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_cavityor are these goodAm I better off going with the Kwok slide 13?http://www.engr.sjsu.edu/rkwok/EE172/Cavity_Resonator.pdfor this guy's method?http://www.chrislmueller.com/studies/Jackson8-6.pdfPlaying catch up here. This page is nifty. http://mathworld.wolfram.com/BesselFunctionZeros.html
...I think it might be time for a spaceflight application minute...
...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....
Quote from: Notsosureofit on 01/17/2015 02:39 pm...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....A minor point. For further clarity you may want to include a prime symbol (or an apostrophe) on X"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).
Space travel applications...If applied, would 'Notsosureofit's mode calculations significantly help or hinder the 'Mulletron Mission to Saturn'?
Quote from: ThinkerX on 01/19/2015 11:55 pmSpace 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.
@RODALA. 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 anywayC. Used square avg to get close to an equivalent vol cylinderD. 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 usingF. 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...
...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 = 2Bradyb => TE213 X'[sub m,n] = 3.05 p = 3Bradyc => TE222 X'[sub m,n] = 6.71 p = 2....
Quote from: ThinkerX on 01/12/2015 01:49 amQuoteAt constant input power, the thrust, and therefore the acceleration, must decrease with time, to ensure that the spacecraft's velocity never exceeds 2*Power/ThrustI 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.
QuoteAt constant input power, the thrust, and therefore the acceleration, must decrease with time, to ensure that the spacecraft's velocity never exceeds 2*Power/ThrustI 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?
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
...TM111 gives me 1.03 GHz ditto for TE011
...I'm thinking the volumetric radius might be the best choice as radiusR^2 = (a^2+a*b+b^2)/3Havn't had time to try recalculations, but if you have Mathematica you can try lots