Quote from: Rodal on 06/12/2015 05:54 PMQuote from: SeeShells on 06/12/2015 05:43 PM...Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?OK, I need your help here Shell.I understand Zeng and Fan's paper analytically, that the smaller the half-angle the greater the attenuation, but I don't have a physical understanding of this to explain it in simple terms. Which means that when it comes down to it, I don't understand really understand it .This is my problem in understanding this:1) A perfect cylinder has no geometrical attenuation: it has no evanescent waves, no progressive cut-off of modes. The modes that are cut-off are the ones that have a bigger diameter than the cylinder, that's it.2) So, how can it be that the biggest geometrical attenuation occurs for the smallest half-angle ? in other words for the geometry that is closest to a cylinder?3) What happens in the limit with the geometrical attenuation as the cone becomes a cylinder (as the half angle becomes zero)? How does one go from the largest geometrical attenuation for a cone to NO geometrical attenuation for a cylinder?In Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.Todd

Quote from: SeeShells on 06/12/2015 05:43 PM...Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?OK, I need your help here Shell.I understand Zeng and Fan's paper analytically, that the smaller the half-angle the greater the attenuation, but I don't have a physical understanding of this to explain it in simple terms. Which means that when it comes down to it, I don't understand really understand it .This is my problem in understanding this:1) A perfect cylinder has no geometrical attenuation: it has no evanescent waves, no progressive cut-off of modes. The modes that are cut-off are the ones that have a bigger diameter than the cylinder, that's it.2) So, how can it be that the biggest geometrical attenuation occurs for the smallest half-angle ? in other words for the geometry that is closest to a cylinder?3) What happens in the limit with the geometrical attenuation as the cone becomes a cylinder (as the half angle becomes zero)? How does one go from the largest geometrical attenuation for a cone to NO geometrical attenuation for a cylinder?

...Also the longer length of the Chinese Frustum makes sense with the sidewalls of the Frustum not attenuating as it (less angle) collapses into the smaller end. Longer distance traveled = higher Momentum gained and speed through less loss, a more focused effervescent wave collapse?

One possible source of drag is eddy currents. As the Aluminum cavity sweeps through the magnetic field created by the levitator any change in the field strength will induce currents in the Aluminum. This can happen because of the wobbles seen in the rotation vs time curves. The Aluminum block moves out of center slightly and the magnetic field strength around it is slightly weaker. That will introduce Ohmic losses and the rotation rate will decrease faster than it would if there was no metal.It's an interesting apparatus; very crafty construction. I hope they collect lots of data from it and see what averages out.

I'm currently in bed, dealing with a few long term health issues. Not much EMDrive build work will happen by me for 4 weeks or so.

Quote from: WarpTech on 06/12/2015 06:13 PMIn Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.ToddSo, your answer boils down to that the explanation is that maximum attenuation occurs when " the cut-off and the wavelength are close to the same value." But what's the problem with a larger cone angle? as I see it the difference is that for a larger cone angle there are extra modes that get cut-off.EXAMPLE:Imagine two different truncated cones, both having the same length and the same small diameter.The only difference is that truncated cone A has a much larger big diameter than truncated cone B. Therefore cone A has a significantly larger cone angle than cone B.In cone A, you have other modes that get cut off. All the modes that get cut off in cone B also get cut of in cone A.So, why is the geometrical attenuation of cone A worse than the geometrical attenuation of cone B?

In Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.Todd

Quote from: Rodal on 06/12/2015 06:25 PMQuote from: WarpTech on 06/12/2015 06:13 PMIn Zeng & Fan, the waves are propagating down a cylinder with little or no attenuation, and little or no phase shift. When they come to a tapered waveguide, then the waves are attenuated because the cut-off wavelength is getting shorter, and shorter. The wave is partially reflected as it contacts the walls at different phases in the cycle, and these reflections propagate backwards causing an interference with the incoming wave. That is what causes the attenuation;alpha = j(k - Beta), Beta is the phase of the reflected wave, k is the phase of the forward wave. if k - Beta, there is no attenuation, that's a cylinder.If the wavelength of the incoming wave is much shorter than the cut-off of the guide, the taper will have very little effect.... until the cut-off and the wavelength are close to the same value.ToddSo, your answer boils down to that the explanation is that maximum attenuation occurs when " the cut-off and the wavelength are close to the same value." But what's the problem with a larger cone angle? as I see it the difference is that for a larger cone angle there are extra modes that get cut-off.EXAMPLE:Imagine two different truncated cones, both having the same length and the same small diameter.The only difference is that truncated cone A has a much larger big diameter than truncated cone B. Therefore cone A has a significantly larger cone angle than cone B.In cone A, you have other modes that get cut off. All the modes that get cut off in cone B also get cut of in cone A.So, why is the geometrical attenuation of cone A worse than the geometrical attenuation of cone B?I think you are misinterpreting Zeng & Fan, when you think the half-angle of a cylinder is zero. My interpretation would be that the half-angle of a cylinder = pi/2. It has a "Flat" end. As you reduce the cone angle, the "end" of the cylinder becomes pointed, and the permitter of the conical end shifts back toward the large end to become a cone. Then the longer the the cone, the smaller the half-angle. A half-angle of zero is not a cylinder, it's a solid rod.Now, for a cylinder of length L, with resonant modes, of N*lambda/2 = L, the allowed wavelengths will be;lambda = 2L/NReflections in this case will have phase Beta = k, and will form standing waves. Attenuation = j(k - Beta) = 0.In the case of a cone at the end of a cylinder, there will be partial reflections from the side walls that are shorter than L. So the allowed wavelengths will be;lamda2 = 2(L-dL)/N,Where, dL is a variable depending on where the wave was reflected along the length of the conical section.Beta =/= k anymore, so now there will be attenuation. Do you see how that causes a phase shift in Beta, of the reflected wave?A larger dL value, coming from a smaller half-angle, will generate larger phase shifts on the returning wave, causing destructive interference moving backwards toward the large end. This relieves the pressure on the big end, allowing the small end to feel a force forward.I'm still working on the hard core mathematics of all this, but you should have a better understanding now, I hope.Todd

It is amazing that Yang achieves record thrust force and record thrust force/powerInput by doing the complete opposite of common wisdom:

...I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.

I'm still fooling with meep, trying to get some 3D images and yesterday I had an interesting accident with the code. Using what I thought was a sealed "Bradycone" cavity in 3D, I continued to see extensive RF energy outside the cone. Of course that can not be as we understand things. I made a Force/Power run and meep measured 10+ times ideal photon rocket.On further investigation i discovered that I had modeled the dielectric shifted in the Y direction - sideways - instead of axially. (Meep uses different coordinates in 3D that in 2D, it seems) So with the dielectric penetrating the center of one of the sidewalls of the cone (still oriented with dielectric and cavity ends parallel) F/P was 10 times higher than typical. OH, and this model won't resonate, that's where my investigation started.I don't know if the above is meaningful but I found it interesting. I wish someone else would do some FDTD runs so I would have someone to discuss wierd resiults with.

...On another note I think my water tank idea for the frustum is trashed. I would need to increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.

As a thought experiment. I've thought of making a Frustum cone collapsing to a point without the small end end plate, but in place of the plate inserting a wire grid where endcap would be. I'm sure it would work just like a normal Frustum until the waveform collapses into the wire grid and effervescent waves. That's where my thought experiment falls apart. On another note I think my water tank idea for the frustum is trashed. I would need to increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.

Quote from: SeeShells on 06/12/2015 08:10 PMAs a thought experiment. I've thought of making a Frustum cone collapsing to a point without the small end end plate, but in place of the plate inserting a wire grid where endcap would be. I'm sure it would work just like a normal Frustum until the waveform collapses into the wire grid and effervescent waves. That's where my thought experiment falls apart. On another note I think my water tank idea for the frustum is trashed. I would need to increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.I'll let you know how it turns out, both my endcaps are mesh, swappable with solid copper plated boards. Expect first test after July 4th holiday weekend.Note: spelling police will get you "Evanescent Waves"

Quote from: SeeShells on 06/12/2015 08:10 PM...On another note I think my water tank idea for the frustum is trashed. I would need to increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.Yes, displacing water acts like a spring that is much stiffer than the torsion of a thin wire for example. Effectively, you would need a liquid with orders of magnitude smaller density than water to have a spring constant small enough such that the displacement for these small forces would be significant.Still, water will work if you do get the magnitude of forces obtained by Prof. Yang (0.3 Newtons).It won't work if you get the forces measured by NASA.

...A cylinder is not at all the limit for θ_{o} =90 degrees . That limit gives you a line. It only defines one line, instead of the two walls of a cylinder.Instead think of what the geometry looks like just before the limit:1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines. However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero....

