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#1560 by TheTraveller on 13 Oct, 2016 13:39
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This argument from you is so fundamentally wrong I do not understand how you can possibly think it makes sense.
And there is you problem as it works and the experimental test data of both big to small Thrust experimental force measurement and small to big accelerative Reaction experimental force measurement supports the theory.
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#1561 by meberbs on 13 Oct, 2016 13:46
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Emphasis mine.This argument from you is so fundamentally wrong I do not understand how you can possibly think it makes sense.
And there is you problem as it works and the experimental test data of both big to small Thrust experimental force measurement and small to big accelerative Reaction experimental force measurement supports the theory.
You finally said what Shawyer's theory predicts correctly. Too bad this is in contradiction to the direction of acceleration found in all independent experiments. See the quote you just posted for what this means. -
#1562 by Tellmeagain on 13 Oct, 2016 14:00
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To explain a bit more about the Differential Radiation Pressure generated Thrust vector (small to big) and how it can be measured to confirm it actually exists, we need only read Roger's Demonstrator EmDrive technical report which details how he measured the Thrust force.
Attached are the Force thrust vectors as measured in the scale based test results and the actual weight change that was measured in each of the force vector directions.
Also attached is the technical report for those who wish to read up on how the tests were conducted.
To measure the accelerative Reaction force that is equal but opposite to the Thrust force requires the EmDrive to be tested on a rotary test rig as was done on 2006, which showed the accelerative Reaction force was in the opposite direction to the Thrust force and of the same level of force as shown in line 5 of the results in the summary attachment.
For DIYers doing force measurements using spring based systems (scales, torsion pendulums, etc), it would seem that what has been measured is the Thrust force and not the accelerative Reaction force.
What this data shows is the EmDrive does work as Roger's theory claims, generating a Thrust force (small to big) from the differential of the radiation pressure due to EmWave momentum change as the frustum diameter varies that can be measured.
Plus the opposite but equal accelerative Reaction force can be measure by using a rotary test rig.
So the test results show there is a small to big Thrust force vector generated which follows the change in frustum diameter and there is an equal but opposite accelerative Reaction force generated that can be used to achieve propollentless but not energy less acceleration.
Should add that you can measure both forces but not at the same time.
Those that desire to develop alternative explanations as to why the Thrust force is generated, need to factor in that the value of the Thrust force follows end plate diameter change vs EmWave momentum change vs excitation mode vs excitation freq vs frustum Qu vs frustum Rf power.
So far there is only one theory that fits all those variables together and that is Roger's theory.
I remembered that Mr. Shawyer once said the test rig must be in moving (?) for the EmDrive to generate thrust. Did that statement contradict to this test? If not, how?
Or did he say in acceleration? If that is the case, may gravity can substitute for the acceleration? -
#1563 by Monomorphic on 13 Oct, 2016 14:06
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Better image of minor end plate and circularization antenna.
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#1564 by TheTraveller on 13 Oct, 2016 14:07
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Did a curve fit of the patent's thrust TC curve to a standard 5x TC curve and they fit as attached.
Which says the 1 Tc cavity time is ~0.2 sec and from that the cavity Q can be calculated as Q = Tc (2 Pi Freq) = ~3.08 billion.
From that the specific accelerative Reaction force can be calculated as (2 Q Pwr Df) / c = (2 * 3,080,000,000 * 1,000 * 0.5 (conservative)) / c = 10,350N/kWrf or 1,056kg/kWrf or 1.056 mt/kWrf.
Not bad.
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#1565 by RotoSequence on 13 Oct, 2016 14:07
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This argument from you is so fundamentally wrong I do not understand how you can possibly think it makes sense.
And there is you problem as it works and the experimental test data of both big to small Thrust experimental force measurement and small to big accelerative Reaction experimental force measurement supports the theory.
EM Drive may work and scale as Shawyer describes, but that doesn't mean the theory of operation is mathematically robust. Mathematical theories that don't stand up to facts are no good, but mathematically inconsistent theories aren't a good stopping point either. -
#1566 by RotoSequence on 13 Oct, 2016 14:11
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Did a curve fit of the patent's thrust TC curve to a standard 5x TC curve and they fit as attached.
