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#2380 by JohnFornaro on 21 Oct, 2014 01:56
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Here's the "small" shawyer device. Correct my titles.
.... Other than that, I really like the way your software presents results.
That software don't do poop. Fornaro Inside!
Every last dimension is eyeballed. I could go to great lengths to geometrically correct for parallax, but without a multiple equational frobnicatory analysis of possible frequencies and resonance, 'twould be a major wast of my time.
There are commonalities in the proportions. I point out that there is no 45 degree cone. -
#2381 by JohnFornaro on 21 Oct, 2014 02:06
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#2382 by ThinkerX on 21 Oct, 2014 02:08
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Found this. Hope the dratted link works:
http://arc.aiaa.org/doi/abs/10.2514/1.32284
Additional 'Photonic Laser Propulsion Paper'
Dunno...is this one relevant?
http://adsabs.harvard.edu/abs/2011SPIE.8164E..06I
http://adsabs.harvard.edu/abs/2008AIPC..997..561H
'Nanonewton thrust measurement of photon pressure propulsion using semiconductor laser'
Most of it seems to be beyond my feeble reach. The parts I can understand and access appear similar to what we are discussing...sort of...maybe...eep...
http://adsabs.harvard.edu/cgi-bin/nph-basic_connect
Other stuff at ADS (Astrophysical Data System) that might be relevant here. Search term I used was 'Photonic Laser Propulsion.'
::Tiptoes quietly away::
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#2383 by RotoSequence on 21 Oct, 2014 02:11
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Not like I can't draw or design or anything.
Did you assume, note from citation, or measure the extruded T-slot aluminum to be standard 1.5" (3.81cm) bars? -
#2384 by JohnFornaro on 21 Oct, 2014 02:18
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Here's the "small" shawyer device. Correct my titles.
But we know from the documentation that the "small" Shawyer device big end was 16 cm diameter.
Like I said, Fornaro Inside....
Dimensions, shmimensions.
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#2385 by JohnFornaro on 21 Oct, 2014 02:21
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Not like I can't draw or design or anything.
Did you assume, note from citation, or measure the extruded T-slot aluminum to be standard 1.5" (3.81cm) bars?We know that the (Faztek) beam (at the bottom of the picture, shown in cross-section) has a square cross section of 1.5 inch by 1.5 inch...
Check out the 3.81cm dimension. If dat ain't right tell me what is.
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#2386 by JohnFornaro on 21 Oct, 2014 02:22
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::Tiptoes quietly away::
I don't have time for that. I'm too busy making incorrectly scaled drawings.
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#2387 by Rodal on 21 Oct, 2014 02:29
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Here's the "small" shawyer device. Correct my titles.
.... Other than that, I really like the way your software presents results.
That software don't do poop. Fornaro Inside!
Every last dimension is eyeballed. I could go to great lengths to geometrically correct for parallax, but without a multiple equational frobnicatory analysis of possible frequencies and resonance, 'twould be a major wast of my time.
There are commonalities in the proportions. I point out that there is no 45 degree cone.
It is a really outstanding job.
Based on the last drawing, that has the dimensions for the Big Diameter based on the 1.5"x1.5" cross section, given to us by Paul March,
I compute the following based on scaling of Fornaro's drawing:
Brady et.al.
Length = 0.33245 m
BigDiameter = 0.39697 m
SmallDiameter = 0.24393 m
The ratios (Length to Big Diameter, and Length to Small Diameter) are pretty close to aero:
L/BigDiameter L/SmallDiameter ( L/SmallDiameter - L/BigDiameter)
Fornaro 0.8375 1.3629 0.525422
aero 0.92 1.3939 0.473939
Mulletrn 0.90538 1.7408 0.835404
Since John used a superior method, using the known dimensions of the cross section of the beam, and it is the median measurement (using the above-mentioned ratios),
Let's take John's measurements for Brady et.al.
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#2388 by Rodal on 21 Oct, 2014 02:31
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Not like I can't draw or design or anything.
Did you assume, note from citation, or measure the extruded T-slot aluminum to be standard 1.5" (3.81cm) bars?
