Quote from: LasJayhawk on 07/28/2015 02:35 AMRandom thoughts:A microwave oven Maggie, operating at 200C in a vacuum may out gas from its radiator. This could cause the appearance of something that looks like working mass, but isn't.Once again, the Edison Effect predates the discovery of the electron by over a decade. Who knows what we could be spitting out the business end... Has anyone considered a fractal antenna, to try and maximize the power dumped into the frustum?Yes a fractal antenna would be a great device to use but normally the ones I've seen have been small and limited in transmitted power. Do you know of larger that could take 1000 watts?Shell

Random thoughts:A microwave oven Maggie, operating at 200C in a vacuum may out gas from its radiator. This could cause the appearance of something that looks like working mass, but isn't.Once again, the Edison Effect predates the discovery of the electron by over a decade. Who knows what we could be spitting out the business end... Has anyone considered a fractal antenna, to try and maximize the power dumped into the frustum?

....Thanks and I took a look at this. Taken at face value it predicts that, in order to maximise the thrust/power ratio, one requires that independently:- Ds/Db to be minimum- Db to be minimum- L to be minimumeven after accounting for the frequency scaling per Appendix I.The thrust predictions seem roughly in line with the experimental data.k = F/P seems to be, for the ranges graphed, about 3*10^{-7} N/W (about 300 uN for 1000 W).I'm looking for a way to get much higher F/P values. I've indicated how this can be done per this formula.I would welcome some concrete suggestions.[A couple of notes on this derivation:1. In Appendix I the n factor due to L is omitted (but the scaling conclusion is correct anyway)2. Whenever I read about "accelerated photons" my toes curl up]

I'm thinking about finesse at a lower power. The whole microwave magnetron seems like the "don't force it Newt, get a bigger hammer" approach. Eagle works was getting "thrust" at 16 watts into the frustum. I'm wondering if attempting to get more at lower power with antenna design and placement is a worthwhile approach?

Quote from: demofsky on 07/28/2015 03:52 AMQuote from: DrLOAC on 07/28/2015 03:21 AM2 comments I'd like to make:First: Solar power. The ISS arrays are about as big as you would want to build. With 80's solar cells they are good for about 30KW at 1 AU and beginning of life. The latest cells are about 3 times more efficient. If you used the same size arrays with the latest high efficiency cells you could eek out maybe 90-100KW per array or roughly 3/4 of a MW for something ISS sized. If you are in orbit, battery charging for night passes reduces usable power by over half. The arrays tend to be rather fragile as well. I think I'd rather hook an EM drive up to a nuclear reactor. ....This is exactly what I would have expected. Remember though that EM (and other electric) Drives are very gentle so it does not matter if a tug uses massive (monstrous?) arrays that are fragile. This thing will change course slowly and things will not fly off. They use chemical rockets on the ISS and so the stress is much higher.Edits: Clarification.Well for attitude control a tug may have more powerful attitude control thrusters. So rates have to be kept down. You also have to be concerned about environments. Atomic oxygen and RCS thruster exhaust can damage or pit cell coverings reducing power generation over time. Finally you have to be concerned about thermal loading. The mast structure has to be flexible to deploy which can leave it susceptible to buckling when thermal stresses are coupled with mechanical stresses from maneuvers. It's not that it can't be done, it has already on the ISS, its just more of a pain in the butt than you might think.

Quote from: DrLOAC on 07/28/2015 03:21 AM2 comments I'd like to make:First: Solar power. The ISS arrays are about as big as you would want to build. With 80's solar cells they are good for about 30KW at 1 AU and beginning of life. The latest cells are about 3 times more efficient. If you used the same size arrays with the latest high efficiency cells you could eek out maybe 90-100KW per array or roughly 3/4 of a MW for something ISS sized. If you are in orbit, battery charging for night passes reduces usable power by over half. The arrays tend to be rather fragile as well. I think I'd rather hook an EM drive up to a nuclear reactor. ....This is exactly what I would have expected. Remember though that EM (and other electric) Drives are very gentle so it does not matter if a tug uses massive (monstrous?) arrays that are fragile. This thing will change course slowly and things will not fly off. They use chemical rockets on the ISS and so the stress is much higher.Edits: Clarification.

