On Apr 2, 7:31=A0am, schoenfeld....@[EMAIL PROTECTED]
wrote:
> In this article we show that "top-down" controlled demolition
> accurately accounts for the collapse times of the World Trade Center
> towers. A top-down controlled demolition can be simply characterized
> as a "pancake collapse" of a building missing its sup****t columns.
> This demolition profile requires that the sup****t columns holding a
> floor be destroyed just before that floor is collided with by the
> upper falling m*****. The net effect is a pancake-style collapse at
> near free fall speed.
>
> This model predicts a WTC 1 collapse time of 11.38 seconds, and a WTC
> 2 collapse time of 9.48 seconds. Those times accurately match the
> seismographic data of those events.1 Refer to equations (1.9) =A0and
> (1.10) =A0for details.
>
> It should be noted that this model differs massively from the "natural
> pancake collapse" in that the geometrical composition of the structure
> is not considered (as it is physically destroyed). =A0A natural pancake
> collapse features a dimini****ng velocity rapidly approaching rest due
> the resistance offered by the columns and surrounding "steel mesh".
>
> DEMOLITION MODEL
>
> A top-down controlled demolition of a building is considered as
> follows
>
> =A0 =A0 =A0 =A0 1. An initial block of j floors commences to free fall.
>
> =A0 =A0 =A0 =A0 2. The floor below the collapsing block has its sup****t
st=
ructures
> disabled just prior the collision with the block.
>
> =A0 =A0 =A0 =A0 3. The collapsing block merges with the momentarily
levita=
ting floor,
> increases in mass, decreases in velocity (but preserves momentum), and
> continues to free fall.
>
> =A0 =A0 =A0 =A0 4. If not at ground floor, goto step 2.
>
> Let j be the number of floors in the initial set of collapsing floors.
> Let N be the number of remaining floors to collapse.
> Let h be the average floor height.
> Let g be the gravitational field strength at ground-level.
> Let T be the total collapse time.
>
> Using the elementary motion equation
>
> =A0 =A0 distance =3D (initial velocity) * time + 1/2 * acceleration *
time=
^2
>
> We solve for the time taken by the k'th floor to free fall the height
> of one floor
>
> =A0 =A0 =A0 =A0 [1.1] =A0 t_k=3D(-u_k+(u_k^2+2gh))/g
>
> where u_k is the initial velocity of the k'th collapsing floor.
>
> The total collapse time is the sum of the N individual free fall times
>
> =A0 =A0 =A0 =A0 [1.2] =A0 T =3D sum(k=3D0)^N (-u_k+(u_k^2+2gh))/g
>
> Now the mass of the k'th floor at the point of collapse is the mass of
> itself (m) plus the mass of all the floors collapsed before it (k-1)m
> plus the mass on the initial collapsing block jm.
>
> =A0 =A0 =A0 =A0 [1.3] =A0 m_k=3Dm+(k-1)m+jm =3D(j+k)m
>
> If we let u_k denote the initial velocity of the k'th collapsing
> floor, the final velocity reached by that floor prior to collision
> with its below floor is
>
> =A0 =A0 =A0 =A0 [1.4] =A0 v_k=3DSQRT(u_k^2+2gh)
>
> which follows from the elementary equation of motion
>
> (final velocity)^2 =3D (initial velocity)^2 + 2 * (acceleration) *
> (distance)
>
> Conservation of momentum demands that the initial momentum of the k'th
> floor equal the final momemtum of the (k-1)'th floor.
>
> =A0 =A0 =A0 =A0 [1.5] =A0 m_k =A0u_k =A0=3D m_(k-1) =A0v_(k-1)
>
> Substituting (1.3) and (1.4) into (1.5)
> =A0 =A0 =A0 =A0 [1.6] =A0 (j + k)m u_k=3D (j + k - 1)m SQRT(u_(k-1)^2+
2gh=
)
>
> Solving for the initial velocity u_k
>
> =A0 =A0 =A0 =A0 [1.7] =A0 u_k=3D(j + k - 1)/(j + k) SQRT(u_(k-1)^2+2gh)
>
> Which is a recurrence equation with base value
>
> =A0 =A0 =A0 =A0 [1.8] =A0 u_0=3D0
>
> The WTC towers were 417 meters tall and had 110 floors. Tower 1 began
> collapsing on the 93rd floor. =A0Making substitutions N=3D93, j=3D17 ,
g=
=3D9.8
> into (1.2) and (1.7) gives
>
> =A0 =A0 =A0 =A0 [1.9] =A0 WTC 1 Collapse Time =3D sum(k=3D0)^93
(-u_k+(u_k=
^2+74.28))/9.8 =3D
> 11.38 sec
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 where
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 u_k=3D(16+ k)/(17+ k )
SQR=
T(u_(k-1)^2+74.28) =A0 =A0 =A0;/ u_0=3D0
>
> Tower 2 began collapsing on the 77th floor. Making substitutions N=3D77,
> j=3D33 , g=3D9.8 into (1.2) and (1.7) gives
>
> =A0 =A0 =A0 =A0 [1.10] =A0WTC 2 Collapse Time =3Dsum(k=3D0)^77
(-u_k+(u_k^=
2+74.28))/9.8 =3D
> 9.48 sec
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 Where
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 u_k=3D(32+k)/(33+k)
SQRT(u=
_(k-1)^2+74.28) =A0 =A0 =A0;/ u_0=3D0
>
> REFERENCES
>
> "Seismic Waves Generated By Aircraft Impacts and Building Collapses at
> World Trade Center
",http://www.ldeo.columbia.edu/LCSN/Eq/20010911_WTC/WTC=
_LDEO_KIM.pdf
>
> APPENDIX A: HASKELL SIMULATION PROGRAM
>
> This function returns the gravitational field strength in SI units.
>
> > g :: Double
> > g =3D 9.8
>
> This function calculates the total time for a top-down demolition.
> Parameters:
> =A0 _H - the total height of building
> =A0 _N - the number of floors in building
> =A0 _J - the floor number which initiated the top-down cascade (the 0'th
> floor being the ground floor)
>
> > cascadeTime :: Double -> Double -> Double -> Double
> > cascadeTime _H _N _J =A0=3D =A0sum [ (- (u k) + sqrt( (u k)^2 +
2*g*h))/=
g | k<-[0..n]]
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 where
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 j =3D _N - _J
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 n =3D _N - j
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 h =3D _H/_N
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 u 0 =3D 0
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 u k =3D (j + k - 1)/(j
+=
k) * sqrt( (u (k-1))^2 + 2*g*h )
>
> Simulates a top-down demolition of WTC 1 in SI units.
>
> > wtc1 :: Double
> > wtc1 =3D cascadeTime 417 110 93
>
> Simulates a top-down demolition of WTC 2 in SI units.
>
> > wtc2 :: Double
> > wtc2 =3D cascadeTime 417 110 77
>
> By Herman Schoenfeld
At the begining there is NO K'th Floor and you are spitting out facts
thata just Nigga ain't true *****!!! HAHA YOU SUCK
Parnell Parnell Parnelll
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