I have some new observations and theory. I lifted the first picture from the Aug. '14 paper, which shows the linear displacement sensor. This device tracks the position of a reflected laser dot on a CMOS image sensor. This is usually done by calculating the center of luminance of the laser dot; a measurement that has an accuracy of a small fraction of the width of a pixel. The second picture describes a typical LDS that has sub-micron accuracy.The only control or "NULL" experiment described in the Aug. 14 paper related to the Eagleworks device was when the dummy load was used instead of sending RF into the cavity. This of course shields the RF very well. The dummy load is 50 Ohms and so the SWR is 1:1. However when the cavity is loaded and the dielectric material is inside the cavity it's possible the SWR is much higher. This would result in RF being reflected back to the amplifier and being radiated from the shield of the RF cable. This is what happens when the SWR is not 1:1. It's possible this RF noise is interfering with the LDS. When the dielectric is not inside the cavity the SWR is lower so no interference takes place. This theory agrees with the results of the cannae test as well.The last picture, also from the Aug. '14 paper shows a negative slope on the baseline position ( no thrust) after each RF pulse. The first one appears to level off just before the final RF pulse. After that pulse it heads down again. I believe the thermally induced change in the CoM of the emdrive causes the balance arm to rotate. This very slight rotation reduces the reflection distance for the laser beam.
What software did you use to do that?I have in the past used Mathematica to transform images like that but one has to write Mathematica code to get it just right like you did.
Dr. Rodal & Notsosureofit:We had an interesting failure in the Eagleworks lab yesterday. That being I was getting ready to pull a vacuum on our copper frustum mounted in its "reverse" or to the right thrust vector position and ran a preliminary data un to see if it was performing in air as it had two weeks ago just before our last RF amplifier died. Sadly it wasn't for it was producing less than half of what it did before and in the wrong direction! I had Dr. White come in and take a look over my latest test article installation last night and he found that the center 1/4"-20 nylon PE disc mounting bolt that holds the second PE disc to the small OD frustum's PCB endplate was no-longer tensioned as it had been before. In fact it had partially melted at the interface between the two PE discs thus relieving the strain induced by its bolts threads and nut. (There are three ~1.00" 1/4-20 nylon bolts mounted on a ~2.00" radius spaced every 120 degrees that hold the first PE disc to the PCB end cap. There is then a layer of 3/4" wide office scotch tape at the interface between the first and second PE discs and the center 1/4"-20 nylon bolt that hold second PE disc to the first PE disc.) Apparently not having the PE discs firmly mounted to the frustum's small OD end cap hindered the thrust producing mechanism that conveys the generated forces in the PE to the copper frustum. And/or the melted nylon was hogging all the RF energy in the PE discs due to its higher dissipation factor in its semiliquid state. Either way it looks like there is a high E-field volume where this center nylon bolt hangs out while running in the TM212 resonant mode. Too bad Teflon bolts are so weak even in comparison to the nylon, for its dissipation factor is at least two orders of magnitude lower than the nylon's. Best, Paul M.
Quote from: zen-in on 02/26/2015 05:04 amI have some new observations and theory. I lifted the first picture from the Aug. '14 paper, which shows the linear displacement sensor. This device tracks the position of a reflected laser dot on a CMOS image sensor. This is usually done by calculating the center of luminance of the laser dot; a measurement that has an accuracy of a small fraction of the width of a pixel. The second picture describes a typical LDS that has sub-micron accuracy.The only control or "NULL" experiment described in the Aug. 14 paper related to the Eagleworks device was when the dummy load was used instead of sending RF into the cavity. This of course shields the RF very well. The dummy load is 50 Ohms and so the SWR is 1:1. However when the cavity is loaded and the dielectric material is inside the cavity it's possible the SWR is much higher. This would result in RF being reflected back to the amplifier and being radiated from the shield of the RF cable. This is what happens when the SWR is not 1:1. It's possible this RF noise is interfering with the LDS. When the dielectric is not inside the cavity the SWR is lower so no interference takes place. This theory agrees with the results of the cannae test as well.The last picture, also from the Aug. '14 paper shows a negative slope on the baseline position ( no thrust) after each RF pulse. The first one appears to level off just before the final RF pulse. After that pulse it heads down again. I believe the thermally induced change in the CoM of the emdrive causes the balance arm to rotate. This very slight rotation reduces the reflection distance for the laser beam.Glad to see people still attempting to come up with null-thrust ideas on here. How does this one account for the measured loss of thrust when the PTFE disk in the resonator cavity came loose after the nylon support bolt melted?
