Quote from: TheTraveller on 05/14/2015 02:05 am....As defined in the attachment dominant circular waveguide cutoff wavelength is 1.7 x diameter.... 1.7 x diameter...Yes of course, but what diameter? A cylinder has only one diameter. It has constant diameter.Shawyer's EM Drive is not a cylinder with constant diameter. Shawyer's EM Drive is a truncated cone.A truncated cone has a variable diameter. The diameters are different at each end.Shawyer's reference (Cullen) does not deal with truncated cones.What diameter ?Look at what I derived in http://forum.nasaspaceflight.com/index.php?topic=36313.msg1373898#msg1373898:...choosing the cutOffWavelength to be TE110,Look at where the factor of 1.7 comes from: it is Pi divided by X'_{11}cOW = gmD *(Pi/1.84118378134065) = gmD *1.70628955426831741.7 times the Geometric Mean of the DiameterscOW = cutOffWavelength gmD = Sqrt[bD * sD]bD = big end diameter (m)sD= small end diameter (m)

....As defined in the attachment dominant circular waveguide cutoff wavelength is 1.7 x diameter....

I'm only asking so as to help out with the maths. It doesn't mean I believe a word about its physicality.For example: if the expression I'm trying to get verified is true, then you can get your blow-up condition. Other conditional relations produce different results (x0 < x1 < x2 or x1 < x2 < x0: x=lambda).Then there's also the case that it blows up exactly when the numerator is also zero. As you know, 0/0 is an indeterminate quantity.I find a nicer way to write Shawyer's Df is like this (again using x=lambda)Df = x0*(x2 - x1) / (x1*x2 - x0^{2})You'll notice that Df=0 when x1 or x2 -> infinity (proof available on request but it's dead simple)Someone said that Df varied between 0 and 1, and should =1 when x1 or x2 -> inf. It doesn't.And obviously Df=0 when x1 = x2. So it can be zero for two separate reasons.Doh!

You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.Therefore 1 - (lambda0^{2}/(lambda1*lambda2)) is always positive definite.Therefore blow-up is impossible

Quote from: deltaMass on 05/14/2015 03:28 amYou missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.Therefore 1 - (lambda0^{2}/(lambda1*lambda2)) is always positive definite.Therefore blow-up is impossibleI agree with you in principle. I have to double check where he puts the cutoff wavelength...

Quote from: Rodal on 05/14/2015 03:37 amQuote from: deltaMass on 05/14/2015 03:28 amYou missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.Therefore 1 - (lambda0^{2}/(lambda1*lambda2)) is always positive definite.Therefore blow-up is impossibleI agree with you in principle. I have to double check where he puts the cutoff wavelength...All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.

Quote from: TheTraveller on 05/14/2015 03:52 amQuote from: Rodal on 05/14/2015 03:37 amQuote from: deltaMass on 05/14/2015 03:28 amYou missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.Therefore 1 - (lambda0^{2}/(lambda1*lambda2)) is always positive definite.Therefore blow-up is impossibleI agree with you in principle. I have to double check where he puts the cutoff wavelength...All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?

Quote from: Rodal on 05/14/2015 04:05 amQuote from: TheTraveller on 05/14/2015 03:52 amQuote from: Rodal on 05/14/2015 03:37 amQuote from: deltaMass on 05/14/2015 03:28 amYou missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.Therefore 1 - (lambda0^{2}/(lambda1*lambda2)) is always positive definite.Therefore blow-up is impossibleI agree with you in principle. I have to double check where he puts the cutoff wavelength...All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation

Df = 0 when lambda1 = lambda2Df = 1 when min(lambda1, lambda2) = lambda0 (free space condition)So it looks like the right Df

Rodal, you seem confused. Lambda0 is the free space wavelength = c/f. Equation (5) tells you the rest.

Quote from: TheTraveller on 05/14/2015 04:14 amQuote from: Rodal on 05/14/2015 04:05 amQuote from: TheTraveller on 05/14/2015 03:52 amQuote from: Rodal on 05/14/2015 03:37 amQuote from: deltaMass on 05/14/2015 03:28 amYou missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.Therefore 1 - (lambda0^{2}/(lambda1*lambda2)) is always positive definite.Therefore blow-up is impossibleI agree with you in principle. I have to double check where he puts the cutoff wavelength...All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equationNo, he doesn't use the standard definition because according to that formula there is only one waveguide wavelength and he is defining two of them.Same question that deltaMass was askingHow does Shawyer define lambdag 1 and lambdag 2 ?He does define lambdag1 not equal to lambdag2, so they are different, we know thatwhat is the difference? In the formula you show the only things that appear are the free space wavelength and the cut off wavelengthBut there is only one free space wavelengthThere is only one cutoff wavelengthSO WHAT IS THE DIFFERENCE between lambdag1 and lambdag2

....The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.