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?

Quote from: QuantumG on 07/27/2015 09:14 pmHey look, someone actually talking about space on this thread.Quote from: Rodal on 07/27/2015 09:10 pmWhich are not a problem for the well-funded Chinese [..] Space Program.What makes you think it's well funded? Last I heard, the best estimates put it below ESA.. i.e., about a third of the US program.What makes you think it's not well funded? After all, it's claimed to be a revolutionary propulsion paradigm. The chief reason that would occur to most people is that it's not funded because it doesn't actually work. I wonder if there'll be a paper about that.

Hey look, someone actually talking about space on this thread.Quote from: Rodal on 07/27/2015 09:10 pmWhich are not a problem for the well-funded Chinese [..] Space Program.What makes you think it's well funded? Last I heard, the best estimates put it below ESA.. i.e., about a third of the US program.

Which are not a problem for the well-funded Chinese [..] Space Program.

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: 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 09:27 pmQuote from: QuantumG on 07/27/2015 09:14 pmHey look, someone actually talking about space on this thread.Quote from: Rodal on 07/27/2015 09:10 pmWhich are not a problem for the well-funded Chinese [..] Space Program.What makes you think it's well funded? Last I heard, the best estimates put it below ESA.. i.e., about a third of the US program.What makes you think it's not well funded? After all, it's claimed to be a revolutionary propulsion paradigm. The chief reason that would occur to most people is that it's not funded because it doesn't actually work. I wonder if there'll be a paper about that.Suppose it does work and they don't publicize it because its such a revolutionary advancement that gives them a major advantage on the final frontier.... By the way, Tajmar's paper has been deemed a "confirmation" by hacked.com https://hacked.com/scientists-confirm-impossible-em-drive-propulsion/

...You may want to put more effort into calculating whether these effects can account for the forces observed (micro-newtons) or conducting something similar to the experiment I described for detecting these particles. Again, I hope I'm wrong! It would be great to see an analyses that says "Yes, these effects are occurring but they cannot account for the thrust because of X."-Joseph KnubleNASA GSFC Code 555Microwave Instrument Technology Branch

Quote from: SeeShells on 07/27/2015 06:57 pmSo if you or one of your friends within that 50 mile radius want to put a couple bucks to help not to make more millions but because they choose to dream, I'll welcome it.Shell*grin* I put a little into your fund the day you started it. Like I said, I want to believe. I'll have a lot more faith in your data (once you have it) than random unsubstantiated statements.

So if you or one of your friends within that 50 mile radius want to put a couple bucks to help not to make more millions but because they choose to dream, I'll welcome it.Shell

Quote from: rfmwguy on 07/27/2015 08:55 pmQuote from: deltaMass on 07/27/2015 07:59 pmAt least an explanation exists for why the Chinese don't already have production lines for churning out EmDrives. Can there be any other excuse?Lack of power at ground level. Lack of millions of potential buyers. Translation; lack of market.Which are not a problem for the well-funded Chinese Air Force and Space Program. Lack of a market has not impeded their on-going Taikonaut and Mini-Space Station programs, Space Defense tests, as well as their long-term programs including ion-drives and Moon program. So, why no deployment of the Yang EM Drive in Space, if it can really do what is claimed?

Quote from: deltaMass on 07/27/2015 07:59 pmAt least an explanation exists for why the Chinese don't already have production lines for churning out EmDrives. Can there be any other excuse?Lack of power at ground level. Lack of millions of potential buyers. Translation; lack of market.

At least an explanation exists for why the Chinese don't already have production lines for churning out EmDrives. Can there be any other excuse?

