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#220 by Notsosureofit on 19 Jan, 2015 00:11
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For c: TE01 doesn't compute but TE022 is a good fit, maybe better than TE222. (w/o dielectric that is)
Edit: Nope, frequency is too high (dielectric might bring it down ?)
For b: TE213 seems the only good fit
For a: Can't tell the difference between TE022 andTM122
Probably as good as this model is gonna get w/o the dielectric (and the COMSOL) -
#221 by Mulletron on 19 Jan, 2015 15:53
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Quick question:
Did someone mention in the past that the equations at the Oracle are no good? Or were they good?
http://en.wikipedia.org/wiki/Microwave_cavity#Cylindrical_cavity
or are these good
Am I better off going with the Kwok slide 13?
http://www.engr.sjsu.edu/rkwok/EE172/Cavity_Resonator.pdf
or this guy's method?
http://www.chrislmueller.com/studies/Jackson8-6.pdf
Playing catch up here. This page is nifty. http://mathworld.wolfram.com/BesselFunctionZeros.html -
#222 by Rodal on 19 Jan, 2015 17:11
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Quick question:
Quick glance: I took a quick gander at them and they all look identical, insofar as being related to X'm,n and Xm,n (the zeros of the derivative of the Bessel function or the zeros of the Bessel function respectively) for TE and TM respectively. What difference do you see between them?
Did someone mention in the past that the equations at the Oracle are no good? Or were they good?
http://en.wikipedia.org/wiki/Microwave_cavity#Cylindrical_cavity
or are these good
Am I better off going with the Kwok slide 13?
http://www.engr.sjsu.edu/rkwok/EE172/Cavity_Resonator.pdf
or this guy's method?
http://www.chrislmueller.com/studies/Jackson8-6.pdf
Playing catch up here. This page is nifty. http://mathworld.wolfram.com/BesselFunctionZeros.html
For example, notice (from http://wwwal.kuicr.kyoto-u.ac.jp/www/accelerator/a4/besselroot.htmlx )
that X'1,1 = 1.84118378134065
And therefore in Kwok slide 13 (http://www.engr.sjsu.edu/rkwok/EE172/Cavity_Resonator.pdf ) with radius of the Coke can a = 1.25
kc = X'1,1 / a = 1.8412 / 1.25 = 1.473
f111=(c/(2*Pi))*Sqrt[(kc)^2+(p*Pi/d)^2]
The expressions for ChrisMueller and Wikipedia look identical except for the factor of 2*Pi due to the fact that ChrisMueller calculates the angular frequency (omega) while Wikipedia calculates the frequency (f). Since omega = 2 *Pi*f, the expressions in Wikipedia and in ChrisMueller are both identical and both are correct. Kwok's expression is missing the factor of the squareRoot of the relative permeability and relative permittivity of the cavity filling because Kwok explicitly wrote that he is calculating the example for air inside the cavity. (For a vacuum they are exactly unity, for air they are approximately unity).
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#223 by Mulletron on 19 Jan, 2015 17:53
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I get the air thing which means I can just drop that part. The part that screwed with me was
f = (c/(2*Pi))*((X[sub m,n]/R)^2+((p*Pi)/L)^2)^.5
silly enough the ^.5 part because I'm used to seeing sqrt typed out, but that works too. Thanks for clearing all that up.
This is all looking pretty good, so far as I've gotten. I think it might be time for @Notsosureofit to put some polish on it, add some prose, and make it a packaged deal.
I think it might be time for a spaceflight application minute. Theory is getting pretty heavy once again. -
#224 by Rodal on 19 Jan, 2015 18:01
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Good idea, I'm ready :-)
I think it might be time for a spaceflight application minute... -
#225 by Rodal on 19 Jan, 2015 18:30
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A minor point. For further clarity you may want to include a prime symbol (or an apostrophe) on X
For TM modes, X[sub m,n] = the n-th zero of the m-th Bessel function.
[1,1]=3.83, [0,1]=2.40, [0,2]=5.52 [1,2]=7.02, [2,1]=5.14, [2,2]=8.42, [1,3]=10.17, etc.
and for TE modes, X[subm,n] = the n-th zero of the derivative of the m-th Bessel function.
[0,1]=3.83, [1,1]=1.84, [2,1]=3.05, [0,2]=7.02, [1,2]=5.33, [1,3]=8.54, [0,3]=10.17, [2,2]=6.71, etc.
...