Which says the 1 Tc cavity time is ~0.2 sec and from that the cavity Q can be calculated as Q = Tc (2 Pi Freq) = ~3.08 billion.
From that the specific accelerative Reaction force can be calculated as (2 Q Pwr Df) / c = (2 * 3,080,000,000 * 1,000 * 0.5 (conservative)) / c = 10,350N/kWrf or 1,056kg/kWrf or 1.056 mt/kWrf.
Not bad.
I'll have to ask that you forgive my incredulity in light of the current lack of public, high impulse test results. This sounds way, way too good to be true. -
#1567 by TheTraveller on 13 Oct, 2016 14:14
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To explain a bit more about the Differential Radiation Pressure generated Thrust vector (small to big) and how it can be measured to confirm it actually exists, we need only read Roger's Demonstrator EmDrive technical report which details how he measured the Thrust force.
Attached are the Force thrust vectors as measured in the scale based test results and the actual weight change that was measured in each of the force vector directions.
Also attached is the technical report for those who wish to read up on how the tests were conducted.
To measure the accelerative Reaction force that is equal but opposite to the Thrust force requires the EmDrive to be tested on a rotary test rig as was done on 2006, which showed the accelerative Reaction force was in the opposite direction to the Thrust force and of the same level of force as shown in line 5 of the results in the summary attachment.
For DIYers doing force measurements using spring based systems (scales, torsion pendulums, etc), it would seem that what has been measured is the Thrust force and not the accelerative Reaction force.
What this data shows is the EmDrive does work as Roger's theory claims, generating a Thrust force (small to big) from the differential of the radiation pressure due to EmWave momentum change as the frustum diameter varies that can be measured.
Plus the opposite but equal accelerative Reaction force can be measure by using a rotary test rig.
So the test results show there is a small to big Thrust force vector generated which follows the change in frustum diameter and there is an equal but opposite accelerative Reaction force generated that can be used to achieve propollentless but not energy less acceleration.
Should add that you can measure both forces but not at the same time.
Those that desire to develop alternative explanations as to why the Thrust force is generated, need to factor in that the value of the Thrust force follows end plate diameter change vs EmWave momentum change vs excitation mode vs excitation freq vs frustum Qu vs frustum Rf power.
So far there is only one theory that fits all those variables together and that is Roger's theory.
I remembered that Mr. Shawyer once said the test rig must be in moving (?) for the EmDrive to generate thrust. Did that statement contradict to this test? If not, how?
Or did he say in acceleration? If that is the case, may gravity can substitute for the acceleration?
An EmDrive generates 2 forces that can be measures but not at the same time nor with the same test setup:
1) Thrust force with a vector small to big that is the result of the differential radiation pressure on the end plates. This is a static force that can be measured with a scale.
2) Reaction force with a vector big to small and is the equal but opposite force to the Thrust force. This is a dynamic force and can only be measured during free acceleration.
Roger has measured both forces and has shown that they are approx equal in force amplitude but opposite in force vector direction.
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#1568 by TheTraveller on 13 Oct, 2016 14:17
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This argument from you is so fundamentally wrong I do not understand how you can possibly think it makes sense.
And there is you problem as it works and the experimental test data of both big to small Thrust experimental force measurement and small to big accelerative Reaction experimental force measurement supports the theory.
EM Drive may work and scale as Shawyer describes, but that doesn't mean the theory of operation is mathematically robust. Mathematical theories that don't stand up to facts are no good, but mathematically inconsistent theories aren't a good stopping point either.
Everyone is free to develop their own theory as long as it matches the experimental data of dual force generation with equal force amplitudes but opposite force vectors. -
#1569 by TheTraveller on 13 Oct, 2016 14:22
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Did a curve fit of the patent's thrust TC curve to a standard 5x TC curve and they fit as attached.
Which says the 1 Tc cavity time is ~0.2 sec and from that the cavity Q can be calculated as Q = Tc (2 Pi Freq) = ~3.08 billion.
From that the specific accelerative Reaction force can be calculated as (2 Q Pwr Df) / c = (2 * 3,080,000,000 * 1,000 * 0.5 (conservative)) / c = 10,350N/kWrf or 1,056kg/kWrf or 1.056 mt/kWrf.
Not bad.
