"Inside Job": Hidden energy in reports by Prof. Bazant, Dr. Greening and D. Thomas

reply to post by remymartin


Interesting interpretation, but I can't agree with it. What I read there is that the towers broke into a million pieces during collapse. Once the collapse front reached the ground, there was no "spring" left.

reply to post by -PLB-


But the stories above the damage remain in one piece. Thats what bazants paper relies on
how can that be

reply to post by remymartin


What Bazants model relies on is that the collision of the top section is concentrated on the columns in the lower section. You could argue that for this to happen, the top section would need to remain undamaged. The reason for this assumption is that a collision concentrated on the columns is favor of arrest. It is unrealistic and we can say for certain that did not happen. However, if we create a more realistic model where the collision of the top is not concentrated on the columns, but on the floors, the top is not required to stay intact at all. This type of failure would offer a lot less resistance than the type of failure in Bazants model.

reply to post by -PLB-


So can we agree that bazants paper
is flawed.

reply to post by remymartin


No, he clearly states in his paper that his assumptions are unrealistic. Had he omitted to state that, I would agree.

The thing is, his paper isn't about what you, and many truthers, want it to be about, which is how the collapse actually happened. That does not make the paper flawed. It just means it is not about the actual collapse. Nevertheless it proves something about the actual collapse, which is that collapse would progress.

To give an analogy: I want to know if an elevator I designed is capable of carrying 10 people. Physically, no more than 20 people fit in the elevator. In my calculations I use a weight that is 5 times higher, equivalent to 50 people. My calculations shows that it can actually carry 50 people without failing. The fact that 50 people do not fit in the elevator in reality does not make my calculations flawed, even though it describes a situation that could never possible happen. It still proves that 10 people would also not make the elevator fail.

Originally posted by -PLB-
reply to post by remymartin

 

No, he clearly states in his paper that his assumptions are unrealistic. Had he omitted to state that, I would agree.

How does saying that not make his paper flawed?

He admits his assumptions, used to compile his report, are unrealistic. I would consider 'unrealistic assumptions' a major flaw by any definition.

reply to post by ANOK


The fact that the paper is not about what you want it to be about does not make it flawed. It would be flawed if there were errors in it. When you say you make a certain assumption, give the reasons for making it, and explain why it is unrealistic, you are not making an error. You do everything how it is supposed to be done.

ANOK, remymartin, the paper is deeply flawed in pretending to make the most optimistic assumptions, then goes on to compute the kinetic energy lost after crushing ONE story and then, instead of input the rest of the kinetic energy for the next iterative step, says: well, that one story broke, so the rest of the kinetic energy will be enough to crush all other floors too. Nothing could hit the nail on the head like PLBs ice-on-a-lake analogy.

PLB, I am sorry for the misunderstanding I caused with my edit. I prefer to edit a post instead of clogging a thread needlessly, because it helps to keep things tidy in case someone wishes to refer to an earlier point in the discussion.

The translation to the WTC is that a support structure that can carry a certain mass for 25 years can fail when this same mass is dropped on it.

Agreed. Now, how do we build a tower with that?

But what exactly is the issue? Do you think the buildings collapsed too fast? Do you think it would not be possible at all? If so, why would it not be possible?

What do you think, according to that which I've written so far, saying that high structures tip, tilt and topple instead of eating themselves up from top to bottom?

In that post you claim that "the experts don't know, they don't have the slightest clue". How did you come to this conclusion? How do you know what experts do or no not understand?

You are right, that was a little far-fetched. Some of them know exactly. But they won't tell you. You wouldn't believe it anyway : )

On what exactly do you base this hypothesis?

PLB, I see no point in continuing this discussion with you if you insist to ignore the explanations I have already given in this very thread in our very discussion as if they were non-existent...

Why should I even take notice of your hypothesis when it is not supported by anything?

...you are therefor free to ignore the hypothesis as much as you ignore the arguments supporting it. I'm not trying to sell you stuff or talk you into something, really.

Can you write in a couple of lines what your most important issue is with Bazants work?

I can and I have. Count the lines:

B&V (07) jumps over one step in the iterative process.

That was one line.

The qualification of random and chaotic is (in this case at least) a subjective one. I has no scientific merit.

You are dissing a whole generation of chaos theorists, be careful what you say ; )

The idea that the collapse needed a lot of planning or extra energy is as of yet a baseless assertion. At least, I have no idea what you base it on. If I were you I would start with providing evidence for this assertion.