It's to bad that we can't find a way that one of the little known or unknown solutions to Maxwell's equations can cause a momentum.
Unless of course it is surface electrons excited by the high power resonant RF, tunnelling through the 35 micron copper ends.
I have some new observations and theory. I lifted the first picture from the Aug. '14 paper, which shows the linear displacement sensor. This device tracks the position of a reflected laser dot on a CMOS image sensor. This is usually done by calculating the center of luminance of the laser dot; a measurement that has an accuracy of a small fraction of the width of a pixel. The second picture describes a typical LDS that has sub-micron accuracy.
Displacement of the pendulum arm is measured via a Linear Displacement Sensor (LDS). The primary LDS components consist of a combined laser and optical sensor on the fixed structure and a mirror on the pendulum arm. The LDS laser emits a beam which is reflected by the mirror and subsequently detected by the optical sensor. The LDS software calculates the displacement (down to the sub-micrometer level) based upon the beam reflection time. Prior to a test run data take, the LDS is positioned to a known displacement datum (usually 500 micrometers) via mechanical adjustments to its mounting platform. Gross adjustments are performed via set screws. Fine adjustments are performed using manually - operated calibrated screw mechanisms and a remotely controlled motorized mechanism that can be operated with the chamber door closed and the chamber at vacuum. The remote adjustment capability is necessary since the LDS datum will change whenever a change to the test facility environment affects the roll - out table or the chamber – e.g., whenever the chamber door is closed or latched and whenever the chamber is evacuated. Once the LDS displacement is adjusted in the final test environment, further adjustment between test run data takes is usually not required.
The only control or "NULL" experiment described in the Aug. 14 paper related to the Eagleworks device was when the dummy load was used instead of sending RF into the cavity. This of course shields the RF very well. The dummy load is 50 Ohms and so the SWR is 1:1. However when the cavity is loaded and the dielectric material is inside the cavity it's possible the SWR is much higher. This would result in RF being reflected back to the amplifier and being radiated from the shield of the RF cable. This is what happens when the SWR is not 1:1. It's possible this RF noise is interfering with the LDS. When the dielectric is not inside the cavity the SWR is lower so no interference takes place. This theory agrees with the results of the cannae test as well.
The last picture, also from the Aug. '14 paper shows a negative slope on the baseline position ( no thrust) after each RF pulse. The first one appears to level off just before the final RF pulse. After that pulse it heads down again. I believe the thermally induced change in the CoM of the emdrive causes the balance arm to rotate. This very slight rotation reduces the reflection distance for the laser beam.
frobnicat I like your diagrams, very clear.Note that as the buckling completes the end plate will be at rest with the frustum. This means that there will be an acceleration of the end plate in the opposite direction (at some time after your diagram 4).
After the buckling completes (and all accelerations are zero) the centre of mass will not have changed. However the position of the centre of mass relative to the connection point to the torsion balance may have changed.
From this I think the torsion balance should see a force in one direction, then some time later a force in the other direction. These forces integrated should be zero.
After the buckling has completed the torsion balance should be showing no force, however as the shape of the frustum + end plate has changed the position of the external surfaces has also changed. If the force is being measured as a change in the position of the frustum then a small force would erroneously be shown (I don't think this is the case, but something to check in the experimental set-ups).