This dimension was given to us by Paul March (a member of Dr. White's team) more than a 100 pages along (time flies) -
#2389 by Rodal on 21 Oct, 2014 02:40
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Found this. Hope the dratted link works:
....
Other stuff at ADS (Astrophysical Data System) that might be relevant here. Search term I used was 'Photonic Laser Propulsion.'
Thanks -
#2390 by JohnFornaro on 21 Oct, 2014 02:44
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(time flies)
Yea. Time flies like the wind, and fruit flies like bananas.
Again. I eyeballed all those dimensions, kinda sorta correcting for parallax. You zoom into the intersecting points of the lines with the JPEG image, and you can see what I mean. We need to add to the Cat's work in getting him to explore the ramifications of the various dimensions and RF frequencies. 'Course, he's French. Prolly atta latteria or whatever they call 'em over there. You know how it is: "Latte" is just the French term for "You paid too much for that coffee". I need to check "News in Slow French", then off to bed. -
#2391 by JohnFornaro on 21 Oct, 2014 02:56
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This dimension was given to us by Paul March (a member of Dr. White's team) more than a 100 pages along (time flies)
Dang it! Neglected to drag one of the dimensions. Rushing to get to bed.
Try this:
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#2392 by Mulletron on 21 Oct, 2014 08:25
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http://www.grainger.com/product/FAZTEK-Framing-Extrusion-5JA96?s_pp=false&picUrl=//static.grainger.com/rp/s/is/image/Grainger/5JA94_AS01?Here's the "small" shawyer device. Correct my titles.
.... Other than that, I really like the way your software presents results.
That software don't do poop. Fornaro Inside!
Every last dimension is eyeballed. I could go to great lengths to geometrically correct for parallax, but without a multiple equational frobnicatory analysis of possible frequencies and resonance, 'twould be a major wast of my time.
There are commonalities in the proportions. I point out that there is no 45 degree cone.
It is a really outstanding job.
Based on the last drawing, that has the dimensions for the Big Diameter based on the 1.5"x1.5" cross section, given to us by Paul March,
I compute the following based on scaling of Fornaro's drawing:
Brady et.al.
Length = 0.33245 m
BigDiameter = 0.39697 m
SmallDiameter = 0.24393 m
The ratios (Length to Big Diameter, and Length to Small Diameter) are pretty close to aero:
L/BigDiameter L/SmallDiameter ( L/SmallDiameter - L/BigDiameter)
Fornaro 0.8375 1.3629 0.525422
aero 0.92 1.3939 0.473939
Mulletrn 0.90538 1.7408 0.835404
Since John used a superior method, using the known dimensions of the cross section of the beam, and it is the median measurement (using the above-mentioned ratios),
Let's take John's measurements for Brady et.al.
Sorry but this solution doesn't work. See the screenshot and link to drawing. The 1.5inch reference is at an arbitrary depth so you can't use it. The same reason I couldn't use the width of the chamber. Also, we want inside dimensions, not outside.
https://drive.google.com/folderview?id=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM&usp=sharing
Stanley Cupric1.dwg
The pic below is the result of the wrong solution.
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#2393 by Mulletron on 21 Oct, 2014 11:14
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I'm chasing a hot lead, which if it holds, may allow emdrive to provide thrust from a sealed cavity without breaking any physical laws, teaching photons new tricks, adjusting C, etc... (dielectric present or not, air is dielectric too anyway)........evanescent wave coupling, quantum tunneling is the QM analogue. Part of my research into near field effects. Anybody want to take a stab at it? I'm at the very early stages of it and I'm desperately trying to prove the idea wrong.
I've had some exposure to this kind of concept in my own nerdery over the years, and I may have actually encountered it at work by accident when I noticed I could pick up the energy inside our waveguides very faintly with a spectrum analyzer and a crude probe I made, even though the waveguides were sealed, no flanges anywhere nearby, and there were no leaks detected by the rf sniffer. I just had to get close to the waveguide with the probe and I could see our transmit carriers.
I have got to find out for certain the wall thickness of the Nasa test article tapered frustum.