2 comments I'd like to make:First: Solar power. The ISS arrays are about as big as you would want to build. With 80's solar cells they are good for about 30KW at 1 AU and beginning of life. The latest cells are about 3 times more efficient. If you used the same size arrays with the latest high efficiency cells you could eek out maybe 90-100KW per array or roughly 3/4 of a MW for something ISS sized. If you are in orbit, battery charging for night passes reduces usable power by over half. The arrays tend to be rather fragile as well. I think I'd rather hook an EM drive up to a nuclear reactor. ....

I stopped looking at the Design Factor as a serious formula as soon as I learnt from TheTraveller that it uses the "cut-off" frequency for an open waveguide of constant cross-section, as it is known that tapered waveguides and cavities do NOT have rigid cut-off. Only constant cross-section waveguides have a rigid cut-off condition.

Quote from: deltaMass on 07/28/2015 12:10 AM....Thanks and I took a look at this. Taken at face value it predicts that, in order to maximise the thrust/power ratio, one requires that independently:- Ds/Db to be minimum- Db to be minimum- L to be minimumeven after accounting for the frequency scaling per Appendix I.The thrust predictions seem roughly in line with the experimental data.k = F/P seems to be, for the ranges graphed, about 3*10^{-7} N/W (about 300 uN for 1000 W).I'm looking for a way to get much higher F/P values. I've indicated how this can be done per this formula.I would welcome some concrete suggestions.[A couple of notes on this derivation:1. In Appendix I the n factor due to L is omitted (but the scaling conclusion is correct anyway)2. Whenever I read about "accelerated photons" my toes curl up]I just figured out that apparently, Dr. McCulloch's formula, Dr. @Notsosurofit's formula and my own formula using the cylindrical approximation, all amount to the same basic force. This force is proportional to, the change in energy from the small end to the big end, divided by the length, i.e, delta_E/delta_z.Where the 3 formula differ, is in how this force is multiplied by the group velocity v/c or (v/c)^{2}, and what formula is used for the group velocity. Dr. McCulloch skips this concept entirely and simply inputs the energy as Power x time, using the speed of light and the length. No consideration of group velocity at all. He simply uses m*c^2 where m is derived by his theory.Dr. @Notsosureofit's formula, after completing the square and factoring the difference between two frequencies squared. The basic force above is multiplied by the average cut-off (I know @Rodal) over the input frequency:(w_{s} + w_{b})/2*wThis was surprising and interesting. Note, that when the frequency is less than the average cut-off, (i.e, becoming evanescent) this factor is > 1.My formula, without Zeng and Fan results in a factor that also depends on the frequency, but has a much larger value near the cut-off: (w + w_{b})/(w - w_{b}),w_{s} and w_{b}, are the resonant or cut-off frequencies at each end respectively.These factors are each multiplied by delta_E/delta_z, where delta_z is the length, and delta_E is the frequency shift from small end to big end. There is one more factor, and that is the impedance plots in Zeng and Fan for a cone. It is somewhere between the "infinite" value of my formula and the subtle value of @Notsosureofit's formula. It's late, I hope I didn't make any errors, I'll be back. Todd

Quote from: birchoff on 07/28/2015 12:47 AMQuote from: rfmwguy on 07/28/2015 12:27 AMSpaceflight. I've avoided useage of the term without enough ground test results, but it might be time to plant a seed of discussion for the future considering Tajmar's paper. We will need to think about electric power for a smallsat. That will not be easy.A magnetron is not 100% efficient, meaning that this type of device would require north of 1kW of electric power. This is not easily attainable on a satellite. RTGs are usually much below this.:https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generatorSolar panels appear to be the best choice at levels above 1kW:https://en.wikipedia.org/wiki/Solar_panels_on_spacecraftSo dreams of deep space travel might look like this: Solar Panels to ? AU, jetison, then RTG takeover when out of solar influence. IOW, a hybrid power design. Moral of story, optimize for max performance to weight ratio/effeciency. Partner with solar panel and/or RTG provider. Agreed on solar power for inside the asteroid belt. However, What do you think abuot fuel cells?A little unclear on those babies. You might have to upskill me a bit.

