WTC Towers: The Case For Controlled Demolition

Le
schoenfeld.one
WTC Towers: The Case For Controlled Demolition
By Herman Schoenfeld

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 support columns.
This demolition profile requires that the support columns holding a
floor be destroyed just before that floor is collided with by the
upper falling masses. 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) and
(1.10) for details.

It should be noted that this model differs massively from a "natural
pancake collapse" in that the geometrical composition of the structure
is not considered (as it is physically destroyed). A natural pancake
collapse features a diminishing velocity rapidly approaching rest due
to the resistance offered by the columns and surrounding "steel mesh".

DEMOLITION MODEL

A top-down controlled demolition of a building is considered as
follows

1. An initial block of j floors commences to free fall.

2. The floor below the collapsing block has its support structures
disabled just prior the collision with the block.

3. The collapsing block merges with the momentarily levitating floor,
increases in mass, decreases in velocity (but preserves momentum), and
continues to free fall.

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

distance = (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

[1.1] t_k=(-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

[1.2] T = sum(k=0)^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.

[1.3] m_k=m+(k-1)m+jm =(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

[1.4] v_k=SQRT(u_k^2+2gh)


which follows from the elementary equation of motion

(final velocity)^2 = (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.

[1.5] m_k u_k = m_(k-1) v_(k-1)


Substituting (1.3) and (1.4) into (1.5)
[1.6] (j + k)m u_k= (j + k - 1)m SQRT(u_(k-1)^2+ 2gh)


Solving for the initial velocity u_k

[1.7] u_k=(j + k - 1)/(j + k) SQRT(u_(k-1)^2+2gh)


Which is a recurrence equation with base value

[1.8] u_0=0



The WTC towers were 417 meters tall and had 110 floors. Tower 1 began
collapsing on the 93rd floor. Making substitutions N“, j , g=9.8
into (1.2) and (1.7) gives


[1.9] WTC 1 Collapse Time = sum(k=0)^93 (-u_k+(u_k^2+74.28))/9.8 =
11.38 sec
where
u_k=(16+ k)/(17+ k ) SQRT(u_(k-1)^2+74.28) ;/ u_0=0



Tower 2 began collapsing on the 77th floor. Making substitutions Nw,
j3 , g=9.8 into (1.2) and (1.7) gives


[1.10] WTC 2 Collapse Time =sum(k=0)^77 (-u_k+(u_k^2+74.28))/9.8 =
9.48 sec
Where
u_k=(32+k)/(33+k) SQRT(u_(k-1)^2+74.28) ;/ u_0=0


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 = 9.8

This function calculates the total time for a top-down demolition.
Parameters:
_H - the total height of building
_N - the number of floors in building
_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 = sum [ (- (u k) + sqrt( (u k)^2 + 2*g*h))/g | k<-[0..n]]
> where
> j = _N - _J
> n = _N - j
> h = _H/_N
> u 0 = 0
> u k = (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 = cascadeTime 417 110 93

Simulates a top-down demolition of WTC 2 in SI units.

> wtc2 :: Double
> wtc2 = cascadeTime 417 110 77
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The Man From Havana
Le #3152561
On Mar 7, 1:48 am, wrote:
WTC Towers: The Case For Controlled Demolition
By Herman Schoenfeld

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 support columns.
This demolition profile requires that the support columns holding a
floor be destroyed just before that floor is collided with by the
upper falling masses. 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)  and
(1.10)  for details.

It should be noted that this model differs massively from a "natural
pancake collapse" in that the geometrical composition of the structure
is not considered (as it is physically destroyed).  A natural pancake
collapse features a diminishing velocity rapidly approaching rest due
to the resistance offered by the columns and surrounding "steel mesh".

DEMOLITION MODEL

A top-down controlled demolition of a building is considered as
follows

        1. An initial block of j floors commences to free fall.

        2. The floor below the collapsing block has its support st ructures
disabled just prior the collision with the block.

        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.

        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

    distance = (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

        [1.1]   t_k=(-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

        [1.2]   T = sum(k=0)^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.

        [1.3]   m_k=m+(k-1)m+jm =(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

        [1.4]   v_k=SQRT(u_k^2+2gh)

which follows from the elementary equation of motion

(final velocity)^2 = (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.

        [1.5]   m_k  u_k  = m_(k-1)  v_(k-1)

Substituting (1.3) and (1.4) into (1.5)
        [1.6]   (j + k)m u_k= (j + k - 1)m SQRT(u_(k-1)^2+ 2gh )

Solving for the initial velocity u_k

        [1.7]   u_k=(j + k - 1)/(j + k) SQRT(u_(k-1)^2+2gh)

Which is a recurrence equation with base value

        [1.8]   u_0=0

The WTC towers were 417 meters tall and had 110 floors. Tower 1 began
collapsing on the 93rd floor.  Making substitutions N“, j , g =9.8
into (1.2) and (1.7) gives

        [1.9]   WTC 1 Collapse Time = sum(k=0)^93 (-u_k+(u_k ^2+74.28))/9.8 =
11.38 sec
                where
                        u_k=(16+ k)/(17+ k ) SQR T(u_(k-1)^2+74.28)      ;/ u_0=0

