Richard Gage Blueprint for Truth Rebuttals on YouTube by Chris Mohr

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Same data - 2 entirely different results.

Incorrect. One more detailed than the other, due to acceleration trend details being more visible and defined on the acceleration graph than is practical to determine from the velocity graph (where differences at lower velocity are very subtle). Again, you are looking at smoothed plots. Use caution. Be conservative. Don't make overreaching conclusions based on poor interpretation.
 
The wiggly line assumes that the data points are accurate. Chandler explains that the data points are not perfectly accurate because they are taken from a grainy video. The deviation from free fall line is that inaccuracy, not a variation in the acceleration.

This is basic high school physics that most high school aged people can understand.

C7, do you have Chandler's data points in tabulated form? As a list of pairs of numerical values "time|drop distance"? This as opposed to "dots" or "lines" drawn in a chart.

Thanks
 
Incorrect. One more detailed than the other
Get serious, they are entirely different.

The acceleration graph has 0.8s of acceleration to FFA.
The velocity graph has ~0.1s of acceleration to FFA, 0.4s of FFA and 0.3s of >g during this same time period.

The acceleration has NO period of FFA [or ~FFA as you claim]
The velocity graph has 1.75s of FFA [or ~FFA as you claim]

These are NOT "details".
 
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C7, do you have Chandler's data points in tabulated form? As a list of pairs of numerical values "time|drop distance"? This as opposed to "dots" or "lines" drawn in a chart.

Thanks
 
C7, do you have Chandler's data points in tabulated form? As a list of pairs of numerical values "time|drop distance"? This as opposed to "dots" or "lines" drawn in a chart.

Thanks

27th June...
NIST - No, though I have digitised their data using a 9000px image of their PDF vector based graph (so the positioning is very accurate)

Chandler - Not fully, though David did upload this during an old discussion...


Some position/time data is there.

There's also...

http://aneta.org/911experiments_com/WTC7/index.htm

...but I think that's digitised, rather than original. Shouldn't make too much difference if it is though, as the data is very sparse anyway.

Oh, and...
73064793.png

...which is a bit better than the screenshots both in clarity and image resolution.
 
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femr2,

all this digitizing of plotted graphs introduces more error on top of the original measurment error inherent in Chandler's raw data.

That's why I asked Christopher7:


C7, do you have Chandler's data points in tabulated form? As a list of pairs of numerical values "time|drop distance"? This as opposed to "dots" or "lines" drawn in a chart.

Thanks
 
femr2,

all this digitizing of plotted graphs introduces more error on top of the original measurment error inherent in Chandler's raw data.
I know.

That's why I asked Christopher7:
Which obviously he doesn't have. Even I don't have the data, despite asking David directly. I managed to get hold of Tony's WTC1 data (generated by David), and the actual Tracker data file. It is also posted within this forum in XML form (as an example of tracker's capabilities)...
http://www.internationalskeptics.com/forums/showpost.php?p=6096940&postcount=120

And in plain-text...
http://www.internationalskeptics.com/forums/showpost.php?p=6096989&postcount=124

(I have the actual video used as well somewhere)

I assume you noticed that the bottom right of the tracker image I've provided you with a few times now does have some of his raw data in it...between 5.6s and 8.4s of his plot, for WTC7.

That's all that is "out there" as far as I'm aware. I don't see the point in repeatedly asking C7 for data you know he doesn't have. He's very unlikely to give you a "no".
 
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I know.

Which obviously he doesn't have.
...
I don't see the point in repeatedly asking C7 for data you know he doesn't have. He's very unlikely to give you a "no".
Oh I simply enjoying showing off how C7 is just like every other truther in that he evades easy questions that have simple answers if those answers may make them look like their argument isn't that strong after all.

The point I am going after is of course: If we don't have the data, how can we numerically assess the quality of the data ("margin of error"...)?

I actually thought initially that C7 might have the raw data somewhere. His reoeated ignoring of my question implies, in trutherese, the answer "no" (same as Tony Sz.)

Even I don't have the data, despite asking David directly. I managed to get hold of Tony's WTC1 data, and the actual Tracker data file. It is also posted within this forum in XML form (as an example of tracker's capabilities)...
http://www.internationalskeptics.com/forums/showpost.php?p=6096940&postcount=120

And in plain-text...
http://www.internationalskeptics.com/forums/showpost.php?p=6096989&postcount=124

(I have the actual video used as well somewhere)
Thanks. May come in handy later.

I assume you noticed that the bottom right of the tracker image I've provided you with a few times now does have some of his raw data in it...between 5.6s and 8.4s of his plot, for WTC7.

