I would like advice on lifetime estimates of the power within the cavity - what is the expected half life? Does a simple expression exist involving the Q-factor? I remind that Q can be defined as the angular frequency multiplied by the ratio of (stored energy) / (rate of energy dissipation).
Continuing with the "Test In Space" theme.We lack cheap space access for unmanned cargo. We have no railgun running from the coast of Ecuador up into the Andes to the east, and we have no Skylon/SABRE SSTO yet. So we must pay many thousands of dollars per launched kilogram rather than what could be only tens of dollars. What we do have is Cubesat and SpaceX. The problem is that the devices under consideration here won't fit even into the largest Cubesat. So let's talk miniaturisation.What we have is photons in an asymmetric cavity. So let's use light instead of microwaves. I'll stop there for now.
Quote from: deltaMass on 05/10/2015 05:39 pmContinuing with the "Test In Space" theme.We lack cheap space access for unmanned cargo. We have no railgun running from the coast of Ecuador up into the Andes to the east, and we have no Skylon/SABRE SSTO yet. So we must pay many thousands of dollars per launched kilogram rather than what could be only tens of dollars. What we do have is Cubesat and SpaceX. The problem is that the devices under consideration here won't fit even into the largest Cubesat. So let's talk miniaturisation.What we have is photons in an asymmetric cavity. So let's use light instead of microwaves. I'll stop there for now.I like where you're going with this...........and I want to talk miniaturization too. @Notsosureofit is working with 10ghz gunn diodes so he might be able to help.Judging from reading (and really reading) the latest tweets from @ElonMusk, I feel that one of these Emdrive tests will find a ride on a rocket real soon. http://forum.nasaspaceflight.com/index.php?topic=36313.msg1368379#msg1368379Note there is also a (very much new) favorable Io9 article released on April 30th,(A day after this: http://www.nasaspaceflight.com/2015/04/evaluating-nasas-futuristic-em-drive/)that wasn't linked to in the tweet. What was linked to was an old negative article from August last year.
Quote from: frobnicat on 05/10/2015 01:31 pmQuote from: Mulletron on 05/10/2015 11:57 amI know that the frustum design that I currently have was originally driven by a high power (and dangerous to most people) magnetron. The frequency range of your average microwave oven magnetron and wifi are the same. I verified the frustum will resonate within this frequency range using a spectrum analyzer and a SNA. Mine works on wifi channel 1 and 10. Given that Eagleworks was able to observe thrust with only 2.6 watts for one of their tests*, I think this is an acceptable risk to take. Besides, it is just money right? So I am literally driving the frustum with the RF from the wifi camera (used to observe and is riding on the experiment) and a 2watt amp. I can scale this up to 20 watts using other amps if needed. Amps are cheap and plug and play.* http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdfI think the frequency hopping aspect of the waveform might end up doing me in though.So this is a gamble. I'm accepting the risk of not getting a successful replication attempt in hopes that if I do, I will have built a ready-made solution for mister tinkerer to easily observe anomalous thrust at home.If all that fails, I'll just shove in a magnetron.
Quote from: Mulletron on 05/10/2015 11:57 amI know that the frustum design that I currently have was originally driven by a high power (and dangerous to most people) magnetron. The frequency range of your average microwave oven magnetron and wifi are the same. I verified the frustum will resonate within this frequency range using a spectrum analyzer and a SNA. Mine works on wifi channel 1 and 10. Given that Eagleworks was able to observe thrust with only 2.6 watts for one of their tests*, I think this is an acceptable risk to take. Besides, it is just money right? So I am literally driving the frustum with the RF from the wifi camera (used to observe and is riding on the experiment) and a 2watt amp. I can scale this up to 20 watts using other amps if needed. Amps are cheap and plug and play.* http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdfI think the frequency hopping aspect of the waveform might end up doing me in though.So this is a gamble. I'm accepting the risk of not getting a successful replication attempt in hopes that if I do, I will have built a ready-made solution for mister tinkerer to easily observe anomalous thrust at home.If all that fails, I'll just shove in a magnetron.
I know that the frustum design that I currently have was originally driven by a high power (and dangerous to most people) magnetron. The frequency range of your average microwave oven magnetron and wifi are the same. I verified the frustum will resonate within this frequency range using a spectrum analyzer and a SNA. Mine works on wifi channel 1 and 10. Given that Eagleworks was able to observe thrust with only 2.6 watts for one of their tests*, I think this is an acceptable risk to take. Besides, it is just money right? So I am literally driving the frustum with the RF from the wifi camera (used to observe and is riding on the experiment) and a 2watt amp. I can scale this up to 20 watts using other amps if needed. Amps are cheap and plug and play.* http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdfI think the frequency hopping aspect of the waveform might end up doing me in though.So this is a gamble. I'm accepting the risk of not getting a successful replication attempt in hopes that if I do, I will have built a ready-made solution for mister tinkerer to easily observe anomalous thrust at home.If all that fails, I'll just shove in a magnetron.