Neat simulation tool plugin for MATLAB:
http://www.problemsinelectrodynamics.com/tools/interactive-fdtd-toolbox -
#2394 by JohnFornaro on 21 Oct, 2014 12:59
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Sorry but this solution doesn't work. See the screenshot and link to drawing. The 1.5inch reference is at an arbitrary depth so you can't use it. The same reason I couldn't use the width of the chamber. Also, we want inside dimensions, not outside.
Not that I'm right or anything, but I would observe that the visual vertical orientation of the device suggests that the line of the cone's vertical diameter passes thru the center line line of the Faztek thingy, virtually in the same plane perpendicular to the camera. The parallax is to be ignored in my analysis, and the distance of the cone and thingy from the camera is immaterial.
There only appears to be two sizes for the Fazteck thingy: 1.5" and 3": http://www.faztek.net/ I'm guessing this isn't the 3" product.
Ipso fatso: Scaling the foto on that piece of evidence is appropriate.
I'm including the CAD file for reference, with a note of appreciation to the foax at AutoCAD, who change their drawing format hourly so as to protect what they think is their intellectual property, while ignoring that were they to have an accessible database, and were to improve the functioning of their CAD products instead of changing the drawing format, they would probably have more sales and greater market penetration, but I digress just a mite.
ACAD R14 is a 1997 product, and they refuse to improve that product. If the whiz bang kids at Mathematica and Comsol cain't read legacy .DWG's, that's their problem.
Point being: Twerk the CAD file. Also, in my eye, color is a distraction, and your drawing is hard on old eyes. That's why I converted the JPEG to a .
Anyhow.....
As to the idea of measuring the inside of the cavity, thou art spot on.
It has been suggested that the inside must be properly mirrored for the desired M/W resonance. The inside of that cone has to be conically flat. Dollars to donuts sez it ain't. Not if it is made of 1/8" thick copper. The inside may have been polished with Turtle Wax, but it has not been polished nor flattened to Hubble specs. The substrate is too thin. The exterior has a mill finish only. Was it formed over a wood mandrel? Who knows?
I employ the doctrine of the fair witness. Also known as calling 'em like I see 'em.
Since i can only follow the addition, subtraction, multiplication and some division of the math, I have trouble following the intricacies of the arguments being presented. I do follow the general gist however.
And. I do know that there's an awful lotta predictions being made without crucial information. Information which is not being shared for obvious reasons.
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#2395 by Mulletron on 21 Oct, 2014 13:18
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Sorry but this solution doesn't work. See the screenshot and link to drawing. The 1.5inch reference is at an arbitrary depth so you can't use it. The same reason I couldn't use the width of the chamber. Also, we want inside dimensions, not outside.
Not that I'm right or anything, but I would observe that the visual vertical orientation of the device suggests that the line of the cone's vertical diameter passes thru the center line line of the Faztek thingy, virtually in the same plane perpendicular to the camera. The parallax is to be ignored in my analysis, and the distance of the cone and thingy from the camera is immaterial.
Notice the 1.5inch dims in yellow in the foreground and background. Because of depth perception, one is behind the test article and the other is in front. That creates a situation where you can't rely on the 1.5inch reference without knowing the viewing angle of the camera and at least one length in the Z axis of the chamber (the direction you're looking). I think I got it pretty close using the 6.25 inch internal PE disk reference from the paper. That allowed me to do 3 transformations in x, y, z to get pretty close but not exact. The z axis of the cavity was a result of x, y transformations, and still ended up being an eyeball job after proper scaling. My CAD drawing is linked to above for peer review. I hear you on Autodesk. I'm using Acad 2000. It is good enough. I'm not buying a new one.
Edit:
I just realized I (or anyone else) can just use both foreground and background 1.5inch references at once and some simple math to nail the drawing. -
#2396 by Notsosureofit on 21 Oct, 2014 13:33
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#2397 by Rodal on 21 Oct, 2014 14:16
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Thank you for being kind to my eyes, and noticing this John. Much appreciated.
Point being: Twerk the CAD file. Also, in my eye, color is a distraction, and your drawing is hard on old eyes. That's why I converted the JPEG to a .
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#2398 by JohnFornaro on 21 Oct, 2014 14:33
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Notice the 1.5inch dims in yellow in the foreground and background. Because of depth perception, one is behind the test article and the other is in front. That creates a situation where you can't rely on the 1.5inch reference without knowing the viewing angle of the camera and at least one length in the Z axis of the chamber...
