tag:blogger.com,1999:blog-4637778157419388168.post2852426820840467815..comments2017-03-25T11:46:41.691-07:00Comments on Physics from the edge: Emdrives, dielectrics, & the Kaporin optimisation.Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-4637778157419388168.post-55727263993709016362017-01-15T02:46:04.385-08:002017-01-15T02:46:04.385-08:00That optimized length equation is interesting. The...That optimized length equation is interesting. The optimum length is 4 times the geometric mean of the small diameter and the large diameter. plshttp://www.blogger.com/profile/09797206871907086579noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-347019301331782492017-01-13T22:18:27.140-08:002017-01-13T22:18:27.140-08:00A question on Rindler horizons. If I have an obse... A question on Rindler horizons. If I have an observer at point B in an inertial frame, that observer would observe a universe with a diameter determined by the speed of light. If I have an observer A under constant acceleration towards point B this observer (A) sees a RIndler horizon behind him that is nearer than the horizon for a stationary or constant velocity (inertial frame).<br /><br />What happens when A passes B? If B can communicate with A then A has 'cheated' the RIndler horizon and made the unobservable directly observable. Or is this OK because any information B supplied would have to be information that A could potentially have observed at an earlier time?<br /><br />Another question on constants. There has been some suggestion that NASA observed possible changes of light speed in their drive. Any change in light speed would dramatically affect the diameter of the observable universe and (presumably) the Rindler horizon proportionally. Yet c is a major player in your equations. This suggests that the behaviour should change on startup. I am not understanding this at all. Can someone point me in the right direction?Trucklehttp://www.blogger.com/profile/11229906370265577613noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-27823275839872588872017-01-11T12:22:15.841-08:002017-01-11T12:22:15.841-08:00It is interesting that proposed optimisation gave ...It is interesting that proposed optimisation gave no result at all. Means that all tried theories are wrong. But there is a hope, despite all problems. Goal is, not achieve mNs, but achieve 40-50 Ns by KW. This force will drive our ships across our galaxy. We should not afraid mistakes, if this drive does not really work, another will be born sometimes...There is should be a way to connect all Universe. Hope that it is. Beloushkin Aleksandrhttp://www.blogger.com/profile/00813647956060798332noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-54901712431692641912017-01-10T09:27:22.184-08:002017-01-10T09:27:22.184-08:00If you speak with me i can only say
No
and that...If you speak with me i can only say<br /><br />No<br /><br />and that pnn origin from maxwell displacement current error<br /><br />http://www.calmagorod.org/pnn-la-sua-genesi/<br /><br />www.asps.it<br /><br />https://www.antennex.com/preview/Nov02/Nov0602/intro-dca1.htm<br />Emidio Lauretihttp://www.blogger.com/profile/07537450348603281511noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-91700598602423117892017-01-10T08:24:56.529-08:002017-01-10T08:24:56.529-08:00http://www.electrogravity.com/STAVROS/Stavros1.pdf...http://www.electrogravity.com/STAVROS/Stavros1.pdfPaulhttp://www.blogger.com/profile/12180228151727984771noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-61637806517661453662017-01-10T08:20:33.429-08:002017-01-10T08:20:33.429-08:00Hello I would like to inquire of your theory is co...Hello I would like to inquire of your theory is compatable with the phenomena published in the links below.<br /><br />http://jnaudin.free.fr/stvdmdoc/prplessp.htm<br /><br />http://jnaudin.free.fr/stvdmdoc/stvcap.htm<br /><br />http://jnaudin.free.fr/html/stvrfpnd3.htmPaulhttp://www.blogger.com/profile/12180228151727984771noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-70961328905300278942017-01-10T03:37:49.005-08:002017-01-10T03:37:49.005-08:00
The PNN (Non Newtonian Propulsion) F242 operative...<br />The PNN (Non Newtonian Propulsion) F242 operative again<br /><br /><br />https://neolegesmotus.wordpress.com/2017/01/08/f242-is-operative-again/Emidio Lauretihttp://www.blogger.com/profile/07537450348603281511noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-13534170735689671812017-01-10T03:12:35.388-08:002017-01-10T03:12:35.388-08:00Joesixpack: As you can see from the data, quantise...Joesixpack: As you can see from the data, quantised inertia also predicts the Cannae drive, which in my opinion works the same way as the emdrive. It's length is 3cm, its width tapers from 22cm to 20cm.Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-39369417779769430712017-01-10T02:23:00.169-08:002017-01-10T02:23:00.169-08:00Sir,
Sorry I misread this at first, the formula w...Sir,<br /><br />Sorry I misread this at first, the formula was staring me in the face. <br /><br />It is absolutely amazing that such a simple formula using an idea that was based in stuff we knew in the 1950s might help to boost us beyond the mire we've been in since the end of the Apollo missions.<br /><br />Rocketry is a 13th century technology. It is absurd to think we will ever get to Tau Ceti, let alone past Mars with rocketry.<br /><br />If the EM drive is superconducting, do you think it will match the results of the superconducting Cannae drive? Do you also believe that quantised inertia can explain the Cannae Drive as well? I can't understand how it would ave the same effect of slowing C in the cavity as the EM drive does.joesixpackhttp://www.blogger.com/profile/08912279232742819732noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-56434454524472632692017-01-09T18:27:22.159-08:002017-01-09T18:27:22.159-08:00Dan, ok, if you want to be all CORRECT about it.