I'll have to ask that you forgive my incredulity in light of the current lack of public, high impulse test results. This sounds way, way too good to be true.
I agree.
The data is based on the calculated Q value from the patents thrust curves vs time.
As I showed, the thrust curves follow a standard 5x TC curve, which allows the 1 TC time to be determined and from that using standard resonant cavity Tc equation to determine what the effective cavity Q would need to be to have a thrust curve that lasts for 1 sec.
Yes the result seems very high but hey Roger has always stated 1ton/kW was his goal. Just maybe he has achieved it? -
#1570 by Rodal on 13 Oct, 2016 14:34
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I note five things in Shaywer's new patent that are blindingly obvious:
Thanks, but another obvious think to notice is that this is not a patent, but just an application at this point in time. Patent applications at this stage are non enforceable since a patent has not been granted at this stage.
....
18 months after the patent application was filed it is automatically published and is available for anyone to look at. This is not a granted patent and the application owner cannot sue anyone yet for using his/her invention.
Within 6 months of publication one must pay a further fee and request examination. During this detailed examination, the UK Intellectual Property Office may write giving reasons why the invention is not new, or is obvious, or challenge it with other arguments. When the UK Intellectual Property Office agrees that the invention is new and inventive, the patent may be granted. The process for obtaining a UK granted patent usually takes about 4 years from the date of the application. However, if one pays some fees early and replies promptly to letters from the UK Intellectual Property Office, it may be possible to reduce this time to as little as 18 months, if the invention is not a complex one. And if the patent is ever granted, it would become, at that point in time, just a UK patent, not a US or an international (PCT) patent. The applicant would have to file a patent application in other jurisdictions to seek protection in other jurisdictions.
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#1571 by andygood on 13 Oct, 2016 14:48
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The e-field strength for this new Bell geometry and circularizing antenna is off the charts. I'm now at 250,000 kV/m.
Does the large end absolutely have to be superconducting YBCO on a quartz substrate? Would it be worthwhile for a DIY-er to fabricate this geometry with a flat plate of copper and test?!
(I appreciate that the small end would be a bit of a challenge)
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#1572 by Gilbertdrive on 13 Oct, 2016 14:49
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That would be saying that the final kinetic energy must always be equal to the chemical energy content delivered by the fuel and no more. Is that your position?
Actually, less: the final kinetic energy of the rocket equals the chemical energy delivered by the fuel burnt less the total kinetic energy gained by the propellant, and less the energy lost as waste heat.
Totally exact.
Bob012345, you can try to give a counter example
It is well known that the upper stage of a rocket can gain more energy that the total energy content of its fuel.
That's because the fuel already has kinetic energy, often far greater than the chemical energy of the fuel. This is a known fact rocket engineers use.
Can you explain how any kinetic energy of the fuel changes how the fuel combusts?
As a rocket moves out of the gravity well of the earth, there is less gravitational resistance to acceleration. The location with respect to the earth's center of mass (gravity) is far more significant than any kinetic energy associated with acceleration.
When the fuel combusts, it goes in the opposite direction that the rocket, at high speed.
For example, if the upper stages goes initially at 3000m/s in the earth reference frame, and that the propellant is ejected to 3000m/s, it means that, after ejection, the propellant goes to 0m/s in earth reference frame. The propellant has given it's Kinetic Energy to the upper stage from the viewpoint of the earth reference frame. That means, that, at this point, the upper stage can really get more Kinetic energy in the earth reference frame that it has spent by it's own propellant chemical properties.
It does not need any gravity wall to be true.
It remains that the claim of Wicoe is still true.
Bob012345, I do not see how your statement would be a counter example.
As Wicoe has written :
the final kinetic energy of the rocket equals the chemical energy delivered by the fuel burnt less the total kinetic energy gained by the propellant, and less the energy lost as waste heat.
You just have to take into account that the kinetic energy gained by the propellant can be a negative number ! in my example it is a negative number.
The fact that an upper stage get more Kinetic Energy that it's own propellant chemical energy is perfectly conform to the statement of Wicoe.
The probem is that your formulation is ambigous.
That is the problem of the man with a wooden leg named smith (see Mary Poppins)
When you write It is well known that the upper stage of a rocket can gain more energy that the total energy content of its fuel.