Quote from: SeeShells on 07/27/2015 06:57 pmQuote from: mittelhauser on 07/27/2015 05:06 pmQuote from: TheTraveller on 07/27/2015 06:58 amQuote from: demofsky on 07/27/2015 06:30 amThe device used by Tajmar looks more like a version of Shawyer's first fustrum than the latest work by Yang, et al. It would be very nice if we could get actual schematics of Tajmar's fustrum rather than squinting at pictures trying to figure out what he did...<clip>So if you or one of your friends within that 50 mile radius want to put a couple bucks to help not to make more millions but because they choose to dream, I'll welcome it.ShellI am game. Maybe you can get a kickstarter togetger. I would contribute and I am sure more people would too

Quote from: mittelhauser on 07/27/2015 05:06 pmQuote from: TheTraveller on 07/27/2015 06:58 amQuote from: demofsky on 07/27/2015 06:30 amThe device used by Tajmar looks more like a version of Shawyer's first fustrum than the latest work by Yang, et al. It would be very nice if we could get actual schematics of Tajmar's fustrum rather than squinting at pictures trying to figure out what he did...<clip>So if you or one of your friends within that 50 mile radius want to put a couple bucks to help not to make more millions but because they choose to dream, I'll welcome it.Shell

Quote from: TheTraveller on 07/27/2015 06:58 amQuote from: demofsky on 07/27/2015 06:30 amThe device used by Tajmar looks more like a version of Shawyer's first fustrum than the latest work by Yang, et al. It would be very nice if we could get actual schematics of Tajmar's fustrum rather than squinting at pictures trying to figure out what he did...

Quote from: demofsky on 07/27/2015 06:30 amThe device used by Tajmar looks more like a version of Shawyer's first fustrum than the latest work by Yang, et al. It would be very nice if we could get actual schematics of Tajmar's fustrum rather than squinting at pictures trying to figure out what he did...

The device used by Tajmar looks more like a version of Shawyer's first fustrum than the latest work by Yang, et al. It would be very nice if we could get actual schematics of Tajmar's fustrum rather than squinting at pictures trying to figure out what he did...

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?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?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?

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?

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?

Quote from: birchoff on 07/26/2015 02:12 am@Rodal or anyone for that matter know how to calculate shawyer's design factor?

...By the way, Tajmar's paper has been deemed a "confirmation" by hacked.com https://hacked.com/scientists-confirm-impossible-em-drive-propulsion/

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?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_HypothesisDo you know or would notsosureofit what grouping of 3 modes are right above > 2GHz at 300 uN?http://emdrive.wiki/File:00_chart21.jpgThanks,Shell

Quote from: jknuble on 07/27/2015 10:40 pm...You may want to put more effort into calculating whether these effects can account for the forces observed (micro-newtons) or conducting something similar to the experiment I described for detecting these particles. Again, I hope I'm wrong! It would be great to see an analyses that says "Yes, these effects are occurring but they cannot account for the thrust because of X."-Joseph KnubleNASA GSFC Code 555Microwave Instrument Technology BranchOutgassing and electrical corona breakdown discussion in the thread go back to very early in early threads, by different individuals. Bringing up multipaction is something that we appreciate your bringing up to the thread. SeeShells has specifically thought about these effects in her design. Rfmwguy also has a perforated shell to reduce most gas effects.SeeShells ( http://www.gofundme.com/yy7yz3k) has a comprehensive program to look at the effects you are describing including a camera to film what's happening inside the EM Drive through her perforated EM Drive. What additional efforts, specifically, would you suggest that SeeShells has not taken to examine the effects you are describing ?

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: zellerium on 07/27/2015 09:44 pm...By the way, Tajmar's paper has been deemed a "confirmation" by hacked.com https://hacked.com/scientists-confirm-impossible-em-drive-propulsion/I give up. people will think whatever they want.

...[snip very long post to minimize bandwitdth]

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 07/27/2015 11:49 pmQuote 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?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_HypothesisDo you know or would notsosureofit what grouping of 3 modes are right above > 2GHz at 300 uN?http://emdrive.wiki/File:00_chart21.jpgThanks,ShellMy understanding is that in:http://emdrive.wiki/images/f/fc/00_chart21.jpgThe three curves represent p=1, p=2 and p=3, where p is the quantum number in the longitudinal direction, for modes TMmnp and TEmnp, for suitable m and n values.Evidently p=1 is best, which is what I would expect (for example: TE011 is better than TE012 which is better than TE013). Similarly TE111 is better than TE112 which is better than TE113. Ditto for TM modes: TM111 is better than TM112 which is better than TM113. Because lower modes have greater amplitude.Outside of Notsosureofit's formula there is the question of whether having a stronger field near the apex is better or worse. Inspection of the Wolfram/Mathematica analysis of Meep runs reveals this story which to me is becoming more clear.