That's all that is "out there" as far as I'm aware.
Yes, I noticed, and the values have some, let's say, "interesting" properties that may allow us to estimate the size of at least one error inherent in that data.

I want to walk through this together with C7 with the intention of estimating a margin of error for each data point in Chandler's WTC7 track, and derive a measure of uncertainty from that for the average velocity and acceleration over some time intervals.



By the way: I may need a little math help further down the road: If you have three data points x1|y1, x2|y2, x3|y3, I am fairly sure that you can fit exactly one parabola of the form y(x) = a + bx + cx2, where a corresponds to an initial distance, b to an initial velocity and c to a constant (per assumption) acceleration. Is there an algorithm that gives us a, b and c, or s0, v0 and a?
 
If you have three data points x1|y1, x2|y2, x3|y3, I am fairly sure that you can fit exactly one parabola of the form y(x) = a + bx + cx2, where a corresponds to an initial distance, b to an initial velocity and c to a constant (per assumption) acceleration. Is there an algorithm that gives us a, b and c, or s0, v0 and a?
The simplest method that comes to mind is to graph the three points in Excel and add a polynomial trendline of order 3, making sure you switch on the display of the equation and R2 value.

For instance:

1 10.258
2.01 15
3 25.332

Excel provides...

y = 2.8707x2 - 3.9456x + 11.333
R2 = 1
 
By the way: I may need a little math help further down the road: If you have three data points x1|y1, x2|y2, x3|y3, I am fairly sure that you can fit exactly one parabola of the form y(x) = a + bx + cx2, where a corresponds to an initial distance, b to an initial velocity and c to a constant (per assumption) acceleration. Is there an algorithm that gives us a, b and c, or s0, v0 and a?
Substituting the x and y values of your three data points into the general quadratic equation gives you three linear equations in a, b, and c. Solving n linear equations in n unknowns is equivalent to solving Au=v where A is an n by n matrix, u is the vector of unknowns, and v is a known vector. The solution is given by u=A-1v where A-1 is the inverse of A, so the problem is nearly equivalent to matrix inversion.

This is the first problem that would be discussed in a typical sophomore-level course on linear algebra. Standard algorithms include Gauss-Jordan eliminationWP and LU decompositionWP.

If the matrix A is singular (meaning its determinant is zero), then there may be infinite solutions or none at all. If the matrix A is almost singular, then the problem is ill-conditioned, which means numerical solutions may contain significant roundoff error. That shouldn't be a problem when the data points come from a short time series.
 
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...

By the way: I may need a little math help further down the road: If you have three data points x1|y1, x2|y2, x3|y3, I am fairly sure that you can fit exactly one parabola of the form y(x) = a + bx + cx2, where a corresponds to an initial distance, b to an initial velocity and c to a constant (per assumption) acceleration. Is there an algorithm that gives us a, b and c, or s0, v0 and a?

I used the Levenberg-Marquardt algorithm a few years ago, tricky but functional. The Excel method above certainly looks easier.
 
Thanks, femr2, for the Excel trick - I'll figure out whether LibreOffice can do the same ^^

WDC: I seem to remember those words from waaaaaaaaaay back when - I was a straight-A+ student in math from first grade to university (business school), but all that was over 20 years ago, never practiced most of that math since :o

DaveThomas, I don't want to do a curve fit, I want to do an exact fit of three parameters for 3 data points. So yes, the Excel method seems to be the one to try. (I can imagine that there are Excel functions that do the kind of matrix operations that WDC talks about)

The idea is about the following: David Chandler (NIST and femr2, too) has a set of measured data points t(1-n)|s(1-n), with s being the drop distance in ft or m. One could pick any three of these points, assume constant acceleration, and compute s0, v0 and aavg for that triplet. For example, if I picked the following data points:
t (s)|s (m)
1,2|-28,2384
1,8|-44,1225
2,4|-63,5364
then the algorithm should give me
s0 = -7,0596 m
v0 = -11,766 m/s
aavg = -9,805 m/s2That would mean an acceleration of almost exactly g (NYC).
(I know this because I computed the s-values with these parameters :p)

