Quote from: SeeShells on 08/02/2015 02:57 PM...Shell1) Concerning estimation of Yang's dimensions:Everybody is welcome to come up with their own estimates for Yang's dimensions. So far, to my recollection, I'm the only one that estimated Yang's dimensions, with two different and clearly stated assumptions: A) assuming the cut-off condition for the small diameter and alternatively B) assuming the angle from the drawings to be approximately correct.To all readers disagreeing: what is your estimate for Yang's dimensions and what is it based on?Let's compare different estimates and the basis for calculating the different estimates._____2) Concerning experiments:The formulas (Shawyer, McCulloch, Notsosureofit), as well as the Meep results and my computations show that the results based on an approximately cylindrical frustum with 6 degree cone angle are going to be inferior to one with a larger cone angle.Concerning the experiments, I very much look forward to have experiments with a frustum having only 6 degree cone half-angle to compare with the experiments with frustum having higher cone angles.An experiment will really clear this. Just like Tajmar's experiment with a Q=50 clarified the relationship to Q.

...Shell

...I even printed off pictures after changing them to line drawings in software and took my handy dandy protractor to measure and I was all over the place in differences in the pictures and the drawings. The best I could come up with was was disagreeing angles. I did go the old fashioned route with pen and paper.Shell

Quote from: SeeShells on 08/02/2015 03:23 PM...I even printed off pictures after changing them to line drawings in software and took my handy dandy protractor to measure and I was all over the place in differences in the pictures and the drawings. The best I could come up with was was disagreeing angles. I did go the old fashioned route with pen and paper.ShellHow much "all over the place"? Did you get an angle of only 6 degrees from any of those pictures?Neither Flyby nor FluxCapacitor or me report an angle of 6 degrees from any of Yang's pictures. Blue is the cavity with 6 degrees: (it looks practically like a cylinder, compared to the other two):

Quote from: Rodal on 08/02/2015 01:02 PMQuote from: flux_capacitor on 08/02/2015 10:18 AMQuote from: TheTraveller on 08/02/2015 07:44 AMQuote from: flux_capacitor on 08/01/2015 03:13 PMYang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492In particular, the big end is smaller than the height, not larger.There is no resonance I can find for those dimensions at 2.45GHz. Can get TE012 resonance at 2.51GHzThey were surely wrong dimensions, based on the assumption Yang's drawings had the correct proportions. ..No, they were the dimensions that I had correctly based on the assumption that the small diameter for Yang was obtained based on the cut-off condition for the small end. And TheTraveller's calculation actually confirms it:<<There is no resonance I can find for those dimensions at 2.45GHz. Can get TE012 resonance at 2.51GHz>>The 2% difference between 2.45 GHz and 2.51 GHz is completely and utterly negligible compared to the other differences we are talking about: compared to the difference between 6 degrees and 15 degrees for the cone half-angle or compared to the uncertainties in having to eyeball the dimensionless ratio equations from Yang's chart.On top of that, the result TE012 resonance at 2.51GHz is obtained by TheTraveller from an Excel spreadsheet using very simplified ad-hoc approximate formulas that do not respect the boundary conditions of the problem while the solution for TE012 2.45GHz resonance uses Wolfram Mathematica to solve the exact solution to the problem using Legendre Associated Functions and Spherical Bessel functions.Taking into account the above considerations, the 2% different ad-hoc TE012 resonance at 2.51GHz means a complete agreement with the dimensions provided, because the 2% difference is completely within bounds.___________________________________________The issue at hand (determination of Yang's geometry) has to do with whether L=24cm provided by Yang in her paper is correct, and if so, whether the other (one and only one available to choose) parameter to consider should be the small diameter (based on cut-off), or the cone half-angle (determined from drawings), or some other parameter like D/L (determined from drawings), etc.One cannot obtain dimensions from Yang's schematic drawings, one can only obtain dimensionless ratios: angles and ratios, that's it.For L=24 cm and f=2.45 GHz, you cannot satisfy both that the diameter of the small base be above cut-off for an open waveguide AND satisfy the cone half-angle from Yang's drawings . They are in violent conflict with each other. Something got's to give. This has to be clearly understood: we have mathematical relations to satisfy here, they are mathematical constraints. You cannot satisfy all the dimensionless ratios from Yang's drawings and simultaneously satisfy the cut-off condition at the small end and simultaneously satisfy L=24 cm and simultaneously satisfy the D/L vs frequency relation.I'd like to insert one other thing here as well, having been in the middle of this as one of the builders. First I understand the why of your calculations and how you arrived at them. Numbers don't lie, but liars and the deceiving will number. That said I think we need to at least make sure that the cone angle for Yang's frustum is correct, better than just taking it off of Yang's drawings which have been proven to be not quite accurate. How do we know that the drawing wasn't modified to fit the page ie: shortening or lengthening the image changing the cone angle? How can we even know who drew this and in what program? PCPaint? The answer is, we truly cant. That throw the angles and ratios under question as well.Unless you are aware of a designed numerical conditions and guides that promotes this anonymously vague thing called thrust (don't think any here are) we still are poking the bear in the dark. How can I design something to optimize thrust when we don't even know what causes it and then throw in the misleading facts and figures from other builders and we have a real corundum on our hands. (ack, Shell runs screaming into the woods here) Numbers don't lie, but confusion in reported numbers or intentional misreporting of them makes for bad designing.I'm not going to throw bricks or mash up my frustum with it but I have much to do in the building of the test stand and I'm gonna sortta float for awhile seeing if better numbers come forth.Shell

