Quote from: Mulletron on 10/08/2014 04:22 PM...The whole device, as a system is biased to one side in symmetry. The Looook harder thing started as a running joke a little bit back.Well, I cannot work mathematically with a description of "The whole device, as a system is biased to one side in symmetry"...

...The whole device, as a system is biased to one side in symmetry. The Looook harder thing started as a running joke a little bit back.

Quote from: Rodal on 10/08/2014 04:25 PMQuote from: Mulletron on 10/08/2014 04:22 PM...The whole device, as a system is biased to one side in symmetry. The Looook harder thing started as a running joke a little bit back.Well, I cannot work mathematically with a description of "The whole device, as a system is biased to one side in symmetry"...I think what he means is that the pillbox is symmetrical radially along its axis, but there is no line of symmetry in a plane 90 degrees from the axis.And it's "looook", not "look".

Quote from: JohnFornaro on 10/08/2014 05:22 PMQuote from: Rodal on 10/08/2014 04:25 PMQuote from: Mulletron on 10/08/2014 04:22 PM...The whole device, as a system is biased to one side in symmetry. The Looook harder thing started as a running joke a little bit back.Well, I cannot work mathematically with a description of "The whole device, as a system is biased to one side in symmetry"...I think what he means is that the pillbox is symmetrical radially along its axis, but there is no line of symmetry in a plane 90 degrees from the axis.And it's "looook", not "look".Darth Vader, there is a great disconnect with the force: what Prof. McCulloch can actually calculate and apparently what Vader thinks can be calculated.

Just a curiosity:The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surfaceQ = c^2/ ( g * lambda )I think.So maybe Shawyer is on to something w/ his deceleration curve ??

Dr. Rodal. I know your expertise on experimental setups.Please could you cast your searching eyes over this please?http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf... and comment in the thread here:http://forum.nasaspaceflight.com/index.php?topic=35805.0Much obliged and thanks.

Quote from: Notsosureofit on 10/08/2014 06:07 PMJust a curiosity:The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surfaceQ = c^2/ ( g * lambda )I think.So maybe Shawyer is on to something w/ his deceleration curve ??Yes, when one says goodbye to "wordy" explanations and just looks at the math, a number of explanations look very similar:Force = ( Q* PowerInput / frequency ) * (1/w_up) - ( Q* PowerInput / frequency ) * (1/w_down) = Q * PowerInput /( frequency * CharacteristicLengthA ) - Q * PowerInput /( frequency *CharacteristicLengthB ) = Q * ( PowerInput / CharacteristicSpeedA ) - Q * ( PowerInput / CharacteristicSpeedB ) = Q * PowerInput * (1/ CharacteristicSpeedA - 1 /CharacteristicSpeedB)Recall that, dimensionally Power = Force * speed

Quote from: Rodal on 10/08/2014 06:49 PMQuote from: Notsosureofit on 10/08/2014 06:07 PMJust a curiosity:The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surfaceQ = c^2/ ( g * lambda )I think.So maybe Shawyer is on to something w/ his deceleration curve ??Yes, when one says goodbye to "wordy" explanations and just looks at the math, a number of explanations look very similar:Force = ( Q* PowerInput / frequency ) * (1/w_up) - ( Q* PowerInput / frequency ) * (1/w_down) = Q * PowerInput /( frequency * CharacteristicLengthA ) - Q * PowerInput /( frequency *CharacteristicLengthB ) = Q * ( PowerInput / CharacteristicSpeedA ) - Q * ( PowerInput / CharacteristicSpeedB ) = Q * PowerInput * (1/ CharacteristicSpeedA - 1 /CharacteristicSpeedB)Recall that, dimensionally Power = Force * speedYes, the problem I'm having that the Q from bandwidth implies an exponential decay time constant:T = c / ( g * pi ) So where is the power going if the cavity is superconductive ?

Quote from: Notsosureofit on 10/08/2014 07:06 PMQuote from: Rodal on 10/08/2014 06:49 PMQuote from: Notsosureofit on 10/08/2014 06:07 PMJust a curiosity:The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surfaceQ = c^2/ ( g * lambda )I think.So maybe Shawyer is on to something w/ his deceleration curve ??Yes, when one says goodbye to "wordy" explanations and just looks at the math, a number of explanations look very similar:Force = ( Q* PowerInput / frequency ) * (1/w_up) - ( Q* PowerInput / frequency ) * (1/w_down) = Q * PowerInput /( frequency * CharacteristicLengthA ) - Q * PowerInput /( frequency *CharacteristicLengthB ) = Q * ( PowerInput / CharacteristicSpeedA ) - Q * ( PowerInput / CharacteristicSpeedB ) = Q * PowerInput * (1/ CharacteristicSpeedA - 1 /CharacteristicSpeedB)Recall that, dimensionally Power = Force * speedYes, the problem I'm having that the Q from bandwidth implies an exponential decay time constant:T = c / ( g * pi ) So where is the power going if the cavity is superconductive ?Superconductivity: a quantum mechanics effect for which people's intuition fails, because our intuition is built around our macro world and not the quantum world. One would have to work out the quantum mechanics math to answer

Quote from: IslandPlaya on 10/08/2014 06:37 PMDr. Rodal. I know your expertise on experimental setups.Please could you cast your searching eyes over this please?http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf... and comment in the thread here:http://forum.nasaspaceflight.com/index.php?topic=35805.0Much obliged and thanks.Hi IslandPlaya, still appreciating you being the first one to welcome me to this forum I wish I could, but unfortunately I can't at the moment embark into another topic. I thank you for the gracious invitation