Me being happy that I am me, I have backed up my statement with logic reasoning, examples, thought experiments, cited the corresponding laws of nature and heard no protest then; now somehow suddenly it isn't true anymore. I have a hard time understanding what exactly you are asking from me that I haven't given you already.

If you wish, however, I will show you my original approach to the collapse mechanics which "inevitably" led me to Professor Bazants papers and how I found the "mistake". It is strictly mathematical, but since you have an engineering background, and we'll only use schoolbook formulas even Professor Bazant and Dr. Greening would agree on, it will not be difficult to follow. Is that okay for you?

reply to post by Akareyon


You can extend the ice analogy of course.

Imagine you have a stack of horizontal ice plates, supported by vertical ice plates in between like this:

__
|_|
|_|
|_|

Now image the exact same principle. When you stand on the top plate, it will not fail. When you jump on it, it will fail. What do you think will happen when the top plate fail and you fall down on the plate below it? Will it stop you? Or will it also fail? And when that one fails, and you fall down on the next, will it stop you? Or will it also fail?

I think you get the point. You now may be asking, but what about the vertical plates? Wouldn't they still be standing? The answer is no. The are kept in place by the horizontal plates. So they will fall over when the horizontal plates fail.

This example is of course a very simplified analogy. It ignores for example a core. Of course you can extend the analogy further. But it should give you insight in how a progressive collapse works, and how it could have worked in the WTC. I think the exact details are less important at this moment, it is more about the mechanics of the basic principle.

I pretty much refutes your argument that the kinetic energy will be less after the first iteration. In fact, it will be larger, as both mass and velocity increases with each failed ice plate.

Originally posted by Akareyon
You are dissing a whole generation of chaos theorists, be careful what you say ; )

I chose my words carefully and added "in this case".

Me being happy that I am me, I have backed up my statement with logic reasoning, examples, thought experiments, cited the corresponding laws of nature and heard no protest then; now somehow suddenly it isn't true anymore. I have a hard time understanding what exactly you are asking from me that I haven't given you already.

I replied to you with Verinage demolition, which proves the possibility of the iterative mechanism. What we are left with is the initiation. I haven't seen any good arguments from you why the initiation in the towers could not happen.

Your argument is basically an argument from incredulity. There is not a law or logic reason why those towers could not have collapsed. That is a bare assertion on your site. I simply do not agree with the assertion, and when you do not further support it, there is nothing left to discuss.

If you wish, however, I will show you my original approach to the collapse mechanics which "inevitably" led me to Professor Bazants papers and how I found the "mistake". It is strictly mathematical, but since you have an engineering background, and we'll only use schoolbook formulas even Professor Bazant and Dr. Greening would agree on, it will not be difficult to follow. Is that okay for you?

I would be very interested in this.

Okay. One thing before we start: we have to agree that altough the first ice plate might not stop me and that even the second might not be able to arrest my fall, that each plate that I crash through will slow me down, so I when I hit the next floor respectively, my velocity will never become as fast it was upon the impact just before and that accordingly (depending on my weight, the height of my first jump and the crushing force of the ice plates) sooner or later my fall will be arrested under the condition that each and every ice plate would be able to support my walking on it carefully - just as a bullet shot through the rain will much sooner drop to the ground than a bullet shot on a sunny day. I will only fall through all floors if none of them was able to support my weight, and I will reach terminal velocity until I hit the basement of our ice palace (I hope the water is warm).

We must also agree that these are only models and thought experiments, no detailed descriptions of what happened. We will not discuss wether it took the towers 11 or 12 seconds to come down, we'll only try to find out the orders of magnitude involved in the collapse.

Let us assume that the weight of the upper block was 58,000 metric tonnes, that the total weight of the towers was 500,000 metric tonnes, that it took the tower 14 seconds to come down, that earth's acceleration is 9.81m/s² and that the towers were 400 meters tall.

I had to start off with something I knew. At high school, I had to make a presentation about friction - the simple sort: dry friction between surfaces, the difference between static friction and kinetic friction.

So I just calculated the "static friction" of the topmost floor under the upper block (Block C) that was able to decelerate Block C to v=0. That one is easy. I took Bazants 58,000,000 kg and g=9.81m/s² for that. If you get something around F_static=569 MN for the upward force of the topmost floor under the upper block, both our calculators are working correctly. It was probably much more than that, but we'll leave the security factor aside for a moment.

