Quote from: WarpTech on 08/06/2015 09:13 PMQuote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.ToddIf F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart. Specifically, as the size of the waveguide decreases Vp increases. An increase in Vp decreases F. That would make the force on the little end --smaller-- than the force on the big end.Is that right?

Quote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.Todd

...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.

Quote from: flux_capacitor on 08/06/2015 10:58 PMI tried to model TheTraveller's EmDrive Mark 2 in 3D, which was a bit difficult since the only dimensions TT provided were end diameters D_{s} and D_{b} and the length (R_{b} - R_{s})^{*}D_{s} = 159 mmD_{b} = 400 mmR_{b}-R_{s} = 240.7 mmR_{s} = D_{s} = 159 mmR_{b} = D_{b} = 400 mmhalf-cone angle = 30°I get slightly different dimensions, on the order of tenths of a millimeter. Which brings up the question, how tight are the tolerances on the manufacturing process, and exactly how tight they need to be to get good results?

I tried to model TheTraveller's EmDrive Mark 2 in 3D, which was a bit difficult since the only dimensions TT provided were end diameters D_{s} and D_{b} and the length (R_{b} - R_{s})^{*}D_{s} = 159 mmD_{b} = 400 mmR_{b}-R_{s} = 240.7 mmR_{s} = D_{s} = 159 mmR_{b} = D_{b} = 400 mmhalf-cone angle = 30°

Quote from: ElizabethGreene on 08/06/2015 10:10 PMQuote from: WarpTech on 08/06/2015 09:13 PMQuote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. <snipp>If F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart. Specifically, as the size of the waveguide decreases Vp increases. An increase in Vp decreases F. That would make the force on the little end --smaller-- than the force on the big end.Is that right?Your thinking is correct. <snip>

Quote from: WarpTech on 08/06/2015 09:13 PMQuote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. <snipp>If F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart. Specifically, as the size of the waveguide decreases Vp increases. An increase in Vp decreases F. That would make the force on the little end --smaller-- than the force on the big end.Is that right?

Quote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. <snipp>

Quote from: Rodal on 08/06/2015 10:18 PMAs to how you can reconcile that the force being larger on the Big End results in motion towards the Small EndThat's why I prefer McCulloch's theory: according to MiHsC, as they travel towards the big end, photons gain momentum. When they bounce back towards the small end, they loose momentum. So to obey CoM, the cavity has to move from the big end towards the small end.

As to how you can reconcile that the force being larger on the Big End results in motion towards the Small End

r2 - r1 = 0.4 - 0.159 = 0.241 exactly

So if it moves, you have chosen to reject Maxwell.And when it moves, you have a further set of choices: reject Noether or reject Einstein.Pretty heady company, if you ask me.

The 3D integration of all photonic forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell.The 3D integration of all Lorentz forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell too.So if it moves, you have chosen to reject Maxwell.And when it moves, you have a further set of choices: reject Noether or reject Einstein.Pretty heady company, if you ask me.I hope you know what you're doing, whoever you are.

The inaccuracy of TheTraveller and Shawyer's code is more evident for this case because of the higher cone angle 30 degrees (compared with other EM Drives that range from 15 degrees to 20 degrees), and therefore the more inaccurate their approximate formulas.

