2=L ??)
Hmmm, let me think about that for a second. What we want to do is isolate the (p*pi/L)^2 term so it will cancel out leaving the Doppler shifts in the accelerated frame. ie. (f0^2 - f2^2) where f0^2 is evaluated at one end and f2^2 at the other. In short, we want to find the frame in which the dispersion from end to end disappears. The complication is the terms from the reflections at the c1,c2 interface.
