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#5681 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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Agreed. The correction is of lower magnitude than the error resulting from defining building scaling metric on the Dan Rather footage, so it's a bit pointless.
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#5682 |
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Graduate Poster
Join Date: Feb 2009
Location: Spain
Posts: 1,320
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The aim I had in mind is not to use a 3D trace for graphing, only for estimating distance to the camera over time for perspective correction. I expect that a heavily smoothed version of the 3D trace would suffice for that purpose. Microscale movements have negligible influence in the perspective correction. It would just be an additional refinement to the vertical position data.
Mostly agreed. That's why I've suggested using the camera position for the correction; the distance between the windows can serve as a secondary check, and also to have an initial idea before going into the more complex calculations. |
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Ask questions. Demand answers. But be prepared to accept the answers, or don't ask questions in the first place. |
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#5683 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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Chandler explains what he did and what the software did. You ask for things you know don't exist and do not need to be known in an attempt to claim that NIST and Chandler don't know what they are talking about when they said WTC 7 fell at free fall for ~100'.
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NIST measured to within 1/10th of 1%. The difference is negligible - to small to be considered. As long as you keep using sophistry to deny the fact that WTC 7 fell at free fall acceleration for about 100' and claiming that NIST and Chandler don't know what they are talking about I will continue to post these facts.
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Originally Posted by C7
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![]() I just moved the line up to show how it fits the data before the moment of >g [which is probably a slight movement in the camera several miles away]. ![]() . |
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#5684 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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Yet you are unable to type them. Big surprise.
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...whilst freely admitting, when asked what "programs" they use. that... That's just dumb. Again, I've TOLD you numerous times... So, comprehension test...have a go at answering my questions again.
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...NIST say their estimation of approximate and average acceleration is equivalent to the acceleration of gravity during that period.
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NIST say their estimation of approximate and average acceleration is equivalent to the acceleration of gravity during that period.
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No-one is denying ~FFA for a period of time, indeed I am ADDING a period >FFA. What part of that are you having a problem with ?
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#5685 |
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#4
Join Date: May 2007
Location: West of Northshore MA
Posts: 14,535
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__________________
Join the team, Show us what your machine can do (or just contribute to a good cause)Join the JREF Folders ! Team 13232 "Remember that the goal of conspiracy rhetoric is to bog down the discussion, not to make progress toward a solution" Jay Windley |
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#5686 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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You hang you hat on the qualifiers and ignore the scientific conclusion.
"the north face descended at gravitational acceleration" The difference is so tiny that it is not worth considering. Only a devout denier would try to imply that it is not FFA.
Originally Posted by C7
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In any case, the first 0.7s are at or greater than FFA. i.e. NO RESISTANCE! |
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#5687 |
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Jellied eel and offal fancier
Join Date: Sep 2006
Location: Arcadia
Posts: 9,213
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And C7 will argue forever and a day about sub-pixel levels of accuracy in video analysis while blithely ignoring the absence of the 000's of explosions required to support his weird theory.
"Straining at gnats" is a good expression, and sums him up well. |
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#5688 |
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#4
Join Date: May 2007
Location: West of Northshore MA
Posts: 14,535
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__________________
Join the team, Show us what your machine can do (or just contribute to a good cause)Join the JREF Folders ! Team 13232 "Remember that the goal of conspiracy rhetoric is to bog down the discussion, not to make progress toward a solution" Jay Windley |
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#5689 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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Incorrect.
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"~" = "approximately".
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11 Incorrect. 1 Correct. 3 Middlin' You do not win a prize. Try harder next time. |
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#5690 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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Originally Posted by C7
Originally Posted by C7
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ETC: FFA for ~0.4s, then >g for ~0.3s. then FFA for 1.8s. ![]() You cannot refute this so you refuse to answer directly. We could quibble about 0.2s of >g but that's not the point. Your velocity graph shows FFA or greater. There is NO <g until 14.9s - 85'. Just saying incorrect without saying why is worthless. |
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#5691 |
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Philosopher
Join Date: Dec 2009
Location: Germany
Posts: 9,930
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#5692 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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Incorrect.
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Stick a ~ in front of "FFA" when you write it, and I'll give you a lot less grief. Oh, but make sure you don't state ridiculous incorrect ~FFA periods suggesting my data supports your nonsense, as it doesn't. Being very generous you could say my data suggests ~1.75s of ~FFA. More than that and I'll be on your case.
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Without your problems above, you could have simply improved your understanding a long time ago, and be discussing the correct details, rather than waste my time going round in circles with the REALLY simplest of friggin things imaginable.
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#5693 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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Here's Chandler's "re-done" WTC7 velocity plot...
![]() In your own words, please explain how the **** you get from that remarkably non-linear (aka wiggly) line, which, of course you understand cannot imply a constant acceleration, "FFA" or not, to saying "indistinguishable from freefall" ?
