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#2120
by
Rodal
on 15 Oct, 2014 22:13
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Those thinking that the results are an experimental artifact should seek artifact explanations that are consistent with the experimental data being linearly dependent on just these three variables.
Yes dr Rodal, working on that.
Better to keep a stiff upper lip, hey !
7 data points is sparse to conduct statistical analysis
I both agree and disagree. Yes, I would prefer to have thousands of data points instead. However, given the fact that this is all the data we have, no statistical analysis of the data would be even worse. Ignoring the data would be worse. Using lots of words and pre-conceived ideas to explain the data would be much worse.
but you do a great job.
Thanks. I wonder what comes next
Though I wonder if it wouldn't be more appropriate to conduct the regressions in log log plane, as the low values dispersions tend to be squashed by the low absolute levels, while their relative dispersion around the (linear) predicted values seem more natural to me : log(experimental/predicted) or equivalently log(experimental)-log(predicted) (then squared as for the least square regression). A linearly scaling formula that predicts 1.0µN for a 1.1µN measure has as much error than predicting 1N instead of actual 1.1N. Maybe this is already the case in your R^2 results ?
In general I don't like log-log regressions because just about everything looks great regressed that way.
You do have a point concerning data squashing however log-log plotting introduces many other problems.
The main problem is the very nonlinear relationship between amplitude and frequency
that has introduced tremendous outliers (particularly in the NASA Eagleworks experiments with extremely small power inputs that also produced extremely small force values). Since the NASA Eagleworks tests were conducted with the smallest Power, the smallest forces, and the smallest differences in diameter, what this does is to exaggerate the outliers.However reluctantly (I favor the linear plots for the aforementioned reasons) I comply with your request in the interest of objectivity and the communal work in this thread
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#2121
by
Rodal
on 15 Oct, 2014 22:19
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dF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).
Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).
P - Power
Q - Quality factor
f - Drive frequency
w_big - diameter of the big end
W_small - diameter of the small end.
My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?
The only way MiHsC enters the picture is because it led Prof. M to the above equation.
As it stands the equation can be written as
dF = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
Does that mean anything helpful?
It is an argument against thermal explanations (stored power instead of loss power)
Anybody seeking explanations on magnetic, artifact or any other explanations should explain the geometrical factor (that came from McCulloch's theory) .
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#2122
by
aero
on 15 Oct, 2014 22:39
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It is an argument against thermal explanations (stored power instead of loss power)
Perhaps, but it is exactly the same equation as always, just arranged differently.
The geometrical factor (that came from McCulloch's theory) needs to be explained by magnetic, artifact or any other explanation
I agree. The terms 1/dia are strange. 1 / big and 1 / small aren't strange at all but why a length? Is that commonly used in radiation calculations / reflections?
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#2123
by
Rodal
on 15 Oct, 2014 22:47
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It is an argument against thermal explanations (stored power instead of loss power)
Perhaps, but it is exactly the same equation as always, just arranged differently.
It is an argument against thermal explanations, because the higher the Q, the less power that it is dissipated (into heat to the walls and elsewhere).
I don't understand <<it is exactly the same equation as always>>. The same equation as what precisely?
Thanks
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#2124
by
aero
on 15 Oct, 2014 22:53
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dF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).
Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).
P - Power
Q - Quality factor
f - Drive frequency
w_big - diameter of the big end
W_small - diameter of the small end.
My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?
The only way MiHsC enters the picture is because it led Prof. M to the above equation.
As it stands the equation can be written as
dF = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
Does that mean anything helpful?
dF = (PQ/f)*((1/w_big)-(1/w_small)) = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
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#2125
by
frobnicat
on 15 Oct, 2014 22:57
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...
Please re-write or withdraw your comment about the Chinese.
Done withdrawal, was silly, but didn't mean anything mean against you or Chinese scientists.
Thank you for the log log plots. Agree that in log log everything looks too good... don't know if it holds for some eminent statistician (which I am not) but I find the log log more convincing as we have very few diversity if the 3 Brady et al entries get squashed. Save the outlier, the other two still look good relative to much lower magnitudes overall.
Still wondering what to do with ranges in the tabulated data...
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#2126
by
Rodal
on 15 Oct, 2014 22:59
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dF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).
Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).
P - Power
Q - Quality factor
f - Drive frequency
w_big - diameter of the big end
W_small - diameter of the small end.
