-
#2800 by Rodal on 30 Oct, 2014 20:46
-
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
-
#2801 by frobnicat on 30 Oct, 2014 20:48
-
.../...
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.
All right, let's stick with this simple right handed system. Let's say the origin O is where the main axis of rotation of the flexure bearings meets the axis of the faztek moving beam.Quotealpha=rotation around x axis
beta=rotation around y axis
gamma=rotation around z axis
OK 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 axis
if you have a force applied at the origin along the y axis, it will be produce a swinging alpha rotation around the x axis
er, a force applied at the origin has no torque... I see a force along the x axis anywhere but on the y axis has torque for beta rotation, but that could be said also of any vector that don't cross the y axis. Only a force vector within Oxz plane has "pure" beta torque and no torque on the others. Is that what you mean by "applied at the origin" ?Quoteif 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:
to be clear, a force applied at x=end_of_arm y=whatever z=thruster_axis_height along the y axis will have a torque relative to gamma and alpha.QuotealphaDot = d alpha /dt
alphaDotDot = d2 alpha /dt2
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 * alphaDotDot
I 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).
So you have 6 dynamical DOF with alpha beta gamma alphaDot betaDot gammaDot, is that it ?
Wouldn't the "centre" xc yc zc of the beam, initially at O, also have some small but significant displacements and coupling with the rotation ? After all, for rotations alpha and beta to be considered means that the 2 flexure bearing can see their axis shift from their rest position. That would add another 6 dynamical DOF, xc yc zc and xcDot ycDot zcDot. Negligible ? More negligible than the first modes of vibration of the beam (if it is no longer considered a perfect solid) ?
I understand you are on the investigation of those complex dynamics, and those aspects need to be addressed to either discard or include them as sources of spurious measures (or wrong interpretations of readings). I can only encourage such endeavour. But I see that it will need a lot more geometric data to incorporate the y oriented forces (or not perfectly y oriented) with their x y z position to yield relevant results from such a sophisticated model.
At the moment, my personal line of inquiry would be happy with a 2 DOF dynamical model gamma and gammaDot and I'm assuming a simple harmonic (under)damped oscillator for now. As to answer my later question "What is the force vs speed function of a magnetic damper ?" I understand from the oracle that we have a linear Fdamp = -K v like a classical simple viscous damping (on first approximation).
I'm still very interested in your further inquiries around the mechanical system as a whole (with more DOF) and in particular if it could illuminate the presence of a significant "background level" of natural freq oscillations in spite of the damper. For instance figure 19, first picture, first calibration pulse, we expect an exponential decay of ringing amplitude but we see the ringing going on much longer... it's like there is a permanent level of excitation of the system. The experiment is sensitive, so it will have some level of "noise" but it would be interesting to know where this noise comes from, and short of that, could we take that explicitly into account to "subtract" this background ringing contribution, how ?
-
#2802 by frobnicat on 30 Oct, 2014 21:10
-
....
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 damper
One can see a porthole that is on a line perpendicular to the beam attached to the magnetic damper
The portholes are on the sides of the chamber
Therefore 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
Why would that give an x axis force more that a z axis force ? This is a non magnetic plunger in a field that's oriented Ox, force are only with movement, so if main movement is along y axis, main force would be y axis (damping main gamma rotation). If small moves along x, I don't see much force from that since you are moving essentially parallel to field lines. Maybe z move would give more z oriented force (torque around, er, beta also, but with a better "lever")
Anyhow, I was talking about the DC spurious force of 9µN or so of current interaction with B of magnets, saying that a wire that's roughly (from the wet contacts to the amplifier) in Oxz, that a magnetic dipole around the x axis would have field lines parallel to Oxz in this plane, and that the force would be directed along y, hence a torque around gamma and alpha, but not beta. But since it all depends on the exact geometry of the 2 wires and of the B gradient, I guess we are stuck : basically we don't know the place of the vector, just its contribution as torque around gamma. So, sad as it is, subtracting this "DC component" is best we can do save a visit at Eagleworks and dismantling of amplifier.
-
#2803 by frobnicat on 30 Oct, 2014 21:31
-
There is no such thing as passive perfect permanent B field shielding, is there ? The magnets are very strong, they are just below (10cm ?) the amplifier that receive the DC wires. I have no precise idea of the leaked B but the gaps in the magnetic circuit are huge, my guess would be around .1T. On a gradient of .001T per distance between + and - plugs, suffice to have 2mm length of unbalanced wire carrying 5 amps to reach a force of 5*2e-3*1e-3 = 10µN of force ! I would be at the place of the experimenters I would go like "well, just subtract that component".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 ?
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.