"for TE modes, X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function"
instead of
"for TE modes, X[subm,n] = the n-th zero of the derivative of the m-th Bessel function"
to differentiate X' (as used for the TE modes) from X (as used for the TM modes).
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#226 by Mulletron on 19 Jan, 2015 19:56
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A minor point. For further clarity you may want to include a prime symbol (or an apostrophe) on X
For TM modes, X[sub m,n] = the n-th zero of the m-th Bessel function.
[1,1]=3.83, [0,1]=2.40, [0,2]=5.52 [1,2]=7.02, [2,1]=5.14, [2,2]=8.42, [1,3]=10.17, etc.
and for TE modes, X[subm,n] = the n-th zero of the derivative of the m-th Bessel function.
[0,1]=3.83, [1,1]=1.84, [2,1]=3.05, [0,2]=7.02, [1,2]=5.33, [1,3]=8.54, [0,3]=10.17, [2,2]=6.71, etc.
...
"for TE modes, X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function"
instead of
"for TE modes, X[subm,n] = the n-th zero of the derivative of the m-th Bessel function"
to differentiate X' (as used for the TE modes) from X (as used for the TM modes).
Perfect timing, in the RF & Microwave Toolbox app I've been using, I found they messed up exactly what you are saying. Emailed developer. See screenshot.
Still trying to think of another spaceflight application. We need more power Scotty!
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#227 by Asteroza on 19 Jan, 2015 22:22
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For applications, I would imagine a key figure is practical thrust/weight and thrust/power. It probably isn't enough for a very low orbit PROFAC atmosphere collector, but a high orbit PROFAC might work. Tether/rotovator reboost is another application.
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#228 by ThinkerX on 19 Jan, 2015 23:55
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Space travel applications...
If applied, would 'Notsosureofits's' mode calculations significantly help or hinder the 'Mulletron Mission to Saturn'? -
#229 by Notsosureofit on 20 Jan, 2015 00:18
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Space travel applications...
If applied, would 'Notsosureofit's mode calculations significantly help or hinder the 'Mulletron Mission to Saturn'?
Might help if it could calculate an "optimum" cavity shape. Such has been done for acoustic cavity refrigerators, etc. What is the maximum asymmetric dispersion you can get ? What is the optimum frequency ? (and the highest Q, of course)
Then too, it's not yet apparent why this calculation works at all.
Maybe that's what the Cannae was about, short cavity w/ high mode numbers ? I could see where they might have been trying to stabilize some mode w/ the fins, sorta like some magnetrons.
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#230 by Mulletron on 20 Jan, 2015 10:49
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Space travel applications...
If applied, would 'Notsosureofit's mode calculations significantly help or hinder the 'Mulletron Mission to Saturn'?
Might help if it could calculate an "optimum" cavity shape. Such has been done for acoustic cavity refrigerators, etc. What is the maximum asymmetric dispersion you can get ? What is the optimum frequency ? (and the highest Q, of course)
Then too, it's not yet apparent why this calculation works at all.
I think the other half of the calculation remains to be completed, which is the part concerning how to, and may provide a way to, conserve momentum. The other side of the momentum vector diagram concerns vacuum radiation pressure; well I think it is anyway. At the bottom of page 91 is a nice mashup. We've discussed this in depth in thread one, yet the math remains elusive, but I don't think it is out of our reach.
Google book link to: The Quantum Vacuum: A Scientific and Philosophical Concept, from Electrodynamics to String Theory and the Geometry of the Microscopic World. Luciano Boi
http://bit.ly/1CKmGWX -
#231 by aero on 20 Jan, 2015 17:27
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@RODAL
A. I'm still having conceptual difficulties w/ the mode numbers trying to resolve the waveguide vs cavity versions of this thing.
B. Don't have access to COMSOL anyway
C. Used square avg to get close to an equivalent vol cylinder
D. Didn't use included dielectric, but assumed it might change the res freq somewhat. Prob the biggest errors. (enough to change mode numbers?)
E. Sounds like what I wound up using
F. Nice !
G. Those still might not be the correct modes given that the dielectric is not taken into account along w/ the dimensional approximations. i didn't try all matches.
...snip...
@Notsosureofit
I encountered similar problems with the mode numbers. I note here http://mathworld.wolfram.com/BesselFunctionZeros.html that the first J'0(x) value is given as non-zero in the table but the curve of J0(x) clearly shows a zero for the derivative at x=0. This is explained in the text, but the text does not resolve the question of which value to use for cavity resonance. That of course will change the TE mode numbers.
