A layman’s open letter to Professor Zdeněk Pavel Bažant


UPDATE, 11/10/11: Professor Bažant and Dr. Greening were so kind to reply to my emails. Please find our correspondence right after the original letter.


CC: Yong Zhou, Mathieu Verdure, Jia-Liang Le, Dr. Frank R. Greening, S. Shyam Sunder (NIST) and f. ASCE.

(a printable PDF of this letter may be downloaded here)

Dear Professor,

dear scientific community!

Thank you for taking the time to consider my thoughts on your works „Why did the World Trade Center collapse?“ and „Mechanics of Progressive Collapse“.

I am writing to you today because I found your publications* to give the only non-conspirational explanations for the 9/11/01 collapses of the Twin Tower buildings. Based on your work (2002), even NIST concluded in NCSTAR 1 (2005) that no further  theoretical or experimental examination of the collapse sequence was needed, because, once the impacted stories began to give in, „the towers were doomed“.

I found this odd ever since I saw the towers collapse for the first time, and although, out of a layman’s natural curiosity, I tried to find explanations other than those outrageous demolition theories that have been postulated by a few outsiders in the past  years, I have found none that made sense to me; so I find no other way than to ask your assistance in understanding your work. Since this is still a matter of great importance to many all around the world, please accept my sincere apologies for choosing to publish this letter openly on my blog so other members of the scientific community may join the discussion.

To be honest, those differential equations scared me off a little, but I remembered from high school that an integral describes the area under a curve in a coordinate system, Energy is Force times Distance, Force is Mass times Acceleration, I remembered Isaac Newton’s lex quarta and so forth, so it wasn’t that hard anymore to decode the equations (1) – (3) in your paper from 2002 and (1) – (6) in 2007.

For better understanding, a copy of Figure. 4 from „Mechanics of Progressive Collapse: Learning from World Trade Center and Building Demolitions“ is included:

Fig. 4 from "Mechanics of progressive collapse"

I understand you have proven that the kinetic energy of a 3.7m drop would be sufficient to overcome the peak of the resisting force curve (Fig. 4a) and, after that, continue to buckle. A quick check on Wikipedia shows the four Euler cases and I understand that from a technical perspective, a steel beam can indeed be regarded as four parts connected by three (plastic) hinges. I understand that at first, it’s hard to push against a match but once it’s broken, it’s broken and the fall will speed up again, driven by gravitation or any other pushing force. In any case, FC, the force needed to crush the story should be much (two or three times) bigger than m*9.81m/s² (as you show in Fig. 4b), otherwise, the building wouldn’t keep upright in the first place. That’s what architecture and engineering is about: putting stuff on top of each other to decelerate gravity’s acceleration of everything in and above it to v=0, which is guaranteed by an FC reasonably greater than the weight of what it’s supposed to hold upwards.

Only when block C comes down at great velocities and a=31*9.81m/s² is m*a >> FC so the story gets „crushed“ (up or down, you choose) as demonstrated by Fig. 4a. But is global collapse really inevitable, logically?

What about the next story? By now, just some of the potential energy of Block C has been converted to kinetic energy (by the energy the planes brought in (deformation and heat, gets a whole story out of the way for Block C’s free fall)), which in turn has been converted to internal energy, because the structure has been deformed. According to what we have learned about the conservation of energy, the sum of all impulses in a closed system stays the same and as we do not assume that any form of energy (as, for example, a lot of unknown mass or gears, levers, momentums or spring energy built in with a watchmaker’s precision, or an upside down rocket on the roof, or chemical energy which could be transformed into thermal energy to cut the colums to provide for better „hinges“ or even  extraterrestrial aliens with superlasers shooting from outer space, which would all serve the same purpose – to assure that g*m(z) > FC) has been smuggled into the towers prior to the attacks to explain why for all stories g*m(z) = m*a > FC (the case you show in Fig 4a), K, as converted to W(uf), should become smaller, i.e. K’=K-W(uf).

Again: of course, in the beginning, a=31*9.81m/s² as you computed from the stiffness C, accordingly, Fig. 4a applies; but g is still only 9.81m/s², so m*9.81m/s² < FC as in Fig. 4b applies for the whole rest of the building which suffered no damage at all (until then), as everyone including you and NIST (2005) agree. So, in theory, F(u) should decrease from story to story, except if someone or something besides earth’s 9.81m/s² is pulling (or pushing, if you will). Accordingly, the deceleration should have resulted in a collapse arrest sooner or later as you show in Fig 4c.

Yet, as we have witnessed on September 11th, 2001, the towers accelerated downwards with an average of aobserved= 2*h/t² = 2*400/(14s)² = 4.08 m/s² ≈ 0.42 g, so the resisting force that accelerated the building upwards was only Ftotalfriction = m*a = 58,000,000 kg * (9.81-4.08) m/s² ≈332 Meganewtons.

