The 9/11 Forum

In lieu of a home for discussions of initiation, I will throw one thing in here.

I am not a structural engineer by a long shot. I understand a lot more than I did before falling in with this crowd, but it is only a thimble-full of knowledge. I am puzzled by the lack of resistance in the collapse continuation but much more so by the abrupt onset of global descent. I've not reviewed the NIST documents regarding the period of time between penthouse collapse and onset, and perhaps some answers lie there. Still, I have some theoretical hurdles to overcome before I can understand how such a thing is possible.

While collapse continuation can be modeled in 1D with the generalized coordinate in the vertical dimension, the initiation can also be modeled in 1D - but with the coordinate horizontal. The issue is lateral progression of instability. The question is, how did it get so far so fast? If the roofline kink is assumed to be downward displacement - and I'm becoming less confident that it is so exclusively - then the failure spread across two-thirds the width in about a second. This makes sense if an over-capacity load is suddenly added at the top but not so much from the perspective of removing strength at the bottom. While not necessarily simultaneous across the span, the velocity of propagation is startling to me in any case.

Dr. G had posted info about progression of instability in CDs over at physorg, I link to it early in this thread but physorg has been down for some time and I can't immediately refer to the post. As I recall, even with a CD, it takes some time for the failure to propagate across the structure and commence global collapse. Not very much time, maybe, but these are buildings that are pre-weakened and conceivably have dozens of columns cut or kicked simultaneously. The Landmark CD, 41 stories, didn't drop this fast. How does a localized point of failure become a global failure in such a short time in such an extensive structure?

I try to imagine a cascading failure, like a zipper opening up. And it seems to me brittle materials may propagate very quickly but ductility ought to slow things down. The stiffer it is, the quicker the propagation, but the less likely to fail in the first place. If a beam can transmit downward deflection laterally so as buckle large numbers of columns below, then it's strong enough not to have failed in the first place, being that it so effectively redistributes load to columns which could hold the same weight in the previous second.

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This is getting longer than I wanted it to without saying much of anything. Let me wrap it up with a hypothetical scenario which perhaps someone can explain to me, to help with my fundamentals. Load redistribution, very basic stuff.

Imagine a slab or beam supported by identical columns evenly spaced underneath. The columns and beam are geometrically ideal and perfectly rigid so do not deform in any way from their ideal shape. The figure labeled #1 below depicts such an arrangement with seven columns supporting the beam.



How much load does each column experience in #1? The OBVIOUS answer is one-seventh the mg of the slab/beam, the total load being distributed equally amongst the columns. Now, magically remove the left three columns without disturbing the system, shown in #2. Since the objects are perfectly rigid, the redistribution of load in the now eccentric arrangement would be instantaneous, but what would that distribution be?

If the columns could deform, then the center column (now leftmost) would deflect downward somewhat from the load which has a cm directly above and would promote a tilt of the beam, which in turn would transfer even more of the load in that direction. But, if the columns can't deflect? In this scenario, the beam is simply resting on the columns and is free to move but won't because it is adequately supported in balance by unyielding columns. Intuition is not such a good guide. My intuition says the three columns on the right are experiencing some load, after all, they're in contact. Perhaps the rightmost sees none, but surely the other two do. I think my intuition comes from real material that deforms, but is not applicable here.

Remove the right three columns and there is only the center column remaining, as in #3. Now I'm back in my comfort zone, it's clear that all the weight is on one column. The beam is balanced so still doesn't move. So, somehow, without displacement, the load has been redistributed from all to one. Since the beam is balanced in #3, I'm led to conclude that, without deformation, all the load is on the center column in #2 as well, and three columns experience no load.

So, then: if columns are removed from the left one at a time, does load come off columns on the right one at a time?

My lay concluson from all of this is that it's meaningless to talk about load distribution without deformation, that idealizations which do not include deflection are of no value in modeling propagation of yield location.