Quote from: Rodal on 10/08/2014 07:10 PMQuote from: Notsosureofit on 10/08/2014 07:06 PMQuote from: Rodal on 10/08/2014 06:49 PMQuote from: Notsosureofit on 10/08/2014 06:07 PMJust a curiosity:The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surfaceQ = c^2/ ( g * lambda )I think.So maybe Shawyer is on to something w/ his deceleration curve ??Yes, when one says goodbye to "wordy" explanations and just looks at the math, a number of explanations look very similar:Force = ( Q* PowerInput / frequency ) * (1/w_up) - ( Q* PowerInput / frequency ) * (1/w_down) = Q * PowerInput /( frequency * CharacteristicLengthA ) - Q * PowerInput /( frequency *CharacteristicLengthB ) = Q * ( PowerInput / CharacteristicSpeedA ) - Q * ( PowerInput / CharacteristicSpeedB ) = Q * PowerInput * (1/ CharacteristicSpeedA - 1 /CharacteristicSpeedB)Recall that, dimensionally Power = Force * speedYes, the problem I'm having that the Q from bandwidth implies an exponential decay time constant:T = c / ( g * pi ) So where is the power going if the cavity is superconductive ?Superconductivity: a quantum mechanics effect for which people's intuition fails, because our intuition is built around our macro world and not the quantum world. One would have to work out the quantum mechanics math to answerOne thing about superconductors most people are not aware of is that they do not work very well with AC. There are specially designed configurations used for 60 Hz AC transmission cables, but at higher frequencies it is my understanding that superconductors are unusable. I don't see any way that superconductors could be used to improve this em thruster.

Quote from: Rodal on 10/08/2014 06:54 PMQuote from: IslandPlaya on 10/08/2014 06:37 PMDr. Rodal. I know your expertise on experimental setups.Please could you cast your searching eyes over this please?http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf... and comment in the thread here:http://forum.nasaspaceflight.com/index.php?topic=35805.0Much obliged and thanks.Hi IslandPlaya, still appreciating you being the first one to welcome me to this forum I wish I could, but unfortunately I can't at the moment embark into another topic. I thank you for the gracious invitation Thank you. However the topic has been nuked (maybe on my advice.) I am currently trying to understand Hubble and Unrhu horizons as I intuit that is the theoretical key to the anomalous thrust.

Quote from: IslandPlaya on 10/08/2014 07:30 PMQuote from: Rodal on 10/08/2014 06:54 PMQuote from: IslandPlaya on 10/08/2014 06:37 PMDr. Rodal. I know your expertise on experimental setups.Please could you cast your searching eyes over this please?http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf... and comment in the thread here:http://forum.nasaspaceflight.com/index.php?topic=35805.0Much obliged and thanks.Hi IslandPlaya, still appreciating you being the first one to welcome me to this forum I wish I could, but unfortunately I can't at the moment embark into another topic. I thank you for the gracious invitation Thank you. However the topic has been nuked (maybe on my advice.) I am currently trying to understand Hubble and Unrhu horizons as I intuit that is the theoretical key to the anomalous thrust.Take another look at http://physicsfromtheedge.blogspot.it/2014/10/mihsc-vs-emdrive-data-1.html, Prof. McCulloch has now incorporated the Chinese data, and all the data [Shawyer, China and NASA] is pretty well calculated by McCulloch's formula except for one experiment (out of 3 in the list) by Brady et.al. that I had pointed out is extremely anomalous (they raised the Q by a factor of 2.5 and the force came down to 1/2). The Unruth/McCulloch formula does a great job [compared to everything else that has been offered, and look at this thread we have considered all kinds of stuff]. What is most interesting again is that McCulloch does not use fudge factors or an excessive number of parameters. Actually McCulloch's formula is bare bones: PowerInput, Q, frequency and the geometry: that's all folks.

Quote from: zen-in on 10/08/2014 07:27 PMQuote from: Rodal on 10/08/2014 07:10 PMQuote from: Notsosureofit on 10/08/2014 07:06 PMQuote from: Rodal on 10/08/2014 06:49 PMQuote from: Notsosureofit on 10/08/2014 06:07 PMJust a curiosity:The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surfaceQ = c^2/ ( g * lambda )I think.So maybe Shawyer is on to something w/ his deceleration curve ??Yes, when one says goodbye to "wordy" explanations and just looks at the math, a number of explanations look very similar:Force = ( Q* PowerInput / frequency ) * (1/w_up) - ( Q* PowerInput / frequency ) * (1/w_down) = Q * PowerInput /( frequency * CharacteristicLengthA ) - Q * PowerInput /( frequency *CharacteristicLengthB ) = Q * ( PowerInput / CharacteristicSpeedA ) - Q * ( PowerInput / CharacteristicSpeedB ) = Q * PowerInput * (1/ CharacteristicSpeedA - 1 /CharacteristicSpeedB)Recall that, dimensionally Power = Force * speedYes, the problem I'm having that the Q from bandwidth implies an exponential decay time constant:T = c / ( g * pi ) So where is the power going if the cavity is superconductive ?Superconductivity: a quantum mechanics effect for which people's intuition fails, because our intuition is built around our macro world and not the quantum world. One would have to work out the quantum mechanics math to answerOne thing about superconductors most people are not aware of is that they do not work very well with AC. There are specially designed configurations used for 60 Hz AC transmission cables, but at higher frequencies it is my understanding that superconductors are unusable. I don't see any way that superconductors could be used to improve this em thruster.Do you know the mechanism behind this? SC are used for NMR and all sorts of hi-freq things. I may be wrong though...