Quote from: flux_capacitor on 08/02/2015 10:18 AMQuote from: TheTraveller on 08/02/2015 07:44 AMQuote from: flux_capacitor on 08/01/2015 03:13 PMYang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492In particular, the big end is smaller than the height, not larger.There is no resonance I can find for those dimensions at 2.45GHz. Can get TE012 resonance at 2.51GHzThey were surely wrong dimensions, based on the assumption Yang's drawings had the correct proportions. ..No, they were the dimensions that I had correctly based on the assumption that the small diameter for Yang was obtained based on the cut-off condition for the small end. And TheTraveller's calculation actually confirms it:<<There is no resonance I can find for those dimensions at 2.45GHz. Can get TE012 resonance at 2.51GHz>>The 2% difference between 2.45 GHz and 2.51 GHz is completely and utterly negligible compared to the other differences we are talking about: compared to the difference between 6 degrees and 15 degrees for the cone half-angle or compared to the uncertainties in having to eyeball the dimensionless ratio equations from Yang's chart.On top of that, the result TE012 resonance at 2.51GHz is obtained by TheTraveller from an Excel spreadsheet using very simplified ad-hoc approximate formulas that do not respect the boundary conditions of the problem while the solution for TE012 2.45GHz resonance uses Wolfram Mathematica to solve the exact solution to the problem using Legendre Associated Functions and Spherical Bessel functions.Taking into account the above considerations, the 2% different ad-hoc TE012 resonance at 2.51GHz means a complete agreement with the dimensions provided, because the 2% difference is completely within bounds.___________________________________________The issue at hand (determination of Yang's geometry) has to do with whether L=24cm provided by Yang in her paper is correct, and if so, whether the other (one and only one available to choose) parameter to consider should be the small diameter (based on cut-off), or the cone half-angle (determined from drawings), or some other parameter like D/L (determined from drawings), etc.One cannot obtain dimensions from Yang's schematic drawings, one can only obtain dimensionless ratios: angles and ratios, that's it.For L=24 cm and f=2.45 GHz, you cannot satisfy both that the diameter of the small base be above cut-off for an open waveguide AND satisfy the cone half-angle from Yang's drawings . They are in violent conflict with each other. Something got's to give. This has to be clearly understood: we have mathematical relations to satisfy here, they are mathematical constraints. You cannot satisfy all the dimensionless ratios from Yang's drawings and simultaneously satisfy the cut-off condition at the small end and simultaneously satisfy L=24 cm and simultaneously satisfy the D/L vs frequency relation.

