Is there anything useful right now, today..... that can be done with an always on 50uN of thrust in space?
Quote from: Mulletron on 02/13/2015 12:12 amIs there anything useful right now, today..... that can be done with an always on 50uN of thrust in space?Why limit oneself to just one EM Drive engine in space?. How about posing the question as:Is there anything useful right now, today..... that can be done with an always on N*50uN of thrust in space where N is an integer multiple (1,2,3,4,...) of 50 watts per engine. For example, 4 engines: total thrust 100uN at total power 200 Watts, 10 engines 500 uN at 500 Watts total.
Quote from: Rodal on 02/13/2015 12:15 amQuote from: Mulletron on 02/13/2015 12:12 amIs there anything useful right now, today..... that can be done with an always on 50uN of thrust in space?Why limit oneself to just one EM Drive engine in space?. How about posing the question as:Is there anything useful right now, today..... that can be done with an always on N*50uN of thrust in space where N is an integer multiple (1,2,3,4,...) of 50 watts per engine. For example, 4 engines: total thrust 200uN at total power 200 Watts, 10 engines 500 uN at 500 Watts total.IIRC, the ISS has 30KWe extra to play with.
Quote from: Mulletron on 02/13/2015 12:12 amIs there anything useful right now, today..... that can be done with an always on 50uN of thrust in space?Why limit oneself to just one EM Drive engine in space?. How about posing the question as:Is there anything useful right now, today..... that can be done with an always on N*50uN of thrust in space where N is an integer multiple (1,2,3,4,...) of 50 watts per engine. For example, 4 engines: total thrust 200uN at total power 200 Watts, 10 engines 500 uN at 500 Watts total.
2) It would be great to get the forces on the center of mass of the EM Drive you compute for the three different cases
Quote from: Mulletron on 02/13/2015 12:20 amQuote from: Rodal on 02/13/2015 12:15 amQuote from: Mulletron on 02/13/2015 12:12 amIs there anything useful right now, today..... that can be done with an always on 50uN of thrust in space?Why limit oneself to just one EM Drive engine in space?. How about posing the question as:Is there anything useful right now, today..... that can be done with an always on N*50uN of thrust in space where N is an integer multiple (1,2,3,4,...) of 50 watts per engine. For example, 4 engines: total thrust 200uN at total power 200 Watts, 10 engines 500 uN at 500 Watts total.IIRC, the ISS has 30KWe extra to play with.That's 600 EM Drives giving 0.03 Newtons total thrust They need to increase the Thrust/Power to decrease the number of EM Drives necessary.
Quote1) You need to give us a contour field mapping rule: what do the colors mean in numerical terms, to further understand the contourplot. For example: what is the intensity of the white areas? of the red areas? of the orange areas? of the blue areas?Attached is the best I can do on short notice. I've looked at the available color palates in HDFView and they are weak, to say the least. Or maybe I just don't know how to scale them. I would like to know paraview better because I think it could display the data and in the format you wanted. Lines of constant value would be good. iso - whatevers.
1) You need to give us a contour field mapping rule: what do the colors mean in numerical terms, to further understand the contourplot. For example: what is the intensity of the white areas? of the red areas? of the orange areas? of the blue areas?
What is white ? What is black?White may be the most positive (off-scale) and black the most negative (off-scale), but I would like you to confirm...