-
#2804 by frobnicat on 30 Oct, 2014 21:48
-
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
Okay. This is not exactly what I expected at first, but it looks like there is a very significant part of the magnitude that is claimed as an "EM thrust" that is lagging quite a lot, and that part would be hardly compatible with any EM thrust interpretation (even the most crazy ones). That part needs to be explained (the fast one also !). That can't be the heating of the flexure bearings by the amplifier, this "drifting baseline" has a much longer time constant. Also this smoothing component is (visually) absent or much weaker (relative to pulse magnitude) from the calibrations pulses and from the "magnetic interaction" DC spurious @ 5.6 Amps (figure 20, with Null load) that are known to be rectangle excitations.
please send complaints to [email protected]
-
#2805 by Rodal on 30 Oct, 2014 22:48
-
....
OK before I answer your posts about the degrees of freedom, etc.
Let's deal with this picture:
What do you make of the first the 70 microNewtons impulsive rise followed by a very long exponentially decaying rise to an extra 60 microNewtons ? The exponentially decaying rise is labeled (in red) RF ON
What's up with that?
They claim that the average thrust for this condition is 55 microNewtons.
What is the response in this case? two separate responses of 70 and 60 ? Why on top of each other? Why is one an impulse and the other one a long slow exponentially decaying rise?
Or is the response 70+60=130 microNewtons?
What's going on with that very well defined exponentially decaying rise to another 60 microNewtons?
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? Well, what they wrote is that they had lots of trouble repeating this run (bold added for emphasis):Quote from: Brady, March, White, et.al.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.
Here we have probably the best well defined rectangular impulse calibration performed, and what is it followed by?
By an instantaneous impulse immediately followed by an exponentially decaying rise that is in proportion 6 to 7 or
0.46 to 1.
-
#2806 by aero on 30 Oct, 2014 23:21
-
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 = 1.025077351
Applying these correction factors gives:
w-b w-s L
246.018292 138.7083271 221 raw chords, pixels
252.187779 139.8234528 226.5420945 factored, pixels
27.92916748 15.48509864 25.0889719 Scaled to cm
Here, w-b factor was used to factor length as they are near in numerical value. -
#2807 by aero on 30 Oct, 2014 23:27
-
Referring to this graphic.
-
#2808 by frobnicat on 30 Oct, 2014 23:31
-
.../...
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?
There are already quite a number of failed comments for a publication with claims of this nature (turning a century of physics on its head). Before I go to sleep : too bad we have only one single pulse to see with this "Brady c" data point. On one hand this is only one data point (but with a lot of points in it !) measured not far away to noise level. I'm not sure it's both fair and statistically relevant to dive in this one at this point. Maybe we could refine the tools ( Fb(t) reconstruction ) on the more voluminous "Brady a" before tackling that one ?
On the other hand, well, the shape is relatively clean in its deviation from an expected rectangle response. I would be tempted to say that the descending slope after the RF ON is also an exp decay rather than just the background drift : note how it is roughly "aiming" back to the pre-pulse level (just before the second calibration pulse. Note also how the ringing amplitude, both on rise and fall, is much much lower than ringings of cal. pulses (of similar magnitude). My instinct would tell me that we have a similar "rectangle + exp_rise_decay" than from "Brady a" but with much longer time constants, to the point that the rectangle part is no longer instantaneous steps but also starts to show signs of limited rise/fall rates and lack the "hit" to ring the bell.
If this comparison holds then we should say this pulse yields more than 70+60(-spuriousDC) about 120 µN magnitude.
What is the main difference between Brady a and c ? Power and Q a lot, but the product of both not that much (a factor of 2, while the difference on the graphs seems more than that)
Possible induction that some effect(s) with time constant(s) depend on Power:
Tau1=K1/Power Tau2=K2/Power
-
#2809 by frobnicat on 30 Oct, 2014 23:49
-
Last word for tonight and I end this frobnicathon :
The magnetic damper is likely very well modelled as Fdamping = -k v where v is velocity and k a damping constant.
For details this paper is full of details (case is for a damper with external circuit but can assume the circuit is closed at the eddy current levels)
https://dspace.mit.edu/bitstream/handle/1721.1/33174/65176850.pdf?sequence=1
page 43 equation 3.27 shows a linear response F = - k v + a dot term for F, that is there is some time delay due to induction in the circuit.
page 44 below equation 3.30 we have a tau for this delay that is the ratio L/R of inductance on resistance. For eddy currents in a block of aluminium 10cm square 1cm thick I get very very roughly R=1e-5 Ohm and L=3e-8 henry that gives a tau of 3ms, way below the main natural period of beam : no phase shifting effect between oscillating velocity and damping force.
-
#2810 by JohnFornaro on 31 Oct, 2014 00:04
-
27.92916748 cm
Correct me if I'm wrong, but would the world come to an end if you called it 28 cm?
