- Dan wrote:

>

First of all I'm sure they believed that they were adequately protected.

> Not really. Both the White House and the Capitol were on the potential

> target list for the planes. There are some arguments that the path of

> the plane that hit the Pentagon indicates that the first target was the

> White

> House, which is surprisingly hard to see from the air....although I think

> "just down Pennsylvania Ave. from the Capitol" should be workable from

> the air.

>

> Unless one assumes that the administration knows something about the

> timing of attacks, which I think we agree is not a reasonable

> assumption, then Bush and Cheney would be at a likely target...and risk

> death. Thus, any

> deliberate looking the other way during such an attack would have to

> involve taking a personal risk.

>

> Just being wrong, of course, requires none of this. For that reason, I

> see it as the simplest explanation.

Secondly, I think you'll find that during that summer Bush was rarely at

the White House, third you may recall how quickly Cheney crawled down a

rat hole at the first sign of danger and finally, you're talking about an

administration that believed that conquering Iraq would be a cakewalk; a

president that was confident that not a single life would be lost in the

effort.

Itâ€™s an administration that just doesn't think these things through very

well.

--

Doug

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http://www.mccmedia.com/mailman/listinfo/brin-l - 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)

(a 9.8))

(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

didn't.

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)

(a 9.8))

(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)

(a 9.8)

(f 6))

(/ (* 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

http://www.rattlesnake.com http://www.teak.cc

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