My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?
The only way MiHsC enters the picture is because it led Prof. M to the above equation.
As it stands the equation can be written as
dF = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
Does that mean anything helpful?
dF = (PQ/f)*((1/w_big)-(1/w_small)) = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
OK, but this is telling a huge amount of information.
Why should the stored power give you the force?
It should not if the force is a thermal artifact.
It should not if the force is an artifact due to losses in the medium.
It does leave the door open for the force being due to being an artifact due to resonance.
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#2127
by
Rodal
on 15 Oct, 2014 23:09
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...
Thank you for the log log plots. Agree that in log log everything looks too good... don't know if it holds for some eminent statistician (which I am not) but I find the log log more convincing as we have very few diversity if the 3 Brady et al entries get squashed. Save the outlier, the other two still look good relative to much lower magnitudes overall.
Still wondering what to do with ranges in the tabulated data...
OK thanks
Yes, although I still favor the linear plots for the aforementioned reason, I agree 100% that it is a great idea to also have the LogLog plots. The more ways that one can look at the data the better. Thank you for bringing it up.
I did not use any averages.
I used the data according to my reading of the papers. My reading of the Chinese papers (there are two of them) is that they used the same geometry as Shawyer's larger device.
Concerning the other ranges I think I used the maximum
but I have to re-check, and get back on what was my justification from reading the papers.
Thanks.
===> We still look forward to your producing an artifact or another physical explanation that explains the data analytically <===
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#2128
by
aero
on 15 Oct, 2014 23:40
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dF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).
Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).
P - Power
Q - Quality factor
f - Drive frequency
w_big - diameter of the big end
W_small - diameter of the small end.
My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?
The only way MiHsC enters the picture is because it led Prof. M to the above equation.
As it stands the equation can be written as
dF = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
Does that mean anything helpful?
dF = (PQ/f)*((1/w_big)-(1/w_small)) = [ Stored power/w_big - Stored power/w_small ] / f
where stored power = Q * Power.
OK, but this is telling a huge amount of information.
Why should the stored power give you the force?
It should not if the force is a thermal artifact.
It should not if the force is an artifact due to losses in the medium.
It does leave the door open for the force being due to being an artifact due to resonance.
Also, look at this formulation.
dF = (PQ/c)*((L/w_big)-(L/w_small)) where L was initially named the Unruh wavelength but was then renamed the RF wavelength.
It happens that the RF wavelength is very nearly equal to w_small, the diameter of the small end of the cavity and perhaps the diameter of the dielectric. By my screen measurements, I measure w_small = 6.6 inches with 6.094 =< wavelength <= 6.278 inches for the RF wave.
Would it be to much trouble to run some statistics on different values of L? In particular the 3 dimensions of the cavity, 6.6, 9.0 and 9.9 inches for the Eagleworks tapered cavity. It may be that L is also related to cavity geometry and not drive power. If that is so it would be very good to know in the design of the next thruster.
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#2129
by
aero
on 16 Oct, 2014 00:00
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OK, let's try to narrow this down.
dF = (PQ/c)*((L/w_big)-(L/w_small))
L= 6.6, 9.0 and 9.9 inches
w_small = 6.6 inches
what do you want for w_big ?
Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.
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#2130
by
aero
on 16 Oct, 2014 00:31
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OK, let's try to narrow this down.
dF = (PQ/c)*((L/w_big)-(L/w_small))
L= 6.6, 9.0 and 9.9 inches
w_small = 6.6 inches
what do you want for w_big ?
Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.
OK, I'm using metric units.
So I will use
L=0.16764, 0.2286, 0.25146 m with the same numbers previously used for w_big and w_small
QUESTION: Do you want me to use this L's only for NASA Eagleworks or also for Shawyer and China?
Or do you want to think what values you want for them UK and China?
[Please assume that the Chinese used the same dimensions as the larger Shawyer device, which makes sense from their text and also because Chinese used 1 KW]
These L's are the dimensions of the NASA Eagleworks device. I don't know the dimensions of the Shawyer device. If you do, then use them consistently. That is, small diameter, Cavity length and Large diameter.
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#2131
by
Rodal
on 16 Oct, 2014 00:39
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OK, let's try to narrow this down.
dF = (PQ/c)*((L/w_big)-(L/w_small))
L= 6.6, 9.0 and 9.9 inches
w_small = 6.6 inches
what do you want for w_big ?
Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.
OK, I'm using metric units.
So I will use
L=0.16764, 0.2286, 0.25146 m with the same numbers previously used for w_big and w_small
QUESTION: Do you want me to use this L's only for NASA Eagleworks or also for Shawyer and China?
Or do you want to think what values you want for them UK and China?
[Please assume that the Chinese used the same dimensions as the larger Shawyer device, which makes sense from their text and also because Chinese used 1 KW]
These L's are the dimensions of the NASA Eagleworks device. I don't know the dimensions of the Shawyer device. If you do, then use them consistently. That is, small diameter, Cavity length and Large diameter.
We might as well do the best job we can at the outset rather than re-visit later on.Here is a link to Shawyer's paper:
http://www.emdrive.com/IAC-08-C4-4-7.pdfplease let me know what L dimensions you want to use for the smaller and the larger Shawyer devices
Thanks
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#2132
by
Rodal
on 16 Oct, 2014 01:41
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...
Thank you for the log log plots. Agree that in log log everything looks too good... don't know if it holds for some eminent statistician (which I am not) but I find the log log more convincing as we have very few diversity if the 3 Brady et al entries get squashed. Save the outlier, the other two still look good relative to much lower magnitudes overall.
Still wondering what to do with ranges in the tabulated data...
This paper by Shawyer has 19 test runs:
http://www.emdrive.com/flightprogramme.html
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#2133
by
Rodal
on 16 Oct, 2014 02:18
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Though I wonder if it wouldn't be more appropriate to conduct the regressions in log log plane, as the low values dispersions tend to be squashed by the low absolute levels, while their relative dispersion around the (linear) predicted values seem more natural to me : log(experimental/predicted) or equivalently log(experimental)-log(predicted) (then squared as for the least square regression). A linearly scaling formula that predicts 1.0µN for a 1.1µN measure has as much error than predicting 1N instead of actual 1.1N. Maybe this is already the case in your R^2 results ?
Here are Log Log plots with the Brady 1937 GHz outlier removed. It improves the R^2, foremost for the Q regression. Still the same conclusion regarding the frequency data having a low R^2, even with the outlier removed.
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#2134
by
JohnFornaro
on 16 Oct, 2014 02:19
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In McCulloch's quantised inertia the Unruh waves are allowed only if they fit exactly within the Hubble horizon (or within a local Rindler horizon). For the formula to apply the EM Drive cavity walls must act like a horizon.
For a theory to work, reality "must" be a certain way. Uhhh... that would be a no.
Overdue emphasis on physical explanations rather than mathematical examination of the data is misplaced, particularly with anomalous results of early experimental results with high uncertainty bars.
An objective, cool, mathematical viewpoint (rather than passionate subjective beliefs) is called for.
I get the principle...
McCulloch's simple formula, without any fudge factors, and with a minimum of parameters, does a much better job at predicting the experimental results than anything else presented so far...
But you're trying to fit an equation to a data set, while depending on hypothetical waves and a "skin effect" that is not supported by theory or experiment.
Me no get it, kemosabe. Me primitive man. Me hung out with mixologist earlier.
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#2135
by
ThinkerX
on 16 Oct, 2014 02:48
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Seeking clarity here.
4) So far, all the experimental data variation in the US (NASA Eagleworks, including the statistical outlier), the UK and China can be explained solely in terms of just three variables:
A) (1/DiameterOfSmallBase-1/DiameterOfBigBase)
B) Q (resonance quality factor)
C) Power Input
So...if I am following this correctly, the logical thing for the research teams to do would be to run thousands (?) of tests in which these three points differ somewhat, ideally in a vacuum. Aka...different 'truncated cone sizes', different frequencies (?), degrees of power input, that sort of thing, right? And then a much clearer picture as to what is going on should emerge. Is that a fair assessment?
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#2136
by
aero
on 16 Oct, 2014 02:57
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OK, let's try to narrow this down.
dF = (PQ/c)*((L/w_big)-(L/w_small))
L= 6.6, 9.0 and 9.9 inches
w_small = 6.6 inches
what do you want for w_big ?
Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.
OK, I'm using metric units.
So I will use
L=0.16764, 0.2286, 0.25146 m with the same numbers previously used for w_big and w_small
QUESTION: Do you want me to use this L's only for NASA Eagleworks or also for Shawyer and China?