Quote from: frobnicat on 02/25/2015 11:37 pm...I like your drawings with the buckled condition idealized as a beam with three hinges: one hinge at the center and a hinge at each end. Thank you very much for taking the time to make those drawings. I should have looked at that initially. I agree with your drawing. To the extent that my prior wording disagreed with your drawing, my prior words were incorrect, when and if they referred to the thermal force Fpr.Let's then address what happens in the opposite case that the plate instead buckles to the right:The flat plate can theoretically buckle towards the left or towards the right (if the copper is thin enough) depending on initial imperfections. If when the buckled plate moves towards the left it gives a force towards the right then if the plate buckles towards the right it would produce a force towards the left, in the same direction as the EM Drive's motion, do you agree?.Then, if this buckling analysis is correct, it gives transient force that is towards the left when the plate buckles towards the right and it gives a transient force to the right when the plate buckles towards the left.Now I have to give further thought to which direction the plate buckles when heated. The real plate has a neutral surface that is not at the middle of the cross-section. It is really a bi-material thermostat: with the epoxy expanding much more than the very thin copper coating. If this unsymmetric laminate would be exposed to a uniform temperature it would expand towards the outside, producing a force towards the left, like the EM Drive force. The question is what happens under the superposed thermal gradient through the thickness. The IR Camera shows very pronounced heating on the outside surface....
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Quote from: Rodal on 02/26/2015 01:14 amQuote from: frobnicat on 02/25/2015 11:37 pm...I like your drawings with the buckled condition idealized as a beam with three hinges: one hinge at the center and a hinge at each end. Thank you very much for taking the time to make those drawings. I should have looked at that initially. I agree with your drawing. To the extent that my prior wording disagreed with your drawing, my prior words were incorrect, when and if they referred to the thermal force Fpr.Let's then address what happens in the opposite case that the plate instead buckles to the right:The flat plate can theoretically buckle towards the left or towards the right (if the copper is thin enough) depending on initial imperfections. If when the buckled plate moves towards the left it gives a force towards the right then if the plate buckles towards the right it would produce a force towards the left, in the same direction as the EM Drive's motion, do you agree?.Then, if this buckling analysis is correct, it gives transient force that is towards the left when the plate buckles towards the right and it gives a transient force to the right when the plate buckles towards the left.Now I have to give further thought to which direction the plate buckles when heated. The real plate has a neutral surface that is not at the middle of the cross-section. It is really a bi-material thermostat: with the epoxy expanding much more than the very thin copper coating. If this unsymmetric laminate would be exposed to a uniform temperature it would expand towards the outside, producing a force towards the left, like the EM Drive force. The question is what happens under the superposed thermal gradient through the thickness. The IR Camera shows very pronounced heating on the outside surface....I have calculated the variables that govern which way the circular plate will buckle.Using S. Timoshenko's classic solution to the bimaterial thermal expansion problem (Journal of the Optical Society of America, JOSA, Vol. 11, Issue 3, pp. 233-255 (1925) ) the radius of curvature can be expressed as:radiusOfCurvature = (t1+t2)*(3(1+m)+(1+m*n)(m^2 +1/(m*n)))/(6*deltaAlpha*deltaT*((1+m)^2))where I have expressed the following variables as non-dimensional ratios (allowing use of any consistent system of units, and revealing the important parameters, instead of the expression in Wikipedia that is awkwardly expressed in dimensional units, and expressed for the curvature instead of its reciprocal):m=t1/t2 Thickness ration=E1(1-nu1^2)/(E2(1-nu2^2)) Plate stiffness-ratioand the differences in thermal expansion and temperature:deltaAlpha=alpha2 - alpha1 deltaT=T - Tot1 and t2 are the thicknesses of the layersE1 and E2 are the Young's moduliinu1 and nu2 are the Poisson's ratiosFrom Paul March we know:t1=0.00138 inchest2=0.063 inchesthe material "1" is copperWikipedia and Engineering Toolbox give E1=17*10^6 psiWikipedia, NBS (Hidnert &Krider's classic article) and Engineering Toolbox give alpha1= 17*10^(-6) 1/degK and Wikipedia gives nu1=0.33and that the material "2" is FR4The following references give:WikipediaECW = 3*10^6 psi alphaCW= 14*10^(-6) 1/degK nuLW=0.136ELW = 3.5*10^6 psi alphaLW= 12*10^(-6) 1/degK nuCW=0.118P-M Services (UK) alphaCW= 15*10^(-6) 1/degK alphaLW= 11*10^(-6) 1/degKLeiton (Germany) alphaCW= 17*10^(-6) 1/degK alphaLW= 12*10^(-6) 1/degKIt is trivial to show that the sign of the curvature (which way the plate is going to buckle) is governed only by the difference in thermal expansion coefficients between the two layers.Bimaterial thermal bending will take place along the anisotropic in-plane direction with the highest thermal expansion and lowest modulus. It is obvious that this is the CW direction.It is immediately obvious, that using Leiton's (Germany) coefficient of thermal expansions, the copper/FR4 laminate circular plate will not experience any bimaterial bending whatsoever because according to Leighton alphaCW of FR4 is exactly the same as the universally accepted alpha1 of copper.The thermal IR camera shows temperature readings ranging from below 79 deg F to a concentrated maximum at certain small spots of 94.3 deg F.Using Wikipedia's properties, the plate will experience an extremely small amount of bimaterial bending towards the inside, for a temperature increase from 68 deg F to 79 deg F, with a huge radius of curvature exceding 7000 inches (practically flat, in relation to the thickness of only 0.064 inches).Also, using Wikipedia's properties, the plate will experience an extremely small amount of bimaterial bending towards the inside, for a temperature increase from 68 deg F to 94.3 deg F, with a huge radius of curvature exceeding 3000 inches (practically flat, in relation to the thickness of only 0.064 inches).CONCLUSION: due to the fact that FR4 has a coefficient of thermal expansion very similar to the one of copper, and that the thickness of the copper is extremely small compared to the thickness of FR4, the circular plate will experience either no bimaterial thermal bending whatsoever, or it will be extremely small (will stay practically flat) under the measured changes in temperature. Therefore, bi-material bending due to a change in temperature is irrelevant to the buckling problem. Buckling is instead governed by the plate's initial imperfect flatness
FYIFound my Gunn Oscillator !!
Quote from: Notsosureofit on 02/27/2015 02:27 pmFYIFound my Gunn Oscillator !!Where was it hidden? Inquiring minds would like to know...
While looking into the tunnelling of electrons through a barrier, I observe that the mathematics for tunnelling of electrons is very similar to the mathematics presented for propagation evanescent waves through a restriction in an RF wave-guide. See page 15, here http://wwwsis.lnf.infn.it/pub/INFN-FM-00-04.pdf for evanescent wave propagation and sheet 8, here http://tuttle.merc.iastate.edu/ee439/topics/tunneling.pdf for tunnelling of electrons through a barrier. (I would paste the math here, but both papers are PDF's, so I can't.)I guess I shouldn't be surprised that the math is similar, if not the same, as tunnelling electrons are evanescent matter waves so the math should be similar. The big difference in the two math formulations is that the author of the first paper, evanescent wave propagating through a restriction in the wave guide, relies on extended special relativity to keep the exp (+term) in the wave equation while the author of the electron tunnelling paper relies on a physical observation to retain that term. This brings up a question. "Since we are concerned with the mass of the thruster, why isn't it made from a light metal, aluminium for example?" The work function of aluminium is very close to that of copper, is all metals. Granted that aluminium oxidized when exposed to air so that its work function more than doubles after exposure, but a very thin layer of gold electro deposited on both sides would eliminate that problem with the result of a much less massive thruster. Can anyone suggest a reason that an aluminium cavity would not work?