There is also the remark that I found (somewhere) while reading up on Bessel functions or resonance cavitys. In communicating mode numbers it is important to specify the order of the subscripts. Some people communicate TE0,1 as TE1,0 and so forth.
Regarding your concern W.R.T. frequency changing with dielectric. The dielectric constant for the Brady cavity dielectric is about 1.76, I think. I obtained this value by iteratively running meep for different values of the dielectric constant and using Harminv to calculate the resonant frequency. The cavity resonated at 1880.5 MHz using a dielectric value of 1.76. I called that close enough. Note that is about 50% density of P.E.
I will check using meep again to see how strong an effect the dielectric constant has on the resonance frequency. I have conflicting recall on that subject at the moment. -
#232 by Rodal on 20 Jan, 2015 19:56
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For TM modes, X[sub m,n] = the n-th zero of the m-th Bessel function.
[1,1]=3.83, [0,1]=2.40, [0,2]=5.52 [1,2]=7.02, [2,1]=5.14, [2,2]=8.42, [1,3]=10.17, etc.
and for TE modes, X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function.
[0,1]=3.83, [1,1]=1.84, [2,1]=3.05, [0,2]=7.02, [1,2]=5.33, [1,3]=8.54, [0,3]=10.17, [2,2]=6.71, etc.
So, using these to identify the frequencies, I chose:
Bradya => TM122 or TE022 X[sub m,n] = 7.02 p = 2
Bradyb => TE213 X'[sub m,n] = 3.05 p = 3
Bradyc => TE222 X'[sub m,n] = 6.71 p = 2
....
1) I have checked your above-given definitions for Xm,n and X'm,n vs. the ones obtained from Mathematica and vs. the ones given by University of Kyoto, Japan in http://wwwal.kuicr.kyoto-u.ac.jp/www/accelerator/a4/besselroot.htmlx . They are all identical, therefore the discrepancies in mode shape definitions that I will discuss below have nothing to do with the definitions of Xm,n and X'm,n . You and I are using identical definitions. (For anybody interested in the reason why Mathematica, the University of Kyoto and others has the J'0(x)@x=0 value as non-zero (as opposed to the old reference: Abramowitz) , see for example Theorem 3.1 in http://www.irjabs.com/files_site/paperlist/r_1259_130901233803.pdf : the value of the derivative at the origin (x=0) is mathematically undefined as the closed-form solution has zero/zero at the origin.)
2) It appears that the exact geometrical dimensions play a paramount role concerning the actual mode shape for a given frequency, particularly for higher modes, where different mode shapes are more likely to give frequencies close to each other. This makes sense.
3) Let's define as "Aero geometry" the following definition for the NASA Brady et. al. cavity:
Aero Best estimate as of 11/9/2014 http://forum.nasaspaceflight.com/index.php?topic=29276.msg1285896#msg1285896
cavityLength = 0.24173 m
bigDiameter = 0.27246 m
smallDiameter = 0.15875 m
4) Let's define as "Fornaro geometry" the following definition for the NASA Brady et. al. cavity:
Fornaro estimate http://forum.nasaspaceflight.com/index.php?topic=36313.msg1302455#msg1302455
cavityLength = 0.332 m
bigDiameter = 0.397 m
smallDiameter = 0.244 m
5) Let's use the GeometricalMeanDiameter=Sqrt[bigDiameter*smallDiameter] as the equivalent diameter of the equivalent cylindrical cavity
6) Given the experimentally reported frequencies, the geometrical dimensions and the value of speed of light in air, one can invert the frequency equation (see: http://en.wikipedia.org/wiki/Microwave_cavity#Cylindrical_cavity ) to obtain Xm,n and X'm values as a function of constants and the longitudinal mode shape number "p". Let's define the error difference between these Xm,n and X'm,n values and actual Xm,n and X'm values as:
error= (value of Xm,n or X'm,n obtained from frequency eqn.)/ (correct value of Xm,n or X'm,n ) -1
where Xm,n is used for TM modes and X'm,n is used for TE modes.