That is not much compared to the force the topmost floor of the lower building block must have had at least before a plane crashed into it: Fload= m*g = 58,000,000 kg * 9.81 m/s² ≈ 569 MN (+some little just to be sure). And that’s just the topmost floor, in theory, all Fload of each of the 110 floors should sum up, according to the superposition principle. Just imagine Fload for the colums on ground level!

This, in turn, means, that – against all logic (except that of a few outsiders), but according to your own explanations – indeed Fig 4a does apply so that m*g >Fc during collapse, distributed evenly among all 110 stories, so, essentially, a huge ΔFa was hidden in the building, ready to be triggered by a relatively small F(u), to make sure that FC< m*g all the way from the top to the bottom. Ergo, we can conclude that, since we do not want to insinuate that m somehow became greater because a lot of mass has been brought up there, and since we may want to exclude hypothesises explaining how g locally became a lot greater than 9.81m/s² during this very hour (or that springs or an upside down rocket attached to the roof accelerated the towers all the way down to Ground Zero) to explain how g (or a, in this case) remains much greater than 9.81m/s² for un=400m, that the only way to bring down the buildings would be for FC to drop way below m*g*security factor on each level one after another – for all remaining 108 floors above and beneath the impact area.

You state that what matters is neither strength nor stiffness, but is energy. So, E_kinplaneimpact= 1/2 * 124,000kg * (225 m/s)² ≈ 3.14 Gigajoules. This plus E_heatkeroseneis what, in our 1D-model, causes E_potBlockC to be released all of a sudden by displacing all columns (which obviously took several minutes to prepare, for until then, the towers were standing – hurt and smoking from 3GJ of deformation energy and burning from E_heatfires, but upright). E_potBlockC now in turn, after falling 3.7 meters and gaining a speed of 8.52 m/s (since v= sqrt(2*g*s) = sqrt(2*9.81m/s² * 3.7m)), becomes E_kincrush= 1/2 * m * v² = 1/2* 58,000,000kg * (8.52 m/s)² ≈ 2.1 GJ (≈K in your paper, I believe).

Compared to the potential energy of each tower, Epotential= m*g*h = 500,000,000 kg * 9.81m/s² * 400m / 2 ≈ 981 GJ, a very small amount of input energy sufficed to bring down the complete structure, while one would expect that the sum of all W(u1) to W(u110) would be a huge lot greater than K, even greater than Epotential, diminishing a*m(z)*un along u in Eq. 3 (2007) as a=31*g approaches a=g, since usually in a building ΔFd> ΔFa, resulting in an increasing Φ(u), so that sooner or later Φ(u) > K (Eq. 5) and the fall is stopped, as with every other static structure we know of.

To be even more precise, Φ(u) is even bigger because of the area over λh in Fig. 4a-c. The „rubble“ and the softened steel would „dampen“ any force, as indicated by the steep rise in the curve for F(u), and so would the acceleration force needed to overcome the inertia of the next floor slow downwards movement.

As the collapse was not arrested for 110 floors and the downwards acceleration was bigger than the deceleration (=upwards acceleration) for 10-23 seconds, a critical review of your own analysis begs the question why m*a was greater than FC throughout the whole structure so that ΔFa>ΔFd and Φ(u) < K for all u from top to bottom of three buildings (WTC 1, 2 & 7) that have been built to withstand storms and earthquakes and faithfully did so for three decades.

The towers were „doomed“ and collapse was „inevitable“ only under the assumption that for each floor, the rules of Fig. 4a apply. It is hard to see why such a building would be allowed to be built in the first place, as its collapse bears features of a metastable system or a 19th century perpetuum mobile mechanism rather than of anything remotely statical; just a small impulse would trigger a huge mechanism that switches from decelerating earth’s 9.81m/s² to 0m/s to accelerating 500.000.000 kg structure of steel and concrete with 0.42 gs so it keeps moving all the way, folding into itself.

This cannot be the result of some chaotic and random process, but requires meticulous planning, as can easily be verified in experiment by alternately stacking weights and paper rings on each other (as seen in http://www.youtube.com/watch?v=caATBZEKL4c and http://youtu.be/rGw58logz0o): no model of what looks anything like a „tower“, even with so much m distributed (as m1, … mn) over h that FC is just a little bigger than m*9.81m/s² (putting just a little more weight on one paper ring than 1.8 Kg would crush it instantly, an extreme form of Fig. 4a). Choosing a generous drop height and a nice mC for a good C-block (picking up a third of the 1825g and letting them drop from 0,2m) will not result in total collapse. Parts of the „tower“ structure will get „crushed“ between earth and falling object. „Crush-up“ and „crush-down“ appear simultaneously, the impulse runs through the structure, „crushes“ the weakest „hinges“ (right down to floor level, where m*g is closest to FC with just a tiny ΔFd), and, as expected, Fig. 4b&c ensue: collapse is arrested (if kept from „toppling“) with most damage where block C and B+A made contact and on floor level, while „rubble“ (crumpled paper rings) dampened the impacting force.