Quote from: rfmwguy on 07/28/2015 12:27 AMSpaceflight. I've avoided useage of the term without enough ground test results, but it might be time to plant a seed of discussion for the future considering Tajmar's paper. We will need to think about electric power for a smallsat. That will not be easy.A magnetron is not 100% efficient, meaning that this type of device would require north of 1kW of electric power. This is not easily attainable on a satellite. RTGs are usually much below this.:https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generatorSolar panels appear to be the best choice at levels above 1kW:https://en.wikipedia.org/wiki/Solar_panels_on_spacecraftSo dreams of deep space travel might look like this: Solar Panels to ? AU, jetison, then RTG takeover when out of solar influence. IOW, a hybrid power design. Moral of story, optimize for max performance to weight ratio/effeciency. Partner with solar panel and/or RTG provider. Agreed on solar power for inside the asteroid belt. However, What do you think abuot fuel cells?

Spaceflight. I've avoided useage of the term without enough ground test results, but it might be time to plant a seed of discussion for the future considering Tajmar's paper. We will need to think about electric power for a smallsat. That will not be easy.A magnetron is not 100% efficient, meaning that this type of device would require north of 1kW of electric power. This is not easily attainable on a satellite. RTGs are usually much below this.:https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generatorSolar panels appear to be the best choice at levels above 1kW:https://en.wikipedia.org/wiki/Solar_panels_on_spacecraftSo dreams of deep space travel might look like this: Solar Panels to ? AU, jetison, then RTG takeover when out of solar influence. IOW, a hybrid power design. Moral of story, optimize for max performance to weight ratio/effeciency. Partner with solar panel and/or RTG provider.

Quote from: SeeShells on 07/28/2015 03:12 AMQuote from: A_M_Swallow on 07/28/2015 01:59 AMQuote from: birchoff on 07/28/2015 12:47 AMAgreed on solar power for inside the asteroid belt. However, What do you think abuot fuel cells?Fuel cells may have a use for powering aircraft on oxygen free places like Venus.http://www.lockheedmartin.com/us/products/compact-fusion.htmlLockheed Martin: Compact Fusion Research & Development at the Skunk Works is working on this and since we're talking about future tech I think this would scale quite well to space.There are other promising fusion projects that might fit the bill as well. The question here is how well will a particular approach scale down to 5-10 MW? The Lougheed approach does not have direct energy conversion so you need the standard turbines, etc. That said, the main thing is once you get something into orbit that is reliable you get a lot of amortization...It will be interesting to do the trades on the different approaches. Solar cells are a surprisingly strong alternative - especially to someone who strongly assumed you would need something nuclear for these power levels.

Quote from: A_M_Swallow on 07/28/2015 01:59 AMQuote from: birchoff on 07/28/2015 12:47 AMAgreed on solar power for inside the asteroid belt. However, What do you think abuot fuel cells?Fuel cells may have a use for powering aircraft on oxygen free places like Venus.http://www.lockheedmartin.com/us/products/compact-fusion.htmlLockheed Martin: Compact Fusion Research & Development at the Skunk Works is working on this and since we're talking about future tech I think this would scale quite well to space.

Quote from: birchoff on 07/28/2015 12:47 AMAgreed on solar power for inside the asteroid belt. However, What do you think abuot fuel cells?Fuel cells may have a use for powering aircraft on oxygen free places like Venus.

Agreed on solar power for inside the asteroid belt. However, What do you think abuot fuel cells?

