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#1440 by rq3 on 08 Jan, 2016 00:05
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Acceleration does give a calculable frequency shift (Need to calculate it...)
Note: Then you have to compensate for velocity relative to your reference frame.
Fortunately, nature sometimes does it for you at non-relativistic velocities. That's why airborne phased array RADAR don't (generally) bother (the reference and return signal can be compensated at base-band), but GPS satellites always bother (they move fast enough, and are in high enough orbits to introduce timing error when looking for sub-meter resolution. The error is both velocity and gravimetric). -
#1441 by TheTraveller on 08 Jan, 2016 00:05
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I wondered about this cutoff point for some time and aero and Dr. Rodal susgested months ago we run a frustum past a cutoff point....
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
I stopped the video when the wavefront reached the part of the frustum that was its cutoff.
After the wave propagates past the cuttoff and hits the small end it bounces back hitting the cutoff area again and the frustum begins to make the cavity resonate in two directions.
For your viewing pleasure and some insight.
Shell
BRB with a gif that runs.
Shell,
This is a rule of thumb. A 1st cut value when designing a frustum. I don't remember Roger ever saying it would not work, but like the use of a dielectric, something to be avoided to get good thrust generation.
Phil -
#1442 by rq3 on 08 Jan, 2016 00:12
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The air bearing this is mounted on is virtually frictionless but he does have fans and pumps running plus the thrumming of the air compressor nearby.
Take a look of Shawyer's original turning table experiment on YouTube. Surely it needs vibration (or tapping) to overcome the initial static friction. But such a vibration is not needed for a torsion balance or a boat on water experiment to overcome static friction that is zero.
Think of an imagined perfect air cushion hovercraft on the land. The land is not perfect, with tiny highs and lows that is not detectable by bear eyes. When stay idle, the perfect hovercraft will find itself a local low position and stay there. now suppose we apply a small force on the hovercraft (say, very light wind). If the force is not enough to overcome the initial climbing out of the local low position, the hovercraft will not move away. Only when a small earthquake shakes the land so the hovercraft overcomes the initial obstacle, will it start to move; and with built-up momentum it will overcome any new small low lands.
I have no experience with air bearing; but I have spent lots of time thinking of magnetic bearing that will suffer from similar problems.
From someone who HAS built air bearings, this is a beautiful analogy! Thank you! -
#1443 by rq3 on 08 Jan, 2016 00:18
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...
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
Inside the frustum, freq is the same as outside. Need to work in wavelengths, Lambda.
If the quide wavelength "turns imaginary" then there is a limitation of some kind in your model. The wavelength cannot "turn imaginary", but the wavequide impedance can go asymptotically large. Perhaps the model cannot deal with that? -
#1444 by Rodal on 08 Jan, 2016 00:24
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Excellent
I wondered about this cutoff point for some time and aero and Dr. Rodal susgested months ago we run a frustum past a cutoff point....
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
I stopped the video when the wavefront reached the part of the frustum that was its cutoff.
After the wave propagates past the cuttoff and hits the small end it bounces back hitting the cutoff area again and the frustum begins to make the cavity resonate in two directions.
For your viewing pleasure and some insight.
Shell
BRB with a gif that runs.
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#1445 by RFPlumber on 08 Jan, 2016 00:47
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I wondered about this cutoff point for some time and aero and Dr. Rodal susgested months ago we run a frustum past a cutoff point....
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
I stopped the video when the wavefront reached the part of the frustum that was its cutoff.
After the wave propagates past the cuttoff and hits the small end it bounces back hitting the cutoff area again and the frustum begins to make the cavity resonate in two directions.
For your viewing pleasure and some insight.
Shell
BRB with a gif that runs.
Ok, awesome. Now what are you all talking about because I do not understand... Oh, and I cannot see anything on that attached gif, just a black rectangle and a grey dot...
So what "cut-off" does this refer to? If this is not the mode cut-off freq for the waveguide when what is it?
It would have been real nice if someone has mentioned this "cut-off / no thrust" thing a few weeks ago when I was starting to build the frustum per dimensions... Now it is kind of late.
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#1446 by TheTraveller on 08 Jan, 2016 00:47
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...
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
Inside the frustum, freq is the same as outside. Need to work in wavelengths, Lambda.