Even if there’s a gradient of FC over h, but so is of m. And of course FC(n) and F(h) is just the weakest points, hinges, connectors and trusses combined, including shearing, buckling and all that which can happen when things go mechanic. However, for a, say, 80% intact steel building, that should be more than just 332 MN resisting a 58,000t-block (distributed over 400 meters and its own load being something between 250,000t to 400,000t) vertically; bearing in mind that the structure hardly swayed visibly when it was hit horizontally by a fast-moving plane and even absorbed the huge fireballs that we saw when half of the kerosene exploded. So, even with another 2.1 GJ being triggered by the ensuing office fires and fireproofing being razored off the columns in the impact zone, global collapse should neither be inevitable nor the most logical thing to expect, as documented by the EMS setting up a triage desk in the lobby of the already burning south tower (NCSTAR 1, p. 44) and hundreds of firefighters still making their way upwards to save lives. Those were heroes, not maniacs with suicidal tendencies.

Dave Thomas was kind enough to explain on his website how the stiffness of a spring can be computed and inserts a value of 71,000,000,000N/m, as you do for (2002) and (2007) as a basic premise for the rest of your calculations. If, however, F=-kx, then, as we observed, the „spring“ was displaced 400 meters by 31 times the force of block C, so k=-F/x and for F=31*FLoadBlockC: K=-31*58,000,000kg*9.81m/s²/-400m=44095950N/m which is 1610 times smaller than 71×109N/m – what a „spring“ the towers must have been then that it lost 99.94% of its stiffness!

Although I am not an expert, I’d like to prefer science, empiric studies and sane logic over blind superstition, ignorance and outlandish conspiracy theories as they were common during the dark ages and, as we have learned from history, inevitably lead to oppression, prosecution of minorities and the torture of innocents. This is the 21st century.

As Galileo Galilei said: „in questions of science, the authority of a thousand is not worth the humble reasoning of a single individual“, I herewith would like to offer my 0.02€ to the discussion over why the towers fell and express my hopes that you as the leading expert in the research of scaling the mechanics of solids will propose a model of the towers featuring the same distinct properties during a – however induced – collapse sequence to shed even more light on what happened on September 11th, 2001 and which still influences the lives of many all around the world.

Kind regards,


*this letter refers to:

Bazant, Zhou (2002): Why Did the World Trade Center Collapse?—Simple Analysis

Bazant, Le, Greening, Benson (2007): Collapse of world trade center towers: what did and did not cause it?

Bazant, Verdure (2007/2008): Mechanics of Progressive Collapse: Learning from World Trade Center and Building Demolitions


Professor Bazants reply:

Dear Mr. Akeyron: It is impressive when a layman grasps phenomena normally requiring higher level of training. We appreciate your expression of support, but we are too busy to have time to engage in a discourse at this level. For us, the analysis of WTC was a small, unfunded and temporary interest.

Dear Professor Bazant,

I understand that you must be very busy, so thank you very much for your fast reply.

I just wanted to point out that according to your 2007 paper, m*g would have to be greater than F_c as in Fig. 4a for all 110 floors of each tower to guarantee global collapse so, if interpreted correctly, regardless on which level, your work inadvertedly SUPPORTS controlled demolition theories instead of refuting them.

What for you may have been a temporal interest, is of great significance to those still riddled with doubt over what happened on 9/11, as policy makers around the world, including those in my country, still base their decision-making mainly on the validity of your work (on which NIST relies, on which, in turn, media rely and shape public opinion), and many suffer from that very policy – for example, when the war against terror is fought in their country.

I must not speak on their behalf, but on my own, for when I went to school (in 1990’s Germany), our teacher wanted us to interview our grandparents about their knowledge of the Holocaust. 9/11 being the „American Holocaust“ which I have witnessed in my lifetime, I must expect my kids to ask me one day why the WTC collapsed and I would feel ashamed to admit that – even after asking the very experts who proved a global collapse under its own weight feasible – I could only „guess“ how and why the towers fell, even if it gives them a bad lesson about science, superstition and conscience.

If, perchance, you or your colleagues ever start a new small and unfunded analysis, they may want to avoid the mistake that I have pointed out in my letter to you, so I thought I better notify you about that pitfall.

Again, thank you for your time and consideration!