Now the question I would ask of C7 is: How uncertain are the s-values? Expressed as a +/- deviation; this could be derived from pixel resolution and other considerations. Or perhaps a guess. C7 should have a rough idea of what this uncertainty is, at a minimum (I am sure the pixel resolution error can be determined from the data itself, as there distances seem to be spaced at whole multiples of some fraction of a meter). If, for example, C7 thinks that the distances are measured with a margin of error of +/- 0.2 m, then each of the 3 data points could deviate by as much, and that would change the computed average acceleration. In the case of monotonously negative velocity, I would get the maximum (negative) acceleration amax from

t (s)|s (m)
1,2|-28,2384 - 0.2
1,8|-44,1225 + 0.2
2,4|-63,5364 - 0.2

and the minimum (negative) acceleration amin from

t (s)|s (m)
1,2|-28,2384 + 0.2
1,8|-44,1225 - 0.2
2,4|-63,5364 + 0.2

If I can find any three data points within Chandler's data set where the interval [amax,amin] does not include -g, and both exceed g, then that would mean that, even according to C7's assumptions of what Chandler's margin of error may be, his data proves a period where the average acceleration was >g.



Several disclaimers in order:
- Sorry if I am confusing people with mins and maxs of negative values, I am imprecise in my wording, but I hope you get my drift
- I am assuming that the t-values are super precise - at a frame rate of 29.97 Hz (or what is it), t-values such as "1.2 s" are of course imprecise, too; but evidently Chandler didn't care
- The whole excercise will be rated a success if C7 admits that he doesn't have Chandler's data and/or admits he has not slightest clue what margin of error Chandler's data has; both would render any claims about what is within or outside the margin of error invalid.
- I could do the same excercise with femr2's data. I am sure that it has been debated elsewhere what margin of error is inherent in his method, and that such estimates have been derived by someone somewhere. But that would be going too far at this moment. My goal really is to have C7 commit to certain quantitative claims about the quality of Chandler's data.
 
C7, do you have Chandler's data points in tabulated form? As a list of pairs of numerical values "time|drop distance"? This as opposed to "dots" or "lines" drawn in a chart.

Thanks
No.
Your question is a diversion to bury the fact that FEMR's graphs disagree with each other - as are the rest of the posts on this page.

FEMR,
Address these discrepancies please.

The acceleration graph has 0.8s of acceleration to FFA.
The velocity graph has ~0.1s of acceleration to FFA, 0.4s of FFA and 0.3s of >g during this same time period.

The acceleration graph has NO period of FFA [or ~FFA as you claim]
The velocity graph has 1.75s of FFA [or ~FFA as you claim]
 
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He has not released it. There's no legitimate reason why he should not do so.

Your question is a diversion to bury the fact that FEMR's graphs disagree with each other - as are the rest of the posts on this page.
Again, one is derived mathematically from the other, utilising the savitzky-golay smoothing method.

By definition they "agree".

FEMR,
Address these discrepancies please.
They are not discrepencies beyond your misinterpretation of what you're seeing.

But, hell, I'll make it clearer for you, though you really should have listened earlier. Hand-holding for you is getting more than a little tedious.

The velocity graph has ~0.1s of acceleration to FFA, 0.4s of FFA and 0.3s of >g during this same time period.
Why don't you look a little closer...


Red - Velocity (Scale on LHS ft/s)
Cyan - Acceleration (Scale on RHS ft/s^2)
Black - NIST Linear Fit (Scale on LHS ft/s)

Your interpretation is nonsense.

Note the correlation between velocity behaviour and acceleration behaviour.

Remember that you're looking at smoothed data, so you must understand that values at each point are determined by curve fitting the points around it. A 20 sample window for each curve fit iirc.
 
Why don't you look a little closer...
http://femr2.ucoz.com/_ph/7/2/886374797.jpg

Red - Velocity (Scale on LHS ft/s)
Cyan - Acceleration (Scale on RHS ft/s^2)
Black - NIST Linear Fit (Scale on LHS ft/s)
Thanks for the composite.

Times rounded to nearest tenth of a second:
Your acceleration line shows acceleration going from release to g between 12.0 and 12.8s while the velocity line shows the release is ~12.3s and g is attained ~12.4s.

The building is not moving up then down, then up again between 11.5 and 12.3s, that is noise. There cannot be less than zero acceleration so the first second of the acceleration graph is also noise.

You acceleration graph has the roofline accelerating downward for 0.4s before your velocity graph shows it starting down.

Between 12.4 and 12.7 your acceleration graph shows less that g while your velocity graph shows g. Starting at 12.8s, both graphs show >g.

If you dispute this interpretation, be specific as to the times of transition as I have done.


You did not include the part where the velocity graph shows g for ~1.8s and the acceleration graph shows <g. Would you show that part please?
 
...
Again, one is derived mathematically from the other, utilising the savitzky-golay smoothing method.

By definition they "agree".
...
Why don't you look a little closer...