Next, of course, I wanted to know what the force was that decelerated the upper block. Normally, a stone thrown down the top of the tower (without the friction of air) would hit the ground after t = sqrt(2*h/g) = 9.03 seconds. From that, I could conclude that the average downwards acceleration of the towers was a = 2*h/t² = 4.08 m/s² ≈ 0.42 g, so the kinetic friction force that decelerated the descent of the upper block was no more than F_dynamic = m*a = 58,000,000 kg * (9,81-4,08) m/s² = 236,640,000 Newtons ≈ 332 Meganewtons.

Until now, it doesn't matter for our model what the lower part of the structure was made of or how much it weighed, but it must be stressed that after initiation, the upwardly working forces of the complete structure were only 58% of the force with which the topmost floor held up Block C for 30 decades.

Since forces are vectors that obey the laws of vector addition, according to Newtons lex quarta (the superposition principle, more accurately the d'Alembert principle that Bazant refers to), we must assume that each and every story lost more than 40% of its static friction force, so that the top could "glide" at terminal velocity through the rest of the structure just like a stone gliding through a fluid, decelerated only by an average of 5.73m/s².

Since I had taken the values for the weight of Block C from Bazants paper, which until then still looked quite greek to me, I was familiar with his model of Block C dropping from 3.7m height on the rest of the building. Again, I calculated the energy which is derived from the momentum the mass had at the velocity it would have gained after a 3.7m fall. So with an impact velocity of v = sqrt(2*g*s) = sqrt(2*9,81m/s² * 3,7m) = 8.5202 m/s, energy of impact would be E_impact = 1/2 * m * v² = 1/2* 58,000,000kg * (8,52 m/s)² ≈ 2.1 Gigajoules (Dr. Greening agrees). That's not the amount of energy I usually have to do with, but I could compare it with the kinetic energy of a plane flying at 503mph with the weight of 124,000kg: E_kin_impakt = 1/2 * 124,000kg * (225 m/s)² ≈ 3.14 GJ.

So, the horizontal impact from the side high up at the "lever" brought more energy into the system than dropping Block C vertivally on the rest of the structure. The first hardly made the towers sway, the second crumblified them completely, releasing a potential energy of something around E_pot = m*g*h/2 = 500,000,000 kg * 9.81m/s² * 400m / 2 ≈ 981 GJ (assuming the centroid is at half the tower's height, it probably was way below that. I got inspiration and confirmation for this calculation from a debunking website btw). Usually, if a small momentum releases a lot of energy, the system is called "metastable". Bombs are metastable, too. You light it with a match and it releases its chemical energy. Were the towers mechanical bombs?

tbc.

Originally posted by Akareyon
Okay. One thing before we start: we have to agree that altough the first ice plate might not stop me and that even the second might not be able to arrest my fall, that each plate that I crash through will slow me down, so I when I hit the next floor respectively, my velocity will never become as fast it was upon the impact just before and that accordingly (depending on my weight, the height of my first jump and the crushing force of the ice plates) sooner or later my fall will be arrested under the condition that each and every ice plate would be able to support my walking on it carefully - just as a bullet shot through the rain will much sooner drop to the ground than a bullet shot on a sunny day. I will only fall through all floors if none of them was able to support my weight, and I will reach terminal velocity until I hit the basement of our ice palace (I hope the water is warm).

I do not agree. In fact, I think this is the culprit where most truthers go wrong.

First lets make some assumptions. Lets say that jumping from 1m height will make the ice plate fail. Lets also say that the ice plates are 2m apart.

1st ice plate: my weight falls on the ice plate from 1m height. Some kinetic energy is transformed into other forms (breaking the ice for example). I have a certain leftover amount of kinetic energy (and thus velocity), as I do not come to a full halt.

2nd ice plate: my wight falls on the ice plate from 2m height. Some kinetic energy is transformed into other forms (breaking the ice for example). I have a certain leftover amount of kinetic energy, as I do not come to a full halt.

In the second iteration, I fall from a greater height, and there is already an initial velocity. The amount of kinetic energy right after falling through the first ice plate is smaller than the kinetic energy right after the second ice plate (of you do not believe this, write down the equations). This continues until you reach a certain velocity. Only then there will be an equilibrium.

The thing you seem to forget is that when an object falls, potential energy is constantly transformed into kinetic energy. There is a constant increase in kinetic energy, and only a momentary decrease (the short moment you are falling through the ice plate).

Also note that in this explanation I exclude the weight of the broken ice that also starts falling. But lets ignore it as it is not of much influence of whether the collapse will come to a halt or not.

