Quote from: Flyby on 06/11/2015 09:17 AMQuote from: deltaMass on 06/11/2015 08:22 AMThat 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...Well, someone might built a one meter frustum and connect it to a couple megawatts (continuous power) klystron at 300 Mhz but no one will spend such amount of money without clear evidence.

Quote from: deltaMass on 06/11/2015 08:22 AMThat 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...

That 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.

Quote from: Paul Novy on 06/11/2015 09:31 AMQuote from: Flyby on 06/11/2015 09:17 AMQuote from: deltaMass on 06/11/2015 08:22 AMThat 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...Well, someone might built a one meter frustum and connect it to a couple megawatts (continuous power) klystron at 300 Mhz but no one will spend such amount of money without clear evidence.what would be the theorized (so far) thrust of such experimented device?

Disturbing to se the EM Drive's strange preference of one direction vs. another (reminds me of the issue with NASA turning it around by 180 degrees).

Quote from: aceshigh on 06/11/2015 03:58 PMQuote from: Paul Novy on 06/11/2015 09:31 AMQuote from: Flyby on 06/11/2015 09:17 AMQuote from: deltaMass on 06/11/2015 08:22 AMThat 720 mN came from about 1 KW of power. The Baby EmDrive uses not kilowatts, not Watts, but milliwatts. You need to adjust your expectations accordingly.It doesn't need to float 1 ton of force for me... no need for floating cars as evidence...just a few hundreds of mN , not to be attributed to any already (thermal, magnetic, etc) known effect, would be enough, i think...Well, someone might built a one meter frustum and connect it to a couple megawatts (continuous power) klystron at 300 Mhz but no one will spend such amount of money without clear evidence.what would be the theorized (so far) thrust of such experimented device?If my Theory of Operation is correct, damping factor = attenuation/frequency. So a lower frequency should have a lower Q but a higher thrust output, due to higher damping on one side. The greater the differential of phase between the forward and backward standing waves, the higher the power factor, the more work that can be done as thrust.Todd

Quote from: Rodal on 06/11/2015 01:13 AMDisturbing to se the EM Drive's strange preference of one direction vs. another (reminds me of the issue with NASA turning it around by 180 degrees). Has anyone checked for a North/South/East/West orientation dependency in the measured thrust? If the quantum vacuum conjectures are correct (or at least on the right track), could it be that the EM drive's measured thrust is affected by frame dragging from the rotation of the Earth?

Quote from: WarpTech on 06/11/2015 04:44 AM...We are "guessing" what Yang did with her design and @Rodal suggested she followed Shawyer's DF making the small end equal to (half?) the input wavelength. I'd be surprised if it wasn't 50% smaller than that, given her thrust measurements and low Q values.What Shawyer suggest for the small end is it should operate just ABOVE cutoff, defined as Guide Wavelength at small end still above zero. That way there is an EM wave to bounce back off the small end plateThat gives max Df as either small end diameter, excitation mode or input frequency will need to be adjusted / chosen to obtain that result.Next he suggests getting big end as big as possible while obtaining desired internal 1/2 wave multiples of the effective overall guide wavelength to fit inside end plate spacing.

...We are "guessing" what Yang did with her design and @Rodal suggested she followed Shawyer's DF making the small end equal to (half?) the input wavelength. I'd be surprised if it wasn't 50% smaller than that, given her thrust measurements and low Q values.

...... Asymmetrical attenuation will cause thrust. Copper losses will cause heat. Reduce the losses, but increase the attenuation by making the small end smaller and the frustum longer, "should" result in much higher thrust.Todd

Quote from: Vix on 06/11/2015 01:00 PMHistory repeats itself. Which reminds me of Leo Szilard in the 1930'es. Skeptics need a proof first. As it looks now, it could easily happen that the Chinese will provide it...Shawyer has stated he will release a new peer reviewed Superconducting EMDrive in 2015, created with the assistance of other companies SPR works with. He normally releases at IAC meetings. Next is mid Oct 2015.

History repeats itself. Which reminds me of Leo Szilard in the 1930'es. Skeptics need a proof first. As it looks now, it could easily happen that the Chinese will provide it...

Quote from: WarpTech on 06/11/2015 04:24 PM...... Asymmetrical attenuation will cause thrust. Copper losses will cause heat. Reduce the losses, but increase the attenuation by making the small end smaller and the frustum longer, "should" result in much higher thrust.ToddIf you continue on that mental path - by making the small end smaller and smaller - you end up with a simple cone shape, no?Or do the front and back plates have a role to play?

I know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations.

Quote from: WarpTech on 06/11/2015 04:24 PMI know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations. I dispute that Yang's maths is correct.Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.

Quote from: deltaMass on 06/11/2015 06:00 AMI'm trying to deduce the thrust from the Baby Emdrive data, and failing.Let q represent dw/dt or the angular acceleration (always < 0), whereq_{0} is the acceleration with the drive off = -29.408q_{+} is the acceleration with the drive prograde (impeding decay) = -24.987q_{-} is the acceleration with the drive retrograde (assisting decay) = -35.007where the numbers have been read off from the interpolated graphical slopes.Let b = the constant frictional force acting against the motion (Newtons)Let F = the drive force (Newtons)Let a = I/R, so that a*q is also in Newtons (torque = moment of inertia * dw/dt)Then we have 3 equations in 3 unknowns:1. a q_{0} = -b2. a q_{+} = -b + F3. a q_{-} = -b - FFor consistency we are forced to have 2 q_{0} = q_{+} + q_{-} and from the interpolated graphical slopes, this is quite nicely the case (58.82 vs. 59.99).However, we cannot solve for F independently of a or b.In order to get the value of F, either the friction needs be known (which is the value 'b'), or the moment of inertia of the cavity and its lever arm needs be known (which yields the value 'a').Note that the numerical values used for q are not yet in the correct (SI) units at this point.Its difficult for me to wrap my head around what the natural anguiar deceleration should be, which is why I would like to see more of the non-thrusting behavior. On the one hand, if it were due primarily to "smooth" aerodynamics, the decelerating torque should be proportional to the square of the angular rate. On the other hand, the airflow in the jar may be completely chaotic or vortical making prediction impossible. On the third hand, is there a magnetic dampening torque?

