Quick question.In my physics education. "Q" stood for heat energy, or charge, or resonance efficiency, depending on what you are talking about.In most of these EMDrive conversations, the distinction hasn't been particularly worrisome, but I find myself asking what does "Q" mean in the context of these latest conversations. Relatedly, is the definition of "Q" interchangeable in this context?Thanks.
Quote from: Rodal on 06/16/2015 02:51 pmQuote from: WarpTech on 06/16/2015 03:43 amI've updated my Theory on the Wiki.http://emdrive.wiki/Todd_Desiato_(@WarpTech)%27s_Evanescent_Wave_Theory#Application_of_Theory_to_the_EmDriveThe EM Drive is intentionally designed to have asymmetrical attenuation. As such, energy is reflected from the large end to be stored at the small end as induced currents. Standing waves store energy and as such, store mass. As the EM drive charges and the Q ramps up, energy from the input source is reflected from the large end and stored in the small end on each successive reflection cycle. This energy is stored as induction currents caused by the near-field effects of evanescent waves. Due to the phase shift, the Power Factor is not zero as it is with standing waves. Therefore, work can be done to move the EM Drive. This dynamic action of storing mass-energy toward the front causes the center of mass to walk forward. The increasing pressure on the small end causes the EM Drive to accelerate forward due to the internal pressure gradient, until the pressure is equalized. Then the cycle builds again. This dynamic implies that a high Q value is not required, but rather how quickly can energy be ramped up under extreme attenuation conditions.ToddGot to find some way that enough mass (or energy) leaks out (somehow) asymmetrically of the EM Drive to justify the claimed self-acceleration of its center of mass without breaking Conservation of Momentum. Either that, or you have to couple to an external directional field.(delete)EDIT: I'm confused as to why putting energy "in" does not result in the same physics as letting energy "out". If the system is gaining energy from the outside, +dm/dt, why is that not the same as expelling it to the outside as -dm/dt in the opposite direction?
Quote from: WarpTech on 06/16/2015 03:43 amI've updated my Theory on the Wiki.http://emdrive.wiki/Todd_Desiato_(@WarpTech)%27s_Evanescent_Wave_Theory#Application_of_Theory_to_the_EmDriveThe EM Drive is intentionally designed to have asymmetrical attenuation. As such, energy is reflected from the large end to be stored at the small end as induced currents. Standing waves store energy and as such, store mass. As the EM drive charges and the Q ramps up, energy from the input source is reflected from the large end and stored in the small end on each successive reflection cycle. This energy is stored as induction currents caused by the near-field effects of evanescent waves. Due to the phase shift, the Power Factor is not zero as it is with standing waves. Therefore, work can be done to move the EM Drive. This dynamic action of storing mass-energy toward the front causes the center of mass to walk forward. The increasing pressure on the small end causes the EM Drive to accelerate forward due to the internal pressure gradient, until the pressure is equalized. Then the cycle builds again. This dynamic implies that a high Q value is not required, but rather how quickly can energy be ramped up under extreme attenuation conditions.ToddGot to find some way that enough mass (or energy) leaks out (somehow) asymmetrically of the EM Drive to justify the claimed self-acceleration of its center of mass without breaking Conservation of Momentum. Either that, or you have to couple to an external directional field.
I've updated my Theory on the Wiki.http://emdrive.wiki/Todd_Desiato_(@WarpTech)%27s_Evanescent_Wave_Theory#Application_of_Theory_to_the_EmDriveThe EM Drive is intentionally designed to have asymmetrical attenuation. As such, energy is reflected from the large end to be stored at the small end as induced currents. Standing waves store energy and as such, store mass. As the EM drive charges and the Q ramps up, energy from the input source is reflected from the large end and stored in the small end on each successive reflection cycle. This energy is stored as induction currents caused by the near-field effects of evanescent waves. Due to the phase shift, the Power Factor is not zero as it is with standing waves. Therefore, work can be done to move the EM Drive. This dynamic action of storing mass-energy toward the front causes the center of mass to walk forward. The increasing pressure on the small end causes the EM Drive to accelerate forward due to the internal pressure gradient, until the pressure is equalized. Then the cycle builds again. This dynamic implies that a high Q value is not required, but rather how quickly can energy be ramped up under extreme attenuation conditions.Todd
Jose,Thanks for the explanation of "Q"! One more quick sophomoric question for anyone if I may: Could the EM Drive be creating an energy allegory to mass (for this narrow slice of frequency)? And it's the allegorical mass that is reacting to the directional field. In other words, E=mc2, so the more "E" we feed into the device, the more our allegorical mass grows until we can detect the space-time distortion with a laser (as Eagleworks did). That then begs the question of whether our allegorical mass is preferential in its formation due to the frustum shape, resulting in perceived thrust to an external viewer, when really it's just "falling" in a preferential direction.