Quote from: TheTraveller on 08/02/2015 07:44 AMQuote from: flux_capacitor on 08/01/2015 03:13 PMYang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492In particular, the big end is smaller than the height, not larger.There is no resonance I can find for those dimensions at 2.45GHz. Can get TE012 resonance at 2.51GHzThey were surely wrong dimensions, based on the assumption Yang's drawings had the correct proportions. ..

Quote from: flux_capacitor on 08/01/2015 03:13 PMYang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492In particular, the big end is smaller than the height, not larger.There is no resonance I can find for those dimensions at 2.45GHz. Can get TE012 resonance at 2.51GHz

Yang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492In particular, the big end is smaller than the height, not larger.

Shell,I think your narrow angled frustum is an ideal case to test Todd's theory, needing a long narrow shape. If it works better then all the rest, then we've all learned what direction to take for better results... If it doesn't, it might indicate that Shawyer's path might yield better results. So, whatever the outcome, your findings and data will bring an important contribution...I too still have mixed feelings about the accuracy of Yang's frustrum+waveguide "technical drawing".Some aspects make it read as a genuine technical illustration, where other aspects turn it "make believe" technical drawing that in essence hold no true dimensioning, scaling or proportion...If only there was more consistency in the drawings and they had measurements on them... sigh...

Well, i have no clear view on the current status of Todd's ideas are, as they tend to evolve rather quickly, but I'm sure that at certain stage it was mentioned in one of his posts...I do think it is very helpful to tailor the DIY projects according some of the theories that float around, as these tests might help to prove or disprove an idea/theory.The faster we're clearing this forest of theories and ideas, the better it will be.. no?

Quote from: Rodal on 08/01/2015 08:39 PMQuote from: Rodal on 08/01/2015 05:51 PMQuote from: Flyby on 08/01/2015 05:07 PMQuote from: Rodal on 08/01/2015 04:50 PMThe difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wikiCompared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !Could you please superpose the geometry of the EM Drive Wiki ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.Blue outline is the Wiki version of Yang...As you said... it's way off..Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a length value that is way off?The only dimension given by Yang in her paper is the Length (also called height) measured perpendicular to the bases. It is 0.240 meters.Look at page 811, Table 1 http://www.emdrive.com/NWPU2010paper.pdfHence it would be best if you superpose all the images so that they all have the same length: 0.240 metersThen one has to use this image:knowing that 1) L=24 cm2) f = 2.45 GHz3) D = (Dbig + Dsmall)/24) TE012 line should be usedOK, using the above and Cone half-angle = 15 degrees I proceed as follows:TE012 Equation (from Yang's graph): y = 13.5 + 8.5 xwhere y = ((f D) ^2)*10^(-20)x = (D/L)^2replacing L=24 cmf=2.45*10^9 Hzand solving the quadratic equation for D, I getD=17.26915 cmand since(Db - Ds)/2 = (24 cm)* tan (15 degrees)and(Db + Ds)/2 =D = 17.26915 cmand solving these coupled equations for Db and Ds, we finally getDb = 23.69993 cm ~ 23.70 cmDs = 10.83836 cm ~ 10.84 cmL = 24 cmNeed to calculate whether these numbers give a natural frequency for TE012 of 2.45 GHzif not, need to modify the diameters in order to get TE012 @ 2.45 GHzThe main imprecision comes from the coefficients of the equation y = 13.5 + 8.5 xObviously, there is a whole family of solutions that satisfy the equation for TE012 in Yang's graph, for different values of the cone angle. The larger the cone angle, the more different are the values of Db and Ds, all we know is their average D.After a 5 mile run, I'm ready to put this to bed.We proceed as follows:We adopt the Cone half-angle value from Flyby 15.44 degrees, which is the Median of the three values obtained (15 degrees from SeeShells and the two values from Flyby 15.44 and 18.27 degrees)Cone Half-Angle = 15.44therefore the constraint isDb - Ds = 2 (24 cm) Tan(15.44) = 13. 25746 cmand vary Db and Ds subject to this constraint to get 2.45 GHz for TE012 from the exact solutiondoing so, we get:Db = 0.247 m Ds = 0.114425 m L = 0.24 mr1= 0.211022 mr2= 0.455515 mCone half-angle = 15.44 degreesthese numbers represent only a small change from the previous numbers, which gives some level of encouragement, as the previous numbers were obtained from Yang's chart, just eyeballing it, without performing any frequency calculationNotice that these diameters for Yang are close to the diameters of Shawyer's Flight Thruster, the only difference with Flight Thruster is that Yang has a greater length to enable resonating at TE012 for Yang at 2.45GHz instead of TE013 for Flight Thruster at 3.85GHz. Recall that at the time that Yang embarked on her project, the Flight Thruster was the latest and highest effective force design by Shawyer. The longer length of Yang being motivated perhaps by the need to resonate at 2.45 GHz instead of 3.85 GHz.We adopt these numbers as the official Yang geometry and we correct the previous numbers I had entered in the EM Drive Wiki accordingly without further ado.***A cone half-angle of 6 degrees does not make any sense to me for an EM Drive, it is like a cylinder ===> bad***