...Dan, ok, if you want to be all CORRECT about it. <br /><br />I'm still not convinced I understand the debate about conservation of energy, though. Take instead a rock at a very high altitude beyond Earth's atmosphere, like L1. But it isn't in orbit; it is simply falling towards the Earth. It also will experience a constant acceleration at a constant mass, so a constant force accelerating it to very high speeds before it hits the Earth. I understand that there is an infinitesimal deflection of the Earth and conservation isn't a factor here, but why isn't this a problem of conservation of energy? It will accelerate to a speed where the kinetic energy is greater than unity, also. The standard explanation of "potential energy" seems strange to me. This is energy that doesn't exist; I don't think the energy heated the the rock and I I don't think it got heavier (does it)? It's seems to me a way of describing a way for energy to change form in way that allows for oscillation. So, the question is not (in my mind) does it violate the principle of conservation of energy, but what is the form of potential energy that it is trading against, which may not be obvious to us right now, just like the potential energy of the rock is undetectable if you don't know how it got there, or that the Earth is nearby. Potentially Dr. MCulloch's Casimir effect is where it comes from, but I'm sure we'll figure it out when we play with it more. But the real quaestion is "does it work," and that's starting to be answered. Jonathan Cardhttp://www.blogger.com/profile/17258349623846497813noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-42996076469367011532017-01-09T05:53:25.575-08:002017-01-09T05:53:25.575-08:00Julien: 1) OK, good point, I had to use a log-log ...Julien: 1) OK, good point, I had to use a log-log plot because of the wide range of values. I will add a table of raw values. 2) I did say the spread of values 'along' the ideal line, not across it.Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-86602254203487934002017-01-09T04:56:05.929-08:002017-01-09T04:56:05.929-08:00I have two questions about your diagram:
1. You ...I have two questions about your diagram: <br /><br />1. You said "the scales are logarithmic". <br />We must pay attention to the linear nature of the plots, because in a log-log graph, the logarithmic scale makes many disparate values look like a straight line:<br />https://en.wikipedia.org/wiki/Logâ€“log_plot#/media/File:LogLog_exponentials.svg<br /><br />2. You also said "The spread of the comparisons along the 'ideal' diagonal line is due to the Q factor."<br />I understand this for the dispersed white square plots, but the superconducting Cannae drive with huge Q > 10^7 as well as the Tajmar cavity with very small Q < 50 are almost exactly on the line, and both slightly on the same left side of the line. So plot spreading on either side of the ideal line does not seem only related to the Q factorâ€¦<br />Julien Geffrayhttp://www.blogger.com/profile/06498085495224042768noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-81800370504327142442017-01-09T02:44:21.618-08:002017-01-09T02:44:21.618-08:00Ho ho. Luckily, since Boltzmann's constant is ...Ho ho. Luckily, since Boltzmann's constant is so small: 1.38x10^-23 J/K, it is an infinitesimal amount! I wonder if, in the far future, information will be taxed?Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-72224274295411946382017-01-09T01:25:06.280-08:002017-01-09T01:25:06.280-08:00"Julien: thanks for the information.."
