Quote from: frobnicat on 10/29/2014 09:44 PM.../...So, in the end, for the principal dynamic activity of the balance around A, at 99% we have a simple (under)damped harmonic oscillator no ? What is the force vs speed function of a magnetic damper ?the arm is axis x positive front (thruster) negative back (RF amplifier and magnets). The y axis is orthogonal, going to the right, the z axis is upward.

.../...So, in the end, for the principal dynamic activity of the balance around A, at 99% we have a simple (under)damped harmonic oscillator no ? What is the force vs speed function of a magnetic damper ?

alpha=rotation around x axisbeta=rotation around y axisgamma=rotation around z axisOK let me answer one important question at the outset: where does the coupling come from.The coupling comes from this nasty fact:if you have a force applied at the origin along the x axis, it will be produce a swinging beta rotation around the y axisif you have a force applied at the origin along the y axis, it will be produce a swinging alpha rotation around the x axis

if you have a force applied at the end of the x arm, oriented along the y axis, it will produce a gamma rotation around the z axis, but also (because of the above facts) one has nonlinear coupling:

alphaDot = d alpha /dtalphaDotDot = d^{2} alpha /dt^{2}In the equations of motion for the gamma rotation around the z axis one also gets contributions from a number of terms, the most important ones being the following rates: alphaDot * betaDot and another contribution from beta * alphaDotDotI obtained the 3-dimensional, nonlinearly coupled equations of motion by solving the Lagrangian.Please notice that while the department of Aeronautics and Astronautics at MIT has an inverted pendulum designed at MIT to eliminate this coupling (only linear x and y motions are allowed for the thruster), NASA Eagleworks neglected to eliminate this coupling. The department of Aeronautics and Astronautics at MIT has been a leader in nonlinear dynamics for the last century (starting with the problems of flutter and divergence and self-excited oscillations).

Quote from: Rodal on 10/29/2014 11:08 PMQuote from: frobnicat on 10/29/2014 09:44 PM....From this line of reasonning the DC spurious force would have a direct torque around the z (mode A) and x (mode C) axis. Just the opposite of what you said ....Based on my memory of their setup, I am not following your line of reasoning here. Let me go back to Brady's report and check whether I remember correctly their setup.Figure 3 shows the magnetic damperOne can see a porthole that is on a line perpendicular to the beam attached to the magnetic damperThe portholes are on the sides of the chamberTherefore the magnetic damper is connected to the beam that runs longitudinally along the length of the chamber. Therefore the magnetic damper is connected to a beam that runs along the x axis.If that beam running along the x axis goes through the center of rotation, the force oriented along direction x cannot produce a torque along the z axis, because the direction of the force will go through the center of torsional rotation.A force directed along the x axis produces a swinging rotation "beta" around the y axis, swinging with largest rotary inertia. <------------- X direction Y axis perpendicular to this page <-- / \ beta rotation l

Quote from: frobnicat on 10/29/2014 09:44 PM....From this line of reasonning the DC spurious force would have a direct torque around the z (mode A) and x (mode C) axis. Just the opposite of what you said ....Based on my memory of their setup, I am not following your line of reasoning here. Let me go back to Brady's report and check whether I remember correctly their setup.

....From this line of reasonning the DC spurious force would have a direct torque around the z (mode A) and x (mode C) axis. Just the opposite of what you said ....

Quote from: frobnicat on 10/29/2014 09:44 PMOnly... isn't a DC current supposed to go and come back ? A twisted pair would in principle suffice to neutralize any net imbalance. Only when the cables separate we have a loop with coupling to magnetic field. So, where the cables separate (at the wet contacts box? inside the amplifier?) and at what angle in what plane ?Since they claim to know they have b field coupling, one presumes they waved some Mu metal between or some such. It wouldn't be the first time a twisted pair produced a noticeable force and these are very small forces.

Only... isn't a DC current supposed to go and come back ? A twisted pair would in principle suffice to neutralize any net imbalance. Only when the cables separate we have a loop with coupling to magnetic field. So, where the cables separate (at the wet contacts box? inside the amplifier?) and at what angle in what plane ?

The actual response of Eagleworks does look something like this response to this impulse + exponentially decaying rise forcing function:Piecewise[{{(80*10^(-6))*(1-ssr)+(80*10^(-6))*ssr*(1-Exp[-t/tau])/(1-\Exp[-30/tau]),t<30},{(80*10^(-6))*ssr*(Exp[-(t-30)/tau]),t>= \30}}]ssr=0.4 tau=4 seconds

....

Prior to the TM211 evaluations, COMSOL® analysis indicated that the TE012 was an effective thrust generation mode for the tapered cavity thruster being evaluated, so this mode was explored early in the evaluation process. Figure 22 shows a test run at the TE012 mode with an operating frequency of 1880.4 MHz. The measured quality factor was ~22,000, with a COMSOL prediction of 21,817. The measured power applied to the test article was measured to be 2.6 watts, and the (net) measured thrust was 55.4 micronewtons. With an input power of 2.6 watts, correcting for the quality factor, the predicted thrust is 50 micronewtons. However, since the TE012 mode had numerous other RF modes in very close proximity, it was impractical to repeatedly operate the system in this mode, so the decision was made to evaluate the TM211 modes instead.

.../...And look at that very long exponentially decaying fall after that. What's up with that?How come the researchers failed to comment on this?

27.92916748 cm

Ok, here are my final estimates of Bradly cavity.The method:1. Extract pixel lengths of chords from the photograph on the left, scale lengths from Frazt beams, and including the visible length of the Frazt beam extended in back.2. Extract pixel length of visible Frazt beam extended in back from photo on right, and verify it's scale. Calculate the added length value in right photo to the length in left photo.3. Calculate distance from edge of base plate to Frazt beam in back using geometry from photo on right. It is 52.30234549 pixels.4. Calculate the scale factor at the center of the base plate. It is 34.40258068 px/cm.5. Calculate distance of camera from center of base plate in left hand photo using pinhole projection formula.6. Calculate angular diameter of base plate viewed from camera.7. Calculate angle between radius of base and base tangent line from camera.8. Calculate radius of small and bases plate using above angles and projected chord length. It is factor times the chord length. w-s factor = 1.008039357, w-b factor = 1.025077351Applying these correction factors gives:w-b w-s L246.018292 138.7083271 221 raw chords, pixels252.187779 139.8234528 226.5420945 factored, pixels27.92916748 15.48509864 25.0889719 Scaled to cmHere, w-b factor was used to factor length as they are near in numerical value.

Arguments about who's the better round-offer might be misunderstood.

John: The camera is very close. Its distance can be calculated by measuring that 1 1/2" FZTK stuff in front and back then estimating/measuring the distance difference between them.

Sorry for asking, but -as the discussion is getting very technical- (1) could someone of you make a quick update for the non-physicists among us (like myself)? (2) is there any tangible progress, (3) or has the device been demistified once for all?thanks!

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