Hi OWE,

Great problem. Your model is theoretical which means that the centre of mass is exactly in the middle of the middle column. You could indeed balance the whole building in a centre column and if it is perfectly in balance I would say you could place a very thin column at the right for example and that shouldn’t feel anything. It’s like you place a symmetrical heavy weight on a table, it almost tips over. Assume it is exactly in the middle then the edge of the table is providing an equal opposite force. The more you move it to the table the more the rest of the table will provide, I’m absolutely not sure (I think the last classical mechanics I took was in 1993…) but expect it is a linear function which becomes a singularity in the theoretical model.

But the problem becomes more important interesting in your case of 7 columns from which you remove one.

One thing is sure, since it doesn’t move, we have

F1+F2+F3+F4+F5+F6-mg=0

Otherwise it would accelerate in the vertical directory. Force vectors can be shifted , but there is something to add! And you know that from a seesaw, if a person is a ratio x heavier than the other person the whole thing is balanced if he sits more in the middle than the lighter person, it is exactly the same ratio.

And that is also the thing that should be added to this problem, Torque! It doesn’t get rotation velocity therefore that should also add up to zero. For a seesaw, where there is a fixed hinge, it is trivial, i.e. F1*h1=F2*h2, but for this problem, where is the hinge ?? I expect it will lead to 6 equations with 6 unknowns or something like that, maybe it is even easier.

Without calculations and further thinking I would say your problem is identical to a book that is on a table and you slowly shift its centre of mass to the edge.

This means that if you remove the first colum the second one provides the most force and it decreases with the third and so on, this will always be the case until the middle column aligns with the centre of mass, in that case it provides the full force. It is like you place your books centre of mass in steps to the edge of the table. but I could be wrong.

Yes, einsteen, I believe you've got it! Embarassingly simple when you put it in terms of sum of forces and torques equal zero! What gets me about this problem is:

- the idealization of contact, that the surfaces meet at a point... mathematically, do they share the same point, or does one surface, by definition, have dibs on a specified point, and the other surface only comes infinitesimally close to that point? If they occupy the same point, doesn't that violate the exclusion principle (excluding bosons?)? If they don't share the location, how does one specify the location of the 'other' surface?
- this is simply a problem in statics... what prevents possible solutions of say, only the center or outer columns providing the upward directed force? By what principle do we assume equipartition of load in #1, since there are an infinite number of solutions that will lead to zero net motion?
- if one of the end columns is raised infinitesimally, the load instantaneously becomes distributed between the end columns only in this perfectly rigid arrangement. Nasty singularities all over this perfectly stable solution!
- if both the columns and beam are allowed to deform, things become much harder analytically, but the discontinuities go away

Hope I didn't look too stupid in the last couple of posts. I really don't have that big a problem with abstractions (and I was sort of joking about the exclusion principle). The point was to show how meaningless or trivial excessive idealization can become. It's easy enough to imagine all sorts of arrangements - think of a camshaft under the columns - in which the dynamic redistribution of load is arbitrarily complex and frequent, all occurring instantaneously without any motion or work done. Sure, it's easy to show how such an ideal system behaves, but the system has no relevance whatsoever to anything real.

Physics engines don't like certain idealizations, either. Configure a system such as I've described in a physics engine, start the simulation, and the first thing that happens: the pieces blow apart, scattering at high velocity. Why? Because the solid bodies overlap at a point value, and only there. Without some adjustments and/or cheats, the solver blows up. To make it work (i.e., behave like a real physical system), the solids must have a small but finite skin width specifying the region in which objects may interpenetrate. As objects contact or collide, their surfaces are allowed to overlap up to this depth, with a repulsive force applied in accordance with the depth of penetration. What's more, the objects must be initially positioned outside this region, allowed to come into contact over time, and settle; otherwise they'll start with false kinetic energy derived from the artificial repulsive potential. In other words, the beam must be gently placed on top of the columns, not dropped onto them, neither invading their space like a bad Star Trek transporter accident.

Thus collisions are not instantaneous, regions are fuzzy, and singularities better avoided (though not entirely of course). The notion of even something as simple as contact is quite complicated in physics engines, much like the real world.

Sure I still look stupid.

