Loading ...
Sorry, an error occurred while loading the content.

Re: Physics Prof Finds Thermate in WTC Physical Samples

Expand Messages
  • Doug Pensinger
    ... First of all I m sure they believed that they were adequately protected. Secondly, I think you ll find that during that summer Bush was rarely at the White
    Message 1 of 112 , Jul 4 2:55 PM
    • 0 Attachment
      Dan wrote:

      >
      > 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.

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

      It’s an administration that just doesn't think these things through very
      well.

      --
      Doug
      _______________________________________________
      http://www.mccmedia.com/mailman/listinfo/brin-l
    • Robert J. Chassell
      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
      Message 112 of 112 , Jul 4 7:39 PM
      • 0 Attachment
        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
        _______________________________________________
        http://www.mccmedia.com/mailman/listinfo/brin-l
      Your message has been successfully submitted and would be delivered to recipients shortly.