Quote from: Prunesquallor on 06/10/2015 03:37 pmYou are absolutely correct - there are no other forces accounted for in that plot other than spacecraft thrust acceleration. As I mentioned in the above quote, I think you need to be clearly above the region where predicted drag deceleration is within an order of magnitude of your expected thrust acceleration. As time permits, I was going to try to do some calculations and include them as a "keep out zone" in that plot.I am actually NOT a big fan of a space test, especially CubeSat. I think the prospect of introducing error sources is much higher than in the lab. Drag is one, but for example most CubeSats do NOT have active attitude control - they just tumble. You have no ability, really to shield thruster electronics from spacecraft electronic and vice versa - you just don't have the space or mass available.Somewhere I saw that the AVERAGE CubeSat deployment altitude was 600 km, which is why I used it in that example, but the analysis was just to answer the question of what kind of altitude change you could expect as a function of predicted thruster performance. I still maintain that an early CubeSat test at this stage that returns a null results tells us nothing.That is indeed a worry. Note, however, that reaction wheels (and, for that matter, micro thrusters) are now available for CubeSats. Seehttp://www.cubesatkit.com/docs/datasheet/DS_CSK_ADACS_634-00412-A.pdf (for example).However, my thoughts have been moving in a different direction, and I would like to present that here. I would suggest that a CubeSat test should be 3U, not 1 U, and should have two thrusters, to enable a spin test, not a thrust test.The proposed test would have two drives, oriented in opposing direction (i.e., 180 deg relative to each other), to enable the drives to spin the satellite up (i.e., in a "pin-wheel" mode). That way, drag becomes much less important. In Ascii Art mode, the directions of thrust would look (from above the long axis) like >| | + | |<where + denotes the center of Mass and <,> the directions of thrust. A 3U CubeSat is 30 cm long (L = 0.3 m) and 10 cm high and wide (h = 0.1 m). Assume the thrust is 2.5 x 10^-5 N per drive, they are separated by 25 cm, are each 12.5 cm from the center of mass , and the total mass is 3 kg. The moment of inertia, I, of a spin perpendicular to the L axis is ~ M * (L^2 + h^2) / 12, or 0.025 kg m^2 (a non-uniform distribution of mass might change that by up to a factor of two, which doesn't matter now but would have to be measured before launch). The torque, T, from two drives would be 2 x 2.5 x 10^-5 N x 0.125 m = 0.625 x 10^-5 N m = 6.25 x 10^-6 kg m^2 / sec^2. The spin up (or down), d Omega / dt = T / I = 6.25 x 10^-6 / 0.025 = 2.5 x 10^-4 radians sec^-2A d Omega / dt = 2.5 x 10^-4 radians sec^-2 means that in 1 day (86,400 seconds, assuming the orbit was in continual sunlight), the spin rate would be 21.6 radians / second, or a spin period of 0.29 seconds (or 206 rpms). With two transmitters on either end of the CubeSat, you could easily see the 3 m/sec relative Doppler shift that 21.6 radians represents, and of course gyroscopes, accelerometers and sun-sensors (all for measuring rotation) are available and routinely used on CubeSats. Even in 8 hours of Sun, the spin would be 68 RPM or 0.87 seconds. I, personally, think that spinning a test satellite up to 200 rpms or even 68 RPMS in a day would be pretty conclusive. (Of course, you would need a "null mode," with power but no expected thrust, as a control.) And, note, this could be done at ISS altitude (or on a "next available" orbit, which is cheapest on commercial flights) - drag at those altitudes does not spin satellites up like that. If these thrusts are at all realistic, then the test would succeed or fail in a very short period of time (2 or 3 days), which is also a plus. Also, the first such test you run, if successful (a big if in my opinion) would establish a market for EM drives for attitude control.
You are absolutely correct - there are no other forces accounted for in that plot other than spacecraft thrust acceleration. As I mentioned in the above quote, I think you need to be clearly above the region where predicted drag deceleration is within an order of magnitude of your expected thrust acceleration. As time permits, I was going to try to do some calculations and include them as a "keep out zone" in that plot.I am actually NOT a big fan of a space test, especially CubeSat. I think the prospect of introducing error sources is much higher than in the lab. Drag is one, but for example most CubeSats do NOT have active attitude control - they just tumble. You have no ability, really to shield thruster electronics from spacecraft electronic and vice versa - you just don't have the space or mass available.Somewhere I saw that the AVERAGE CubeSat deployment altitude was 600 km, which is why I used it in that example, but the analysis was just to answer the question of what kind of altitude change you could expect as a function of predicted thruster performance. I still maintain that an early CubeSat test at this stage that returns a null results tells us nothing.
