Via two simple models, Kausel was able to determine that the fall of the upper building portion down onto a single floor must have caused dynamic forces exceeding the buildings design loads by at least an order of magnitude. He also performed some computer simulations that indicate the building material fell almost unrestricted at nearly the speed of free-falling objects. "The towers' resistive systems played no role. Otherwise the elapsed time of the fall would have been extended," he noted. As it was, the debris took about nine seconds to reach the ground from the top.
http://www.public-action.com/911/jmcm/sciam/The two towers of world trade center in New York were impact by two aeroplanes on Sep. 11 of 2001. The two towers were collapsed completely and more than 3 thousand people died in this affair. This affair obtained everyones focus and many explanations are put forward for the reason of this collapse, including fire damage, heaped load, second-damage and so on. Because the collapse of the towers is very complex and it is difficult to recur by test, most explanations are based on qualitative analysis. However, the computer technology applied us a chance to simulate the collapse in the computer, so that the reason for the collapse can be discussed, as well as the methods can be raised to avoid this disaster.
http://www.luxinzheng.net/publications/english_WTC.htmHeh! Heres one from a physics blogger..
When I did the calculations, what I got for a thousand feet was about nine seconds- let's see,
d = 1/2at^2
so
t = (2d/a)^1/2
a is 9.8m/s^2 (acceleration of gravity at Earth's surface, according to Wikipedia),
d is 417m (height of the World Trade Center towers, same source)
so
t = (834m/9.8m/s^2)^1/2 = 9.23s
OK, so how fast was it going? Easy enough,
v = at
v = (9.8m/s^2 x 9.23s) = 90.4m/s
So in the following second, it would have fallen about another hundred meters. That's almost a quarter of the height it already fell. And we haven't even made it to eleven seconds yet; it could have fallen more than twice it's height in that additional four seconds. The time for it to fall from start to finish was three and a half times its own height.
Let's see:
KE = 1/2mv^2
The mass of the towers was about 450 million kg, according to this. Four sources, he has. I think that's pretty definitive. So now we can take the KE of the top floor, and divide by two- that will be the average of the top and bottom floors. Then we'll compare that to the KE of a floor in the middle, and if they're comparable, then we're good to go- take the KE of the top floor and divide by two and multiply by 110 stories. We'll also assume that the mass is evenly divided among the floors, and that they were loaded to perhaps half of their load rating of 100lbs/sqft. That would be
208ft x 208ft = 43,264sqft
50lbs/sqft * 43264sqft = 2,163,200lbs = 981,211kg
additional weight per floor. So the top floor would be
450,000,000 kg / 110 floors = 4,090,909 kg/floor
so the total mass would be
4,090,909 kg + 981,211 kg = 5,072,120 kg/floor
Now, the velocity at impact we figured above was
90.4m/s
so our
KE = (5,072,120kg x (90.4m/s)^2)/2 = 20,725,088,521J
So, divide by 2 and we get
10,362,544,260J
OK, now let's try a floor halfway up:
t = (2d/a)^1/2 = (417/9.8)^1/2 = 6.52s
v = at = 9.8*6.52 = 63.93m/s
KE = (mv^2)/2 = (5,072,120kg x (63.93m/s)^2)/2 = 10,363,863,011J
Hey, look at that! They're almost equal! That means we can just multiply that 10 billion Joules of energy by 110 floors and get the total, to a very good approximation. Let's see now, that's
110 floors * 10,362,544,260J (see, I'm being conservative, took the lower value)
= 1,139,879,868,600J
OK, now how much is 1.1 trillion joules in tons of TNT-equivalent? Let's see, now, a ton of TNT is 4,184,000,000J. So how many tons of TNT is 1,139,879,868,600J?
1,139,879,868,600J / 4,184,000,000J/t = 272t
Now, that's 272 tons of TNT, more or less; five hundred forty one-thousand-pound blockbuster bombs, more or less. That's over a quarter kiloton. We're talking about as much energy as a small nuclear weapon- and we've only calculated the kinetic energy of the falling building. We haven't added in the burning fuel, or the burning paper and cloth and wood and plastic, or the kinetic energy of impact of the plane (which, by the way, would have substantially turned to heat, and been put into the tower by the plane debris, that's another small nuclear weapon-equivalent) and we've got enough heat to melt the entire whole thing.