If the quide wavelength "turns imaginary" then there is a limitation of some kind in your model. The wavelength cannot "turn imaginary", but the wavequide impedance can go asymptotically large. Perhaps the model cannot deal with that?
Not my equations.
Standard microwave equations for determining guide wavelength in a cylindrical waveguide. Approaching cutoff, as the diameter drops, the guide wavelength becomes longer and longer, until it can't get any longer and the wave can't propogate. -
#1447 by TheTraveller on 08 Jan, 2016 00:52
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It would have been real nice if someone has mentioned this "cut-off / no thrust" thing a few weeks ago when I was starting to build the frustum per dimensions... Now it is kind of late.
It is not a secret. I have stated it many times. -
#1448 by Rodal on 08 Jan, 2016 00:54
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I wondered about this cutoff point for some time and aero and Dr. Rodal susgested months ago we run a frustum past a cutoff point....
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
I stopped the video when the wavefront reached the part of the frustum that was its cutoff.
After the wave propagates past the cuttoff and hits the small end it bounces back hitting the cutoff area again and the frustum begins to make the cavity resonate in two directions.
For your viewing pleasure and some insight.
Shell
BRB with a gif that runs.
Ok, awesome. Now what are you all talking about because I do not understand... Oh, and I cannot see anything on that attached gif, just a black rectangle and a grey dot...
So what "cut-off" does this refer to? If this is not the mode cut-off freq for the waveguide when what is it?
It would have been real nice if someone has mentioned this "cut-off / no thrust" thing a few weeks ago when I was starting to build the frustum per dimensions... Now it is kind of late.
First of all, this cut-off rule does not prevent resonance of a truncated cone (facts are facts).
As to preventing "thrust", then ask those who impose this rule, how does the rule accommodate for the use of dielectrics at the small end?
Because the fact is that NASA has been claiming thrust with the EM Drive, both in air and in vacuum (Shawyer and Yang have never reported a single test in vacuum). All of NASA's tests claiming thrust have dielectric inserts at the small end that lower the natural frequencies of the resonating cavity below the natural frequency of the cavity without a dielectric.
So, how does the "cut-off" rule apply to NASA's claimed thrust according to those imposing this rule ? Do they have a modified cut-off rule accommodating for "thrust" for dielectric inserts? Where is that cut-off rule for dielectric inserts written ? In what report?
NASA truncated cone dimensions (final result in meters, intermediate in inches (") )
bigDiameter = (11.01")*2.54 cm*(1"/(100 cm));
smallDiameter = (6.25")*2.54 cm*(1"/(100 cm));
axialLength = (9")*2.54 cm*(1"/(100 cm));
According to TT the small diameter of NASA's EM Drive (6.25 inches) is smaller than allowed for the natural frequency for TE012 without a dielectric measured at 2.168 GHz by NASA, and that's why NASA registered no thrust force without a dielectric insert.
With the dielectric insert, NASA reported its highest force/PowerInput, at a frequency of 1.880 GHz with the same mode shape TE012
So, what happened to the cut-off rule, when the dielectric insert is used? the frequency was even lower (1.880 GHz instead of 2.168 GHz) yet, NASA not only reported thrust at the lower frequency, but the highest thrust per input power in the NASA report -
#1449 by VAXHeadroom on 08 Jan, 2016 01:00
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Uhhhh...I mean an in-house frustum/cavity tool...LOL! Ya got me there, Glenn. Well done...cavity spreadsheet modeling tool in my brain, emdrive on my keyboard...forgive me.
Seems we are moving towards an "in-house" emdrive modeling tool.
uh oh
Note to self - try to minimize posting when actually doing work elsewhere
You didn't quite read my uh oh thought. I was trying to calculate how many "free" hours I have in 2016 in my copious free time to devote to writing some simulation software given some empirical data and builder requirements that forces it to be written.
There are a few other coders in this forum... an open source development project was where my brain was at, and how much pain it is to make that work. I wasn't correcting your word usage.
I was calculating my personal pain threshold at doing lots of billable work for free. Have to do some market research to see if I could license it to someone in the RF industry without violating an existing NDA....
I'm a DAMN good coder and can write code for any equations someone can give me. Here's a link to my C++ CSV to POV-Ray converter (and the POV-Ray source for an animation from mid-November): http://www.untiedmusic.com/emory/emdrive/CE3-copper-64cycles-2015-11-13.zip
Let's see what we can work up!