Quote from: SeeShells on 05/10/2015 02:03 pmNot a single thing I built defied the laws of physics or the formulas of the trade. . . Maxwell, Ohms law, etc. If something didn't work for some weird reason, it still followed the basic laws and formulas when it ended up.It doesn't really matter to me what is happening inside of the EM Chamber it must follow the principals of physics and conservation of energy and momentum is one of them. If I have a Air Tank pressurized with 200psi of air and a audio speaker inside that can blast at 100 watts any frequency range no matter what mixture of sound or what mixture of harmonics I crank, the tank will not move, but put a hole in one end and stand back. The second law of thermodynamics states that the entropy of an isolated system never decreases and the EM Chamber is an isolated enclosed system, we think. If we are getting thrust that, thrust must be acting outside the chamber in some form. This is why I asked the simple question if smoke was used in the tests, it wasn't to detect thermal air currents but to see if it was moving away from any thrust from the EM Chamber. Smoke is small .5 to 2 um and might be be directly effected. If not then look for other forms of accelerated energy, providing thrust emanating out of the EM Chamber. I too would like to see a smoke test. I can't see conservation of momentum being violated. It just goes against everything we know both empirically and theoretically. I think that even in the off chance that the EmDrive is not experimental error, conservation of momentum will still hold albeit in a more subtle manner than the classical analysis would expect.You gave the illustrative example of a closed container with different traveling and standing waves of different frequencies and amplitudes bouncing around inside. There is a very neat quantum mechanical reason that such a container is not truly closed. Even in an infinite potential well, the wave function can extend outside the walls of the well, leading to effects such as tunneling. Another great example of the wavefunction extending beyond barriers that appears to be somewhat related to the possible effect seen here is the Aharanov Bohm effect: http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect. This is due to the wavefunction of a particle outside of a container extending past the barrier of the container and interacting with the EM field on the inside of the container.Now I leave this paper to ruminate upon:http://arxiv.org/abs/0708.0681Perhaps the EmDrive is acting as an evanescent mode photon rocket where momentum is carried away outside the cavity via this mechanism.
Not a single thing I built defied the laws of physics or the formulas of the trade. . . Maxwell, Ohms law, etc. If something didn't work for some weird reason, it still followed the basic laws and formulas when it ended up.It doesn't really matter to me what is happening inside of the EM Chamber it must follow the principals of physics and conservation of energy and momentum is one of them. If I have a Air Tank pressurized with 200psi of air and a audio speaker inside that can blast at 100 watts any frequency range no matter what mixture of sound or what mixture of harmonics I crank, the tank will not move, but put a hole in one end and stand back. The second law of thermodynamics states that the entropy of an isolated system never decreases and the EM Chamber is an isolated enclosed system, we think. If we are getting thrust that, thrust must be acting outside the chamber in some form. This is why I asked the simple question if smoke was used in the tests, it wasn't to detect thermal air currents but to see if it was moving away from any thrust from the EM Chamber. Smoke is small .5 to 2 um and might be be directly effected. If not then look for other forms of accelerated energy, providing thrust emanating out of the EM Chamber.
Quote from: deltaMass on 05/07/2015 02:52 amQuote from: Einstein79 on 05/07/2015 02:48 amThe reason for the confusion over the violation of classical physics is because this system has nothing to do with classical physics. Moreover, the “thrust” that is being calculated is not thrust at all but space moving the drive from one position to another which can merely be related to thrust but is not, per se, thrust. The controlling factor here is, of course, the resonant frequency. If you match the resonant frequency that space uses to “hold” the object you will develop a “cavity” that the “object will move towards”. The reason why the device cannot be “pushed off of” for conservation of momentum to hold true is because space is already pushing on it satisfying the law. A couple of postulates to keep in mind that will help with these experiments are:1. Space creates light.2. Space itself is a resonating chamber.Interesting! Would you then be prepared to write down the equations of motion so that we can play with them?The equations of motion do not exist from an inertial reference frame. We must assume that the object is not moving and that space is moving around and through the object. I am trying to develop the Hamiltonian for space but having difficulty because we have always assumed space to be a virtual plasma and it is not virtual at all, but real.