About that spinning idea. T = I dw/dt is the rotational equivalent of F = m a. T = F r and I is the moment of inertia of the thing being accelerated. So for a given F, what r maximises dw/dt?dw/dt = F r / I, so naively maximum r results in maximum dw/dtBut I itself depends on r; I = m r2, the result of some integral and averaging, sodw/dt = F / (m r), so here minimum r results in maximum dw/dt.Which is it?
Quote from: deltaMass on 06/10/2015 05:10 pmAbout that spinning idea. T = I dw/dt is the rotational equivalent of F = m a. T = F r and I is the moment of inertia of the thing being accelerated. So for a given F, what r maximises dw/dt?dw/dt = F r / I, so naively maximum r results in maximum dw/dtBut I itself depends on r; I = m r2, the result of some integral and averaging, sodw/dt = F / (m r), so here minimum r results in maximum dw/dt.Which is it?Suppose you had a uniform (very thin) rod, of (linear!) density rho, so that M = rho L and I = M L^2 / 12 = rho L^3 / 12dw/dt = F L / I = 12 F L / rho L^3 = 12 F / (rho L^2) so you are correct, a large lever arm will have a smaller d Omega / dt. However, if your structure's size is fixed, then you want the separation of the thrusters to be as large as possible, as then I is fixed (to first order) and T depends linearly on the separation. A 1U test would be better in terms of a smaller I, maybe, but I don't think all of this would fit in 1U, and I worry about doing a rotation test on a small cube (the thrusters would not be symmetrically placed with respect to the face of the cube they were pointing at). (I would have the same response to a rotational test _about_ the long axis of a 3-U CubeSat.)But, these are all implementation details, and FWIW I could certainly be convinced differently if there was a good reason for it.
Quote from: TMEubanks on 06/10/2015 05:31 pmQuote from: deltaMass on 06/10/2015 05:10 pmAbout that spinning idea. T = I dw/dt is the rotational equivalent of F = m a. T = F r and I is the moment of inertia of the thing being accelerated. So for a given F, what r maximises dw/dt?dw/dt = F r / I, so naively maximum r results in maximum dw/dtBut I itself depends on r; I = m r2, the result of some integral and averaging, sodw/dt = F / (m r), so here minimum r results in maximum dw/dt.Which is it?Suppose you had a uniform (very thin) rod, of (linear!) density rho, so that M = rho L and I = M L^2 / 12 = rho L^3 / 12dw/dt = F L / I = 12 F L / rho L^3 = 12 F / (rho L^2) so you are correct, a large lever arm will have a smaller d Omega / dt. However, if your structure's size is fixed, then you want the separation of the thrusters to be as large as possible, as then I is fixed (to first order) and T depends linearly on the separation. A 1U test would be better in terms of a smaller I, maybe, but I don't think all of this would fit in 1U, and I worry about doing a rotation test on a small cube (the thrusters would not be symmetrically placed with respect to the face of the cube they were pointing at). (I would have the same response to a rotational test _about_ the long axis of a 3-U CubeSat.)But, these are all implementation details, and FWIW I could certainly be convinced differently if there was a good reason for it.And it still begs the question, why not just do it in the lab on a low friction turntable or magnetic suspension?
...Well, that should be done first, ideally in vacuum. Any idea what the torque drag is on a good low friction turntable? I was thinking of a torsion pendulum for ground tests, which I think can be made very low drag,Even on the ground, I would think rotation tests would have the great advantage that static perturbations (of the "the drive is attracted to that hunk of metal over there" type) will average out in a rotational test. And, if this was ever flown, the actual flight hardware should be tested this way too.
WallofwolfstreetWould you care to tell us about Your theory on the em-drive...
Quote from: TMEubanks on 06/10/2015 05:51 pm...Well, that should be done first, ideally in vacuum. Any idea what the torque drag is on a good low friction turntable? I was thinking of a torsion pendulum for ground tests, which I think can be made very low drag,Even on the ground, I would think rotation tests would have the great advantage that static perturbations (of the "the drive is attracted to that hunk of metal over there" type) will average out in a rotational test. And, if this was ever flown, the actual flight hardware should be tested this way too.NASA has been using a low torsional stiffness torque pendulum for their measurements. The forces they are reporting are extremely small (from a few microNewtons to 100 microNewtons). They have an inclination problem, discussed up by @frobnicat in very detailed posts. The main issue has been that they have measured in vacuum chamber a 5*10^-6 Torr partial vacuum, forces of 55 microNewtons with 50 watts power input, but when they turn it around by 180 degrees pointing in the opposite direction, they only measured 9.9 microNewtons under an input power of 35 watts.See:http://emdrive.wiki/Experimental_ResultsNASA Glenn has offered to do measurements but they have a minimum threshold of 100 microNewtons for their measuring equipment, so they need NASA Johnson (Eagleworks) to up their thrust to over 100 microNewtons to repliate the experiments.