(edit: should really have said "algorithm" instead of "equations" above
)
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#1450 by rq3 on 08 Jan, 2016 01:07
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Shell,
This is a rule of thumb. A 1st cut value when designing a frustum. I don't remember Roger ever saying it would not work, but like the use of a dielectric, something to be avoided to get good thrust generation.
Phil
[/quote]
A rule of thumb may be a great way to lose a digit. An example. On one of the ***classified*** RADAR sytems I was privileged to work on, one of the requirements was that the 9 Megawatt transmitter be able to withstand various Voltage Standing Wave Ratios (VSWR) into it's launch wavequide. Many meetings were held, great ponderance was provided, and great barrels of water and dummy loads were rolled about for many weeks. Upon the intial test, the detector circuit declared foul, rotated the 20 foot long antenna to the "park" position, and threw the Techtronics oscilloscope resting upon it through the lab wall. One tech lost a finger.
I asked if the waveguide could be pressurized to 60 PSI with sulfur hexafluoride. Yes. It could. Can I place a rotary waveguide switch here? Yes. You can. Well, then, folks, given THIS position of the switch, you get VSWR "A". Given THIS position of the switch, you get VSWR "B", etc.
Much blustering. Much finger pointing. I volunteered to rotate the switch by hand. What had they to lose? $3M had already been spent on this silly test. I had faith in my calculations, and all they had to lose was a couple of very expensive klystrons and a snotty EE.
I'm still here, the wavequide switch is now stepper motor driven under feedback from a vector network analyzer, and we never lost an EE or a transmitter. Even the switches didn't care.
A rule of thumb can be a time-saver. Or it can be a life-saver. Or it can be a waste of time. In this case, it was a waste of time. No-one ever questioned whether you can choke a high powered wavequide without fatal consequences. In all the history of RADAR engineering. Ever.
You all be careful out there!
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#1451 by SeeShells on 08 Jan, 2016 01:07
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I find it interesting that meep which employs electromagnetic simulations by the FDTD method, can calculate resonate modes and the travel of those modes within this cavity. It shows that there does appear to be a impedance "wall" but it also shows it somehow tunnels past it to generate another set of modes in the small end and as the energy of the small end increases it bounces of the cutoff point to reform another traveling mode. This may be a very small amount of energy that is showing but it's there....
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
The guide wavelength turns imaginary when the small end is <= cutoff.
Need to do the cutoff calc 1st and then calc the guide wavelength.
Inside the frustum, freq is the same as outside. Need to work in wavelengths, Lambda.
If the quide wavelength "turns imaginary" then there is a limitation of some kind in your model. The wavelength cannot "turn imaginary", but the wavequide impedance can go asymptotically large. Perhaps the model cannot deal with that?
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#1452 by SeeShells on 08 Jan, 2016 01:26
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One more time.
Ok. You can see the cutoff about 2/3rds down from the large end.
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#1453 by Rodal on 08 Jan, 2016 01:47
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One more time.
Notice
Ok. You can see the cutoff about 2/3rds down from the large end.
1) that it does not prevent resonance.
2) the mode shape distribution looks similar to what they look in my "no cut-off" report. What happens at the small end is similar to what I showed in my report, based on the exact solution for the problem.
By the way, in my report I list a number or peer-reviewed references where this fact is discussed, and known in the literature, as electromagnetic resonance of a frustum of a cone, as well as resonance of cones was well investigated both analytically, numerically and experimentally much prior to R. Shawyer's reports
What's "anomalous" about Shawyer's EM Drive (and Yang's and NASA's) claim is the "anomalous force" being claimed.
Electromagnetic resonance of frustums of a cone and complete cones was well understood much prior to R. Shawyer. -
#1454 by SeeShells on 08 Jan, 2016 02:09
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Excellent.One more time.
Notice
Ok. You can see the cutoff about 2/3rds down from the large end.
1) that it does not prevent resonance.
2) the mode shape distribution looks similar to what they look in my "no cut-off" report. What happens at the small end is similar to what I showed in my report, based on the exact solution for the problem.
By the way, in my report I list a number or peer-reviewed references where this fact is discussed, and known in the literature, as electromagnetic resonance of a frustum of a cone, as well as resonance of cones was well investigated both analytically, numerically and experimentally much prior to R. Shawyer's reports
What's "anomalous" about Shawyer's EM Drive (and Yang's and NASA's) claim is the "anomalous force" being claimed.