Quote from: Einstein79 on 05/07/2015 02:48 amThe reason for the confusion over the violation of classical physics is because this system has nothing to do with classical physics. Moreover, the “thrust” that is being calculated is not thrust at all but space moving the drive from one position to another which can merely be related to thrust but is not, per se, thrust. The controlling factor here is, of course, the resonant frequency. If you match the resonant frequency that space uses to “hold” the object you will develop a “cavity” that the “object will move towards”. The reason why the device cannot be “pushed off of” for conservation of momentum to hold true is because space is already pushing on it satisfying the law. A couple of postulates to keep in mind that will help with these experiments are:1. Space creates light.2. Space itself is a resonating chamber.Interesting! Would you then be prepared to write down the equations of motion so that we can play with them?
The reason for the confusion over the violation of classical physics is because this system has nothing to do with classical physics. Moreover, the “thrust” that is being calculated is not thrust at all but space moving the drive from one position to another which can merely be related to thrust but is not, per se, thrust. The controlling factor here is, of course, the resonant frequency. If you match the resonant frequency that space uses to “hold” the object you will develop a “cavity” that the “object will move towards”. The reason why the device cannot be “pushed off of” for conservation of momentum to hold true is because space is already pushing on it satisfying the law. A couple of postulates to keep in mind that will help with these experiments are:1. Space creates light.2. Space itself is a resonating chamber.
Quote from: Einstein79 on 05/08/2015 03:42 pmQuote from: deltaMass on 05/07/2015 02:52 amQuote from: Einstein79 on 05/07/2015 02:48 amThe reason for the confusion over the violation of classical physics is because this system has nothing to do with classical physics. Moreover, the “thrust” that is being calculated is not thrust at all but space moving the drive from one position to another which can merely be related to thrust but is not, per se, thrust. The controlling factor here is, of course, the resonant frequency. If you match the resonant frequency that space uses to “hold” the object you will develop a “cavity” that the “object will move towards”. The reason why the device cannot be “pushed off of” for conservation of momentum to hold true is because space is already pushing on it satisfying the law. A couple of postulates to keep in mind that will help with these experiments are:1. Space creates light.2. Space itself is a resonating chamber.Interesting! Would you then be prepared to write down the equations of motion so that we can play with them?The equations of motion do not exist from an inertial reference frame. We must assume that the object is not moving and that space is moving around and through the object. I am trying to develop the Hamiltonian for space but having difficulty because we have always assumed space to be a virtual plasma and it is not virtual at all, but real.Bumping this because of its elegance.
I'm asking about the energy in the cavity (for reasons that may be clearer later). Here's a really dumb way to calculate it. This is so ugly, I hope someone shoots it down. It goes like this:We know the input power, so it remains only to estimate the time to "fill" the cavity. We can think of the Q factor as the number of bounces inside the cavity (!). Taking Q=6,000 and L = 0.25 m, we get t = 5 us.Thus for an input power of 50 W, the cavity energy is 2.5*10-4 Joules.Pretty bad hunh?Another way to calculate it is to take the cavity volume and the average E-field from the simulation, and then the cavity energy is 0.5*epsilon0*E2*V Joules.
Quote from: deltaMass on 05/10/2015 07:50 pmI'm asking about the energy in the cavity (for reasons that may be clearer later). Here's a really dumb way to calculate it. This is so ugly, I hope someone shoots it down. It goes like this:We know the input power, so it remains only to estimate the time to "fill" the cavity. We can think of the Q factor as the number of bounces inside the cavity (!). Taking Q=6,000 and L = 0.25 m, we get t = 5 us.Thus for an input power of 50 W, the cavity energy is 2.5*10-4 Joules.Pretty bad hunh?Another way to calculate it is to take the cavity volume and the average E-field from the simulation, and then the cavity energy is 0.5*epsilon0*E2*V Joules.Resonate circuit has a time constant. 5 time constants to fill.TC = Q / (2 Pi Fr). Shawyer did comment on this and gave example.
Quote from: Rodal on 03/10/2015 01:42 amQuote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.The previous expression is only valid approximation for a "uniformly varying dielectric". There is no L1 and L2 in that case.What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )
Quote from: Notsosureofit on 03/10/2015 01:29 am@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !I find your previous expressiondel f = ( f/(2*c^2)) * (c1^2-c2^2)more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length, while on the other hand del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))goes to zero for equal dielectric lengths, regardless of their dielectric constants.
@ RODALJust got a minute but from your p expression;If L1/c1 = L2/c2del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))might be a solution ??Got to check the thinking later.Night !
Sure, but I'm just pointing out what the lack of cheap and readily available space access costs us in terms of either proving or disproving new propulsion theories. It would have saved Woodward 20 years of mucking about, for instance. It will cost this endeavour years also.