I would just interject at this point that none of this matters, as you still have to get a small chunk of building through a larger chunk of building in under 15 seconds, "naturally", whether the columns are meeting up, or whether they're mangled and look like dried steel noodles, it's all still mass and the upper mass is not stronger than the lower. It may have an initial advantage if it can actually "fall" through the plane impact zone (which it can't) but the advantage stops quickly as that initial impact is shared and transferred between the two colliding bodies. There is no further advantage to be had gravitationally because the lower block easily arrests the upper.
Totally, fully, wholly WRONG. Not a bit of actual understanding of physics involved here.
Most science-ignorant post in recent history.
Listen. An intact column can, by desing, excert a maximum
force upwards of about
2 x m x g
Where m is the mass resting on that column, g is gravitational acceleration, and 2 is the safety factor (you may wiggle that to 3, if you like, won't change the argument much)
It does normaly excert a force of 1 x m x g - the simple static load.
The column must be intact for that - unbent, vertical, and laterally braced. In other word: Part of an undamaged assembly.
Now, if that mass m is not at rest, but moving downards at a speed of v, then our so far undamaged column has to excert more force to arrest that fall:
- It still must bear the mass' weight of 1 x m x g
- In addition, it must decelerate that mass according to delta-v= a x t, where a is the (negative) maximum acceleration associated with the remaining force the column has to spare, and t is the time, measured from beginning of impact. Since the remaining force is (2 x m x g) - (1 x m x g) = 1 x m x g, it follows that a = g
It will take some time until the falling mass has decelerated to 0, and during that time, the mass will travel down a certain distance h = 1/2 a t
2.
It is easy to show (high school physics) that this distance (height) is equal to the height that the mass fell freely just before impact. So, if the mass had fallen only 3 feet at g, or twice as long by 1/2 g, then our intact column would arrest the fall after another 3 feet.
This would imply that the column gets shortened by 3 feet!
Which it can't.
It will seize being an
intact column after a few inches. More likle, the bolts which conncet it to the column piece below will break.
In any case, after a few inches of fall, the force our column can still apply up, will diminish greatly, and pretty soon something will break, and force goes down to 0.
Long before the fall is arrested.
The falling mass, plus the newly added floor, will then fall freely until they hit the next floor.
Where the game starts over.
In an overall picture, it can be shown that the energy needed to break all structural steel elements on all floors is only a small fraction of the potential energy contained in the building's mass on account of its height. So only a fraction of the potential energy will be dissipated by buckling. Another fraction will be dissipated by inelastic collisions with non-structural material (breaking concrete floors etc). But 2/3 of the energy will be available to be converted into kinetic energy. And that translates into a fall at 2/3 of g.
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