7) Then I obtain the following mode shapes:
BRADY "A"
Fornaro Geometry
Best result: TM022 error=+0.956%
2nd best: TM310 error= - 1.206%
Aero Geometry
Best result: TE310 error=+0.255%
2nd best: TM111 or TE011 error= + 4.11%
BRADY "B"
Fornaro Geometry
Best result: TM310 error= - 0.997%
2nd best: TM022 error= + 1.230%
Aero Geometry
Best result: TE310 error=+0.468%
2nd best: TM111 or TE011 error= + 4.36%
BRADY "C"
Fornaro Geometry
Best result: TE412 error= + 1.166%
2nd best: TE313 error= + 1.259%
Aero Geometry
Best result: TE212 error=+0.859%
2nd best: TE011 or TM111 ; error= + 0.971%
CONCLUSIONS
1) The mode-shape to frequency relation is very sensitive to the exact geometrical dimensions of the cavity. The above-given Fornaro and Aero guesses of the dimensions of the Brady et.al cavity give very different mode shapes (all other parameters being the same).
2) Before these calculations the consensus was that Aero's latest estimates of the geometry were superior (for a number of reasons). These calculations give further confirmation that Brady et.al. actual geometry may be closer to Aero's estimates:
2a) In all cases examined above, the errors are smaller using the Aero estimate of geometry.
2b) Aero's estimate of geometry (predicting a smaller Brady et.al. cavity than Fornaro) lead to more stable values of mode shape with frequency: Aero's estimate gives the same mode shape (TE310 for both Brady "A" and Brady"B", which differ very little in frequency). Fornaro's geometry gives different mode shapes for Brady "A" and "B".
2c) Aero's estimate of geometry gives more discrimination between mode shapes for Brady cases "A" and "B": the mode shape (TE310) with the smallest error has an error off less than 1% while the next closest mode shapes have errors exceeding 4%. Fornaro's geometry has errors much closer together which do not provide as much power to discriminate between the actual mode shape.
3) It is interesting that I obtained for Brady case "C" (NASA's experiment which gave the largest by far thrust/PowerInput) mode TE01 as the second best mode shape, with an error of less than 1%: just 0.971%. This coincides with NASA's Brady et.al. mode shape first two quantum numbers (circumferential and radial). I propose that TE01 may indeed be the actual mode shape for Brady "C" (the difference in errors between TE01 and TE21 is insignificant) because of the effect of the dielectric polymer in the cavity (which we do not take into account). The dielectric polymer is a circumferential doughnut-shape polymer that must have a TE01 mode shape - obviously (see this picture). This dielectric polymer must force the cavity into the TE01 mode shape. As to why I calculate TE011 while NASA Brady et.al give TE012 that also can be readily explained by the polymer dielectric as the dielectric will produce an extra longitudinal full wave in the dielectric polymer, resulting in two full waves in the longitudinal direction of the whole cavity: 1) one full wave within the doughnut-shaped polymer dielectric itself, along the thickness of the doughnut and 2) the other full wave within the longitudinal direction of the rest of the empty cavity that has no dielectric polymer.
4) The Aero geometry estimate of the Brady et.al. cavity (that appear to be the best geometrical estimate) gives TE (transverse electric) mode shapes as the best mode shape estimates for all frequencies tested by NASA Brady et.al. Actually, the TE01 mode is a close mode shape for all the Brady experiments. It just happens that the frequency that best excites TE01 happens to be Brady C, that provided the highest thrust/PowerInput. This is critical, because TE01 is also the mode excited by the dielectric polymer. This is important also if indeed the thrust force is mainly a result of the internal, centrally located magnetic field (contained within the Transverse Electric (TE) circumferential field) being responsible for the experimental results (as previously argued either as thermal buckling artifact or whether as a result of the central longitudinal magnetic field interacting with the Quantum Vacuum providing quantum vacuum radiation pressure, for example).
5) I have to double-check my Mathematica program for any errors and my above transcription for any errors. Thank you for your patience
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#233 by JasonAW3 on 20 Jan, 2015 21:18
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Quote
At constant input power, the thrust, and therefore the acceleration, must decrease with time, to ensure that the spacecraft's velocity never exceeds 2*Power/Thrust
I was going to start a thread, but there is a photon rocket variant idea I have been batting around for a few months now:
take a long hollow cylinder, closed at one end, open on the other. Probably several hundred meters long, by three or four meters in diameter. Running the length of this cylinder, spaced at even intervals are low beams - probably no more than ten or twenty centimeters high. So the inner edge of the cylinder has...call it a dozen shallow troughs. At regular intervals - maybe a meter - these beams have specially designed reflective points. Open end of the cylinder, you have a powerful high frequency laser (or something emitting a focused photon beam) aimed at a 45 degree angle into each trough. One laser per trough, call it twelve total.
Now, a laser, like a military searchlight, is also a photon rocket.
Photons, as pointed out in the previous thread are durable little critters, and can bounce around a good 50,000+ times before going wherever it is expired photons go. And photons can transfer momentum with each bounce.