Electromagnetic resonance of frustums of a cone and complete cones was well understood much prior to R. Shawyer. -
#1455 by glennfish on 08 Jan, 2016 02:09
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Let's see what we can work up!
(edit: should really have said "algorithm" instead of "equations" above )
I've always wondered about the VAX in your moniker. Dare I ask if you know RT-8, RT-11, TSX, RSX-11, RSTS? I suspect VMS is on the list too? -
#1456 by TheTraveller on 08 Jan, 2016 02:59
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Roger shared some advise with us about real world effects of trying to get good thrust.
1) don't use a dielectric in the frustum.
2) avoid making the small end too narrow such that a constant diameter waveguide of that diameter, excited in the mode of choice, at the frequency of choice, would be at or below cutoff using standard microwave industry equations. His advise is that ideally the small end diameter should be just a bit bigger diameter that cutoff diameter.
You are here building EmDrives because of his work. Your choice to take his advise notice or not.Quote from: Roger Shawyer emailHi Phil
Your proposals sound fine to me.
Note that the Q you achieve will also be dependent on how well you tune and match the impedance of the input antenna. We have used probe, loop and waveguide iris plates as input circuits. All have their own problems, but you should first calculate the wave impedance of the cavity at the input position. Standard text book equations work, as they always do. You can then design your chosen input circuit to match the wave impedance at the cavity resonant frequency.
All successful EmDrive thrusters that I know of have incorporated a tuning element of some sort at the input.
Also no successful design used COMSOL without correction, as the software does not seem to cope with conditions close to cut-off, as NASA should have realised.
Best regards
Roger -
#1457 by Emmett Brown on 08 Jan, 2016 03:03
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Question for the COMSOL experts..
What would be the recommended way to model a magnetron antenna source in these two scenarios:
1) Direct insertion in the frustum
2) Magnetron in a rectangular waveguide attached to the frustum
(similar to a typical microwave oven, eg. a WR340 waveguide)
COMSOL's microwave-oven-heating-a-potato tutorial used a rectangular Port on the end of the waveguide (scenario 2), but is that an over-simplification given that we are interested in the best simulation of the EM modes?
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#1458 by RFPlumber on 08 Jan, 2016 03:27
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...
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
Ok, sorry, never mind. I take this back. The "a" in those formulas is actually a radius, not a diameter. So, 158mm is indeed a tad too small a diameter for a cylindrical waveguide to pass 2312 MHz as the cut off freq is ~2314 MHz.
Will need to cut ~4 mm alongside the circumference at the bigger end to move the mode frequency up ~20 MHz to formally comply with the requirement. This is a good catch.
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#1459 by TheTraveller on 08 Jan, 2016 03:34
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...
BTW at the indicated 2.312 GHz, the small end is below cutoff when calculated as a constant diameter circular waveguide. According to what Roger has taught me, there should be no thrust. This is not to say the small end is actually cutoff, just a rule of thumb situation to avoid.
...
Umm.. How do you calculate the cut-off freq? I am using attached, and so for 0.158 m smaller end one would get:
P01 = 3.832
Cut-off Freq = 3.832 * c / (6.28 * 0.158 * sqrt(1)) ~= 1,159 MHz.
The TE012 frustum mode at 2.312 GHz seems to be way above the cut-off for the smaller end... What am I missing?
Ok, sorry, never mind. I take this back. The "a" in those formulas is actually a radius, not a diameter. So, 158mm is indeed a tad too small a diameter for a cylindrical waveguide to pass 2312 MHz as the cut off freq is ~2314 MHz.
Will need to cut ~4 mm alongside the circumference at the bigger end to move the mode frequency up ~20 MHz to formally comply with the requirement. This is a good catch.
Please note Rogers advise about COMSOL.QuoteAlso no successful design used COMSOL without correction, as the software does not seem to cope with conditions close to cut-off, as NASA should have realised.
As I don't use it, I never asked him for more information on this.
The NASA reference is in relation to their use of too low a freq, voiding the small end cutoff rule, when they did the non dielectric test, which had no thrust.
I was calculating my personal pain threshold at doing lots of billable work for free. Have to do some market research to see if I could license it to someone in the RF industry without violating an existing NDA.... 
)