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The null force testing indicated that there was an average null force of 9.6 micronewtons present in the as tested configuration. The presence of this null force was a result of the DC power current of 5.6 amps running in the power cable to the RF amplifier from the liquid metal contacts. This current causes the power cable to generate a magnetic field that interacts with the torsion pendulum magnetic damper system. The null test data is also shown in Fig. 20.

Quote from: frobnicat on 10/28/2014 12:49 amEagleworks inverted torsional pendulum response to exponentially decaying forcing functions (force in Newtons) © Rodal 2014 Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{0,t>= 30}}],tau=0,0.5,1,2,3Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{(80*10^(-6))*(Exp[-(t-30)/tau]),t>= 30}}],tau=0,0.5,1,2,3Piecewise[{{(80*10^(-6))*(1-Exp[-t/tau]),t<30},{(80*10^(-6))*(Exp[-(t-30)/tau]),t>= 30}}],tau=2

For what it's worth, I measure the FRP board at 0.060" and the copper cladding at 0.002". (the stuff I have here anyway)

I need a little help here. I'm trying to get an accurate estimate of the Brady cavity dimensions so that frobnicat will have the data he needs to evaluate dimensionally accurate equation formulations for Force. Using the attached left side photo, I have used my screen pixel ruler to extract dimensions as follows. x -pixels y -pixels pixel dist w-b 3 246 246.018292 w-s 14 138 138.7083271 L 221 0 221------------------------------------------top slope -221 57 0.257918552bot slope 213 54 -0.253521127taper 0.511439679Using the right side photo I have sketched what I think the camera sees. That is, it sees a chord of the big and small ends, and foreshortened length. To estimate the degree of foreshortening, I measured the cross section of the Faztek beams circled in red on the photos. The beam near the camera measures 41 pixels on a side, and the far beams measure 25 pixels wide. I measured both diagonals of the near beam end, and both widths of the far beams. Only one measurement differed by 1 pixel.Assuming the center of the base is equidistant from the near and far beams, (Might need to adjust this slightly)I calculate the beam width of the Faztek beam supporting the cavity corresponding to the distance from camera to axial center of the cavity to be 33.125 pixels = 1.5 inches = 3.81 cm, or 0.115018868 cm/pixel. Using this conversion I calculate the chord lengths illustrated in the attached drawing to be w-big chord = 28.29674543 cmw-small chord = 15.95407475 cmand foreshortened length = 25.41916981 cmBut I need some help calculating the actual diameters and real length. It is not a huge factor but it is probably the largest error source remaining in the estimated cavity size.

Quote from: frobnicat on 10/29/2014 01:01 pm...We should also model thisQuote from: Brady, March, White, et.al.The null force testing indicated that there was an average null force of 9.6 micronewtons present in the as tested configuration. The presence of this null force was a result of the DC power current of 5.6 amps running in the power cable to the RF amplifier from the liquid metal contacts. This current causes the power cable to generate a magnetic field that interacts with the torsion pendulum magnetic damper system. The null test data is also shown in Fig. 20.And the awful drift up and down of their "baseline".Frobnicat, do you have any suggestion on how best to model this? In which direction (A) torsional, B) swinging with largest rotary inertia, C) swinging with lowest rotary inertia) is their "9.6 micronewton null force" ?I don't think it could be torsional. It looks (from their arrangement) like it is a B) swinging motion of the Faztek beams that gets measured as a torsional displacement because of the coupling between swinging and torsion.If so, it should be entered into my model, rather as they do (in a sort of clumsy way) by subtracting it from their measured response (they assume linearity).

Well, there is a lot to answer there. But what do you think of just modeling the impulse as a trapezoid ?That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.

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 ?