Or do you want to think what values you want for them UK and China?
[Please assume that the Chinese used the same dimensions as the larger Shawyer device, which makes sense from their text and also because Chinese used 1 KW]
These L's are the dimensions of the NASA Eagleworks device. I don't know the dimensions of the Shawyer device. If you do, then use them consistently. That is, small diameter, Cavity length and Large diameter.
We might as well do the best job we can at the outset rather than re-visit later on.
Here is a link to Shawyer's paper: http://www.emdrive.com/IAC-08-C4-4-7.pdf
please let me know what L dimensions you want to use for the smaller and the larger Shawyer devices
Thanks
I get.
Shawyer Experimental Demonstrator
Dimension - meters meters
w_big, in. 0.1600 0.2800
w_small, in 0.1050 0.0778
height, in 0.1700 0.3811
w_small, external view. 0.1711
The Demonstrator has a constant external small end diameter for about half the length. I speculate that the taper continues internally with the constant diameter section there for whatever reason. That constant diameter section is about 0.1711 meter. Did the Chinese do both sizes?
Edit: I don't think you should use demonstrator numbers. W_big is OK, it's from the text but w_small is probably to small and height to large. I expect the taper stops in the cylindrical section at a height of about 300 mm based on the Experimental and Eagleworks device ratios.. Point is, I can't tell the dimensions of the demonstration device cavity because of the construction. Garbage in, garbage out so just don't use the above Demonstrator numbers I guess.
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#2137
by
Rodal
on 16 Oct, 2014 03:10
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Seeking clarity here.
4) So far, all the experimental data variation in the US (NASA Eagleworks, including the statistical outlier), the UK and China can be explained solely in terms of just three variables:
A) (1/DiameterOfSmallBase-1/DiameterOfBigBase)
B) Q (resonance quality factor)
C) Power Input
So...if I am following this correctly, the logical thing for the research teams to do would be to run thousands (?) of tests in which these three points differ somewhat, ideally in a vacuum. Aka...different 'truncated cone sizes', different frequencies (?), degrees of power input, that sort of thing, right? And then a much clearer picture as to what is going on should emerge. Is that a fair assessment?
No, it looks like I didn't get my point across well at all. On the contrary, data is very well explained in terms of those three variables in McCulloch's equation. No need to run thousands of data points.
Main experimental need is to be able to tune precisely to the (initially unknown) frequency that gives highest amplitude. And then for the device to stay tuned to that frequency and not deviate from it while very high resonance with very low damping is achieved. This was very difficult for researchers to accomplish on a reliable basis. Once that is accomplished the need is then for NASA to make EM Drive that runs at 1000 watts (like in the UK and China) instead of 20 watts to produce much larger forces.
Secondary need is to explain precisely what is the physical effect that is producing the thrust measurements.
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#2138
by
aero
on 16 Oct, 2014 03:28
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While stumbling through Shawyer's papers, I reached the conclusion that his thrust equation, after substituting the parameter names that Prof. M used, is:
T = 2*So * P*Q/c *(RF wavelength/w_big - RF wavelength/w_small) or
T = 2*So *P*Q/f *(1/w_big - 1/w_small)
where So = (1 - (RF wavelength^2)/(w_big*w_small))^-1
In other words, Shawyer's thrust model differs from Prof. M's thrust model by a multiplicative factor of 2*So .
That of course assumes that I interpreted Shawyer's definition of terms correctly.
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#2139
by
ThinkerX
on 16 Oct, 2014 05:07
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Many thanks for your clarification, Doctor Rodal.
Main experimental need is to be able to tune precisely to the (initially unknown) frequency that gives highest amplitude. And then for the device to stay tuned to that frequency and not deviate from it while very high resonance with very low damping is achieved. This was very difficult for researchers to accomplish on a reliable basis. Once that is accomplished the need is then for NASA to make EM Drive that runs at 1000 watts (like in the UK and China) instead of 20 watts to produce much larger forces.
Ah ha...sounds very tricky, especially given the at times erratic results. This is looking like one very sensitive and touchy mechanism. But the potential payoff...wow!
Secondary need is to explain precisely what is the physical effect that is producing the thrust measurements.
If this thread is anything to go by, the theory people will still be arguing this one while the test craft is enroute to the moon or Mars or someplace.