Well, I took anudder look at your sketch, and still stick with my interpretation. The camera appears to be damn near perpendicular to the centerline of the cone, and perpendicular to the z-axis of the large end diameter which is nearly centered on the lens of the camera. If anything, the rear of the cone rim is slightly to the left of center, but I ignored that.
I note that the support is not at right angles to the plane of the large diameter of the cone, and not at all at right angles to its own vertical support Faztek, but I don't think that matters.
What matters to the camera is that the Faztek horizontal support is centered on the cone's axis. If that is the case, the dimension that I show should be spot on within a plus or minus. The support is clearly not vertically off center, but it may be nearer or further from the camera than the centerline of the cone's axis. I can't tell, and I can't help that.
The other support that you dimension appears to be a 3" chunk of Faztek, at least to this pair of retinas.
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#2399 by Rodal on 21 Oct, 2014 14:33
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No, it was not a suggestion. It is not based on words, or intuition. It follows from Maxwell's equations.
It has been suggested that the inside must be properly mirrored for the desired M/W resonance. The inside of that cone has to be conically flat. Dollars to donuts sez it ain't. Not if it is made of 1/8" thick copper. The inside may have been polished with Turtle Wax, but it has not been polished nor flattened to Hubble specs. The substrate is too thin. The exterior has a mill finish only. Was it formed over a wood mandrel? Who knows?
...
It is not based on looking at Wikipedia articles. It is not based on scouring the Internet looking at blogs.
It is based on looking at MIT Electrical Enginering course work material.
What was stated in http://forum.nasaspaceflight.com/index.php?topic=29276.msg1273676#msg1273676
It is based on this:
Q =( Length / skinDepthForResonantFrequency) / (1 + ( 2*Length / InnerDiameter ))
skinDepthForResonantFrequency = 1/Sqrt[Pi*ResonantFrequency*μ*σ]
The statement was:
1) Both inner surface ends of the truncated cone must have been made out of copper. Otherwise the cavity would not have the correct boundary conditions to be a resonant cavity: it would be a waveguide. One wouldn't be able to have high Q and resonance if the inner surface of the ends wouldn't be copper.
2) The inner surface of the copper must have been pretty well polished, in order to get Q~50 000, given the calculated skin depths
3) There cannot be any significant irregularities on the inside surface of the truncated cone.
This follows from Maxwell's equations. If Shawyer reports for his demo Q=45000, then it must follow that either:
A) the inner surface of Shawyer's demonstration EM Drive must have been pretty well polished in order to obtain Q=45000, without significant irregularities on its inside surface
or
B) the reported Q=45000 is an incorrect reported value
Also observe that NASA Eagleworks did not report as high Q's as the Q=45000 reported by Shawyer for the demonstrator EM Drive. NASA Eagleworks reported Q's ranging from 7000 to 22000, which is significantly less (Q=7000 is 7 times less). Therefore, it must follow (from the solution to Maxwell's equations) that:
A) NASA Eagleworks truncated cone's inner surface was not as well polished and free of irregularities as Shawyer's demonstration EM Drive
or
B) the reported Q's are incorrect
John, you just cannot have a high Q with a poor, irregular surface on the inside, when the skin depth is a couple of micrometers or less. You don't need to exaggerate about Hubble specs or similar exaggeration not present in the statement that you are objecting to.
John, if there is no inner copper on the ends of the cavity (showing PCB board on the outside) then you cannot have a resonant cavity, so the inner surface ends must have been made out of highly-conductive material like copper.
There is no way out of this: reported Q's are giving us information about the inner surface and material of the cavities. If you claim that the inner surface of the ends was not made out of a highly conductive material like copper, then you are stating that the reported Q's are completely incorrect, if you state that the inner surface of the cavities was significantly irregular, then Q cannot reach 45000, and effectively you are stating that you think that the reported Q is incorrect.
Similarly regarding Shawyer's statement about Q's several orders of magnitude higher: the only way to obtain such Q's is to make the inner surface of the cavity out of superconducting material and have the cavity at temperatures where superconductivity can take place.