Quote from: SeeShells on 06/10/2015 05:11 pmBy placing one of my tests in water, any and all abnormalities can show up and I'm looking to address any of them. Things you'll not see on a bench strapped down to a air bearing or simply pushing on a force gauge or hanging in free air with a wire. Being submerged with the device water cooled I can reduce the issues with the thermal expansion coefficients of the case. See any unknown effects and log them, measuring in real time the EmDrive to move freely in XYZ & T. It's all about the data and getting data in 3D is better than in 2D.ShellIt will be very interesting. My prediction is that you will see a random walk
By placing one of my tests in water, any and all abnormalities can show up and I'm looking to address any of them. Things you'll not see on a bench strapped down to a air bearing or simply pushing on a force gauge or hanging in free air with a wire. Being submerged with the device water cooled I can reduce the issues with the thermal expansion coefficients of the case. See any unknown effects and log them, measuring in real time the EmDrive to move freely in XYZ & T. It's all about the data and getting data in 3D is better than in 2D.Shell
...Iulian was able to build a model in what, a few weeks, and with a maybe a few hundred dollars? A bit more time and money and he might have quite a nice setup. How can Shawyer justify 15 years and exponentially more money to produce a worse result? All we have is a few confused papers and a 30 second youtube video! Strange, very strange. (By now you should realize I'm a cynical person, so I acknowledge that those Shaywer criticisms might be unfounded)
Quote from: wallofwolfstreet on 06/10/2015 06:17 pm...Iulian was able to build a model in what, a few weeks, and with a maybe a few hundred dollars? A bit more time and money and he might have quite a nice setup. How can Shawyer justify 15 years and exponentially more money to produce a worse result? All we have is a few confused papers and a 30 second youtube video! Strange, very strange. (By now you should realize I'm a cynical person, so I acknowledge that those Shaywer criticisms might be unfounded)I share your skepticism.What compounds the issue is that Prof. Yang and her team at a Chinese University publishes in a Chinese peer-reviewed journal her results that exceed Shawyer's claims, achieving 1 N/Kw.Her Force/InputPower results are thousands of times greater than the one reported by NASA Eagleworks.And her papers claim low errors between the tests (look at the curves in the figure above). Very difficult to reconcile all of this disparate data...See:http://emdrive.wiki/Experimental_Results#cite_note-design_factor-3
...It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.Sorry to ask but I've tried to dig it out on my own and nadda.Shell
...I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind.
Quote from: SeeShells on 06/10/2015 06:35 pm...I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind. md2z/dt2 + ρgAz = 0if you use an encasing cylinder, it will have a constant cross-section A, and it will act like a spring with stiffness ρgAhttp://www.codecogs.com/library/engineering/fluid_mechanics/floating_bodies/the-oscillation-of-floating-bodies.phphttp://physics.stackexchange.com/questions/64154/shm-of-floating-objectshttp://www.okphysics.com/1-oscillations-of-a-body-floating-in-a-liquid/
Quote from: Rodal on 06/10/2015 07:01 pmQuote from: SeeShells on 06/10/2015 06:35 pm...I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind. md2z/dt2 + ρgAz = 0if you use an encasing cylinder, it will have a constant cross-section A, and it will act like a spring with stiffness ρgAhttp://www.codecogs.com/library/engineering/fluid_mechanics/floating_bodies/the-oscillation-of-floating-bodies.phphttp://physics.stackexchange.com/questions/64154/shm-of-floating-objectshttp://www.okphysics.com/1-oscillations-of-a-body-floating-in-a-liquid/Perfect! Thnaks!!!!
Quote from: SeeShells on 06/10/2015 06:53 pm...It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.Sorry to ask but I've tried to dig it out on my own and nadda.ShellBecause (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.
Quote from: Rodal on 06/10/2015 06:55 pmQuote from: SeeShells on 06/10/2015 06:53 pm...It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.Sorry to ask but I've tried to dig it out on my own and nadda.ShellBecause (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.And that's it? Thanks. I thought it might be it but I could never know if I missed something.
Because (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.