Quote from: Rodal on 10/29/2014 03:59 pmQuote from: frobnicat on 10/29/2014 01:01 pm...We should also model thisQuote from: Brady, March, White, et.al.The null force testing indicated that there was an average null force of 9.6 micronewtons present in the as tested configuration. The presence of this null force was a result of the DC power current of 5.6 amps running in the power cable to the RF amplifier from the liquid metal contacts. This current causes the power cable to generate a magnetic field that interacts with the torsion pendulum magnetic damper system. The null test data is also shown in Fig. 20.And the awful drift up and down of their "baseline".Frobnicat, do you have any suggestion on how best to model this? In which direction (A) torsional, B) swinging with largest rotary inertia, C) swinging with lowest rotary inertia) is their "9.6 micronewton null force" ?I don't think it could be torsional. It looks (from their arrangement) like it is a B) swinging motion of the Faztek beams that gets measured as a torsional displacement because of the coupling between swinging and torsion.If so, it should be entered into my model, rather as they do (in a sort of clumsy way) by subtracting it from their measured response (they assume linearity).At the moment I do ask a lot of questions and don't answer a lot. Have to say I still don't have a good picture of the torsion pendulum geometry and where gizmos are relative to pivot (save the thruster).Time to get a better understanding : from figure 1, left picture, we have a gantry that's static, along the vertical left "leg" of this gantry there are two flexure bearing with vertical axis, on the right picture we can only see the upper one, as two dark grey bloc and I guess the "spring foils" are hidden in the cylindrical space between them. Is that it ? So essentially we have a vertical pivot on the left of the diamond shaped plate (second picture figure 17) that's fixed on the right of the horizontal rotating arm, so that pivot axis is roughly centered in the middle of the faztek arm. This pivot has a linear restoring torque proportional to angular deviation from rest position. On the right of the upper part of vertical plate that links the arm to the flexure bearings there is a connecting box, this box will move with the arm. The static gantry is on three height adjustable (and vibration isolating ?) platforms so that the axis can be set a good vertical I guess. The oblique horizontal faztek linking the arm to the right leg of the gantry is seen on some pictures, not others; I guess it is used to fasten the rotating arm when mounting things and then removed for measures.Figure 1 picture left : the liquid contacts system is the thing on top of the gantry, on the left, above the flexure bearings (aligned with axis of rotation ?). On figure 6 the upper white board is fixed to the gantry, the white board below (with cut angles) is the part that moves with the arm. It is somehow mechanically and electrically connected to the connecting box below.We see the damping system on the third picture of figure 17 : it is situated at the back of the arm, below the amplifier. When the arm rotates, the "fin" enter and leaves the space between the strong magnets : figure 3 second picture the view is lateral to the arm, the plunging "fin" is fixed to back end of arm, it goes forward backward relative to the view. The permanent magnets are enclosed in a ferromagnetic U trying to close the circuit.It's tempting to see the leaks of this magnetic circuit as a dipole with axis roughly aligned with arm. Hard to tell from the pictures (maybe with an added hour of eye straining ...) where is the cable that goes from the liquid contacts, above the axis, to the RF amplifier, back of the arm. Could be in the same plane, vertical plane defined by the arm. And magnetic field lines would be parallel to this plane (that needs to be checked, the U closing the magnetic circuit is not symmetric relative to that plane). That would give a cross product ILxB force directed orthogonal to that vertical plane above the arm.Ok lets say, 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. And origin at the middle of the arm, at the axis of rotation. I would say the spurious DC force is along the y direction, applied somewhere between the wet contacts above the origin at x=y=0 z=+something and the amplifier at x=-something y=0 z=+not_muchIf you could confirm my x y z link to your A B C :A) torsional : rotation around z axis ?B) swinging with largest rotary inertia : rotation around y axis ?C) swinging with lowest rotary inertia) : rotation around x axis ?The B and C modes would be very stiff (similar stiffness)The A and B modes would have similar moment of inertiaThe flexure bearings would introduce some level of coupling between the 3 angles + 3 displacements, is that where you get the nonlinearities ?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 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 ? Short of those answers, best guess is to take into account only what goes into A mode and discard the chaotic aspects of A B C coupling.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 ?

Quote from: Rodal on 10/29/2014 01:22 pmWell, there is a lot to answer there. But what do you think of just modeling the impulse as a trapezoid ?That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me. From my "more complicated picture" that will take many hours to just utter (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like : ______ / |____| \_______I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.

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 10:01 pmQuote from: Rodal on 10/29/2014 01:22 pmWell, there is a lot to answer there. But what do you think of just modeling the impulse as a trapezoid ?That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me. From my "more complicated picture" that will take many hours to just utter (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like : ______ / |____| \_______I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.f=0 @ t=0f=f1 @ t=t1f=f2 @ t=t2f=f2 @ t=t3f=f3 @ t=t4f=0 @ t=t5Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would likef1 = f2 - f3 therefore f3 = f2 - f1 ?t1 = t4 - t3 therefore t4 = t1 + t3 ?t2 - t1 = t5 - t4 therefore t5 = t2 + t3 ?therefore, need f1, f2, and t1,t2,t3

Ok...time for the rest of you to post the cute videos demonstrating how deep of a hole I dug for myself this time...