What difference does a better perspective correction make in the CBS video? Turns out, not much. I searched through Dr. G's posts until I found where he did the trig to show consistency between:

- 186m total height
- 6m parapet height
- 3.96m floor height
- 630m ground distance to camera
- ~17- 18 floors visible

I didn't find the reference to a 38 degree camera angle that I thought I remembered. But the above figures indicate an initial NW roofline elevation angle of about 16 degrees and small traversal of view angle over the displacement, so intuition says a scale approximation assuming delta_height = constant x delta_angle should be pretty good. And it is.

By having not only the distance between two points, but also their actual elevation, a better perspective correction than simple linear scaling can be done. Based on the above numbers, such a transformation for the CBS video at the location of the upper NW corner is:

y = 630 * tan(0.287 - 0.0002588 * delta_P(t))

where y is the elevation in meters of the tracked location and delta_P(t) is the set of acquired data points in pixel units. By contrast, the simplified linear scaling that came out of my last verification of scale correctness is:

y = 186 - (0.174 * delta_P(t))

They don't look anything alike mathematically, but the resulting values are very close over the entire range of data. So close that when plotted together over the whole range, they almost appear to lay on top of each other. Green is better and red is simplified scaling:


http://tinypic.com/view.php?pic=2cqi55h&s=4

This shows the simplified correction undershooting the better value over most of the range, but only by a tiny bit, after which there is a crossover. Remember these are elevation values and not deltas, so the red line being above means it underestimates the drop. The plot of the difference, or residual error, is more revealing:


http://tinypic.com/view.php?pic=i5677l&s=4

Here we see the undershoot then overshoot. The peak magnitude is so small it's tempting to say it shouldn't matter at all. But I can't predict how this would affect a specific process like a quadratic fit, so I did a sample analysis on generated data contrived to appear as a function y = c*t^2, a simple parabola passing through the origin, when using the simplified scaling. Doing a quadratic fit on the simplified scaling returned the expected coefficients, a0 ~ 0, a1 ~ 0, a2 = c. However, when the better scaling was applied and subjected to the same fit, the result was an a2 value 1.7% higher than c. No surprises, proper direction and small magnitude.

Less than 2% is pretty insignificant, and piecewise fits will only reduce the error. Conclusion: the camera position and view angle in this video are such that a simplified correction is probably sufficient for most circumstances. This sort of check needs to be done for each video where possible, though, because the effects will only become more pronounced for closer camera positions. And I will eventually use the better correction with more carefully obtained values just to keep the error as small as possible.

OneWhiteEye:

Nice post!

And fear not, ....... you did not hallucinate a 38 degree camera angle!

It was a number I gave in a post on this thread on Oct 29th:

Quote: "The Camera 3 (CBS) footage was taken from a location that was about 38 degrees off perpendicilar to the north face."

I based this estimate on the useful map of camera locations given in the NIST Draft Report on page 263 of NCSTAR 1-9.

The map also shows why Camera 6 was shooting almost directly into the sun which was low and in the west southwest at ~ 5:20 pm on Sept 11th, 2001.

We also need to mark the location of the newly released WTC 7 collapse video which I would place near River Terrace and North End Avenue on NIST's map.

Dr. G wrote:Nice post!

Thank you. That one's been screaming to get out for a while.

And fear not, ....... you did not hallucinate a 38 degree camera angle!

It was a number I gave in a post on this thread on Oct 29th:

Quote: "The Camera 3 (CBS) footage was taken from a location that was about 38 degrees off perpendicilar to the north face."

OK, it all makes sense now; I'm not crazy! Thanks for digging up that info. I need to check out that map. I don't suppose they have a camera map for the WTC1 and 2 videos they used, too?

We also need to mark the location of the newly released WTC 7 collapse video which I would place near River Terrace and North End Avenue on NIST's map.