Quote from: Rodal on 08/01/2015 05:51 PMQuote from: Flyby on 08/01/2015 05:07 PMQuote from: Rodal on 08/01/2015 04:50 PMThe difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wikiCompared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !Could you please superpose the geometry of the EM Drive Wiki ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.Blue outline is the Wiki version of Yang...As you said... it's way off..Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a length value that is way off?The only dimension given by Yang in her paper is the Length (also called height) measured perpendicular to the bases. It is 0.240 meters.Look at page 811, Table 1 http://www.emdrive.com/NWPU2010paper.pdfHence it would be best if you superpose all the images so that they all have the same length: 0.240 metersThen one has to use this image:knowing that 1) L=24 cm2) f = 2.45 GHz3) D = (Dbig + Dsmall)/24) TE012 line should be usedOK, using the above and Cone half-angle = 15 degrees I proceed as follows:TE012 Equation (from Yang's graph): y = 13.5 + 8.5 xwhere y = ((f D) ^2)*10^(-20)x = (D/L)^2replacing L=24 cmf=2.45*10^9 Hzand solving the quadratic equation for D, I getD=17.26915 cmand since(Db - Ds)/2 = (24 cm)* tan (15 degrees)and(Db + Ds)/2 =D = 17.26915 cmand solving these coupled equations for Db and Ds, we finally getDb = 23.69993 cm ~ 23.70 cmDs = 10.83836 cm ~ 10.84 cmL = 24 cmNeed to calculate whether these numbers give a natural frequency for TE012 of 2.45 GHzif not, need to modify the diameters in order to get TE012 @ 2.45 GHzThe main imprecision comes from the coefficients of the equation y = 13.5 + 8.5 xObviously, there is a whole family of solutions that satisfy the equation for TE012 in Yang's graph, for different values of the cone angle. The larger the cone angle, the more different are the values of Db and Ds, all we know is their average D.

Quote from: Flyby on 08/01/2015 05:07 PMQuote from: Rodal on 08/01/2015 04:50 PMThe difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wikiCompared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !Could you please superpose the geometry of the EM Drive Wiki ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.Blue outline is the Wiki version of Yang...As you said... it's way off..Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a length value that is way off?The only dimension given by Yang in her paper is the Length (also called height) measured perpendicular to the bases. It is 0.240 meters.Look at page 811, Table 1 http://www.emdrive.com/NWPU2010paper.pdfHence it would be best if you superpose all the images so that they all have the same length: 0.240 metersThen one has to use this image:knowing that 1) L=24 cm2) f = 2.45 GHz3) D = (Dbig + Dsmall)/24) TE012 line should be used

Quote from: Rodal on 08/01/2015 04:50 PMThe difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wikiCompared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !Could you please superpose the geometry of the EM Drive Wiki ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.Blue outline is the Wiki version of Yang...As you said... it's way off..Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a length value that is way off?

The difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wikiCompared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !Could you please superpose the geometry of the EM Drive Wiki ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.

...So i picked a dimensionality that came from calculations (yours and me following them and I didn't catch the booboo either and that's on me) and reported high thrusts and built it. I was wrong and you were wrong, it happens. ...

...Quote from: flux_capacitor on 08/01/2015 03:13 PMYang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492......the cutoff dimension, just below the guide wavelength for TE01 mode at 2.45GHz = 148.7 mm..

Yang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492...

Quote from: SeeShells on 08/02/2015 03:49 PM...So i picked a dimensionality that came from calculations (yours and me following them and I didn't catch the booboo either and that's on me) and reported high thrusts and built it. I was wrong and you were wrong, it happens. ...Nope. There was no "booboo" and there was not something "wrong" in my previous selection of dimensions for Yang. Those dimensions are the correct dimensions if one assumes that Yang selected her small diameter to be at the waveguide cut-off frequency, following Shawyer's prescription. Quote from: flux_capacitor on 08/02/2015 10:18 AM...Quote from: flux_capacitor on 08/01/2015 03:13 PMYang's frustum has the following dimensions:- cavity length (m): 0.24- big diameter (m): 0.201- small diameter (m): 0.1492......the cutoff dimension, just below the guide wavelength for TE01 mode at 2.45GHz = 148.7 mm.. Proof that I selected those Yang's dimensions based on the assumption that she followed Shawyer's prescription for the small diameter to be at the cut-off frequency:Known input:Mode Shape: TE012Frequency: 2.45 GHzTable of Bessel zeros: http://wwwal.kuicr.kyoto-u.ac.jp/www/accelerator/a4/besselroot.htmlx____________________________________________________Proof:Equation for cut-off diameter:Ds = (X'01 *c)/(Pi*frequency) =(3.83170597020751*299792458 m/s)/(Pi*2.45*10^9 1/s) =0.1492 mQED. Thus the cut-off diameter for TE012 at 2.45GHz is exactly 0.1492 m and that's where the previous dimensions I selected came from.________________________________________________________________________________________________________The dimensions for Yang's EM Drive that I have placed now in the Wiki are based instead on the cone half-angle in Yang's drawings. There is only free parameter one can choose: for example that she chose Ds based on the cut-off frequency, or that the cone half-angle is known. One free-parameter. That's it. The cone half-angle in her drawings is in conflict with the assumption that her small diameter was chosen based on cut-off frequencies. The present dimensions based on the cone half-angle in her drawings are in agreement with:1) similar cone-half angles as NASA and Shawyer 2) in much closer agreement with the formulas from Shawyer. McCulloch and Notsosureofit for thrust

How sure are you about the 0.240m length?If you think that's it we have all together

...How sure are you about the 0.240m length?If you think that's it we have all together

...Did ever someone ask for the dimensions (for verification) personally? [email protected]It isn't helpful to poke in the dark, IMHOPS: I am not the one who want to do

...If you think that's it we have all together SD=0.1492mBD=0.200mL=0.240mf_res(TE012)=~2,4537GHzangle=6,041degwith SD 0.15m i get 2,4490GHzangle=5,946deg

Quote from: X_RaY on 08/02/2015 07:03 PM...Did ever someone ask for the dimensions (for verification) personally? [email protected]It isn't helpful to poke in the dark, IMHOPS: I am not the one who want to do Several people have asked (Ref: NSF threads 1 and 2). No answers have been reported. Ever.It has been reported Prof. Yang told a Scientific News reporter that "publicity about her EM Drive research is most unwelcome"