Originally posted by Akareyon
Until now, it doesn't matter for our model what the lower part of the structure was made of or how much it weighed, but it must be stressed that after initiation, the upwardly working forces of the complete structure were only 58% of the force with which the topmost floor held up Block C for 30 decades.

From the looks of it your calculations are all ok. What you are doing wrong here is comparing an average force over distance with a force in equilibrium.

In this case figure 3 posted earlier gives a lot of insight. I already talked about this is an earlier post. You need to look at the instantaneous force, which is the curved line. At one moment this force is much greater than the static force (mg), and another moment it is much smaller. At the moment it is greater, it is destroying the resistance, at the moments it is smaller, it is almost in free fall, most resistance of the floor is already destroyed. So when you take the average force, it can well be below mg.

I think the example with the ice plates can help here. At the moment you are breaking through the ice plate, the forces are large. Once you broke through it, you are in free fall. During this moment you gain a lot of kinetic energy. Until you hit the next plate. Again there will be a small moment of large forces (breaking the ice), and a long moment of negligible force (falling to the next plate).

Since I had taken the values for the weight of Block C from Bazants paper, which until then still looked quite greek to me, I was familiar with his model of Block C dropping from 3.7m height on the rest of the building. Again, I calculated the energy which is derived from the momentum the mass had at the velocity it would have gained after a 3.7m fall. So with an impact velocity of v = sqrt(2*g*s) = sqrt(2*9,81m/s² * 3,7m) = 8.5202 m/s, energy of impact would be E_impact = 1/2 * m * v² = 1/2* 58,000,000kg * (8,52 m/s)² ≈ 2.1 Gigajoules (Dr. Greening agrees). That's not the amount of energy I usually have to do with, but I could compare it with the kinetic energy of a plane flying at 503mph with the weight of 124,000kg: E_kin_impakt = 1/2 * 124,000kg * (225 m/s)² ≈ 3.14 GJ.

So, the horizontal impact from the side high up at the "lever" brought more energy into the system than dropping Block C vertivally on the rest of the structure. The first hardly made the towers sway, the second crumblified them completely, releasing a potential energy of something around E_pot = m*g*h/2 = 500,000,000 kg * 9.81m/s² * 400m / 2 ≈ 981 GJ (assuming the centroid is at half the tower's height, it probably was way below that. I got inspiration and confirmation for this calculation from a debunking website btw). Usually, if a small momentum releases a lot of energy, the system is called "metastable". Bombs are metastable, too. You light it with a match and it releases its chemical energy. Were the towers mechanical bombs?

At first sight I don't see much wrong with this. But I also don't see how this is very relevant. You can't just look at the energy and say something about the damage done. I can hit you with my fist and you say auch. I can stab you with a dagger and you are dead. All while using the same amount of energy.

I think calling the towers mechanical bombs is a bit exaggerating. But sure, there was an enormous amount of potential energy stored in it.

I also made a few experiments with the law of conversation of momentum. Actio=reactio, the sum of all momentums in a system stay the same. So, since p=p', and p=m*v, in an inelastic collision m_1*v_1=(m_1+m_2)*v'_2-m_2*v_2. From that, I could derive the velocity v_1 necessary for 58,000 tonnes (m_1) to accelerate 442,000 tonnes (m_2) to 400 metres in 14 seconds (v'_2), assuming they can move horizontally and frictionlessly on a plane, no "bouncing" occurs and no energy goes into the deformation of either block. Just try this for yourself.

When I presented these calculations on my blog, I was confronted with B&V ('07) again. This time, with all the thought experiments I've already undertaken, the basics didn't scare me at all and although the formulas presented there were more complex than mine and some of the variables are a little misleading, it was all still just math. For clarification, I'll leave the compaction ratio aside as it adds little to understanding what is going on.

In the first step, presented in B&Z ('02) we compute the force exerted on the spring by 58,000t dropping from 3.7m height based on the spring equation and an assumed stiffness of 71 GN. Now comes the critical part, Eq 2 in B&V. Please read it again and make sure you understand.

After impact, the column resistance causes mass m(z) to decelerate, but only until point u_c at which the load-deflection diagram intersects the line F=gm(z) [Figs. 3 and 4(a)]. After that, mass m(z) accelerates until the end of column crushing.

So, upon impact, the mass is decelerated. When it has crushed the floor beneath, no upwards force is exerted any more, and the fall continues. Quite logical.