@RodalLet me know if your mind has changed on which question should be asked. I'll be checking in on the NSF throughout the day up to the talk. -I

As expected for the last couple of months, based on knowledgeable sources that Tajmar measurements in vacuum were even lower than NASA's, Tajmar's presentation instead of confirming the EM Drive is being used to make fun of it (in the same publication that previously had the headline about reaching Pluto in 18 months):http://www.wired.com/2015/07/really-propellantless-space-drives-still-not-thing/

As expected for the last couple of months, based on knowledgeable sources that Tajmar measurements in vacuum were even lower than NASA's, Tajmar's presentation instead of confirming the EM Drive is being used to make fun of it (in the same publication that previously had the headline about reaching Pluto in 18 months):http://www.wired.com/2015/07/really-propellantless-space-drives-still-not-thing/Quote . A new publication purports to test the drive’s magical thrust-making abilities. This time, the news is coming from a team at the Dresden University of Technology. They presented their results (thrust signatures of +/-20 microNewtons, if you must know) at a conference today, the Propulsion and Energy Forum and Exposition held by the American Institute for Aeronautics and Astronautics.To be fair, these researchers constructed their version of the device so they could try to eliminate potential sources of error or interference, and they don’t say that they’ve validated the drive—just that they can’t explain where their teeny tiny thrust signatures are coming from. Despite what the Internet is saying, nobody has confirmed anything, and those silly physical laws still say propellantless space drives are impossible. If you want a physics primer and a refresher on the history of this crazy hype-machine of a device, here you go. Sorry to crush your dreams, space cadets. I fail to see how constructing and testing an EM Drive with a Q below 50 (gross overcoupling ?) , with a huge waveguide entering its side asymmetrically (hence not surprising to get a huge side-force) and reporting its dimensions incorrectly by a factor of 2 (error-checking and proof-reading ??, what other errors are present in their unreported procedures ? ) qualifies as "constructing their version of the device so they could try to eliminate potential sources of error."There has been more attention to detail in constructing a device to eliminate potential sources of error by DIY experimenters, frankly speaking...

. A new publication purports to test the drive’s magical thrust-making abilities. This time, the news is coming from a team at the Dresden University of Technology. They presented their results (thrust signatures of +/-20 microNewtons, if you must know) at a conference today, the Propulsion and Energy Forum and Exposition held by the American Institute for Aeronautics and Astronautics.To be fair, these researchers constructed their version of the device so they could try to eliminate potential sources of error or interference, and they don’t say that they’ve validated the drive—just that they can’t explain where their teeny tiny thrust signatures are coming from. Despite what the Internet is saying, nobody has confirmed anything, and those silly physical laws still say propellantless space drives are impossible. If you want a physics primer and a refresher on the history of this crazy hype-machine of a device, here you go. Sorry to crush your dreams, space cadets.

<attached: a promise not to pollute this incredible thread>regarding the eagleworks laser interferometer results.. should this be a standard feature of all EM drive experiments going forward? i noted they seem to have created an EM drive just for this test with design differences. i wonder also if some fine particulate could be introduced into the fustrum and a camera to view the thrust effects internally.