Quote from: ElizabethGreene on 08/06/2015 10:10 PMQuote from: WarpTech on 08/06/2015 09:13 PMQuote from: ElizabethGreene on 08/06/2015 07:22 PM...I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ......Disclaiming again, I probably broke the math.The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.ToddIf F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart. Specifically, as the size of the waveguide decreases Vp increases. An increase in Vp decreases F. That would make the force on the little end --smaller-- than the force on the big end.Is that right?Pretty much... It is that the momentum of an EM wave depends on the phase velocity and Planck's constant. It is the phase velocity, Vp = w/k that give us the momentum per photon, p=h*k. So if Vp is increasing, p is decreasing, and if Vp is decreasing, p is increasing. Either way, the force on the frustum is in the same direction in order to conserve momentum. The momentum is "forward rectified" by the gradient. It is only the reflection at the big end that is opposing this force. Anything that reduces the force on the big end, will result in forward thrust. In order for the big end to 100% cancel these forces, the force on the big end would have to be larger than the force at the small end. Definitely not smaller. If it is smaller by hook or by crook, or by leakage or heat, then the frustum will feel a thrust forward. The real question is still, how does it get thrust greater than a photon rocket? I see 3 possibilities:1. A group velocity that is faster than light in free space.2. A phase velocity that is slower than light in free space.3. A means to change the speed of light such that Vg*Vp = (c/K)^{2}, in which case the group velocity could be faster than light in a frustum, but slower than light in free space and result in an amplified ratio;Vg/c^{2} -> K*Vg/c^{2}, where K > 1 This way, nothing ever exceeds c in free space, but it reduces both the phase velocity and the group velocity to be much less than c. It is consistent with the PV Model of gravity. I just don't know how to derive it yet. I have 2 key pieces to the puzzle. Squeezed light behaves this way, and what Zeng and Fan show for an impedance curve looks almost exactly like a Reissner Nordstrom "charged" metric refractive index. Somehow, this all seems to make sense to me, but not well enough to write down the correct equations. Todd

QuoteThe 3D integration of all photonic forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell.The 3D integration of all Lorentz forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell too.So if it moves, you have chosen to reject Maxwell.And when it moves, you have a further set of choices: reject Noether or reject Einstein.Pretty heady company, if you ask me.I hope you know what you're doing, whoever you are.Key word in this to me is 'averaged.'Lately, the thought has crossed my mind more than once that maybe something about this device breaks or twists the averages somehow. There are such things as loaded dice. And coins do land on edge now and again. Maybe the EM Drive represents one such?

As the wave moves from the small end toward the big end it expands, but it's wavelength is getting shorter because the phase velocity is getting slower. So the momentum of the wave is increasing when moving from the small end toward the big end. Therefore, to conserve momentum, the frustum must move the other way. It feels the force that moves it in direction toward the small end. This is thrust, just like a rocket.When the wave reflects from the big end, the amount of momentum it transfers depends on the impedance match and the angle of reflection. If the impedance doesn't match exactly, or the angle of reflection is not perpendicular, then the force will vary too. Don't ask me for the equations please, I don't have them quite right yet. So this is just hand waving, based on my understanding. When the wave is traveling toward the small end, its momentum is decreasing and therefore, to conserve momentum the frustum must again feel a force in the "forward" direction, like a sail on a boat. This is due to the gradient, which is unique to the cone geometry. According to Zeng & Fan's paper, if the phase constant Beta goes to zero, there is a perfect impedance match and all the power and momentum is attenuated into the frustum with no reflection.So regardless if the wave is traveling forward or backwards inside, the frustum will be pushed "forward" toward the small end. The only component that is in the other direction, opposing this thrust is the reflection from the big end. From what I've seen in the meep models, the wave is not perpendicular to the big end. The angle of reflection produces a glancing blow. Anything that reduces the momentum transferred to the big end plate, results in thrust forward. Todd

Any way, whether using 30 degrees or 30.0412 degrees this geometry is going to resonate at TE012 at a lower natural frequency than the 2.45 GHz TheTraveller calculated.

Quote from: Rodal on 08/07/2015 02:47 AMAny way, whether using 30 degrees or 30.0412 degrees this geometry is going to resonate at TE012 at a lower natural frequency than the 2.45 GHz TheTraveller calculated.So what do your exact solutions give for each value of the possible angles we found (30° and 30.0412°) for TE012 and TE013? Are they really different with such a slight difference?I think TheTraveller may have accounted for that lower natural frequency, as he also noted on his drawing: "Min operational frequency 2.3 GHz".

It doesn't matter if there is a more momentum at one vs the other. This theory of operation will not result in the frustum moving because....The small end and the large end are physically connected, so the total momentum is ZERO.This mechanism of action is a no go. It is just like pushing your car from the inside.

Oh, I see. That was supposed to be physics.Carry on.