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#5694 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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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. |
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#5695 |
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Illuminator
Join Date: Mar 2009
Location: Moss Vale, NSW, Australia
Posts: 3,656
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#5696 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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The "wiggly line" is not sentient, and assumes nothing.
The line simply connects the actual data-points. The RESULT is a "wiggly line".
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All you achieve is an average, an estimation, an approximate result that is easier to "read". The question you must ask, is how you KNOW that variation is "noise" and not "real" ? The answer, as I know you'll fluff it, is that you don't know. There are ways of "cutting through noise", but a linear fit is FAR from the best or most accurate way to do that.
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My data shows that some of that variation is indeed "real". My acceleration profile is much closer to actual, and reveals detail not presented by either Chandler nor NIST. You should update your position. It's not wildly different from your previous position, as it's still valid to suggest ~1.75s of ~FFA during which the NW corner descends ~83ft. Being more specific than that, or omitting any of the approximate qualifiers is not supported by my data, let alone the lower quality and inferior data from either Chandler or NIST.
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If so, then why are you trying to defend your eronious position ? If not, then why are you trying to defend your eronious position ? |
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#5697 |
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Muse
Join Date: Aug 2011
Posts: 917
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#5698 |
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Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles
Join Date: Jan 2007
Location: In a world lit only by fire.
Posts: 17,907
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Bollocks. Basic high school physics says that, unless you've analysed and understood the measurement error of the individual points, you can't tell what level of detail in the data is a genuine measurement and what part is that inaccuracy. As usual, C7 is starting from a conclusion and reasoning backwards to an unfounded interpretation of the data. If he weren't, he'd be able to state the width of the error bar on each point, which he won't because he can't.
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#5699 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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The software computes the average and draws the line. It is designed to compute the velocity of objects in a video and that's the way it's done.
In the debate, Dave acknowledged "there was a 2 er 1 3/4 second period of FFA of this outer shell". He's probably referring to your graph and he understands that it confirms at least 1.75s of FFA. Your data points are closer to the free fall line than Chandlers or NIST's. Nice work. Since data taken from a video is imprecise, a straight line thru the average of the data points is the logical scientific conclusion. That is the accepted way it is done. Your interpretation that WTC 7 was varying around FFA is wrong. |
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#5700 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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You used the same data for your acceleration and velocity graphs but got completely different results.
If you wish to claim that I am wrong, point to where I am wrong. On your acceleration graph, you have ~0.8s of acceleration from zero to FFA Then ~1 s of >g. Then 0.5s of g or >g ??? Then <g ![]() But on your velocity graph you have ~0.1s to 0.2s of acceleration from zero to FFA Then ~0.4s of FFA Then ~0.3s of >g Then ~1.75s of g Then <g
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#5701 |
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Graduate Poster
Join Date: Feb 2011
Location: SE Michigan
Posts: 1,368
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Give him some slack
Every other one of his pet theories has been slapped around like a red headed step child. His "free fall" issue is the only one with fuzzy enough data to survive. ![]() And of course he will continue to ignore the fact that even with "free fall" it does nothing to advance his absurd claim of explosive devices. |
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#5702 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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It's a best fit, not the average as such. It's still a straight line placed over a wiggly one.
You do not create new accuracy that was not there in the first place by placing a straight line over the top of a wiggly one. You can eliminate some noise using a variety of techniques, but you cannot remove it all. All a best-fit linear regression gives you is an estimated average between two points in time. All manner of real variation can occur during that time.
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Derivation can be performed in a number of ways. The Savitzky-Golay smoothing method I use performs a mathematical differentiation of each curve fit function (which is very handy). There is one curve fit per sample. Hundreds of 'em.
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I have no great issue with folk suggesting ~1.75s of ~FFA during which the NW corner descended ~83ft. Not one "~" is optional.
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You do have a problem with that, but it's YOUR problem. Add the approximation qualifiers and you'll be fine.
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You actually think that a non-spherical object outside of a vacuum will "fall" AT FFA on the surface of this planet ? Did you know an apple dropped from the roof would take >8s to hit the ground (In a vacuum it would be <6s) ![]() You have NO WAY to determine acceleration with any more accuracy than I do. You can estimate. About FFA for a bit. Around FFA for a bit. Roughly FFA for a bit. Some >g |
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#5703 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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#5704 |
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Graduate Poster
Join Date: Mar 2011
Posts: 1,364
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On chrismohr911.com, I have finished all 44 of my first points in the re-re-rebuttal section except for some links. Am open to suggestions on any corrections there. Thanks, Chris
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20 videos rebutting Blueprint for Truth YouTube keyword chrismohr911 http://www.youtube.com/watch?v=jC3JgWkNNIQ Playlists http://www.youtube.com/playlist?list...eature=viewall and http://www.youtube.com/playlist?list...eature=viewall WTC Dust study http://dl.dropbox.com/u/64959841/911...12webHiRes.pdf Hundreds more links and info both sides: http:www.chrismohr911.com |
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#5705 |
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New Blood
Join Date: Jun 2012
Location: Eastern Finland
Posts: 5
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Thanks for the videos, Chris!