I'm trying to deduce the thrust from the Baby Emdrive data, and failing.Let q represent dw/dt or the angular acceleration (always < 0), whereq_{0} is the acceleration with the drive off = -29.408q_{+} is the acceleration with the drive prograde (impeding decay) = -24.987q_{-} is the acceleration with the drive retrograde (assisting decay) = -35.007where the numbers have been read off from the interpolated graphical slopes.Let b = the constant frictional force acting against the motion (Newtons)Let F = the drive force (Newtons)Let a = I/R, so that a*q is also in Newtons (torque = moment of inertia * dw/dt)Then we have 3 equations in 3 unknowns:1. a q_{0} = -b2. a q_{+} = -b + F3. a q_{-} = -b - FFor consistency we are forced to have 2 q_{0} = q_{+} + q_{-} and from the interpolated graphical slopes, this is quite nicely the case (58.82 vs. 59.99).However, we cannot solve for F independently of a or b.In order to get the value of F, either the friction needs be known (which is the value 'b'), or the moment of inertia of the cavity and its lever arm needs be known (which yields the value 'a').Note that the numerical values used for q are not yet in the correct (SI) units at this point.

...We seem to be the only two so far discussing the Baby EmDrive data - thanks for the valuable feedback.If I take your suggestion that "the decelerating torque should be proportional to the square of the angular rate", then we solve by integration to end up with w(t) = w_{0} / (b w_{0} t + 1)where b is a frictional constant of the motion. Two comments:1. The test was done in vacuum, so I don't think this applies. That's why I adopted a friction model that was independent of w.2. We see to 1st order a constant negative slope of w(t) in all 3 cases (drive off, drive on prograde, drive on retrograde), which is the observation that leads to my equations.Still, I have provided an w(t) based on your hypothesis. Do you think you can fit it to the data?

Quote from: deltaMass on 06/11/2015 05:27 PM...We seem to be the only two so far discussing the Baby EmDrive data - thanks for the valuable feedback.If I take your suggestion that "the decelerating torque should be proportional to the square of the angular rate", then we solve by integration to end up with w(t) = w_{0} / (b w_{0} t + 1)where b is a frictional constant of the motion. Two comments:1. The test was done in vacuum, so I don't think this applies. That's why I adopted a friction model that was independent of w.2. We see to 1st order a constant negative slope of w(t) in all 3 cases (drive off, drive on prograde, drive on retrograde), which is the observation that leads to my equations.Still, I have provided an w(t) based on your hypothesis. Do you think you can fit it to the data?I was not participating since so little is known at this point.In this message you state "The test was done in vacuum" yet in prior messages others state that it was not done in vacuum. How do you know it was done in vacuum ? What was the partial vacuum pressure?You state you use a friction model. I thought they had the Baby EM Drive levitated with magnets, so if there is no air, where is the friction coming from in your model ?

...That's not a vacuum chamber? Are you saying that it was not evacuated?

Would be nice to create enclosure from which you'd be able to remove most of the air. Next, initiate spin using external electromagnet pulses. Record the curve of the decay (chart #1). Next, Restart the spinning, this time in the middle of recording (chart #2), turn on the EMD via laser beam (you probably would have to install sensor and indicator for that). Comparison of curve #1 and #2 would be great!Great work!

Thanx. We got a vacuum chamber, but the levitator is too high, so we need to modify it.The thruster can be turned on by radio.

Quote from: deltaMass on 06/11/2015 05:14 PMQuote from: WarpTech on 06/11/2015 04:24 PMI know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations. I dispute that Yang's maths is correct.Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.Prof. Yang did not publish the actual equations she (may have?) used to model thrust. How she can get thrust from the equations she shows in her paper is a mystery...Therefore: no basis to tell whether her equations for thrust are correct or incorrect ...She invites the reader to think about ions inside the cavity and she displays equations including J terms. Then she switches the story saying that there are no ions inside the cavity but that the conclusions still apply...I would have expected the journal referees in her "peer reviewed paper" to further inquire as to where the thrust comes from. The word "evanescent waves" or geometrical attenuation are not found in her paper, to my recollection.

Quote from: Rodal on 06/11/2015 05:26 PMQuote from: deltaMass on 06/11/2015 05:14 PMQuote from: WarpTech on 06/11/2015 04:24 PMI know, but Shawyer's mathematics is definitively wrong. Yang's is definitively correct, using Maxwell's equations. I dispute that Yang's maths is correct.Maxwell predicts zero thrust. Maxwell doesn't predict that group velocity differences contribute to thrust.Prof. Yang did not publish the actual equations she (may have?) used to model thrust. How she can get thrust from the equations she shows in her paper is a mystery...Therefore: no basis to tell whether her equations for thrust are correct or incorrect ...She invites the reader to think about ions inside the cavity and she displays equations including J terms. Then she switches the story saying that there are no ions inside the cavity but that the conclusions still apply...I would have expected the journal referees in her "peer reviewed paper" to further inquire as to where the thrust comes from. The word "evanescent waves" or geometrical attenuation are not found in her paper, to my recollection.In the attached paper Prof. Yang get thrust from Eq. 13, i.e., by integrating the Maxwell's tensor over the thruster's boundary.