Next come Eq 3+4. What's happening? We compute the kinetic energy that is "used up" in the crushing of one floor. Energy is the product of force and distance, hence, we define it as the integral (the area under a curve) of the force exerted on the floor over the distance u. The load displacement curves shown in Fig. 3&4 are of course merely models and depend on the characteristics of the material used. Something that breaks easily, like glass, will have a thin peak at the left hand side and touch the ordinate for the rest of u, while a rubber column might have a long, increasing slope from left to right. The maxwell line defines the force needed to "crush" a column, and "m*g" denotes the force that acts on the column, which is a little misleading here but we'll come back at that later. Now comes the important part:

Clearly, collapse will get arrested if and only if the kinetic energy does not suffice for reaching the interval of accelerated motion, i.e., the interval of decreasing Φ(u), i.e., Fig. 4, right column. So, the crushing of columns within one story will get arrested before completion [Fig. 4(c)] if and only if K < W_c

In BASIC terms: this is an IF...THEN condition.

Graphically, this criterion means that K must be smaller than the area under the load-deflection diagram lying above the horizontal line F=g*m(z) [Figs. 3 and 4 right column].[...] The next story will be impacted with higher kinetic energy if and only if W_g > W_p, where W_g=g*m(z)*u_f = loss of gravity when the upper part of the tower is moved down by distance u_f; [...] and W_p=W(u_f)=area under the complete load-displacement curve F(u) [Fig. 3]. [...] For the WTC, it was estimated by Bažant and Zhou (2002a) that K≈8.4*W_p > W_p for the story where progressive collapse initiated.

It is clear that upon initiation, g*m(z), the force exerted by 58.000 t after a 3.7m free fall is huge, therefor W_g = F*u_f = g*m(z)*u_f = K is quite big too. We find out that it is 8.4 times greater than W_p, the energy needed to crush one floor.

This is our first input into the little program we have just written. From that, we derive W_c, W_b, dF_d, dF_a and u_c for the next step, depending on the shape of the curve for F(u) as shown in Fig. 3, using Eq. 3&4.

Now comes the next floor. Clearly, W_g (or K) is not 8.4*W_p any more, because some of the kinetic energy went into the deformation of the first floor which means that our next K = W_g-Φ(u). In other words, we draw a new line in Fig. 3 for F=m*g [g*m(z), that is] below the old one, because the force that now acts on the second floor has become smaller (one could also say that the "free fall drop" now isn't 3.7m anymore, but u-u_c). That means that W_c or (Φ(u)) becomes bigger, and, consequently, W_b becomes smaller. We check IF K<W_c THEN RETURN "collapse arrested!", ELSE we GOTO [start].

We will assume that all curves for F(u) as shown in Fig. (3) of each floor look the same, (although in reality, peak F_0 will move up and u_c shift a few metres to the right because the columns downstairs are a little stronger). So, with every step, we derive a new g*m(z) and draw the new line into the curve shown in Fig. 3, and on each step, we substract Φ(u) from W_g for the next K.

That's my point.

I was so busy writing I didn't notice your replies... you know, I really wonder why in discussions, people transform their agression into funny thought experiments. On another forum, I've been thrown out of a helicopter through a tree already and had to stand with one foot on a coke can, you have me falling through ice plates, stab me and punch me. Interesting observation.

It was the other way round, btw... the stab didn't hurt, the punch on the head (if there was one) brought the structure down.

I think calling the towers mechanical bombs is a bit exaggerating.

Wouldn't you also call a 2.1GJ to 981GJ ratio a bit "exaggarated" for something that is meant to be statical? : )

And about Fig. 3, see my other post :-)

Originally posted by Akareyon
Now comes the next floor. Clearly, W_g (or K) is not 8.4*W_p any more, because some of the kinetic energy went into the deformation of the first floor which means that our next K = W_g-Φ(u).

This is where you go wrong. W_g is the released potential energy. At the first floor, W_g is:

W_g1 = mg•2h = 2•2.1GN = 4.2GN

At the next floor it is

W_g2 = mg•3h = 3•2.1GN = 6.3GN

At the floor after that it is

W_g3 = mg•4h = 4•2.1GN = 8.4GN

You get the drift. According to Bazant, the energy that is used up to crush the columns is 0.5GN. This is the value that we plug into the iteration. Here I denote Kx as the kinetic energy before a floor impact and Φ_t(u) the total energy consumed in crushing columns.