Quote from: Star-Drive on 02/11/2015 12:44 pm... Just as a note, we've already tried re-enforcing the frustum endplates with angle aluminum mounted on their outside surfaces and we didn't notice any marked change in its thrust response. ...This was expected not to make any significant difference.As I wrote in my report:Quote from: Dr. J. RodalCotterell and Parkes (based on Cotterell's Ph.D. thesis at the University of Cambridge) correctly point out that the distribution of the heat flux "is not significant in the problem" of thermal buckling of a circular plate, whether the heating takes place uniformly over the whole circular plate or is concentrated in a central region. Cotterell chose a distribution with a heatedDiameterRatio =1/0.3=3.333 instead of the heatedDiameterRatio=1 analyzed by Noda et.al. The fact that the exact distribution is not significant for the deltaT that will produce buckling or for the buckling displacement follows from equilibrium: the membrane stress (=E*alpha*deltaT) force resultant (the integral of the membrane stress through the thickness) is reacted at the simply supported edges (that constrain the in-plane displacement). The membrane force resultant is uniform and it is equal in the polar radial and angular (azimuthal) directions. If only a central area is heated, the membrane stress is still equilibrated throughout. If the plate has uniform thickness and isotropic material properties, the strain in the non heated area prior to buckling is the same as in the heated area.See:Cotterell, B., and Parkes, E. W., Thermal Buckling of Circular Plates, (United Kingdom's) Aeronautical Research Council, Ministry Of Aviation, Reports and Memoranda No. 3245, September, 1960QUESTION: Why did you use a glass-fiber-reinforced polymer printed circuit board as the end plate ? << 0.063 inch thick FR4 printed circuit board with 1.0 oz copper, (~35 microns thick of Cu epoxied to the FR4 fiberglass)>>1) The IR measurement was done from the outside, with the IR camera looking at the composite polymer surface of the circuit board surface they had on the exterior of the big diameter flat end. Since this composite polymer has much lower thermal conductivity and much lower thermal diffusivity than copper, please take into account that these IR measurements represent a temperature and temperature gradients significantly lower than those present on the inner (copper) surface of the big diameter flat end. In other words, the composite polymer circuit board surface being measured with the IR camera acts like an insulating surface concealing the higher temperature of the inner copper surface. Moreover, due to very low thermal diffusivity of the glass-fiber-reinforced polymer printed circuit board, measurement of its exterior surface presents a considerable time delay of the interior temperature vs. time profile (as it takes time for the heat to conduct through the thickness of the very low diffusivity of the glass-fiber-reinforced polymer printed circuit board).2) The modulus of elasticity of the glass-fiber-reinforced polymer printed circuit board is much lower than the modulus of elasticity of the copper. The glass-fiber-reinforced polymer printed circuit board has orders of magnitude lower thermal conductivity and thermal diffusivity than the copper. (Comparison noted below).3) Why not get rid of the fiber-reinforced-polymer printed circuit board and just simply use a 1/4 inch thick (0.25 inches) copper plate for flat ends to prevent this thermal instability, and hence eliminate this artifact from consideration ?As to your questions, I would need some time to give them the analytical consideration they deserve and to calculate, rather than give you an impulsive, reflexive, answer that may be incorrect.NOTE: FR-4 is a composite material made with woven fiberglass cloth embedded in an epoxy resin (polymer) matrix. The in-plane Young's modulus of FR4 is 3.0×10^6 psi , about six times smaller than Copper's Young modulus of 17.0×10^6 psi. The modulus of elasticity in the thickness direction is much lower, practically as low as the modulus of elasticity of epoxy. FR4's coefficient of thermal expansion - x-axis 1.4×10^(−5) 1/K, Coefficient of thermal expansion - z-axis 7.0×10^(−5) 1/KThe thermal conductivity is a tiny 0.29 W/m·K in the thickness direction, due to the low thermal conductivity of the epoxy resin. Copper has a thermal conductivity of 401 W/m·K, that is 1400 times higher than the thermal conductivity of FR4
... Just as a note, we've already tried re-enforcing the frustum endplates with angle aluminum mounted on their outside surfaces and we didn't notice any marked change in its thrust response. ...
Cotterell and Parkes (based on Cotterell's Ph.D. thesis at the University of Cambridge) correctly point out that the distribution of the heat flux "is not significant in the problem" of thermal buckling of a circular plate, whether the heating takes place uniformly over the whole circular plate or is concentrated in a central region. Cotterell chose a distribution with a heatedDiameterRatio =1/0.3=3.333 instead of the heatedDiameterRatio=1 analyzed by Noda et.al. The fact that the exact distribution is not significant for the deltaT that will produce buckling or for the buckling displacement follows from equilibrium: the membrane stress (=E*alpha*deltaT) force resultant (the integral of the membrane stress through the thickness) is reacted at the simply supported edges (that constrain the in-plane displacement). The membrane force resultant is uniform and it is equal in the polar radial and angular (azimuthal) directions. If only a central area is heated, the membrane stress is still equilibrated throughout. If the plate has uniform thickness and isotropic material properties, the strain in the non heated area prior to buckling is the same as in the heated area.