So, turn the lasers on. The initial 'thrust' is backward. Actually, 'backward and sideways' because of the angle.
That thrust gets negated at the first bounce point. Photons hit that (reflective) point, transfer momentum, and head over to the next bounce point, set at a 45 degree angle to the first.
At the second bounce point, the whole thing is moving forward. Repeat for the length of the cylinder. Because the photons are hitting at an angle, the cylinder might start rotating as well as moving forward, but I don't see that as a major issue. At the end of the cylinder, the photons hit a shaped surface and bounce back along the tubes center and out into space.
Did a bit of reading on laser propulsion systems. A Doctor Bae ran some laboratory tests on this: bouncing laser beams multiplied the 'thrust' by a factor of 3000+ - into EM Drive territory without the physics headache. He proposed two linked spacecraft, with laser beams between them, something NASA is supposed to be looking into for near earth applications. My idea is one spacecraft (the cylinder) and a multiplier of about 1500, if the cylinder is long enough. Not sure, but that's should be on a par with the Brady EM drive model.
Alter the angles a bit, test different lasers/emitters, might get a lot more work out of the photons, increasing thrust further.
Would this violate the paradox?
You seem to misunderstand the fundamentals of mechanics.
If a photon hits a mirror at a 45 degree angle and reflects off it, the mirror will receive an impulse perpendicular to the plane of the mirror only. It will not be pushed in the direction of the other component of the photon at all.
It's not just with photons. If you have a billiard ball and you bounce it off another billiard ball that was stationary so that the original ball end up leaving at a 90 degree angle to its initial direction of travel, the other ball will end up traveling at a 45 degree angle to the path of the original ball.
It's non-intuitive because our intuition is shaped by friction tending to pull things along, but such friction is not a part of purely elastic collisions, and photons bouncing off mirrors are purely elastic.
So, every time your photon bounces off the wall, the momentum it imparts will only be to push outward perpendicular to the axis of the tube. And it will be cancelled by the next bounce off the opposite wall.
The only effect of the net momentum of the tube is the opposite of whatever momentum the photo has when it finally leaves the tube. Whatever it does as it bounces around in the tube will have no net effect.
What about a severe conewith a 5 degree opening? Lasers shot up into th mirrorwould glance off at an approximate 7 degerr angle, still proceeding upwards to the conical apex, where they would then be relected directly outwards. Obviously the photons would loose energy with each bounceand you'd have to cool the conical mirror, as at best, you MIGHT get a 97% reflectivity from it, but could THAT work as a propulsive system? (BTW the 7 degree is an approximation of the angle. Your milage may vary). -
#234 by Rodal on 20 Jan, 2015 23:19
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For TM modes, X[sub m,n] = the n-th zero of the m-th Bessel function.
[1,1]=3.83, [0,1]=2.40, [0,2]=5.52 [1,2]=7.02, [2,1]=5.14, [2,2]=8.42, [1,3]=10.17, etc.
and for TE modes, X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function.
[0,1]=3.83, [1,1]=1.84, [2,1]=3.05, [0,2]=7.02, [1,2]=5.33, [1,3]=8.54, [0,3]=10.17, [2,2]=6.71, etc.
So, using these to identify the frequencies, I chose:
Bradya => TM122 or TE022 X[sub m,n] = 7.02 p = 2
Bradyb => TE213 X'[sub m,n] = 3.05 p = 3
Bradyc => TE222 X'[sub m,n] = 6.71 p = 2
....
OK, let's use your terminology to check the labeling of the modes.
The lowest modes that have exactly the same numerical values, according to your definition above are:
For TM modes, X[sub m,n] = the n-th zero of the m-th Bessel function.
[1,1]=3.83 Therefore m=1,n=1 and TM11 corresponds to X1,1=3.83
and for TE modes, X'[subm,n] = the n-th zero of the derivative of the m-th Bessel function.
[0,1]=3.83 Therefore m=0,n=1 and TE01 corresponds to X'0,1=3.83
* Actually, in general X1,n = X'0,n for any n *
*
EDIT:
therefore
frequency of TM1np = frequency of TE0np for given radius and cavity length, for any value of n and any value of p
*
Therefore the modes that have the same numerical value are TM11 and TE01 (and not TM12 or TE02).