Some can understand
1)It is well known that the upper stage of a rocket can gain more energy that the total energy content of the fuel included in the upper stage.
other can understand :
2)It is well known that the upper stage of a rocket can gain more energy that the total energy content of the fuel included in the entire Rocket at lift off.
The problem is that we have to guess what the "its" refers to. "The rocket" or "the upper stage"
That is why you get so different answers. Some will understand 2), and tell that it is false, and why. Others will understand 1), and tell that it is true, and why.
Also when you write "the total energy content of the fuel" it is not clear if the Kinetic Energy is already included, or if you means "the chemical energy content of the fuel"
You need to choose a better formulation.
For example : It is well known that the upper stage of a rocket can gain more energy that the total chemical energy content of the fuel included in the upper stage"
With this formulation, there is no more misunderstanding.
It seems that Wicoe formula was about the entire rocket, because he takes directly into account the final kinetic energy of the rocket, not an increase of Kinetic Energy. That supposes that the Kinetic Energy start to 0. It is not the case for the upper stage.
The formula would become in a upper stage only version
"the increase of Kinetic energy of the upper stage equals the chemical energy delivered by it's own fuel burnt less the total kinetic energy gained by the propellant, and less the energy lost as waste heat.
This increase of Kinetic energy of the upper stage can be superior to the chemical energy spent, if the propellant onboard the upper stage has lost more kinetic energy than the losses as waste heat."
I assume that I neglected the potential gravity energy between the upper stage, and it's own propellant
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#1573 by oyzw on 13 Oct, 2016 14:51
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Better image of minor end plate and circularization antenna.
Can you tell me S11？ -
#1574 by TheTraveller on 13 Oct, 2016 14:54
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I note five things in Shaywer's new patent that are blindingly obvious:
Thanks, but another obvious think to notice is that this is not a patent, but just an application at this point in time. Patent applications at this stage are non enforceable since a patent has not been granted at this stage.
....
18 months after the patent application was filed it is automatically published and is available for anyone to look at. This is not a granted patent and the application owner cannot sue anyone yet for using his/her invention.
Within 6 months of publication one must pay a further fee and request examination. During this detailed examination, the UK Intellectual Property Office may write giving reasons why the invention is not new or is obvious. When the UK Intellectual Property Office agrees that the invention is new and inventive, the patent may be granted. The process for obtaining a UK granted patent usually takes about 4 years from the date of the application. However, if one pays some fees early and replies promptly to letters from the UK Intellectual Property Office, it may be possible to reduce this time to as little as 18 months, if the invention is not a complex one. And if the patent is ever granted, it would become, at that point in time, just a UK patent, not a US or an international (PCT) patent.
Roger would have a priority filing date over most of the world that would allow other patents to be applied for and enforced in other countries.
Would suspect by publication date, Roger is well advanced on further developments and bringing this to market with the the new JV partner Giles, aka Gilo Cardozo of the Gilo Industries Group who are a UK aerospace firm.
Please remember Roger did say he was developing a drone with a Uk aerospace company? Well guess what Gilo group supplies drone motors. Easy to use a 1t/kWrf EmDrive to replace a fossil fuel drone motor or even the backpack motor of a ParaJet para glider or to build Gilo's desired Sky Car.
Interesting future ahead. -
#1575 by Rodal on 13 Oct, 2016 14:57
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...Roger would have a priority filing date over most of the world that would allow other patents to be applied for and enforced in other countries.
With the publication of this application the clock has started ticking for the applicant to file such applications in other jurisdictions if the applicant wishes to do so, since there is a well-specified small amount of time for the applicant to seek such protection in other jurisdictions if the applicant wishes to do so, after the application has been made public and part of the open literature. After that time runs out, it means that nobody will be able to seek identical claim protection as published, since now the information on those claims has become public information. -
#1576 by TheTraveller on 13 Oct, 2016 15:00
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...Roger would have a priority filing date over most of the world that would allow other patents to be applied for and enforced in other countries.
With the publication of this application the clock has started ticking for the applicant to file such applications in other jurisdictions if he wishes to do so, since there is a well-specified small amount of time for the applicant to seek such protection in other jurisdictions if he/her wishes to do so, after the application has been made public and part of the open literature.