Yes, as the realtors say, "location, location, location." You've got a pretty good feel for the (3D) layout in the area. As if there weren't enough to do, are you amenable to working on assembling relevant spatial data - building dimensions, foorprints, camera locations? Someday, not any time soon probably. I've got the beginnings of a 3D representation of the world of interest: LERA plans, WTC plaza footprints, a huge sewn-together Google map of Manhattan, the approximate Sauret camera location for the North Tower collapse, etc. Renderings and animations can be done. The camera can move and zoom to anywhere... could be useful.


Edit: this is the kind of thing I'm talking about: http://i5.tinypic.com/6asq8bc.png

Like this one, too: http://i17.tinypic.com/52gbqkk.gif

OneWhiteEye:

I certainly try to collect useful photographs for "calibration" purposes.

Here's a great one.

http://www.nycfoto.com/showPicture.php? ... 86&start=3

It shows the new WTC 7 building (face on), which is 226 meters tall.

The white 101 Barclay St building, which is 100 meters tall, is on the left ...

And the stone-grey New York Telephone building, which is 152 meters tall, is in the middle.

I very useful exercise is to try to figure out a vertical calibration scale for this photograph using a print-out of the picture.

I get 1 cm = 17 meters for the 101 Barclay building and 1 cm = 18 meters for the new WTC 7 building.

This nicely shows the effect of perspective on vertical scaling.

Dr. G wrote:Here's a great one.

http://www.nyctoto.com/showPicture.php? ... mID=108&...

I want to see this, but the board mangled your link. It's done it to me, too. It seems certain addresses must be enclosed in the URL tags, otherwise part of it gets chopped out. I didn't have to enclose the two links above, but others have been truncated.

I get 1 cm = 17 meters for the Irvine Trust building and 1 cm = 18 meters for the new WTC 7 building.

This nicely shows the effect of perspective on vertical scaling.

Haven't seen it yet, but I know what you mean.

OWE:

The link to the picture was ok initially, because I tested it after I did the post. Then I did an edit for spelling and it mangled the link to the photo.

I have done a second edit, (re-entering the link), and it appears to have fixed the problem...

Please try it again!

Dr. G wrote:Please try it again!

Yes, works fine now. Great shot. Indeed, it's a nice example of perspective effects. Quite a large difference, considering it seems to be at quite a distance, the buildings are close, etc.

OWE:

I have done some more research on the location of the new WTC 7 collapse video. Triangulation on some of the buildings that are identifiable in the video indicates it was taken from the large high-rise apartment building at 22 River Terrace, about 500 meters northwest of building 7.

In addition to the white building, 101 Barclay, and the grey New York Telephone Building, also on Barclay, that are visible near the centre of the WTC 7 collapse segment of the new video, three other buildings are noteworthy in the “zoom-out” shots near the end of the video, namely WFC 3 and the Embassy Suites Hotel buildings on the right, and the tall and slender Woolworth building in the background on the left. With the help of Google Maps it is pretty easy to determine from the relative alignment of the major buildings noted above that the new WTC 7 collapse video was shot near River Terrace and Park Place West, (this is a little further north than Murray St and River Terrace – my first guesstimate of the filming location.)

What is most interesting in the new video is the location of the filming of the North Tower collapse sequence, also included in the newly released footage, in comparison to the WTC 7 sequence. It is clear from this comparison that the North Tower sequence was photographed from a significantly higher location than the WTC 7 sequence. This is best seen from comparisons of the visibility, or non-visibility, of the roof of the Embassy Suites building – the “low-rise” building in front of WFC 3 - in the two sequences.

Interestingly, 22 River Terrace is a two-tiered high-rise building that opened in the spring of 2001. I believe the North Tower collapse sequence was filmed from the 100-meter high roof of this building while the later, (5 hours +), WTC 7 collapse sequence was shot from a balcony about 50 meters lower down in the same building. Why this would be so, one can only speculate …………..

Fabulous work, Dr. G! Thanks for that info.

Oh, I saw the post where you quoted "In order to do the type of analysis that he is talking about we first have to know where he got the video and the camera" and all I can say is wow. Wow. Did you realize we were wasting our time over here? I feel so foolish. I doubt we'll ever be able to determine where the camera was purchased, might as well give it up. But thanks for all your effort, Dr. G.