Re: More on progressive collapse at WTC.
- On 05/07/2006, at 12:10 AM, Robert Seeberger wrote:
> Yet kerosene in an semi-enclosed space likely doesn't burn much hotterThere is plenty of draught in a high-rise with smashed windows.
> than 350C.
> You have to have free access to oxygen to even get it to burn at 500C.
> 800C is a best case temp and WTC was not even close to a best case.Plot of an office fire with and without sprinklers here:
You'll see the real fire reached 1400 degrees for a while.
>Individually, no. Together, they show a myriad of conditions which
> I'm not saying there was a grand conspiracy, but that the
> explanantions so far are incomplete and do not fully explain the
combined to weaken the structure enough to bring the thing down.
>Molten Alaska? ;-)
> I need to ask you for a cite on that one. I can't find any information
> that posits molten AL at the Pentagon.
I remember a report from the time, but can't find a cite right now,
so disregard it for the moment.
>>>It is. But it's hardly a "lack of a hole", it's a big hole.
>>> And then there is the question of the hole in the Pentagon, or
>>> rather the lack of a hole.
>> Ask Matt Rhodes about the hole.
> That's a fairly easy one actually.
- But hang on, 110 stories, 0.9 seconds per story? isn't that at
least 99 seconds for the whole thing? And yet the total time is
more like 10 seconds isnt it?
It takes approximately 10 seconds for an object not resisted by
anything causing friction to fall 110 stories.
In Emacs Lisp, the calculation is:
(let ((s 400)
(sqrt (/ (* 2 s) a)))
--> 9 seconds, which is `approximately 10 seconds' and presumes
the falling object is not subject to any kind of friction
My memory, from seeing the second building collapse, is that it took
something around that length of time.
If it takes say 10 seconds to fall 400m, each floor is maybe more
like 0.1 of a second.
I do not understand you at all. The velocity of a falling object
increases until external friction produces enough drag. I don't see
air causing that much drag on a floor. Other floors might, but
Also, you don't need to find out the stress-strain curves for the
bolts. Instead you can figure that conservative engineers design a
resting structure, like the floor of a building, to take no more than
three times its maximum expected load. To design for more is to cost
the investors extra money and to reduce their profit. (Some buildings
were designed to higher limits, of course; but I remember the `safety
factor of three' from a long time ago.)
So we should calculate how far the bolts would have to give in order
to prevent a more than 3 g deceleration of the load falling on it.
(It does not matter how many bolts there are. Actually, we should
take into account the weight of the floor itself, but we are not, so
the situation is worse since the floor itself would add to the total
weight its bolts must bear.) And to tell us more, we can presume a
`safety factor of six' as well.
In one g, the floor above weights roughly as much as the floor below.
Let us find the distance needed to decelerate three times that weight
(or six); both of those distances will be less than the distance
needed to dedecelerate at one gravity, which is the acceleration at
which the top floor fell.
The first drop is slow since the floor is initially standing still.
It takes a bit more than 3/4 second to fall the first three meters. I
figure that three meters is small for a floor; Andrew Paul figures
that 110 stories take up 400 meters, or about 3.6 meters per floor or
about 12 feet. But the distance needed for minimal deceleration is
smaller the less distance the falling floor falls, so let's presume 3
meters rather than 3.6 meters. And let's presume the velocity reached
is 7 meters per second. By calculation, it is a little more than that
for the first drop, but a higher velocity is worse, so 7 meters per
second it is.
With a safety factor of 3, the deceleration takes more than 80 cm.
With a safety factor of 6, the deceleration takes more than 40 cm.
The bolts will not stretch either distance. So the lower floor falls.
Here are the calculations in Emacs Lisp (I am using floating point,
hence the appearance sometimes of erroneous accuracy):
(/ 400.0 110.0)
;; --> 3.6 meters for each floor, presuming they are equal, which
;; they weren't.
(let ((s 3.0)
(sqrt (/ (* 2 s) a)))
;; --> 0.78 second for floor to fall 3 meters in a frictionless space
(* 9.8 0.78)
;; --> 7.64 meters per second, the speed of the falling floor after
;; falling 3 meters in a frictionless space
;; s = 0.5at^2
;; t = v/a
;; s = 0.5 a v^2 / a^2
;; = v^2 / 2a
;; therefore, s = v^2 / 2a where a is one gravity deceleration
;; safety factor of 3
(let ((v 7.0) ;; velocity of falling floor
(a 9.8) ;; gravitational acceleration on Earth's surface
(f 3)) ;; safety factor
(/ (* v v) (* 2 a f)))
;; --> 0.83 meters, the distance needed for 3 gravity deceleration
;; to reduce the speed of the falling floor from 7 meters
;; per second to zero
;; safety factor of 6
(let ((v 7.0)
(/ (* v v) (* 2 a f)))
;; --> 0.42 meters, the distance needed for 6 gravity deceleration
This provides you a way to convince yourself through your own thinking
that the `pancake theory' is highly suggestive. You do not need to
depend on anyone else, on good or bad hearsay, to reach your conclusion.
Robert J. Chassell
bob@... GnuPG Key ID: 004B4AC8