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#5706 |
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Illuminator
Join Date: Apr 2011
Location: Not America.
Posts: 4,739
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#5707 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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Originally Posted by C7
On your acceleration graph, you have ~0.8s of acceleration from zero to FFA Then ~1 s of >g. Then 0.5s of g or >g ??? Then <g But on your velocity graph you have ~0.1s to 0.2s of acceleration from zero to FFA Then ~0.4s of FFA Then ~0.3s of >g Then ~1.75s of g Then <g |
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#5708 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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The wiggly line is a figment of your imagination, a deliberate misinterpretation of the data. The data points are not accurate because they are taken from a video and you know it.
Originally Posted by C7
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Originally Posted by C7
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#5709 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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#5710 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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Incorrect.
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![]() ...and anyway, that's not AT, which is what YOU keep saying ![]()
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#5711 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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Interpret?
0.8s is quite different from 0.1 or 0.2s etc. How do you interpret the data? Give numbers please. Say what each part is and how long it lasted as I have done. You have stated that the >g lasted ~1s but you have also said that it was ~0.5s. Indeed, that is close to what your graphs show. You cannot have two different sets of results from one set of data. |
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#5712 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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#5713 |
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Graduate Poster
Join Date: Feb 2011
Location: SE Michigan
Posts: 1,368
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#5714 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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Originally Posted by C7
NIST didn't, Chandler didn't, and neither does anyone outside the denoir choir. Talk to yourself much? "approximately" "estimation" "equivalent" [ignore the conclusion of "at gravitational acceleration" after the qualifiers about the data] Quote mine. The context is clear, and I've told you many times. Your continued use of such simply highlights that either you're too stupid to realize that repeating the words as explicit fact on their own is erroneous, or you do understand and do it deliberately. Either way, shame on you. ![]()
Originally Posted by C7
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#5715 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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Acceleration graph
a. release to somewhat over-g in approximately 1s. b. reduction in acceleration to approximately g over approximately 1.5s. c. reduction in acceleration to [<g] There is NO ~1.75s of ~FFA Velocity graph You don't give values for: a. ? b. ? c. ? Just: ~1.75s at ~FFA Same data - 2 entirely different results. |
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#5716 |
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Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles
Join Date: Jan 2007
Location: In a world lit only by fire.
Posts: 17,907
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I realise that there's little point trying to push back the vast Sahara of ignorance that blankets C7's understanding of this topic, but as usual that statement is such a colossal over-simplification that it's reduced to an absurdity. The purpose of a straight line fit is to determine a rate of change over a period of time based on the assumption that the said rate of change is constant over that period of time. C7's insistence that the straight line fit proves that the acceleration was constant over the time covered by the fit is classic circular reasoning.
Dave |
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"We will punish the murderer together. Our punishment will be more generosity, more tolerance and more democracy." - Fabian Stang, Mayor of Oslo SSKCAS, covert member |
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#5717 |
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Philosopher
Join Date: Aug 2006
Posts: 6,535
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#5718 |
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Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles
Join Date: Jan 2007
Location: In a world lit only by fire.
Posts: 17,907
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Chandler is an idiot, so I care very little what he concludes. I would suggest you post the exact wording of what you imagine to be NIST's conclusion, so that I can explain to you which bit you don't understand. And make sure you get the wording right, because someone's sure to notice if you change it.
Dave |
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"We will punish the murderer together. Our punishment will be more generosity, more tolerance and more democracy." - Fabian Stang, Mayor of Oslo SSKCAS, covert member |
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#5719 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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======
Figure 12–76 presents a plot of the downward displacement data shown as solid circles. A curve fit is also plotted with these data as a solid line. A function of the form z(t) = A{1 – exp[–(t/λ)k]} was selected because it is flexible and well-behaved, and because it satisfies the initial conditions of zero displacement, zero velocity, and zero acceleration. The constants A, λ, and k were determined using least squares fitting. The fitted displacement function was differentiated to estimate the downward velocity as a function of time, shown as a solid curve in Figure 12–77. Velocity data points (solid circles) were also determined from the displacement data using a central difference approximation1. The slope of the velocity curve is approximately constant between about 1.75 s and 4.0 s. To estimate the downward acceleration2 during this stage, a straight line was fit to the open-circled velocity data points using linear regression (shown as a straight line in Figure 12–77). The slope of the straight line, which represents a constant acceleration, was found to be 32.2 ft/s2 (with a coefficient of regression R2 = 0.991), equivalent to the acceleration of gravity g. Note that this line closely matches the velocity curve between about 1.75 s and 4.0 s. For discussion purposes, three stages were defined ====== |
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#5720 |
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Illuminator
Join Date: Apr 2009
Location: UK
Posts: 3,627
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