Tower 2 began collapsing on the 77th floor. Making substitutions Nw,
j3 , g=9.8 into (1.2) and (1.7) gives

        [1.10]  WTC 2 Collapse Time =sum(k=0)^77 (-u_k+(u_k^ 2+74.28))/9.8 =
9.48 sec
                Where
                        u_k=(32+k)/(33+k) SQRT(u _(k-1)^2+74.28)      ;/ u_0=0

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 = 9.8


This function calculates the total time for a top-down demolition.
Parameters:
  _H - the total height of building
  _N - the number of floors in building
  _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  =  sum [ (- (u k) + sqrt( (u k)^2 + 2*g*h))/ g | k<-[0..n]]
                      where
                        j = _N - _J
                        n = _N - j
                        h = _H/_N
                        u 0 = 0
                        u k = (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 = cascadeTime 417 110 93


Simulates a top-down demolition of WTC 2 in SI units.



wtc2 :: Double
wtc2 = cascadeTime 417 110 77- Hide quoted text -


- Show quoted text -



Who cares , get over it.


fish..
Le #3152551
In article <48d4a5cd-f1dd-4253-9519-
,
says...



Who cares , get over it.




problem is, they're so fixated on it, to give it up would be an
admission of their fuckwittery.

They're the new Xtians.

starbuck
Le #3152501
oh for fucks sake we dont give a fucking rats arse ok

news:
WTC Towers: The Case For Controlled Demolition
By Herman Schoenfeld

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 support columns.
This demolition profile requires that the support columns holding a
floor be destroyed just before that floor is collided with by the
upper falling masses. 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) and
(1.10) for details.

It should be noted that this model differs massively from a "natural
pancake collapse" in that the geometrical composition of the structure
is not considered (as it is physically destroyed). A natural pancake
collapse features a diminishing velocity rapidly approaching rest due
to the resistance offered by the columns and surrounding "steel mesh".

DEMOLITION MODEL

A top-down controlled demolition of a building is considered as
follows

1. An initial block of j floors commences to free fall.

2. The floor below the collapsing block has its support structures
disabled just prior the collision with the block.

3. The collapsing block merges with the momentarily levitating floor,
increases in mass, decreases in velocity (but preserves momentum), and
continues to free fall.

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

distance = (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

[1.1] t_k=(-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

[1.2] T = sum(k=0)^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.

[1.3] m_k=m+(k-1)m+jm =(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

[1.4] v_k=SQRT(u_k^2+2gh)


which follows from the elementary equation of motion

(final velocity)^2 = (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.

[1.5] m_k u_k = m_(k-1) v_(k-1)


Substituting (1.3) and (1.4) into (1.5)
[1.6] (j + k)m u_k= (j + k - 1)m SQRT(u_(k-1)^2+ 2gh)


Solving for the initial velocity u_k

[1.7] u_k=(j + k - 1)/(j + k) SQRT(u_(k-1)^2+2gh)


Which is a recurrence equation with base value

[1.8] u_0=0



The WTC towers were 417 meters tall and had 110 floors. Tower 1 began
collapsing on the 93rd floor. Making substitutions N“, j , g=9.8
into (1.2) and (1.7) gives


[1.9] WTC 1 Collapse Time = sum(k=0)^93 (-u_k+(u_k^2+74.28))/9.8 > 11.38 sec
where
u_k=(16+ k)/(17+ k ) SQRT(u_(k-1)^2+74.28) ;/ u_0=0



Tower 2 began collapsing on the 77th floor. Making substitutions Nw,
j3 , g=9.8 into (1.2) and (1.7) gives


[1.10] WTC 2 Collapse Time =sum(k=0)^77 (-u_k+(u_k^2+74.28))/9.8 > 9.48 sec
Where
u_k=(32+k)/(33+k) SQRT(u_(k-1)^2+74.28) ;/ u_0=0


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 = 9.8


This function calculates the total time for a top-down demolition.
Parameters:
_H - the total height of building
_N - the number of floors in building
_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 = sum [ (- (u k) + sqrt( (u k)^2 + 2*g*h))/g |
k<-[0..n]]
where
j = _N - _J
n = _N - j
h = _H/_N
u 0 = 0
u k = (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 = cascadeTime 417 110 93


Simulates a top-down demolition of WTC 2 in SI units.

wtc2 :: Double
wtc2 = cascadeTime 417 110 77




SAM
Le #3152481
In article <48d4a5cd-f1dd-4253-9519-
,
says...


Who cares , get over it.




problem is, they're so fixated on it, to give it up would be an
admission of their fuckwittery.

They're the new Xtians.


et en frenchie ça fait quoi ?


Horry
Le #3152431
On Fri, 07 Mar 2008 11:28:16 +0100, SAM wrote:

In article <48d4a5cd-f1dd-4253-9519-
,
says...


Who cares , get over it.




problem is, they're so fixated on it, to give it up would be an
admission of their fuckwittery.

They're the new Xtians.


et en frenchie ?a fait quoi ?


bonjour! ?a va?



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