All of known physics is based on CoM. All of it. Therefore, you are not going to get anywhere with known physics as an explanatory method for EmDrive.Therefore, either EmDrive is an artifact of poor experimental technique, or new physics is required.There is no third way.
Quote from: SeeShells on 06/16/2015 02:19 pmI look at it this way and correct me if I'm wrong. The EM Drive could be thought of like this. You are trapped in an enclosed tank with a fire extinguisher and you turn on the fire extinguisher expecting it to move the tank, it doesn't because it's a enclosed system, but if you run up the side of the tank changing the local enclosed gravity profile of the tank, then you can move it in the direction you are running. And you haven't violated any laws just changed the local enclosed profile.ShellIt's an interesting idea, but the flaw is the concept of an enclosed system. In moving up the tank wall you are working against the local gravity which is part of the working system. If your tank was in orbit, you could run around the inside all day and you'd get the tank spinning, but not change your orbit, as opposed to rolling across the ground, constantly applying a torque and lifting yourself against gravity.
I look at it this way and correct me if I'm wrong. The EM Drive could be thought of like this. You are trapped in an enclosed tank with a fire extinguisher and you turn on the fire extinguisher expecting it to move the tank, it doesn't because it's a enclosed system, but if you run up the side of the tank changing the local enclosed gravity profile of the tank, then you can move it in the direction you are running. And you haven't violated any laws just changed the local enclosed profile.Shell
This has been my thought experiment...a point or points coupling to a natural entropic force. Surfing the wave so to speak or riding the wind. Sorry for the basic description...I cannot visualize particle or wave ejection/leakage thrusting forward nor thermal radiation of some sort. If emdrive works I'm convinced (without the math to prove it yet) that its being coupled to an elemental, natural force we have yet to measure directly. I'll leave it as an entropic force for the time being...the natural tendency for all energy and matter to disperse.
If I understand correctly your intent to attribute cavity momentum to a field property, then the field will carry off equal and opposite momentum. So are you describing radiation reaction? If you are, then because the system is closed, there is net zero momentum expressed externally.You also need to account for forces that are 1000x greater than that expected of a photon rocket, and you seem intent on doing that using electromagnetism. Known physics says that this is impossible.
Quote from: deltaMass on 06/16/2015 06:29 pmIf I understand correctly your intent to attribute cavity momentum to a field property, then the field will carry off equal and opposite momentum. So are you describing radiation reaction? If you are, then because the system is closed, there is net zero momentum expressed externally.You also need to account for forces that are 1000x greater than that expected of a photon rocket, and you seem intent on doing that using electromagnetism. Known physics says that this is impossible."Known" does not mean "understand", since I "know" that gravity can be mimicked by an EM effect, and this is not well understood by others. So this response does not surprise me. In a Newtonian gravitational field, an object falls to lower its potential energy. It loses energy and gains mass in the process. It does not eject mass to conserve momentum;E => E/sqrt(K)m => m*K^3/2In what I've proposed, the frustum "falls" forward to lower it's potential energy as it gains mass. The frustum is charged by a magnetron to raise it's potential energy and input more mass. It doesn't lose mass if it gains velocity.The "New Physics" you want is right in front of you! I'm showing you how to mimic gravity using magnetic flux as a gauge-gravity potential. This is about as NEW as it gets! Gauge potentials are simply the potential for a phase shift. The phase shift caused by magnetic flux is indistinguishable from the phase shift caused by a gravitational field, acting on the identical wave function. Most people do not understand this clearly.Todd
Df = 0.7311 @ 24,100,700,000Hz (from their freq meter) @ TE01x as per attached
PIn=3 - 10mW Q=5e3 - 10e3therefore Thrust=0.073uN - 0.488uN