Quote from: Rodal on 07/27/2015 10:36 PMQuote from: deltaMass on 07/27/2015 10:28 PMQuote from: Rodal on 07/27/2015 09:41 PMQuote from: deltaMass on 07/27/2015 09:37 PMQuote from: deltaMass on 07/26/2015 07:07 AMQuote from: TheTraveller on 07/26/2015 06:13 AMQuote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.Writing x0,x1,x2 for the 3 lambdas, this can be expressed as D = [(1-a)/sqrt(a)] * [sqrt(b)/(1-b)], where a = x1/x2, b = x0^{2}/(x1*x2)Notice that D is a separable function of a,b and so can be readily optimised by inspection.Dmax -> infinity when a->0 and/or b->1.Do other relations between x0,1,2 exist to prevent D becoming infinite?Obviously if a > 0 and b < 1 then Dmax when a is min, b is maxI would like to know the maximum theoretical value of Df.Based on the expression I derived above, it corresponds to - a min, i.e. (x1/x2) min- b max, i.e. x0^{2}/(x1*x2) max.What are the values of aMin and bMax, and why?I stopped looking at the Design Factor as a serious formula as soon as I learnt from TheTraveller that it uses the "cut-off" frequency for an open waveguide of constant cross-section, as it is known that tapered waveguides and cavities do NOT have rigid cut-off. Only constant cross-section waveguides have a rigid cut-off condition.Fair enough. So what formula (if any) can you recommend for calculation of the thrust-to-power ratio?The one of Dr. Notsosureofit, a formula by a Ph.D. in Physics, knowledgeable of General Relativity and Radar, and a formula that is explicitly dependent on the mode shape:http://emdrive.wiki/@notsosureofit_HypothesisThanks and I took a look at this. Taken at face value it predicts that, in order to maximise the thrust/power ratio, one requires that independently:- Ds/Db to be minimum- Db to be minimum- L to be minimumeven after accounting for the frequency scaling per Appendix I.The thrust predictions seem roughly in line with the experimental data.k = F/P seems to be, for the ranges graphed, about 3*10^{-7} N/W (about 300 uN for 1000 W).I'm looking for a way to get much higher F/P values. I've indicated how this can be done per this formula.I would welcome some concrete suggestions.[A couple of notes on this derivation:1. In Appendix I the n factor due to L is omitted (but the scaling conclusion is correct anyway)2. Whenever I read about "accelerated photons" my toes curl up]

Quote from: deltaMass on 07/27/2015 10:28 PMQuote from: Rodal on 07/27/2015 09:41 PMQuote from: deltaMass on 07/27/2015 09:37 PMQuote from: deltaMass on 07/26/2015 07:07 AMQuote from: TheTraveller on 07/26/2015 06:13 AMQuote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.Writing x0,x1,x2 for the 3 lambdas, this can be expressed as D = [(1-a)/sqrt(a)] * [sqrt(b)/(1-b)], where a = x1/x2, b = x0^{2}/(x1*x2)Notice that D is a separable function of a,b and so can be readily optimised by inspection.Dmax -> infinity when a->0 and/or b->1.Do other relations between x0,1,2 exist to prevent D becoming infinite?Obviously if a > 0 and b < 1 then Dmax when a is min, b is maxI would like to know the maximum theoretical value of Df.Based on the expression I derived above, it corresponds to - a min, i.e. (x1/x2) min- b max, i.e. x0^{2}/(x1*x2) max.What are the values of aMin and bMax, and why?I stopped looking at the Design Factor as a serious formula as soon as I learnt from TheTraveller that it uses the "cut-off" frequency for an open waveguide of constant cross-section, as it is known that tapered waveguides and cavities do NOT have rigid cut-off. Only constant cross-section waveguides have a rigid cut-off condition.Fair enough. So what formula (if any) can you recommend for calculation of the thrust-to-power ratio?The one of Dr. Notsosureofit, a formula by a Ph.D. in Physics, knowledgeable of General Relativity and Radar, and a formula that is explicitly dependent on the mode shape:http://emdrive.wiki/@notsosureofit_Hypothesis

