Re: DATA on collapsing WTC

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

MY REPLY FOLLOWS AT THE END.

"Glenn A. Carlson" wrote:

> I've posted my graph of the position v. time data including best fit power
> law curves.  http://www.stchas.edu/faculty/gcarlson/physics/wtc/index.htm
>
> I'm getting very different results.  For all the data (start of collapse =
> 0 s), I find that the tower is nearly in free fall, but during the first
> second the tower is falling at nearly a constant speed.
>
> It seems reasonable to me that early in the collapse the tower might not
> be in free fall, because the lower intact structure would exert an upward
> force on the falling floors.  But later in the collapse as the falling
> floors gain momentum (more floors and more speed) any upward force
> from the intact structure would have a negligible effect.  The building
> would fall as if the lower floors were not even there.
> Glenn
>
> At 02:00 AM 9/19/01, you wrote:
> > Date: Tue, 18 Sep 2001 22:04:27 -0400
> > From: Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>
> > Subject: Re: GLEN'S DATA on collapsing WTC
> >
> > There is something suspicious with data I used to get the second
> > line of my table. Perhaps it was my clerical error, I will check it
> > later. Everything else looks good. The v versus t plot is not a
> > straight line. As one might expect the slopes fluctuate, reflecting
> > what was happening below. At t close to 1.1s the acceleration
> > was nearly constant and large (7 m/s^2) but at t close to 2.6 s it
> > was nearly zero. The overall acceleration (the best straight line
> > through all points was about 2.7 m/s^2. At no time was it a free
> > fall because the acceleration was never larger than 7 m/s^2. In
> > other words, the net force down was never as large as m*g.
> >
> > Ludwik Kowalski wrote:
> >
> > > SEE MY TABLE BELOW.  FIRST I WANT TO THANK
> > > Glenn A. Carlson,  [from St. Charles Community College,
> > > St. Peters, MO  USA] for sharing his digital data on the
> > > collapsing tower with the list; and for allowing us to use the
> > > data, noncommercially. I am sure that many physics teachers
> > > will appreciate what he did. Glen's e-mail address is:
> > > gcarlson@chuck.stchas.edu
> > > ***************************************
> > > For convenience I decided to group data into sets of nine
> > > frames. For each set I calculated the mean distance and
> > > converted it into meters. I ignored the first 9 frames;
> > > but "invented" the first line (y=0 at t=0). Is it OK?
> > >
> > > For many of us averaging nine distances is acceptable.
> > > In my table times are as specified by Glen for the middle
> > > frame of each set. Do not take all the digits shown as
> > > significant; I prefer to exaggerate. The last line in my
> > > table is from the set of last 6 frames.
> > >
> > > Here is my table:
> > >
> > > t (s)         y (m)
> > > ********************
> > > 0.000         0.000
> > > 1.108        -1.561   <-- Here I must look at Glen's data again
> > > 1.712        -3.444
> > > 1.981        -5.359
> > > 2.283        -8.577
> > > 2.585       -13.245
> > > 2.887       -17.839
> > > 3.189       -23.505
> > > 3.491       -29.630
> > > 3.760       -35.129
> > > 4.062       -42.110
> > > 4.331       -48.541
> > > 4.633       -54.059
> > > 4.884       -61.215
> > >
> > > Ludwik Kowalski

I checked and found no clerical error in the way of reducing
nine Glen's data points into one line (t=1.108s  y=-1.561m).

Here are last lines of the above table showing calculations of v.
I have no time to check other values of accelerations now.

 t(s)              y(m)
4.062       -42.110
4.331       -48.541  |v|=(48.54-42.11)/(4.33-4.06)=4.63/0.27=17.15 m/s
4.633       -54.059  |v|=(54.06-48.54)/(4.63-4.33)=5.52/0.30=18.40 m/s
4.884       -61.215

On the basis of the two speeds above a=(18.40-17.15)/0.30=4.16 m/s^2
Unless my times and distances are wrong (I do not see where) the above
case shows that at t~4.5 seconds "a" is below 1/2 of 9.8m/s^2.
Where did I goof?
Ludwik Kowalski