Let the "input energy" be that of 12 floors dropping from h = (98+1)*3.7m height right from the start, in other words: let's lift the 12 floors 3.7
metres, but pretend for a moment that the rest of the tower isn't there anymore. From bottom to ground, there are 366.3 metres. K would be huge upon
impact on the ground, wouldn't it? There would be dust and debris all over the place, steel coloums flying everywhere and nothing left of the Top Of
The World. Let us say that K_start=0 before it drops and K_ground=m*g*h when it crushes on the ground. The
potential energy of block C is
K_ground, therefor: E_pot=m*g*h. This is the potential energy of the upper 12 floors, Block C.
Now let's put the rest of the tower back under it. Those 98 stories have suffered nothing yet. We could just put a new roof on it (and be glad we
have a wise president who doesn't scream "worldwide war" over a destroyed building). On each floor, m*g is
still only half the maxwell line,
accordingly, each Φ(u) is still big enough for each floor to support all the floors above. Imagine all the 98 curves for F(u) from Fig. 4c, a
perfect, stable building the way it should be, lined up next to another. It would look a little like a twisted sine wave with 98 peaks. This new
curve we call F(s), with s=98*u. To keep things neat and easy, we don't account for the gradient in strength, so the line for m*g stays the same, it
doesn't move, and so doesn't the maxwell line at 2*m*g. We have changed nothing, we have only lined them up all next to another. Now we can go ahead
and calculate Φ(s) - again, it is the area between F(s) and m*g, it is also the area between the maxwell line and m*g, it is: (2*m*g-m*g)*s=m*g*s.
Φ(s)=98*Φ(u).
There is one catch, though: we have added one floor between the 98th and the 99th floor consisting of air. So, to the leftmost side of F(s), we must
add another u with an F(air) that nearly touches the ordinate. Needless to say that Φ(air) is negative, because its maxwell line is waaay below
m*g, it's almost not there. We have a negative area, but right now, we're mathematicians, we could even sqrt(-1) and have imaginary numbers if we
need them
No, seriously, this is were Block C gains its momentum and K is increased from zero as it drops through the very first u, diminishing
E_pot.
Therefor, now, as we release the upper block, it first gains momentum (a lot of it!) while going through the first u, rolling down F(air). K is here
"diminished" by Φ(air), but as Φ(air) is negative, it isn't diminished, but greatly increased (and, of course, E_pot is greatly
diminished). After that first u, however, it must climb all those little peaks over s. It will gain momentum when it "rolls" down a peak, and climb
another peak, until all K is "gone". When will that be? Easy. When the energy "gone" will be
E_gone = E_pot - (Φ(s)+Φ(air))
E_gone is the energy that goes into deformation, and it is equal to the [potential energy of Block C] MINUS [the area between F(s) and the maxwell
line of the 98 stories below] PLUS [the NEGATIVE area between m*g and the "maxwell line" that represents air resistance F(air) for the leftmost part
of our 98 lined-up displacement curves].
Let's check: if Φ(air) is zero (no drop), the potential energy of Block C is just as big as Φ(s), the "containing" energy that keeps it up,
so their difference is zero, so no energy goes into deformation. If there is a little drop, Φ(air) (the area between F(air) and m*g) becomes
"negative", so the part between the parantheses gets SMALLER than E_pot, so E_gone increases - energy goes into deformation.
I hope I haven't lost you already because I'm talking of negative energy. I know it's not professional, but I'm trying to explain as good as a
layman can, and I enclose a diagram to show what I mean:
Let's try and make K=0. E_gone = E_pot - (Φ(s)+Φ(air)). So, collapse will arrest WHEN [the energy "gone" (into deformation)] IS EQUAL TO
[the potential ("potentially kinetic") energy] MINUS [the area between the looong sinus-like load displacement curve and the maxwell line] PLUS [the
"negative" area at the leftmost part of our curve where it drops in freefall].
Of course, all this is utterly false. It is even complete rubbish! And in the next post, I'll explain why.