Point was the patent application filing date has locked down a worldwide priority date. -
#1577 by Gilbertdrive on 13 Oct, 2016 15:13
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Dear Traveller
You claimed several times that the experiments were in favour of the constant thrust for constant input power.
I personally claim that there is no experimental evidence for thrust bigger than Pi/V where Pi is the input Power, and V the speed of the frustum in the laboratory reference frame. I do not claim that it is a true limit, I just claim that there is still no experiment that shows more thrust.
For example, for a speed V=3km/s the maximum thrust predicted by my formula is 333mN/Kw (this one is also indicated by Shawyer)
For a speed of V=300m/s the maximum thrust is 3,333N/Kw
For a speed of V=3m/s the maximum thrust is 333,33 N/Kw
For a speed of 1 m/s the maximum thrust is 1000N/Kw (much, is not it ?)
For a speed of 1cm/s the maximum thrust is 100 000N/Kw...
For a speed of 0 m/s there is no limit to the maximum thrust.
Does any experiment show a stronger force that the maximum predicted by my formula ?
What was the higher force measured by a rotary rig ? at what speed ?
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#1578 by TheTraveller on 13 Oct, 2016 15:24
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Dear Traveller
You claimed several times that the experiments were in favour of the constant thrust for constant input power.
I personally claim that there is no experimental evidence for thrust bigger than Pi/V where Pi is the input Power, and V the speed of the frustum in the laboratory reference frame. I do not claim that it is a true limit, I just claim that there is still no experiment that shows more thrust.
For example, for a speed V=3km/s the maximum thrust predicted by my formula is 333mN/Kw (this one is also indicated by Shawyer)
For a speed of V=300m/s the maximum thrust is 3,333N/Kw
For a speed of V=3m/s the maximum thrust is 333,33 N/Kw
For a speed of 1 m/s the maximum thrust is 1000N/Kw (much, is not it ?)
For a speed of 1cm/s the maximum thrust is 100 000N/Kw...
For a speed of 0 m/s there is no limit to the maximum thrust.
Does any experiment show a stronger force that the maximum predicted by my formula ?
What was the higher force measured by a rotary rig ? at what speed ?
In his later papers, Roger stated his earlier statements on velocity reducing thrust were incorrect and he withdrew them.
As far as I know, no rotary test rig has even started to get to those values. Long way away. -
#1579 by Rodal on 13 Oct, 2016 15:28
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...Roger would have a priority filing date over most of the world that would allow other patents to be applied for and enforced in other countries.
With the publication of this application the clock has started ticking for the applicant to file such applications in other jurisdictions if he wishes to do so, since there is a well-specified small amount of time for the applicant to seek such protection in other jurisdictions if he/her wishes to do so, after the application has been made public and part of the open literature.
Point was the patent application filing date has locked down a worldwide priority date.
Date of filing patent application April 7, 2015 (07/04/15) (*)
https://forum.nasaspaceflight.com/index.php?action=dlattach;topic=40959.0;attach=1379060;sess=45576
Example: if filing international (PCT) application has to be done within one year of filing the original national application, the deadline for that PCT patent application is long past: April 7, 2016 (**)Quotefor filing internationally is to file a patent application in accordance with the Paris Convention for the Protection of Industrial Property. Taking this route, the applicant files a patent application in a single Paris Convention member country (usually required to be the country of residence of at least one of the inventors), which establishes a first or priority filing date for the application. The applicant can then delay filing in other Paris Convention countries for up to 12 months after the priority filing date. Member countries of the Paris Convention agree to recognize the priority date of a patent application filed in one member country and to give the benefit of that priority date to corresponding applications in all member countries. This approach delays the costs associated with international patent procurement for one year. Procurement costs initially accrue in the country of first filing, and then, up to one year later, the costs associated with filing applications in the other Paris Convention countries begin to accrue
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(*) Date format in the US is mm/dd/yyyy , whereas in Europe (including the UK) the format is dd/mm/yyyy
(**) The priority date may be earlier than the actual filing date of the application. If a patent application is an original, non-provisional patent application, not a continuation application, and not previously filed in another country, its filing date is usually the same as its priority date.
(I appreciate that the small end would be a bit of a challenge)