All those digits to the right of the decimal point are as arbitrary as the scaling factors I derived.
Not trying to be difficult or anything, but absent actual measurements, all of us are making judgement calls as to what we're measuring. Nobody as yet has made a case regarding possible frequency resonances with specific copper frustrum dimensions. Or izzat what you all are doing and I, well, flat out do not get it?
If you all got complaints, our complaint manager is Helen Waite. If you gotta complaint, go to Helen Waite. -
#2811 by Rodal on 31 Oct, 2014 01:03
-
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 = 1.025077351
Applying these correction factors gives:
w-b w-s L
246.018292 138.7083271 221 raw chords, pixels
252.187779 139.8234528 226.5420945 factored, pixels
27.92916748 15.48509864 25.0889719 Scaled to cm
Here, w-b factor was used to factor length as they are near in numerical value.
bigDaero = 27.92916748;
smallDaero = 15.48509864;
lengthaero = 25.0889719;
bigDJohn = 39.7;
smallDJohn = 24.4;
lengthJohn = 33.2;
Ratio of geometrical factor used in Shawyers and McCulloch1 formulas:
((1/smallDaero)-(1/bigDaero))/((1/smallJohn)-(1/bigDJohn)) = 1.82
(aero gives 82% higher results than Fornaro)
Ratio of geometrical factor used in McCulloch2 formula:
((lengthaero/smallDaero)-(lengthaero/bigDaero))/((lengthJohn/smallJohn)-(lengthJohn/bigDJohn)) = 1.38
(aero gives 38% higher results than Fornaro)
-
#2812 by aero on 31 Oct, 2014 01:04
-
The spread sheet gave 27.92916748 15.48509864 25.0889719 but the world would continue to turn at 28 , 15.5, and 25. Of course the measurement depends on the corner points chosen. The sensitivity to a single pixel one way or the other is 0.113516284 cm/pixel or 1.135 mm/pixel and you can round just as easily as I can.
Arguments about who's the better round-offer might be misunderstood. Someone cowboy would take it as a round-up offer, quite a different thing.
-
#2813 by JohnFornaro on 31 Oct, 2014 01:28
-
Well, here's Aero's sketch and mine as well.
The one dimension we know for sure is the 1 1/2" FZTK stuff. We can arbitrarily assume that the focal plane of the camera is the verticle radius of the copper cone. Looking at the image on the left, if the camera is 8 feet away then the "pinhole effect would indicate that the diameter of the cone would be some 26.abc-xyz cm in diameter.
If you don't know where the camera is or where the FZTK is, you don't know the correct angle to project.
There's not quite enough info in the images to get the cone diameter more accurately than around 28 cm, because every time you turn around, you have to make an arbitrary assumption.
I mean, that's how I On the dimensional aspect, people.
-
#2814 by JohnFornaro on 31 Oct, 2014 01:31
-
Arguments about who's the better round-offer might be misunderstood.
Aero: Nobody's saying they're better at rounding off. Take me, for zample. I insist that there's too many unknowns to get a better estimate of those dimensions.
Plus, if in my previous sketch, I divided wjhen I should have multiplied, lemme know. don't be shy. -
#2815 by aero on 31 Oct, 2014 02:07
-
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. And that distance can also be estimated/measured very closely from features in the two images. That distance difference and the pinhole projection formula is all that is needed, knowing the FZTK stuff is 1 1/2". It is a lot of work though and I'm not about to do it again. But it happens that the camera is only about 68 cm from the center axis of the big end base.
You can tell that the camera is close by looking at the diameter of the vacuum chamber which is 30 inches. In the 36 inches from the front to the back, the apparent diameter shrinks markedly. I didn't use that information because knowing the distance of the camera from the chamber door is not helpful. -
#2816 by JohnFornaro on 31 Oct, 2014 11:23
-
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.
Good point. Lemme look at it again over the weekend, after I see my mixologist. -
#2817 by francesco nicoli on 31 Oct, 2014 11:49
-
Sorry for asking, but -as the discussion is getting very technical- could someone of you make a quick update for the non-physicists among us (like myself)? is there any tangible progress, or has the device been demistified once for all?
thanks!
-
#2818 by JohnFornaro on 31 Oct, 2014 12:29
-
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!
1) Second that emotion.
2) No, per the experimentors. Hopefully, per the math whizzes here.
3) No. -
#2819 by JPLeRouzic on 31 Oct, 2014 14:40
-
If you all got complaints, our complaint manager is Helen Waite. If you gotta complaint, go to Helen Waite.
As American-English culture is something that I learn painfully, I am happy to capture here yet another language joke nearly undecipherable for foreigners, thanks :-)