Yours sincerely



Professor Bazants reply:

With this I disagree. Briefly, out have to consider dynamic equilibrium (d’Alembert principle). in which all the inertia forces must be superposed on the gravity forces and the resisting forces. You need to look at some elementary text on structural dynamics. Our argument refutes rather then supports the fairy tale of controlled demolition.


that’s exactly my argument. m*g in Fig. 4 cannot be 31*m*9.81m/s² for all 110 floors, because kinetic energy m*g*u = m*g’*u‘ + W_deformation, so g should become smaller and smaller while proceeding through u until m*g < F_C as should be in a static building. The superposed resisting forces of the lower block as a structure that has not suffered any serious damage yet should be much greater than block C’s „feeble“ 31*m*9,81m/s², unless something besides earth’s 9,81m/s² keeps accelerating block C+rubble layer.



(no reply so far)



Dr. Greenings reply:

Dear Mario,
Perhaps the attached material will help answer your questions:
Frank Greening

(attached was a Word document with chapter 6.2 from The Pulverization of Concrete in WTC 1 During the Collapse Events of 9-11, with the addition of „An Assessment of the Time Delays Involved in the WTC Collapse Events“ made which, according to google, hasn’t been published yet so I really feel honored :-)

Dear Doctor Greening,

thank you very much for your time and consideration!

However, the attached excerpt from your work „The Pulverization of Concrete in WTC 1 During the Collapse Events of 9-11“, riddled me with questions very similar to those in my original letter, even though your approach is slightly different from that of Bazant/Zhou (2002) and Bazant/Verdure (2007).

Let me quote:

„Thus, setting Mn to 5.8 * 107 kg, M1 to 0.39 * 107 kg and with vi equal to 8.52 m/s, we readily determine that the first impacted floor of WTC 1 moved off with a velocity vf equal to 5.4 m/s; that is 3.1 m/s or 36 % slower than the impact velocity. Nevertheless, this reduced velocity was more than sufficient to guarantee a self-sustaining global collapse of WTC 1.“

With all due respect: non sequitur. You can’t just crush one story and say, just like NIST and Bazant/Verdure do: „after that, the towers were doomed“, „global collapse ensued“, „global collapse was inevitable“. It does not match experimental observation, it is not the most logical thing to happen. We still have another 108 floors to go! (110-1 for the impact damage which causes freefall and -1 we’ve just crushed).

Because, for the next story, the same equation applies:

½ Mn vi2 = ½ [Mn + M1] vf2 + Ed

This time, with our second vi only vf from the first step (first step’s vi * 64/100 = first step’s vf = 5.4m/s = second step’s vi) and according to the law of conservation of energy, another Ed (603 MJ) must again be substracted from the second ½ Mn vi2 to derive vi for the third step. Only one plane impact, so only one free fall, and of course only one energy input of 2.1 GJ, sorry. Thus, with a total input energy of only 2.1GJ (which I agree upon given a 3.7m freefall of 58.000t near earth’s surface), the collapse should get arrested after crushing three or four floors, if each floor’s Ed = 603 MJ, as you calculated.

According to Table 1 in your report, [„An Assessment of the Time Delays Involved in the WTC Collapse Events“] the structure’s „Force action distance“ is only 11 centimetres, which might be the case in a perfect scenario given the fact that the impacting forces found their least energy-consuming way through the weakest parts of the structure, i.e. bolts, „plastic hinges“ and so on, and afterwards, there’s another small „free fall“. But I checked your numbers according to the equations given [d = (v2 – u2)/(2g – Fy/M] and [Δt = 2d/(u + v)] and found that you must have derived each Δt and d in your table for each story from Fy = 2 * 1010 Newtons – while conservation of momentum, Newton’s lex quarta and the d’Alembert principle (which Professor Bazant was so kind to inform me of) and every observation I have made in my lifetime dictate that Fy diminishes with each story crushed, d be 11 centimeters or 3.59 meters!

In essence, what you have done, mathematically, is just what Bazant/Verdure (2007) had to do in their report to make the towers collapse as fast and as completely as they did in reality. In your report, Fy = 33 * Fst for all 110 floors. In their work, m*g > FC (where g = 31*9,81m/s²) as shown in Fig. 4a applies to all 110 floors (and as I have explained in my open letter and which is no mistake, as Professor Bazant insists). 500.000.000kg of mass distributed over 400 metres‘ height, yet the retarding force acting against earth’s acceleration of 9.81m/s² is only m*a = 58.000.000 kg * (9,81-4,08) m/s² ≈ 332 Meganewtons where a = 2*h/t² = 4,08 m/s² ≈ 0,42 g. for h=400m and t=14s.

In other words, your math is correct, as the graphs in your Figure 1 show, but begs for a different interpretation, as it also proves an „inside job“ for the collapse of the Twin Towers – literally: there must be a lot of force, i.e. 110*Fy, that is, a lot of energy hidden inside the buildings (other than that of the planes and the fires and the resulting 3.7m freefall) to ensure global collapse. I am desperately out of jokes by now about what that could have been – springs, gears, levers, martians with superlasers or an upside down rocket on the roof, there are many forms of energy around – so I must suggest in all seriousness, for this is not a laughing matter, that you and your colleagues rethink the official „gravitational collapse“ hypothesis, blow the whistle and inform the public about your – inadvertent – findings.