Conclusion: it appears that the radial quantum number "n" in your mode number is higher than the correct value by one. (For some reason it also appears that your longitudinal "p" is also inflated by one: it should be TM111 instead of TM122 and it should be TE011 instead of TE022). -
#235 by Notsosureofit on 20 Jan, 2015 23:45
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X[sub 1,a] = X'[sub 0,a]
These are the columns that correspond.
http://mathworld.wolfram.com/BesselFunctionZeros.html
TM111 gives me 1.03 GHz ditto for TE011
I couldn't find a formula for a tapered cavity.
I'm thinking the volumetric radius might be the best choice as radius
R^2 = (a^2+a*b+b^2)/3
Havn't had time to try recalculations, but if you have Mathematica you can try lots -
#236 by Rodal on 21 Jan, 2015 00:02
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TM111 gives me 1.03 GHz ditto for TE011
From http://en.wikipedia.org/wiki/Microwave_cavity#Cylindrical_cavity, the frequency equation gives:
frequencyTM111 =( cAir/(2*Pi))*Sqrt[((X1,1 /(aeroGeometricMeanDiameter/2))^2) +(1*Pi/aeroLength)^2]
= 1.8643 GHz instead of 1.03 GHz
ditto for frequencyTE011
using
m=1
n=1
p=1
X1,1 = 3.83170597020751
cair = 299705000 m/s = (299792458 m/s) / (Sqrt[mur*epsilonr] = c / (Sqrt[mur*epsilonr])
aeroGeometricMeanDiameter = 0.207974 m = Sqrt[ (0.27246 m) * (0.15875 m)]
aeroLength = 0.24173 m
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#237 by Rodal on 21 Jan, 2015 00:26
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TM111 gives me 1.03 GHz ditto for TE011
If, were one to use the diameter instead of the radius in the frequency equation one would obain:
frequencyTM111 =( cAir/(2*Pi))*Sqrt[((X1,1 /(aeroGeometricMeanDiameter))^2) +(1*Pi/aeroLength)^2]
= 1.07 GHz which is very close to your calculated 1.03 GHz
==> Please consider whether your mode shape calculations may be incorrect perhaps because you inadvertently may have used the diameter instead of the radius in the expression for frequency <===
Remember:
Aero Best estimate as of 11/9/2014 http://forum.nasaspaceflight.com/index.php?topic=29276.msg1285896#msg1285896
cavityLength = 0.24173 m
bigDiameter = 0.27246 m
smallDiameter = 0.15875 m -
#238 by Notsosureofit on 21 Jan, 2015 01:50
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We are using different radii. I'm thinking of going to the volumetric, might be closest, still looking for an actual f solution.
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#239 by Rodal on 21 Jan, 2015 01:59
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Are you thinking this way?
I'm thinking the volumetric radius might be the best choice as radius
R^2 = (a^2+a*b+b^2)/3
Havn't had time to try recalculations, but if you have Mathematica you can try lots
Defining these symbols for the radii of the truncated cone (frustum of a cone):
r=smallRadius=smallDiameter/2
R=bigRadius=bigDiameter/2
The volume of a truncated cone and the volume of a cylinder are:
Volume of the frustum of a cone =Height*Pi*(r^2+r*R+R^2)/3
Volume of a cylinder= Pi*(EquivalentR^2)*Height
Equating these volumes one arrives at an expression for the Equivalent Radius of a cylinder having the same volume as the volume of the frustum of a cone:
Volume of a cylinder = Volume of the frustum of a cone
Pi*(EquivalentR^2)*Height = Height*Pi*(r^2+r*R+R^2)/3
hence
EquivalentCylindricalRadius = Sqrt[(r^2+r*R+R^2)/3]
= Sqrt[(smallDiameter^2+smallDiameter*bigDiameter+bigDiameter^2)/12]
See this example: it makes very little difference what mean to use:
aeroLength=0.24173 meter;aeroBigDiameter=0.27246 meter;aeroSmallDiameter=0.15875 meter;
aeroGeometricMeanDiameter=Sqrt[aeroBigDiameter*aeroSmallDiameter]
= 0.207974 meter
aeroMeanDiameter =(aeroBigDiameter+aeroSmallDiameter)/2
= 0.215605 meter
aeroVolumetricDiameter=Sqrt[(aeroSmallDiameter^2+aeroSmallDiameter*aeroBigDiameter+aeroBigDiameter^2)/3]
= 0.218089 meter
The differences between these different means (just a few % between them) is less than the level of uncertainty we have for the geometry. The difference between these means is much less than the difference between Aero and Fornaro's estimates of the geometry. It would be useful to know what dimensions did you use.
