Quote from: Rodal on 06/11/2015 05:31 PMQuote 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?

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 ?

...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: Prunesquallor on 06/11/2015 11:05 AMQuote 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?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 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.

...So, I can't figure out a way to apply any analytics to the deceleration. What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing. THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope. Then copilot everything normalizing start time to the thruster-on angular velocity. THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.

Quote from: Prunesquallor on 06/11/2015 06:32 PM...So, I can't figure out a way to apply any analytics to the deceleration. What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing. THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope. Then copilot everything normalizing start time to the thruster-on angular velocity. THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube): http://arxiv.org/pdf/1201.6402Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see Eq. 3.4http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf

Quote from: Rodal on 06/11/2015 06:43 PMQuote from: Prunesquallor on 06/11/2015 06:32 PM...So, I can't figure out a way to apply any analytics to the deceleration. What I would like to see is the experimenters run their system unpowered for about a dozen trials and determine the angular rate vs. time envelope that they are seeing. THEN power up the thruster at a consistent angular rate and do a dozen runs with it on and determine THAT envelope. Then copilot everything normalizing start time to the thruster-on angular velocity. THEN we would be able to see if the variability in the powered and unpowered cases are overlapping, or whether there clearly a distinct deceleration time history envelope with the thruster turned on.Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube): http://arxiv.org/pdf/1201.6402Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see Eq. 3.4http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdfPower = Torque * wSo if P = k_{1} w^{3}, thenF = k_{2} w^{2}using constants k.

Quote from: Rodal on 06/11/2015 06:43 PMDisk drag power goes (empirically) like the RPM to the 2.8 power (almost cube): http://arxiv.org/pdf/1201.6402Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see Eq. 3.4http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdfor Eq. 13 here http://www.dtic.mil/dtic/tr/fulltext/u2/a365512.pdfThe "drag power" is typically D x V IIRC. Since D is proportional to V^{2}, that makes sense. I still think the change in angular acceleration is driven by drag (at least, dimensional analysis would say so).

Disk drag power goes (empirically) like the RPM to the 2.8 power (almost cube): http://arxiv.org/pdf/1201.6402Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see Eq. 3.4http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdfor Eq. 13 here http://www.dtic.mil/dtic/tr/fulltext/u2/a365512.pdf

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?

...... 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: Prunesquallor on 06/11/2015 07:05 PMQuote from: Rodal on 06/11/2015 06:43 PMDisk drag power goes (empirically) like the RPM to the 2.8 power (almost cube): http://arxiv.org/pdf/1201.6402Drag power losses on helicopter blades goes like the cube of the RPM (or angular speed), see Eq. 3.4http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdfor Eq. 13 here http://www.dtic.mil/dtic/tr/fulltext/u2/a365512.pdfThe "drag power" is typically D x V IIRC. Since D is proportional to V^{2}, that makes sense. I still think the change in angular acceleration is driven by drag (at least, dimensional analysis would say so).I agree. Perhaps it should be power of 1.8 (2.8 for the power, then 1.8=2.8-1 for the drag force dependence on velocity) but 2 is a better model than assuming a linear coefficient of friction (power of 1).

How does she calculate the energy density? If she calculates the energy densities based on the standing waves solution, then the results (over an integer number of periods) should be zero.

Quote from: Rodal on 06/11/2015 06:17 PMHow does she calculate the energy density? If she calculates the energy densities based on the standing waves solution, then the results (over an integer number of periods) should be zero.How can Integral[E^{2}] = 0 under any circumstances? (for nonzero E)

This means Egan, who showed his work, integrated the Maxwell stress-energy tensor correctly, and it therefore means that Yang, also using the Maxwell stress-energy tensor, but not showing her work, did it wrong.Which is kinda what I said to WarpTech.

Assuming you're right, can you write down your equation for thrust as a function of parameters that can be engineered?

Quote from: deltaMass on 06/11/2015 09:02 PMAssuming you're right, can you write down your equation for thrust as a function of parameters that can be engineered?The equation is already written by Egan and by Yang. What I don't have is the expressions for E and H, that are dependent on the design of the frustum. I can only say, the function will be of the form E_{0}*e^{-t/T}, where T will be a function dependent on the direction of the unit Normal vectors and the shape of the frustum. If you have software that can crunch Hankel functions, we could determine what they are.Todd

Evanescent waves are a wholly natural phenomenon - for example, the skin depth calculation for e/m waves incident upon a conductor shows a negative exponential decay of field strength inside the conductor. So as such, they cannot violate CoM. How then are they supposed to explain the EmDrive's violation of CoM?