Otherwise, one would have to explain how an input energy of 2.1 GJ could trigger 110* Ed + Epotential > 1 TJ [where Ed = 603 MJ, as you calculated, and Epotential = m*g*h = 500.000.000 kg * 9,81m/s² * 400m / 2 ≈ 981 GJ] UNLESS the towers were built as amplification machines (with gears, springs and levers in them) rather than good old buildings, as those working in it for 30 years believed and as we know them ever since our forefathers erected the first pyramids.

Yours sincerely



Dr. Greenings reply:


No amplification machine required! That’s why it’s called disproportionate collapse…..

Dr. Greening,

Frankly, what it’s called is not of primary interest because it has been pancake, card house, crush-up-crush-down, trusses, global collapse, progressive collapse, gravitational collapse, piledriver, controlled demolition, implosion, explosion and telescope mast already. In religion, opinions may differ about how to name things, in science, two plus two make four. When science says that bumblebees can’t fly, bumblebees don’t stop flying. Instead, science seeks to explain and found that bumblebees have those wings to actually flap them. If Fy = 33 * Fst, as you say (or g*m=31*9,81m/s²*m, as Professor Bazant demonstrates), and if Fy = 20,000,000,000 N and h=110*0.11m (your „force action distance“) then W=242GJ – that’s not unlike our 2,105 GJ *110 floors = 231 GJ. But there was only one plane, one fire and a 80-90% unhurt solid structure. So mathematically, you’re dropping Block C again and again and again. Or pulling. Or pushing. Or getting something out of the way. Please re-read my original letter, third-to-last paragraph: „If F=-kx, then, as we observed, the ’spring‘ was displaced 400 meters by 31 times the force of block C, so k=-F/x and for F=31*FLoadBlockC: K=-31*58,000,000kg*9.81m/s²/-400m=44,095,950N/m which is 1610 times smaller than 71×109N/m“ – where are those 99.94% of its stiffness gone and whence come those additional > 230 GJ?



Dr. Greenings reply:

The „spring“ you refer to was not compressed 400 meters! It was broken into a million pieces long before that, so your calculation is meaningless. Try calculating how much a column can be compressed before it reaches its elastic limit. If you work with Prof Bazant’s values you will see his k is for one floor height, but the pile-driver model is not to be taken too literally because the building ceases to act as a spring after a few milliseconds…..

(emphasis mine)

Thank you, Doctor, now I understand what I’ve been trying to say. Have a nice 11.11.11 :-)

9 Kommentare

9 Kommentare

  1. psikeyhackr  •  Nov 20, 2011 @05:30

    A simple thought experiment which our engineering schools should have been able to simulate some time ago would be to merely remove five simulated levels from the north tower, 91, 92, 93, 94 and 95. That would leave a 60 foot gap with 15 stories floating in the air and 90 intact simulated stories below. Then let gravity take its usual immutably boring course. The bottom of the 15 stories would impact the top of the 90 in just under 2 seconds at 44 mph or 65 feet per second.

    The 90 stories should be 1080 feet tall so if the 15 stories could maintain a constant 65 ft/sec while destroying them the collapse would take 16.6 second plus the 2 seconds totaling 18.6 seconds. But that is significantly longer then most estimates of collapse time therefore the 15 stories would have to accelerate while crushing stories heavier and stronger than themselves.

    Now completely eliminating 5 stories to make that 2 seconds of acceleration possible is more damage than the airliner impact and fire could have done so we know that 60 feet of empty space never existed. But that thought experiment eliminates all argument about how hot the fires got because they could not instantaneously disappear five stories.

    The levels had to get stronger and heavier going down and lighter and weaker going up. So how could 15 stories destroy all 90? Even assuming a 3 to 1 ratio of destruction, which I regard as unlikely, that would leave 45 stories standing which is not what happened on 9/11. So if that simulation is done and it comes nowhere near complete collapse then what is this nonsense that has been going on for more than TEN YEARS?

    And of course how could the simulation be done without knowing the tonas of steel and tons of concrete on every level? You can’t compute the potential energy without that. How could you do the conservation of momentum without that information? So what does it say about the entire physics profession not demanding that information within months of 9/11?

    So why hasn’t any engineering school done such a simple simulation?

    9/11 is the Piltdown Man incident of the 21st century. Maybe when this is settled we should pass a law requiring everyone with a degree in physics to have a picture from 9/11 on the wall in his/her office for the next 100 years. LOL


  2. Akareyon  •  Nov 20, 2011 @12:28

    Psikey, welcome!

