Quote from: WarpTech on 08/01/2015 09:20 PMQuote from: Rodal on 08/01/2015 08:39 PMOK, 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.With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.ToddInteresting. The small plate is getting close to 50% of the large. I need to calculate out for my Yang-Shell model where would I be in L with a 6 degree angle...That's for later now out to sawdust and a effervescence cold something.Shell

Quote from: Rodal on 08/01/2015 08:39 PMOK, 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.With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.Todd

OK, 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.

.....and solving these coupled equations for Db and Ds, we finally getDb = 23.69993 cm ~ 23.70 cmDs = 10.83836 cm ~ 10.84 cmL = 24 cm

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.

...With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.Todd

Quote from: Rodal on 08/01/2015 05:51 PMknowing that 1) L=24 cmHem… taking the longer length L1 = 43.34 mm for the coupling window, according to latest Yang's drawing I get a cavity internal height of 14.87 cm The size of the coupling window may be wrong.However if I follow my first idea of setting the waveguide height as the length of a WR340 waveguide used vertically as in Tajmar's experiment (86.36 mm) I get a cavity length = 22.35 cm, much closer to the 2010 data of 24 cm.

knowing that 1) L=24 cm

Quote from: WarpTech on 08/01/2015 09:20 PM...With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.ToddWhich is very good my friend because k r rules the waves ! This agrees with my computationsThis agrees with all the formulae: McCulloch, Notsosureofit: the larger the difference between the diameters the better

.........and 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 cm Ds = 0.114425 cm L = 0.24 cmr1= 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.

Quote from: Rodal on 08/01/2015 10:20 PM.........and 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 cm Ds = 0.114425 cm L = 0.24 cmr1= 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.Well, the problem i got now is that on 1 side you use the angle from the drawing as being correct, but when I scale the drawing to match the height, my Db and Ds are seriously different that what you have calculated....So...You seem to have an agreement on your calculations and a disagreement with the drawings you based your initial calculus parameters on....doesn't that sound a bit ... odd? I'm not comfortable with that.... either the drawing matches the calculated values or it doesn't...

Quote from: Rodal on 08/01/2015 10:35 PMQuote from: WarpTech on 08/01/2015 09:20 PM...With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.ToddWhich is very good my friend because k r rules the waves ! This agrees with my computationsThis agrees with all the formulae: McCulloch, Notsosureofit: the larger the difference between the diameters the betterUpdated: Her's is still the smallest k*r but only slightly smaller now. I also updated the Tajmar size, and it is "closer" now too.Todd

I am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.There are many constraints to satisfy. Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012As in the movie: something's got to give The most important geometrical constraint from the drawing is the cone half angle. The others you mention are second order and not as important.If it is any comfort to you remember that you thought that Yang's drawings were schematic. There are more important things to base the dimensions on that this drawing. The most important thing to get from her schematic drawings is the cone-half angle.

Quote from: Rodal on 08/01/2015 10:55 PMI am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.There are many constraints to satisfy. Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012As in the movie: something's got to give The most important geometrical constraint from the drawing is the cone half angle. The others you mention are second order and not as important.If it is any comfort to you remember that you thought that Yang's drawings were schematic. There are more important things to base the dimensions on that this drawing. The most important thing to get from her schematic drawings is the cone-half angle.I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.The angle is tied to the dimensions of the big and small angle in relation to the height.How can the angle be correct (for the given height) and not to the big and small bases?After all, all i did is just rescale to the given height...Let me add your new results into a new drawing, so we can see better....brb

Quote from: Rodal on 08/01/2015 10:35 PMQuote from: WarpTech on 08/01/2015 09:20 PM...With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.ToddWhich is very good my friend because k r rules the waves ! This agrees with my computationsThis agrees with all the formulae: McCulloch, Notsosureofit: the larger the difference between the diameters the betterWhat stops one from wanting to take this statement to its maximum (probably wholly illogical) extreme?

I think I found out what's going on here...magenta : original technical drawingred : newly calculated YANG valuesyellow : abstracted frustum that does not take the top/bottom cylindrical shapes into count.Note how, conveniently, that this yellow cone-half angle almost resembles the magenta cone-half angle.I suspect that your calculations do not take those top/bottom edges into account...I do not have enough insight in the matter to understand if those are important or not.

Quote from: Flyby on 08/01/2015 11:08 PMQuote from: Rodal on 08/01/2015 10:55 PMI am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.There are many constraints to satisfy. Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012As in the movie: something's got to give The most important geometrical constraint from the drawing is the cone half angle. The others you mention are second order and not as important.If it is any comfort to you remember that you thought that Yang's drawings were schematic. There are more important things to base the dimensions on that this drawing. The most important thing to get from her schematic drawings is the cone-half angle.I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.The angle is tied to the dimensions of the big and small angle in relation to the height.How can the angle be correct (for the given height) and not to the big and small bases?After all, all i did is just rescale to the given height...Let me add your new results into a new drawing, so we can see better....brbLook at it this way: plot Yang/Shell dimensions vs. these new dimensions and show us which one is closer to Yang's schematic drawing: no contest !!!and the proof of the pudding will be Yang/Shell test, let's see what thrust she gets from a cone angle of only 6 degrees______<<I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.>>I strongly disagree:The cone half-angle is an invariant under internal length-adjustmentThe small and the big bases and the length will vary with reentrant ends, and they will vary with length adjustment

actually, look at my new drawing...Dr Rodal's version of the new YANG cavity is in yellow and has a cone half angle is 15.45° (where the initial magenta design had 15.44°). No idea if it is pure luck or if there is a correlation....

No, of course that I don't take those cylindrical corners into account. What matters is the cone half-angleOn purpose I don't take those into account.The cylindrical extensions make the geometry into an irregular geometry and they break the spherical waves.

ExcellentThanks for the outstanding job

@Notsosureofit,EDIT: Never mind. I answered my own dumb question. I shouldn't post before I finish my coffee. The bandwidth used for Q is not the same, but f is. Therefore, it results in;acceleration g = (c^{2}/L)*(delta_f/f), wheredelta_f = (2pi/f)*(fs^{2} - fb^{2})Q = f/delta_b (b for bandwidth)N*T = (P/2pi*L*f)*(delta_f/delta_b)This implies a lower frequency, large delta_f/L implies a short length, wide half-angle. Small delta_b implies narrow bandwidth. So a wide stubby frustum with both a smaller small end and a larger big end? Using a narrow band RF amplifier rather than a Magnetron. Just as @TT and Shawyer have said.Thank you.Todd