Quote from: Rodal on 06/12/2015 07:31 PM...A cylinder is not at all the limit for θ_{o} =90 degrees . That limit gives you a line. It only defines one line, instead of the two walls of a cylinder.Instead think of what the geometry looks like just before the limit:1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines. However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero....No, you are completely misinterpreting the angle theta. Theta is measured at the "Vertex" of the cone, not the difference between the cone wall and a cylinder wall. See attached....Todd

Quote from: WarpTech on 06/12/2015 08:32 PMQuote from: Rodal on 06/12/2015 07:31 PM...A cylinder is not at all the limit for θ_{o} =90 degrees . That limit gives you a line. It only defines one line, instead of the two walls of a cylinder.Instead think of what the geometry looks like just before the limit:1) for the cone angle going to zero the lateral walls of the cone approach two parallel lines, a cylinder has parallel lines. However as the cone half angle goes to zero, the diameters of the bases become smaller, going to zero....No, you are completely misinterpreting the angle theta. Theta is measured at the "Vertex" of the cone, not the difference between the cone wall and a cylinder wall. See attached....ToddGee, we are not in a good mood today are we (I am completely misrepresenting ?)I never defined the angle θ as "the difference between the cone wall and a cylinder wall. " I even included the figures.This is their figure for θAs θ -> 0 the lines of the cone become parallel, that's what I said. Look at the figure to see what happens as θ -> 0,To describe what happens in the limit you need to consider the spherical radius r.Look at the fact that attenuation is a function of both θ and r in Zeng and Fan's paper, and that for r -> Infinity, the attenuation goes to zero in their figures.Also look at their equations, and what happens for r -> Infinity.You cannot ignore the fact that attenuation is a function of r for large r going to Infinity.

I once during an experiment encountered some non-symmetry that also baffled me. We had a high resistance volt meter connected to a capacitor that was outside and concentric around a large solenoid. It had, I think around 180 picofarads and was aluminum. I could apply DC current through the solenoid in one direction and the voltage would rise on the capacitor and stay that way but decay as charge flowed off slowly. If I discharged it then reversed the voltage wires so current flowed the other way through the Edit:(solenoid) then give it current the magnitude of increase in the voltage on the capacitor was about an order less.

...I apologize if I appear to be rude, it wasn't intended that way. Just trying to squeeze this in "fast" while I'm working. When I look at the diagram above, I see the following:As theta => 0, R => 0, there is no waveguide anymore, the opening closes and it becomes a solid bar.What you are saying is:As theta => 0, R(z_{in}) = R(z_{0})To me, that's not what the diagram shows. To me, as R(z_{0}) => R(z_{in}), theta => pi/2. The vertex of the cone doesn't just "disappear" it gets pulled inward toward the center of the end plate of the cylinder and the angle 2It *theta => pi. Large values of r, are located at the big end. The bigger it is, the less effect the waveguide has on the free-space wavelength. Todd

Quote from: Rodal on 06/12/2015 08:15 PMQuote from: SeeShells on 06/12/2015 08:10 PM...On another note I think my water tank idea for the frustum is trashed. I would need to increase the power to be able to measure above the noise and at the threshold of my home brew laser detector. Than the water turbulence overcomes any meaningful measurement. It was a good idea though.Yes, displacing water acts like a spring that is much stiffer than the torsion of a thin wire for example. Effectively, you would need a liquid with orders of magnitude smaller density than water to have a spring constant small enough such that the displacement for these small forces would be significant.Still, water will work if you do get the magnitude of forces obtained by Prof. Yang (0.3 Newtons).It won't work if you get the forces measured by NASA.It is easy to lower the apparent "stiffness" of a floating object by immersing the main body and letting emerge just a thin cylinder, as in an alcoholmeter :Compared to your previous apparent stiffness calculations, there is a gain in sensitivity as the ratio of section of the main cylinder by the section of the small emerging cylinder.Obviously, the more sensitive it gets (the thinner the emerging cylinder) the more stringent gets the requirement to balance the overall weight so that it floats midwater (or mid alcohol). Needs tuning with leads or other ballast. The sensitivity could in principle be made arbitrarily high (within the limits of surface tension effects), but at some point it will require precise temperature control of water bath to avoid change of density and, ahem, thermal drift of baseline (again) ...Also the move of a large volume underwater will not only add apparent inertia to the mass of the tested rig, but it will stir the whole bath, with hard to predict delayed currents bouncing back from the container walls. High thermal inertia of the bath is a good point to mitigate high temperature at cavities wall (good dissipation), but higher density and viscosity of water means also stronger convection currents. Maybe using a thermal blanket around ? Could be wet, like open cell foam (neoprene for diving...) so that it's heavy enough not to float like cork.So I would say, this could be made sensitive enough as far as static displacement/force at equilibrium goes, but I'm afraid this is going to have poor dynamics, high mass (thing has to have same density as water) so high inertia => slow to reach equilibrium, which for an EM drive means running the power for long time, so lots of heat, with a lot of potential spurious delayed effects hard to foretell. Unless explicitly dealing with liquids from the start (characterisation of surface tension...), precision force measurement are usually done in dry set-ups, vacuum ideally, don't they ? Let's stay sober.