    I just googled the „Piltdown Man“ and I really don’t see many differences between 9/11 and that fakery :-)

    As a movie geek, I have found many resemblances to „Shutter Island“, „Inception“, „The Matrix“ and even „Saw“ on a psychological level. I mean, those towers worked the same way the inverted bear trap does: you put it under tension and then trigger it. Surely left a few mouths open…

    It is plain obvious that the demise of the Twin Towers can not be attributed to „slipshod architecture“ and was surely not „inevitable“. I reproduced your paper loop experiments with vinyl singles, stacking them at least one meter up, and built Jenga towers to understand what’s goin on and I came to the conclusion that if steel is so brittle, we should use paper instead. Maybe the collapse could have been averted with a stack of paper loops? Next time an airliner crashes into a building, we could try. Just yesterday, I was called an uneducated layman for not believing nothing could have stopped the collapse on ATS…

    However, someone (whom you may know) did a great job explaining Bazants „trick“ to me from another perspective. He had me falling through ice slabs (that could support my weight) each 2m apart. Then it came to me why in Bazants calculations, each floor resists the collapse with only 0.5GJ: he’s just throwing something heavy through the roof so it drops (sinks) through all the floors into the basement at terminal velocity. I thought Bazant had just put something very heavy on top of the towers to compress them, at least so it looks from his column buckling displacement diagrams (with m*g above the Maxwell line for all 110 floors). Of course it also works the other way round: dragging the Maxwell line down…

    So right now, the discussion has progressed to the point that I had to bring up E=p*V and am trying to explain the difference between a katana and a hug – it’s in the area affected, the pressure or tension. I’m really just a layman, but it is clear that even lifting the top 12 floors, like Bazant does, would increase the overall „tension“ about just 7% – and that makes the balloon pop! So it must have been stressed to the max already.

    That gave me the idea of a time independent energy evaluation.

    The whole building is undergoing a volume change of little less than 64*64*400m³ – how much pressure must be applied, how much „inner tension“ has to go for that if we know the (average) tension of each floor? Looks it’s quite the potential energy that is converted to kinetic energy, the energy that Bazant refuses to talk about when he’s approching E from m*g*h :-) So if we assume that there was a Factor of Safety around 2 or 3, somebody is „secretly“ putting the structure under huge stress to make sure that the upper part meets as little resistance as possible, but I have a hard time getting my point across, it seems.

    Thank you for unplugging me :-) I’ve been a „conspiracy theorist“ for a long time and just doubted the official explanation, especially with that „Shock Doctrine“ background, knowing people were waiting for something like this to happen and grabbing the opportunity, but your experiments with the tooth picks really gave me that „Heureka!“ experience. Someone made it happen, and they were sure they’d get through with it. Damn, they did. In broad daylight and plain view, and it has become the blind spot for sane logic and institutional analysis although the basic principles could be explained by a bright 14-year old school kid.

    Welcome back to the dark age of superstition, war and torture. Archimedes is sitting in his bathtub, slitting his wrists, while Galileo weeps and Newton spins in his grave.

  3. Akareyon  •  Dez 1, 2011 @09:59

    One more approach based on Figs. 3 & 4a-c in B&V,’07, as resulted from discussion on the 9/11 Forum and ATS.

    If somebody builds something, he will make sure that the maxwell line (F_c) is way above any m*g that can be expected. If it is below m*g, it will crush. With a FoS of 2, the maxwell line will even be two times m*g. The area between the maxwell line and m*g is the energy needed to crush one floor. This E_crush=(maxwellline-m*g) * u. So, if maxwelline=2*m*g, then E_crush=(2*m*g-m*g)*u, so E_crush=m*g*u. This is the energy needed to crush one floor. It is no different from Φ(u) in Eq. 3. The only difference between Φ(u) and E_crush is that Φ(u) is the integral of the load-displacement curve from F(u) minus m*g*u, that is, the area between the curve for F(u) and the line of the load force, while E_crush is simply the „average“ rectangle that stands for the same phenomenon. The greater the load force m*g, the smaller both E_c and Φ(u) – the energy needed to crush one floor. Why? Because the area between the maxwell line and m*g gets „compressed“, it becomes smaller. Therefor, it does not matter if the curve for F(u) has a huge peak at the left hand side and then touches the ordinate (as F(u) for pasta or steel columns would), or if it takes a slope up at the right hand side (as F(u) for a paper loop would). In Bazants words: „what matters is energy, not the strength, nor stiffness.“ What matters is the area between F(u) and m*g.

    If now something impacts one floor with kinetic energy (K) greater than E_crush (or Φ(u)), K will be diminished by E_crush. E_crush will diminish K so long until K=0. The only way for K not to become 0 is to make sure that E_crush (or Φ(u)) is very small. There are two ways to achieve that: either drag the maxwell line down or push m*g up.