Remember, we haven't added the energy of four floors of burning wood, plastic, cloth and paper, at- let's be conservative, say half the weight is stuff like that and half is metal, so 25lbs/sqft? And then how about as much energy as the total collapse again, from the plane impact? And what about the energy from the burning fuel? You know, I'm betting we have a kiloton to play with here. I bet we have a twentieth of the energy that turned the entire city of Nagasaki into a flat burning plain with a hundred-foot hole surrounded by a mile of firestorm to work with. - Schneibster
Despite repeated calculations showing that the energy released simply from the kinetic collapse is on the close order of a small nuclear weapon, without even mentioning the energy contents of the millions of tons of paper, wood, plastic, etc. that were on the floors and a large percentage of which would be in the rubble pile and heated to ignition point by the heat from the kinetic energy dissipated by the collapse.
My best estimate at 13 psf by 35,000 sf/floor by 110 floors by about 30% combustibles, 60% metal and other non-combustible items, by the energy content of common garbage, gives a lot more energy than the energy of the collapse. The insulation provided in that debris pile was apparently pretty good, and thats not surprising. Rock and concrete really are bad heat conductors, air isnt much better, and steel while capable isnt all that good, as you can tell from the fact that the jaws of the shovel arent melting. Ever hear of rock wool? Its insulation; look it up. Youll get the idea pretty quick.
Theres two more factors Ill throw in: first, a certain amount of the office materials didnt make it into the debris pile, perhaps as much as 10% of it just got scattered all over lower Manhattan island. Second, a few floors worth had already burned. So when the time comes, Ill take three floors out, and then another 10%. Youll be surprised, I think, at how much energy there is involved.
This, by the way, is a place where Jim Hoffman makes a serious mistake; in his paper on the dust cloud, he fails to note that he has to ADD THE HEAT BACK IN when hes totaling things up at the end. This is a violation of conservation of energy, the First Law of Thermodynamics (and a foundational law of physics). The energy dissipated during the fall is about 250 or 300 GJ, and the leftover energy at impact is about 600 GJ. So its about a quarter kiloton of TNT for the North tower and about a fifth of a kiloton for the South tower; thats still a hell of a lot of energy, more than sufficient to liquefy a pretty healthy chunk of steel, and it doesnt change the fact that theres a lot more energy in the office contents.
You should be aware that anytime you do mechanical work, the energy you do it with doesnt just go away or get used up. Energy that does work gets dissipated, and when that happens, it turns to heat. This is a well known fact of physics, specifically thermodynamics, that was proven early (or maybe it was late? no, Im pretty sure it was EARLY) in the nineteenth century by the gentleman for whom the SI unit of energy is named, James Prescott Joule. Go look him up on Wikipedia, or elsewhere if youre a newbie and believe what you read in the newspapers about Wikipedia. He did this experiment where he stirred water in buckets and showed it got hotter.
This, by the way, is a place where Jim Hoffman makes a serious mistake; in his paper on the dust cloud, he fails to note that he has to ADD THE HEAT BACK IN when hes totaling things up at the end. This is a violation of conservation of energy, the First Law of Thermodynamics (and a foundational law of physics).
What distance do you drop the load from? The floor of initial collapse: 79 for the South tower, 97 for the North. Its a variable in the program, you can change it for yourself and run it yourself, its a perl. Interestingly, going from a 39-story to a 13-story falling section doesnt make a great deal of difference in the energy, and makes even less difference in the energy thats left over when the building hits the ground.
A falling building is not like a bomb or a laser beam. But it makes heat all the same- just like all work makes heat. Feel the bottom of the bicycle pump after youve pumped the tire up. Where does that heat come from? Same place as this does.
While a 600GJ bomb would take out ten blocks in any direction, the WTC collapse obviously did not. While thats true, you need to know that conservation of energy says that energy NEVER disappears. It ALWAYS winds up SOMEWHERE, and if this is energy capable of knocking buildings over for many blocks in all directions, and it didnt knock them over, then where did it go and what did it do? Answer: it went into the rubble pile, and it melted and burned stuff in there.
