# 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

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

Inscrivez-vous!The Man From HavanaWho cares , get over it.

fish..,

says...

problem is, they're so fixated on it, to give it up would be an

admission of their fuckwittery.

They're the new Xtians.

starbucknews:

SAMet en frenchie ça fait quoi ?

Horrybonjour! ?a va?

Publicitéinscrire, c'est gratuit !.ici.