    In the event of 12 floors dropping from 3.7m height, it is obvious that m*g must be huge and well above the maxwell line. So the area between F(u) and m*g, Φ(u) that is, is very small and K is diminished just a little. So, what is K? It is m*g*h. Usually, h is zero – nothing dropping, so the kinetic energy is zero. If it drops from 3.7m, it is much greater. But – is it actually dropping from 3.7m?

    [One member at ATS] opened my eyes. K is – potentially! – much greater, it is constantly increased as it crushes through the floors – potential energy being transformed into kinetic energy.

    Let therefor the „input energy“ be that of 12 floors dropping from h = (98+1)*3.7m height right from the start, in other words: let’s lift the 12 floors 3.7 metres, but pretend for a moment that the rest of the tower isn’t there anymore. From bottom to ground, there are 366.3 metres. K would be huge upon impact on the ground, wouldn’t it? There would be dust and debris all over the place, steel coloums flying everywhere and nothing left of the Top Of The World. Let us say that K_start=0 before it drops and K_ground=m*g*h when it crushes on the ground. The potential energy of block C is K_ground, therefor: E_pot=m*g*h. This is the potential energy of the upper 12 floors, Block C.

    Now let’s put the rest of the tower back under it. Those 98 stories have suffered nothing yet. We could just put a new roof on it. On each floor, m*g is still only half the maxwell line, accordingly, each Φ(u) is still big enough for each floor to support all the floors above. Imagine all the 98 curves for F(u) from Fig. 4c, a perfect, stable building the way it should be, lined up next to another. It would look a little like a twisted sine wave with 98 peaks. This new curve we call F(s), with s=98*u. To keep things neat and easy, we don’t account for the gradient in strength, so the line for m*g stays the same, it doesn’t move, and so doesn’t the maxwell line at 2*m*g. We have changed nothing, we have only lined them up all next to another. Now we can go ahead and calculate Φ(s) – again, it is the area between F(s) and m*g, it is also the area between the maxwell line and m*g, it is: (2*m*g-m*g)*s=m*g*s. Φ(s)=98*Φ(u).

    There is one catch, though: we have added one floor between the 98th and the 99th floor consisting of air. So, to the leftmost side of F(s), we must add another u with an F(air) that nearly touches the ordinate. Needless to say that Φ(air) is negative, because its maxwell line is waaay below m*g, it’s almost not there. We have a negative area, but right now, we’re mathematicians, we could even sqrt(-1) and have imaginary numbers if we need them :D No, seriously, this is were Block C gains its momentum and K is increased from zero as it drops through the very first u, diminishing E_pot.

    Therefor, now, as we release the upper block, it first gains momentum (a lot of it!) while going through the first u, rolling down F(air). K is here „diminished“ by Φ(air), but as Φ(air) is negative, it isn’t diminished, but greatly increased (and, of course, E_pot is greatly diminished). After that first u, however, it must climb all those little peaks over s. It will gain momentum when it „rolls“ down a peak, and climb another peak, until all K is „gone“. When will that be? Easy. When the energy „gone“ will be

    E_gone = E_pot – (Φ(s)+Φ(air))

    E_gone is the energy that goes into deformation, and it is equal to the [potential energy of Block C] MINUS [the area between F(s) and the maxwell line of the 98 stories below] PLUS [the NEGATIVE area between m*g and the „maxwell line“ that represents air resistance F(air) for the leftmost part of our 98 lined-up displacement curves].

    Let’s check: if Φ(air) is zero (no drop), the potential energy of Block C is just as big as Φ(s), the „containing“ energy that keeps it up, so their difference is zero, so no energy goes into deformation. If there is a little drop, Φ(air) (the area between F(air) and m*g) becomes „negative“, so the part between the parantheses gets SMALLER than E_pot, so E_gone increases – energy goes into deformation.

    I hope I haven’t lost you already because I’m talking of negative energy. I know it’s not professional, but I’m trying to explain as good as a layman can, and I enclose a diagram to show what I mean:

    (the red area is Φ(s), negative and positive part, in each of these four images, sorry the designation is cut off)

    Let’s try and make K=0. E_gone = E_pot – (Φ(s)+Φ(air)). So, collapse will arrest WHEN [the energy „gone“ (into deformation)] IS EQUAL TO [the potential („potentially kinetic“) energy] MINUS [the area between the looong sinus-like load displacement curve and the maxwell line] PLUS [the „negative“ area at the leftmost part of our curve where it drops in freefall].

    Of course, all this is utterly false. It is even complete rubbish! Let me explain why.