There was energy spent pancaking or snapping supports if you believe those theories (I do not). Whether it was explosives or whether it was sheer mass and momentum that snapped them (and I have excellent reason to believe it was nothing but mass- youll see shortly), it STILL made heat, and that heat STILL went into the debris pile at the bottom. Heat is energy and energy NEVER just goes away.
All the collapse theories say that the weight of the top of the building is what caused the collapse well that is HALF true. It was also pushing UP WITH EQUAL FORCE. This force was largely transmitted into the ground during the collapse, not the rubble afterwards. The STATIC FORCE of the building pushes down and the ground pushes up, when the DYNAMIC FORCE of the collapse occurs, it is local to whatever is moving; this is because its the MOTION that causes the DYNAMIC force, and that force is (and must be, to collapse the building) many times the static forces of the building just standing there.
Now, for the program:
**BEGIN PROGRAM**
#!/usr/bin/perl
$m = 4285500; # mass of one floor (kg)
$mt = 0; # mass of falling section
$fc = 39; # floor count of falling section (39 floors for 2 WTC)
$v1 = 0; # beginning velocity for the current step
$v2 = 0; # velocity at impact
$v3 = 0; # ending velocity for prior step
$p = 0; # current momentum
$ke1 = 0; # kinetic energy at impact
$ke2 = 0; # kinetic energy after impact
$de = 0; # total energy dissipated so far
$a = 9.80665; # acceleration of gravity (constant)
$t = 0; # cumulative time taken
$t1 = 0; # time taken for this step
$d = 3.8; # distance between floors (418m/110 stories)
$mt = $fc*$m; # initialize mass of falling section
$rfc = 110 $fc;# initialize remaining floor count of uncollapsed floors
while($rfc > 0) {
$v1 = $v3; # starting velocity is ending for last step
$v2 = (($v1*2)+((2$a)*$d))**0.5; # impact velocity for this step by formula
print(Impact velocity for story , $rfc, was , $v2, n);
$p = $mt*$v2; # momentum at impact
print(Impulse delivered for story , $rfc, was , $p, n);
$ke1 = ($mt*($v2**2))/2; # kinetic energy at impact
print(Impact kinetic energy for story , $rfc, was , $ke1, n);
$fc++; # increment falling floor count
$mt = $fc*$m; # update mass of falling section
$v3 = $p/$mt; # new velocity
print(Velocity after impact for story , $rfc, was , $v3, n);
$ke2 = ($mt*($v3**2))/2; # kinetic energy after impact
print(Remaining kinetic energy for story , $rfc, was , $ke2, n);
$de += $ke1 $ke2; # add dissipated kinetic energy to total
print(The kinetic energy dissipated for story , $rfc, was , $ke1 $ke2,
n);
$t1 = $d/(($v2 + $v1)/2); # time for this step by formula
print(The time spent collapsing story , $rfc, was , $t1, n);
$t += $t1; # add step time to running total
$rfc; # decrement remaining floor count
}
print(The total time was , $t, n);
print(The total energy dissipated during the collapse was , $de, n);
print(The remaining kinetic energy at the end of the collapse was , $ke2, n
);
**END PROGRAM**
Its a perl, you can download perl for just about anything from www.perl.org or somewhere they point. If youre going to get involved in CS, somewhere youre going to encounter perl, and nows as good a time to learn it as any. I highly recommend the OReilly Press perl book which happens to be by the inventors of the language. Just so you can muddle your way through and derive the equations from the code above, * is multiplication, ** is raising to a power (and dont forget that a fractional power is a root; so **0.5 is the square-root operation). The rest of the symbols are obvious, and the parentheses work the same way as they do in standard math notation. You should be aware that the single = in most languages simply ASSIGNS the value of whats on the right to the thing on the left; usually, youre required to put a single variable on the left of an =. The double == TESTS whether one value is equal to another, returning 1 or TRUE if it is, and 0 or FALSE if it is not.
Now what...