Author Topic: EM Drive Developments Thread 1  (Read 796268 times)

Offline JohnFornaro

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Re: EM Drive Developments
« Reply #1720 on: 10/08/2014 05:22 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".
« Last Edit: 10/08/2014 05:37 PM by JohnFornaro »
Sometimes I just flat out don't get it.

Offline Rodal

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Re: EM Drive Developments
« Reply #1721 on: 10/08/2014 05:41 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.  ;)

Offline Rodal

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Re: EM Drive Developments
« Reply #1722 on: 10/08/2014 05:43 PM »
[Hey, kernosabe, ya giv me da material, I work with it]

The Curious Case of Benjamin Button John Fornaro   :)   getting younger and younger









« Last Edit: 10/08/2014 06:07 PM by Rodal »

Offline Notsosureofit

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Re: EM Drive Developments
« Reply #1723 on: 10/08/2014 05:46 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.  ;)

Sawyer worked out an optimization scheme for the truncated cone which would work for McCullough's.

The Cannae drive did have asymmetry in one of the pillboxes, just made no difference.  So that is a math nightmare etc etc

(You are both young pups as far as I'm concerned. Gosh durn it !)
« Last Edit: 10/08/2014 05:51 PM by Notsosureofit »

Offline Notsosureofit

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Re: EM Drive Developments
« Reply #1724 on: 10/08/2014 06:07 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = c^2/ ( g * lambda )

I think.

So maybe Shawyer is on to something w/ his deceleration curve ??
« Last Edit: 10/08/2014 06:18 PM by Notsosureofit »

Offline Mulletron

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Re: EM Drive Developments
« Reply #1725 on: 10/08/2014 06:36 PM »
A factor I didn't apply to the (poorly informed) 3d model I made was the effect of length contraction at relativistic speeds. I don't think there is a way to do this. Also I didn't give any treatment to the Rindler sphere approaching you, to optimize anything. In the end I dropped this problem and didn't try to solve it because it seems trying to solve it would be tantamount to trying to invalidate GR and SR, which you can't. Length contraction is the cosmic speed brake that keeps you from breaking C.

Also on the subject of asymmetries, I never fully solved the questions I brought up previously about chirality in PTFE. I haven't had time to focus on all the aspects of this subject.
« Last Edit: 10/08/2014 06:39 PM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline IslandPlaya

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Re: EM Drive Developments
« Reply #1726 on: 10/08/2014 06:37 PM »
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=35805.0
Much obliged and thanks.

Offline Rodal

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Re: EM Drive Developments
« Reply #1727 on: 10/08/2014 06:49 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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
« Last Edit: 10/08/2014 06:59 PM by Rodal »

Offline Rodal

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Re: EM Drive Developments
« Reply #1728 on: 10/08/2014 06:54 PM »
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=35805.0
Much 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 :)

Offline Notsosureofit

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Re: EM Drive Developments
« Reply #1729 on: 10/08/2014 07:06 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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

Yes, 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 ?

Offline Rodal

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Re: EM Drive Developments
« Reply #1730 on: 10/08/2014 07:10 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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

Yes, 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

Offline zen-in

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Re: EM Drive Developments
« Reply #1731 on: 10/08/2014 07:27 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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

Yes, 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

One 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.

Offline IslandPlaya

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Re: EM Drive Developments
« Reply #1732 on: 10/08/2014 07:30 PM »
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=35805.0
Much 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.

Offline Notsosureofit

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Re: EM Drive Developments
« Reply #1733 on: 10/08/2014 07:34 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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

Yes, 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

Right on!  That gives + and - 90 degree phase shifts that cancel out, so the power goes round-n-round, so to speak.

(must have the uncertainty principle in there some where ?)
« Last Edit: 10/08/2014 07:35 PM by Notsosureofit »

Offline IslandPlaya

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Re: EM Drive Developments
« Reply #1734 on: 10/08/2014 07:37 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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

Yes, 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

One 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...

Offline Rodal

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Re: EM Drive Developments
« Reply #1735 on: 10/08/2014 07:40 PM »
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=35805.0
Much 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.  :)
« Last Edit: 10/08/2014 07:45 PM by Rodal »

Offline IslandPlaya

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Re: EM Drive Developments
« Reply #1736 on: 10/08/2014 07:51 PM »
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=35805.0
Much 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.  :)
Thanks!
I felt in my bones that Prof. McCulloch had a handle on this.
I will try to embiggen my knowledge further.

Offline zen-in

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Re: EM Drive Developments
« Reply #1737 on: 10/08/2014 07:58 PM »
Just a curiosity:

The Q of a perfect superconducting cylindrical cavity of lambda vertical dimension at the earths surface

Q = 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

Yes, 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

One 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...

Low temperature superconductors are used in NMR machines.  However the magnet is a very powerful DC magnet. 

Offline IslandPlaya

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Re: EM Drive Developments
« Reply #1738 on: 10/08/2014 08:04 PM »
What is the mechanism that destroys SC for AC currents?
Please enlighten me.

Offline Mulletron

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Re: EM Drive Developments
« Reply #1739 on: 10/08/2014 08:08 PM »
Challenge your preconceptions, or they will challenge you. - Velik

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