    It’s plain obvious. Φ(s) (the area between the sinus-like F(s) and g*m), is so huge (even with all the negative intervals where F(s) sinks below the m*g line representing the brittleness) that Φ(air), the energy „gained“ (substracted from E_pot) during freefall, would fit tenfold into it. Collapse would be arrested after just a few stories, and that’s not what we want. We want to bring the towers down. How do we achieve that?

    There are numerous ways.

    We could drag the line for m*g up for F(s) so that the area of the 98 peaks of F(s) above m*g are almost as huge as the area of the valleys beneath m*g. That would make sure that Φ(s) is so small that when added with the negative area from Φ(air), there’s almost nothing left in the parantheses of the E_gone = E_pot – (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone – complete destruction. How do we drag up the line for m*g, closer to the maxwell line?

    Another way could be to „cap“ the peaks above m*g. That would drag the maxwell line down, closer to m*g, thus diminish Φ(s) until almost nothing is left in the parantheses of the E_gone = E_pot – (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone – complete destruction. How do we cap the peaks of F(s) above m*g?

    Another way could be to „dig“ the valleys deeper below m*g. That would drag the maxwell line down, closer to m*g, thus diminish Φ(s) until almost nothing is left in the parantheses of the E_gone = E_pot – (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone – complete destruction. How do we dig the valleys of F(s) below m*g deeper?

    Another way could be to increase Φ(air) so that the „negative“ area between m*g and air’s maxwell line is so huge that it is larger than Φ(s) until almost nothing is left in the parantheses of the E_gone = E_pot – (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone – complete destruction. How do we increase Φ(air)?

    Now one may argue that steel is very brittle and more like pasta than like paper, that is, the curve looks different. To explain, Fig. PASTA is enclosed. Here, I left m*g where it is and only changed the shape of the load displacement curve. Accordingly, the maxwell line drops down closer to m*g.

    As we can see, Φ(air) remains the same, only Φ(s) is getting real small now because of all those deep valleys under m*g. Suddenly, global collapse becomes much inevitablerer! It will take a lot of s until all Φ(air) is used up in Φ(s).

    However, a good engineer would account for the the brittleness of the pasta, wouldn’t he? Of course he would, he wants a FoS of 2 at least, and so he’ll make sure that m*g, the forces acting, remain within the boundaries of sanity. Otherwise, everything in the building would be screaming for destruction! Just a little breeze and it would all be crumbling down, but we’re expecting cyclones and storms and office fires! So, obviously, he would pull m*g down by putting less mass on top, which, again, would look like this:

    (Or he would just use more pasta).

    This is what I think I understand and therefor believe that global progressive disproportional collapse is not a new law of nature, but has to be planned.

    One objection: when Block C impacts the lower structure, it won’t be steel columns on steel columns, but a chaotic process. Accounting for that is difficult in these diagrams, so let’s just „flatten“ the first peaks for the moment.

    Another (possible) objection: the mass of Block C seems not to be accounted for. That is not so correct, because there are another 12 peaks on the negative side of (s), but m*g will stay the same throughout.

    A more precise rendering would include the gradient in strength and mass, which would be superposed on the F(s) curve and, of course, m*g, resulting in a constant rise of F(s) (and an even greater Φ(s)).

    The only way for 98 floors to collapse under the impact of 12 floors, that is, to explain a domino effect, would be to view the complete building as one load-displacement curve as shown in Fig. 3, or with 110 load-displacement curves as shown in Fig. 3, all next to another, with an m*g way above the maxwell line:

    But that would require for the building to be built that way – like dominos – either out of complete stupidity or on purpose.

    I would like to exclude a random and chaotic process, too, for a simple reason. Chaos and randomness can be quantified. One last and simple experiment: put up a lot of Jenga blocks on a given area A any way you wish. Make one tip. Only a few will drop. In order to bring them all down, you have to line them up „intelligently“, in ORDER to fall.

  4. Albert  •  Jul 3, 2016 @08:18

    Tony Szamboti, mechanical engineer, hat ebenfalls einen offenen Brief an Bazant gesandt und veröffentlicht.

    Da wird nicht nur nachgewiesen, dass die Annahme des freien Falls für das erste fallende Stockwerk des Nordturms z.B. völlig falsch ist und nicht der Realität entspricht, sondern er zeigt detailliert, dass Bazant und seine Mitarbeiter bei den Masse Ansätzen ebenfalls völlig falsch zu hohe Ansätze in seiner Milchmädchen Rechnung vornahm., seine Berechnungen insgesamt also nicht taugen den Totalkollaps der Türme (wie auch von WTC 7) zu erklären.
    Der von Gläubigen der offiziellen Verschwörungstheorie seither gebrauchte Begriff: progressiver (Total) Kollaps von Gebäuden (aufgrund von asymmetrischen Beschädigungen und von Bränden oder von Bränden allein (WTC 7) hat also keine wissenschaftliche Grundlage.

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