...the large OD end of the Eagleworks copper frustum, the mass of the entire frustum assembly without the PE discs is listed as 1.606 kg. Your 0.25" thick solid copper end plates would add over 3.0 kg of dead mass to this figure just for the small and large OD ends plates and cost us $333.50 for a 12"x 24"x0.25" copper plate stock from McMaster-Carr needed to make them.
we have tried aluminum angles and even 0.090" thick AL plates across the existing PCB end caps and noticed no change in its performance except for the increase in seismic noise pickup that the extra mass such payload-mass increasing modifications always bring to the table.
We are on the edge of a golden age of spaceflight with EM drives and superconducting capacitors. Or maybe not, but at least it is fun to discuss it.
Quote1) You need to give us a contour field mapping rule: what do the colors mean in numerical terms, to further understand the contourplot. For example: what is the intensity of the white areas? of the red areas? of the orange areas? of the blue areas?Attached is the best I can do on short notice. I've looked at the available color palates in HDFView and they are weak, to say the least. Or maybe I just don't know how to scale them. I would like to know paraview better because I think it could display the data and in the format you wanted. Lines of constant value would be good. iso - whatevers.Hopefully this will be more usableQuoteWhat is white ? What is black?White may be the most positive (off-scale) and black the most negative (off-scale), but I would like you to confirm...I don't know the answer to your question but I think that your guess is right. I know that the black boarders of the cavity are perfect metal which is confusing as it is not field strength. But the colors are not single valued. The palates seem to be designed for "Pretty" and not for information. There are two single valued palates in hdfview, both gray and both wash out the differences in field strength. The h5topng color palates also wash out the color, and don't even show the evanescent waves. So I'm not using that program. Again, I need to explore paraview's capabilities.If it were easy, everyone would do it, no?
What I need is more hands and more skills, more skills for sure. I tried to attach the digital output source of the final fields for the copper kettle as designed. It is only 125.4 MB, I would like to share the much more interesting file showing the evolution of the fields from start-up, but it is 4.4 GB so it might not be something Chris would want me to do. Actually, Chris doesn't want people uploading .h5 files at all. I discovered after waiting through the complete upload that .h5 is not an allowed format.
Aero, From what I am reading you keep mentioning "superluminal velocity proposition of evanescent waves". Would I be correct in assuming that you are talking about the apparent velocity of teh wavform themselves and not the matter or energy that the waveforms themselves are composed of? Otherwise, this put's a whole new slant on this EM drive debate.
Quote from: JasonAW3 on 02/13/2015 07:41 pmAero, From what I am reading you keep mentioning "superluminal velocity proposition of evanescent waves". Would I be correct in assuming that you are talking about the apparent velocity of teh wavform themselves and not the matter or energy that the waveforms themselves are composed of? Otherwise, this put's a whole new slant on this EM drive debate.I'm referring to superluminal velocity as presented here http://wwwsis.lnf.infn.it/pub/INFN-FM-00-04.pdfI need an interpretation from someone more knowledgeable than I, in order to know exactly what it is that I'm talking about. Perhaps you can tell me?
.....*** http://arxiv.org/pdf/1411.5359v1.pdf (the QVPT model has officially "got served")
Although the extraction of a net momentum has been postulated in inhomogeneous vacuums [35] (due to a different mechanism than that discussed in this paper), the effect was found to immeasurably small, and it remains unclear whether this small non-zero value is an artefact of the field regularization techniques used.Regardless though, if a continuous net force is indeed being produced in the experiments in Refs. [38–41], one might expect that this should also imply an anomalously high power loss from the electric circuits of the thruster device. This power loss would be in addition to any standard power losses associated with such things as: ohmic heating, eddy current losses, dielectric and ferrite heating, radiation losses, etc. Consequently, a dedicated effort to isolate known power losses from the total input power of the device should identify any additional anomalous losses.