Red - Velocity (Scale on LHS ft/s)
Cyan - Acceleration (Scale on RHS ft/s^2)
Black - NIST Linear Fit (Scale on LHS ft/s)

Your interpretation is nonsense.

Note the correlation between velocity behaviour and acceleration behaviour.

Remember that you're looking at smoothed data, so you must understand that values at each point are determined by curve fitting the points around it. A 20 sample window for each curve fit iirc.

I must agree here with C7 that the correlation isn't really obvious.

For example:

  • Between 12.56 s and 12.59 s, the red curve is flat. That would indicate acceleration = 0. So obviously, the cyan curve in that area is not a mathematical derivation of the slope of the red curve in that interval.

At ~30 frames per seconds, each fram represents ~0.033 s, right? So if you smooth over 20 samples, and if each sample is a frame, we are talking about 0.667 s.

  • Between 12.59 s and 13.31 s, velocity changes by ~30.00 ft/s, that's an average acceleration of -41.81 ft/s2. Your cyan curve never drops below -40.*)
  • No later than ~12.62 s, the red curve gets steeper than the black curve, and it mostly stays above g until ~13.24 s, but the cyan curve drops below -g only at ~12.76 s, that's at least 0,14 s too late.
  • From ~13.24 s to 13.5 s, the red line is flatter than the black line, but the cyan line stays below g, and even goes down during the final 0,08 s.

It seems the cyan curve lags behind by at least 0.15 seconds.
And while the red curve definitely exhibits a major deviation above g for a significant period of time, your smoothing whatever blurs that, and makes the cyan line almost useless, in my opinion.




*) Here's my work:
In your graphic, 1 s = 432 pixels on the x-axis from 12.5 to 13.5 s, and 1 ft/s = 9.4 pixels on the left y-axis from 0 ft/s to -30 ft/s.
The red line, at 543x174 pixels, is at 5.00 ft/s; at 853x456 pixels it crosses the -35 ft/s mark. That's a delta-x of 310 pixels = 0,7176 s (310/432, from 12.59 s to 13.31 s), and a delta-v of -30 ft/s.
Average acceleration is -30/0,7176 ft/s2 = -41.81 ft/s2.
This interval is longer than 20 frames at 29.97 Hz. If the average is below -41 for such a long interval, instantaneous must be below that value somewhere.
 
Thanks for the composite.
You're not particularly welcome, as I see it's done nothing to abate your uninformed interpretations.

As I said, be cautious, conservative.

Have you bothered to compare velocity and acceleration behaviour to displacement behaviour ?

Do you not think it might be an idea to base your opinion on where T0 may be by use of all three of displacement, velocity and acceleration graphs ?

Do you not think that (or realise that) once unrecoverable downward motion is underway, you're already past T0 ?

Do you understand what I'm talking about here ?

What the hell. Here's all three on one graph :)



Your acceleration line shows acceleration going from release to g between 12.0 and 12.8s while the velocity line shows the release is ~12.3s and g is attained ~12.4s.
Nope. Release is <12s.

If you dispute this interpretation, be specific as to the times of transition as I have done.
Your interpretation is, how do I say it nicely, a tad flawed.

I think your problem is that you see things as "literal fact", just like you've done with NIST placing the words "acceleration of gravity" on a piece of paper. You are incapable of taking into account all the other words around them, choosing to only see what you want to. Blinkered. Deliberately.

The graphs do not show instantaneous velocity or acceleration. They reveal the trend beneath the unavoidable noise. The data is good. Best out there. I'm confident that magnitude is pretty accurate also.

The acceleration data is derived from the velocity data in the same way the velocity data is derived from the displacement data. There is no nefarious tweaking going on. You can do it yourself if you like. My raw data is freely available.

You did not include the part where the velocity graph shows g for ~1.8s and the acceleration graph shows <g.
lol. How the bleedin'ell are you going to see detail over a short timespan by including a longer timespan. Plank.

Would you show that part please?
Maybe. I imagine it will simply give you more scope to make "poor" interpretation. I'd rather shake my head in your direction until you improve your understanding of "part 1". Perhaps after that you can look at "part 2".
 
You're not particularly welcome, as I see it's done nothing to abate your uninformed interpretations.

If you wouldn't mind, I can actually see what he's looking at here:

The building is not moving up then down, then up again between 11.5 and 12.3s, that is noise. There cannot be less than zero acceleration so the first second of the acceleration graph is also noise.

I'm thinking the building (point you chose to track) moved back and appeared to move up? (Yes, I could be totally misunderstanding what this means).

If you covered this before a simple link will do.

:)
 
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