..."Julien: thanks for the information.."<br /><br />@mike <br />Now you have to pay back his energy bill!<br />8PAlain Coetmeurhttp://www.blogger.com/profile/08352476615242858677noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-65438450797536193732017-01-09T01:25:02.612-08:002017-01-09T01:25:02.612-08:00ShardPheonix: The minus sign should make no differ...ShardPheonix: The minus sign should make no difference to the optimisation since the formula is differentiated. You are right that eq. 14 does not include the effect of the cut-off width, which is very real and should also be considered: ws > cutoff width.Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-62838738525457969202017-01-09T01:19:39.384-08:002017-01-09T01:19:39.384-08:00Julien: thanks for the information..Julien: thanks for the information..Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-7451936939086348742017-01-09T01:13:48.518-08:002017-01-09T01:13:48.518-08:00Dan: in response to a student who is worried about...Dan: in response to a student who is worried about energy conservation, I would mention the well-known Casimir effect, which is extracting a tiny amount of energy from the zero point field by making a gradient in it.<br /><br />When they complain that this is a tiny effect, you can point out that a door that is even slightly ajar, can open wide, and point them at cosmic acceleration, which is a well-observed huge violation of conservation of energy, which quantised inertia predicts in the same manner: in this case, a horizon (this time at the cosmic edge) is making a gradient in the zero point field that stars are accelerating down. More generally, quantised inertia (MiHsC, horizon mechanics) replaces the old conservation of energy with conservation of mass-energy plus information (a horizon hides information behind it, releasing mass-energy).<br /><br />Indeed, there are other thermodynamic and relativistic complications that will limit the performance of the emdrive..Mike McCullochhttp://www.blogger.com/profile/00985573443686082382noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-73333977519953791482017-01-08T18:17:13.055-08:002017-01-08T18:17:13.055-08:00Dan, not a physicist, but one thing I see is that ...Dan, not a physicist, but one thing I see is that the Emdrive falsifies the idea that Newtonian mechanics are always valid at our size scale, by not conserving momentum. This means all bets are off: You can't use the falsified theory's relation between energy and momentum to say it will violate conservation of energy, though it could violate that too. Luke Bradleyhttp://www.blogger.com/profile/16527503365793312465noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-30528190538124436362017-01-07T19:30:19.617-08:002017-01-07T19:30:19.617-08:00Sir, where can we find the optimised cavity shape ...Sir, where can we find the optimised cavity shape formula/locus? Have any other shapes been devised? Does the EM drive increase C beyond any QM effect?joesixpackhttp://www.blogger.com/profile/08912279232742819732noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-89240465425184698242017-01-07T02:52:57.957-08:002017-01-07T02:52:57.957-08:00Looking at the equation again, I don't think w...Looking at the equation again, I don't think what I posted above can be right. But it would be good if someone less rusty could do the substitution and figure out what the ideal small end size is given a fixed large end size (the assumption being that space and fabrication contrainsts would apply most top the large size and/or length).ShardPhoenixhttp://www.blogger.com/profile/06980692479838159896noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-19624492965519560922017-01-07T02:45:49.938-08:002017-01-07T02:45:49.938-08:00My maths is a bit wobbly, but I only get sensible ...My maths is a bit wobbly, but I only get sensible results from equation 14 if I remove the minus sign at the beginning?<br /><br />Also, given that, it seems that for a fixed-size big end, the force is maximized as the small end (and also the length) goes to zero. Does this contradict the truncated cone design?<br /><br />I may well have made a simple mistake here...ShardPhoenixhttp://www.blogger.com/profile/06980692479838159896noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-5171725886183263102017-01-06T18:52:48.108-08:002017-01-06T18:52:48.108-08:00JC in your example, consider the total time the bi...JC in your example, consider the total time the bikers are spending on that hill; it's approximately 10x less for the second biker.Dan's Test Bloghttp://www.blogger.com/profile/11241554417665894721noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-7657809682037785382017-01-06T18:47:24.503-08:002017-01-06T18:47:24.503-08:00Mike, you noted the main difference between the di...Mike, you noted the main difference between the dielectrics in Shawyer's first experimental prototype thruster and in the Eagleworks cavity is their location (small end for EW, big end for Shawyer).<br /><br />But in fact, besides having quickly dropped dielectrics in all subsequent cavities, Shawyer never experimented with the kind of dielectrics Eagleworks used, which were two non-resonant flat polyethylene discs stacked together. Shawyer's first dielectric was a 2.45GHz ceramic resonator rod with a silver-plated end and a very high Q of around 100,000, whose axial position was tuned to wave front interface to the frustum standing wave at the TE012 internal guide wavelength. This is two very different uses of a dielectric in a resonant cavity. Julien Geffrayhttp://www.blogger.com/profile/06498085495224042768noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-15390624781422943772017-01-06T18:33:14.108-08:002017-01-06T18:33:14.108-08:00I mean, think of it thus way: if you are biking at...I mean, think of it thus way: if you are biking at 1 m/s and coast up a hill that bring you to a stop right at the top, something else with the same mass going at 10 m/s coasting up the hill would go to 9 m/s at the crest. It wouldn't come to 9.5 or something because the last m/s second had more energy in it. It would have a lot more energy to dispel if they both hit a wall, because the change in velocity is bigger, that's all. Jonathan Cardhttp://www.blogger.com/profile/17258349623846497813noreply@blogger.comtag:blogger.com,1999:blog-4637778157419388168.post-28222073758979903502017-01-06T18:28:39.714-08:002017-01-06T18:28:39.714-08:00I don't think that's how it works. Obvious...I don't think that's how it works. Obviously, I will defer to the Professor, but in the scenario you describe, you are adding the energy to go .01 m/s more only. It doesn't take more energy to go from 100 m/s to 101m/s than it does to go from 1 m/s to 2 m/s. The kinetic energy equation does not have relativistic properties. Jonathan Cardhttp://www.blogger.com/profile/17258349623846497813noreply@blogger.com