So:

K1 = W_g1-Φ_t(u) = 4.2GN - 0GN = 4.2GN
K2 = W_g2-Φ_t(u) = 6.3GN - 0.5GN = 5.8GN
K3 = W_g3-Φ_t(u) = 8.4GN - 1.0GN = 7.4GN

etc.

As you can see, K is increasing with each floor.

Originally posted by Akareyon
I was so busy writing I didn't notice your replies... you know, I really wonder why in discussions, people transform their agression into funny thought experiments. On another forum, I've been thrown out of a helicopter through a tree already and had to stand with one foot on a coke can, you have me falling through ice plates, stab me and punch me. Interesting observation.

I was thinking exactly the same when I read it again. I was wondering why I was hitting and stabbing you . I don't know, maybe we unconsciously add some drama to in an attempt to make the point more clear.

It was the other way round, btw... the stab didn't hurt, the punch on the head (if there was one) brought the structure down.

Sure, different things react differently when exerting different types of energy to them in different ways. I can also stab you non fatally, and you will not fall over, or hit you on your head making you go knockout and fall over.

Wouldn't you also call a 2.1GJ to 981GJ ratio a bit "exaggarated" for something that is meant to be statical? : )

I am not sure what you mean to say here.

Originally posted by -PLB-
This is where you go wrong. W_g is the released potential energy. At the first floor, W_g is:

W_g1 = mg•2h = 2•2.1GN = 4.2GN

At the next floor it is

W_g2 = mg•3h = 3•2.1GN = 6.3GN

At the floor after that it is

W_g3 = mg•4h = 4•2.1GN = 8.4GN

You get the drift. According to Bazant, the energy that is used up to crush the columns is 0.5GN. This is the value that we plug into the iteration. Here I denote Kx as the kinetic energy before a floor impact and Φ_t(u) the total energy consumed in crushing columns.

So:

K1 = W_g1-Φ_t(u) = 4.2GN - 0GN = 4.2GN
K2 = W_g2-Φ_t(u) = 6.3GN - 0.5GN = 5.8GN
K3 = W_g3-Φ_t(u) = 8.4GN - 1.0GN = 7.4GN

etc.

As you can see, K is increasing with each floor.

Waitwaitwait. I assume you meant to say Gigajoules, not Giganewtons. But still, 0.5 GJ/3.7m is just 135 Meganewtons per floor. It was more than 569 MN before that for the first floor impacted, even almost 5 GN for the columns on the first floor...?

Originally posted by -PLB-
This is where you go wrong. W_g is the released potential energy. At the first floor, W_g is:

W_g1 = mg•2h = 2•2.1GN = 4.2GN

At the next floor it is

W_g2 = mg•3h = 3•2.1GN = 6.3GN

At the floor after that it is

W_g3 = mg•4h = 4•2.1GN = 8.4GN

You get the drift. According to Bazant, the energy that is used up to crush the columns is 0.5GN. This is the value that we plug into the iteration. Here I denote Kx as the kinetic energy before a floor impact and Φ_t(u) the total energy consumed in crushing columns.

So:

K1 = W_g1-Φ_t(u) = 4.2GN - 0GN = 4.2GN
K2 = W_g2-Φ_t(u) = 6.3GN - 0.5GN = 5.8GN
K3 = W_g3-Φ_t(u) = 8.4GN - 1.0GN = 7.4GN

etc.

As you can see, K is increasing with each floor.

First off how did you calculate the Pe?

Secondly where do you account for the FoS?

Thirdly, you do not account for the loss of Ke to deformation, sound, heat etc.

Originally posted by Akareyon
Waitwaitwait. I assume you meant to say Gigajoules, not Giganewtons. But still, 0.5 GJ/3.7m is just 135 Meganewtons per floor. It was more than 569 MN before that for the first floor impacted, even almost 5 GN for the columns on the first floor...?

edit on 17-11-2011 by Akareyon because: to remove editby tag from quote

Yes, of course joules, my bad. I explained the issue you have with the forces in a previous post:

What you are doing wrong here is comparing an average force over distance with a force in equilibrium. In this case figure 3 posted earlier gives a lot of insight. I already talked about this is an earlier post. You need to look at the instantaneous force, which is the curved line. At one moment this force is much greater than the static force (mg), and another moment it is much smaller. At the moment it is greater, it is destroying the resistance, at the moments it is smaller, it is almost in free fall, most resistance of the floor is already destroyed. So when you take the average force, it can well be below mg.