Quote from: Rodal on 07/27/2015 09:41 PMQuote from: deltaMass on 07/27/2015 09:37 PMQuote from: deltaMass on 07/26/2015 07:07 AMQuote from: TheTraveller on 07/26/2015 06:13 AMQuote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.Writing x0,x1,x2 for the 3 lambdas, this can be expressed as D = [(1-a)/sqrt(a)] * [sqrt(b)/(1-b)], where a = x1/x2, b = x0^{2}/(x1*x2)Notice that D is a separable function of a,b and so can be readily optimised by inspection.Dmax -> infinity when a->0 and/or b->1.Do other relations between x0,1,2 exist to prevent D becoming infinite?Obviously if a > 0 and b < 1 then Dmax when a is min, b is maxI would like to know the maximum theoretical value of Df.Based on the expression I derived above, it corresponds to - a min, i.e. (x1/x2) min- b max, i.e. x0^{2}/(x1*x2) max.What are the values of aMin and bMax, and why?I stopped looking at the Design Factor as a serious formula as soon as I learnt from TheTraveller that it uses the "cut-off" frequency for an open waveguide of constant cross-section, as it is known that tapered waveguides and cavities do NOT have rigid cut-off. Only constant cross-section waveguides have a rigid cut-off condition.Fair enough. So what formula (if any) can you recommend for calculation of the thrust-to-power ratio?

Quote from: deltaMass on 07/27/2015 09:37 PMQuote from: deltaMass on 07/26/2015 07:07 AMQuote from: TheTraveller on 07/26/2015 06:13 AMQuote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.Writing x0,x1,x2 for the 3 lambdas, this can be expressed as D = [(1-a)/sqrt(a)] * [sqrt(b)/(1-b)], where a = x1/x2, b = x0^{2}/(x1*x2)Notice that D is a separable function of a,b and so can be readily optimised by inspection.Dmax -> infinity when a->0 and/or b->1.Do other relations between x0,1,2 exist to prevent D becoming infinite?Obviously if a > 0 and b < 1 then Dmax when a is min, b is maxI would like to know the maximum theoretical value of Df.Based on the expression I derived above, it corresponds to - a min, i.e. (x1/x2) min- b max, i.e. x0^{2}/(x1*x2) max.What are the values of aMin and bMax, and why?I stopped looking at the Design Factor as a serious formula as soon as I learnt from TheTraveller that it uses the "cut-off" frequency for an open waveguide of constant cross-section, as it is known that tapered waveguides and cavities do NOT have rigid cut-off. Only constant cross-section waveguides have a rigid cut-off condition.

Quote from: deltaMass on 07/26/2015 07:07 AMQuote from: TheTraveller on 07/26/2015 06:13 AMQuote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.Writing x0,x1,x2 for the 3 lambdas, this can be expressed as D = [(1-a)/sqrt(a)] * [sqrt(b)/(1-b)], where a = x1/x2, b = x0^{2}/(x1*x2)Notice that D is a separable function of a,b and so can be readily optimised by inspection.Dmax -> infinity when a->0 and/or b->1.Do other relations between x0,1,2 exist to prevent D becoming infinite?Obviously if a > 0 and b < 1 then Dmax when a is min, b is maxI would like to know the maximum theoretical value of Df.Based on the expression I derived above, it corresponds to - a min, i.e. (x1/x2) min- b max, i.e. x0^{2}/(x1*x2) max.What are the values of aMin and bMax, and why?

Quote from: TheTraveller on 07/26/2015 06:13 AMQuote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.Writing x0,x1,x2 for the 3 lambdas, this can be expressed as D = [(1-a)/sqrt(a)] * [sqrt(b)/(1-b)], where a = x1/x2, b = x0^{2}/(x1*x2)Notice that D is a separable function of a,b and so can be readily optimised by inspection.Dmax -> infinity when a->0 and/or b->1.Do other relations between x0,1,2 exist to prevent D becoming infinite?Obviously if a > 0 and b < 1 then Dmax when a is min, b is max

Quote from: birchoff on 07/26/2015 02:12 AM@Rodal or anyone for that matter know how to calculate shawyer's design factor?Shawyer's Df equation is attached. Have verified with Shawyer that it is correct.

@Rodal or anyone for that matter know how to calculate shawyer's design factor?

It's a shame that Tajmar, being Director of an institute with access to fine equipment and trained staff, did such a shoddy job. 1. Not self-contained2. Not correctly impedance matched3. Thermal effects not analysed4. Bloody great hole in the side of the frustum.He's a good theoretician, but struggles as an experimenter. We all have our strengths and weaknesses.