Quote from: Rodal on 10/29/2014 11:29 pmQuote from: frobnicat on 10/29/2014 10:01 pmQuote from: Rodal on 10/29/2014 01:22 pmWell, there is a lot to answer there. But what do you think of just modeling the impulse as a trapezoid ?That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me. From my "more complicated picture" that will take many hours to just utter (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like : ______ / |____| \_______I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.f=0 @ t=0f=f1 @ t=t1f=f2 @ t=t2f=f2 @ t=t3f=f3 @ t=t4f=0 @ t=t5Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would likef1 = f2 - f3 therefore f3 = f2 - f1 ?t1 = t4 - t3 therefore t4 = t1 + t3 ?t2 - t1 = t5 - t4 therefore t5 = t2 + t3 ?therefore, need f1, f2, and t1,t2,t3Not much time, a mountain is waiting for my feet :f=0 @ t=0-f=80(1-ssr) @ t=0+ (instantaneous rise)f=80 @ t=tauf=80 @ t=30-f=80ssr @ t=30+ (instantaneous fall)f=0 @ t=30+tauonly two free running parameters : tau for the slow component and ssr = smooth/sharp ratiolet's try ssr = .4 .2 .1together withtau = 0.25 0.5 1 2 4more later about x y z

Quote from: ThinkerX on 10/30/2014 04:04 amOk...time for the rest of you to post the cute videos demonstrating how deep of a hole I dug for myself this time...We aim to please:

Quote from: frobnicat on 10/30/2014 08:29 amQuote from: Rodal on 10/29/2014 11:29 pmQuote from: frobnicat on 10/29/2014 10:01 pmQuote from: Rodal on 10/29/2014 01:22 pmWell, there is a lot to answer there. But what do you think of just modeling the impulse as a trapezoid ?That means: a linear rise from zero at t=0 to f1 at t=t1, then a slower linear rise from f1 to f2 at t2, and then a linear fall from f2 to zero at t3?then we can plot several trapezoids, essentially I agree that the rise to f1 is fast, followed by a slower rise to f2From your writing I think you are seeing actually a more complicated picture, but both of us are patient (unlike others in this forum) so we could try to understand the behavior to this trapezoidal impulse first.I understand you see a piecewise linear function as the default way to introduce more (not too much) parameters to fit the target data. It's ok for me. From my "more complicated picture" that will take many hours to just utter (a chance that you are patient !) I still suggest rectangle plus exp "charge/discharge". If piecewise linear it would rather look like : ______ / |____| \_______I'm a big fan of chaotic swing too, but right now I'm on an agenda with lower Lyapunov exponent.f=0 @ t=0f=f1 @ t=t1f=f2 @ t=t2f=f2 @ t=t3f=f3 @ t=t4f=0 @ t=t5Please give me f1, f2,f3 and t1,t2,t3,t4,t5 you would likef1 = f2 - f3 therefore f3 = f2 - f1 ?t1 = t4 - t3 therefore t4 = t1 + t3 ?t2 - t1 = t5 - t4 therefore t5 = t2 + t3 ?therefore, need f1, f2, and t1,t2,t3Not much time, a mountain is waiting for my feet :f=0 @ t=0-f=80(1-ssr) @ t=0+ (instantaneous rise)f=80 @ t=tauf=80 @ t=30-f=80ssr @ t=30+ (instantaneous fall)f=0 @ t=30+tauonly two free running parameters : tau for the slow component and ssr = smooth/sharp ratiolet's try ssr = .4 .2 .1together withtau = 0.25 0.5 1 2 4more later about x y zResults for Piecewise[{{(80*10^(-6))*(1 - ssr (1 - t/tau)), t <= tau}, {(80*10^(-6)), t < 30}, {(80*10^(-6))*(ssr/tau)*(30 + tau - t), t <= 30 + tau}}]We see the effect of tau and ssr modifying mainly the dynamic magnification factor, as expected.Varying tau at constant ssrssr=0.4 tau = 0, 1,2,3,4,5ssr=0.4 tau = 4Note: there are diminishing returns on this effect for tau >3Varying ssr at constant taussr=0, 0.25, 0.50, 0.75, 1.00 tau = 4ssr=0.5 tau = 4Note: ssr close to 1 effectively "kills" the dynamic magnification factor (the "first ring" in Fornaro vernacular) without much affecting the second ring