Stephen L. Keeling

Institute for Mathematics and Scientific Computing
Karl-Franzens University of Graz
A-8010 Graz, Austria

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Mathematical Basis of the Model used in wtc.nlogo

Introduction. First consider the simple example of determining the trajectories of two masses m1 and m2 which collide with the respective velocities v1 and v2 in the absence of gravitational forces. Let the respective masses after the collision be denoted by m1 and m2 and the respective velocities by v1 and v2.

From the conservation of mass, the total mass is the same before and after the collision:

m1 + m2 = m1 + m2

From the conservation of momentum, the total momentum is the same before and after the collision:
m1 v1 + m2 v2 = m1v1 + m2v2
Note that the velocity vc of the center of mass is given by dividing each side of the last equation by its corresponding side in the previous equation. Thus, vc remains unchanged by the collision.

From the conservation of energy, the total (kinetic) energy is the same before and after the collision:
½ m1v12 + ½ m2v22 = ½ m1v12 + ½ m2v22

Now assume that no mass is exchanged in the collision:
mi = mi
Also since the masses have no volume, the collision process is necessarily one-dimensional in the inertial frame of the center of mass: vi - vc ~ [vi - vc], i=1,2. Then the velocities after the collision are given in terms of the masses and the velocities before the collision by solving the above system of equations to obtain the following:
vi = vc - [vi - vc],   i=1,2
For instance, these formulas show that in case m1=m2, then v1=v2 and v2=v1 hold, as is observed roughly on the billiard table.

If mass is exchanged in the collision to the extent that:
m1 =m1+m2
then the velocity after the collision is determined by the conservation of momentum alone:
v1 = vc.

Now suppose that the mass m1 is in the free fall of gravity starting with velocity u1<0 at time t1. From the assumption that the acceleration due to gravity is a constant -g, the velocity v1(t) increases in magnitude linearly with the time t according to:
v1(t) = u1 - g (t - t1)
and the height x1(t) above the ground decreases quadratically according to:
x1(t) = (X + h) + u1 (t - t1) - ½ g (t - t1)2
when X+h is the starting height. The time t2 at which the height x1(t) has dropped to X is given by solving the equation x1(t2)=X to obtain:
t2 = t1 + [u1 + √(u12 + 2gh)]/g
Thus, the velocity at time t2 is given by v1=v1(t2) or:
v1 = -√(u12 + 2gh)]

(As an alternative to these algebraic formulations, let the positions xi(t) of masses mi above be calculated by integrating Newton's Law mi xi″(t) = -mi g + Δpiδ(t-t2), where -mi g is the force due to gravity and Δpiδ(t-t2), Δpi = mi(vi-vi), is the force due to the colliding particle given in terms of the Dirac delta function δ(t).)

The WTC Model. Now assume that the masses m1 and m2 move on a vertical axis in a constant gravitational field. At some time t1 the mass m1 is located at height X+h with velocity u1<0 while the mass m2 remains supported and stationary at height X with velocity u2=0. At a later time t2>t1 the two masses collide, where t2 is given in the Introduction:
t2 = t1 + [u1 + √(u12 + 2gh)]/g
Just before the collision, the masses m1 and m2 have velocities v1<0 and v2=0. As a result of the collision, parts of the masses m1 and m2 join to form m1, and the combined mass continues with the velocity v1, where v1<0 provided the mass m1 attains a momentum sufficient to overcome forces holding m2 intact and in place. The rest of the mass m1+m2 turns to dust particles ml (l>1) with velocities vl, where the respective momenta ml vl are randomly distributed and sum to zero:
Σl > 1 ml vl = 0.

It is assumed that m1>m1 holds, and therefore the masses after the collision are written as:
m1 = m1 + (1-σ) m2,    0 < σ < 1
and:
Σl > 1 ml = σ m2
After the lowest point of the falling mass strikes the ground, all of the falling mass is assumed to pulverize and to continue in free fall to the ground. Until then, σ is referred to for simplicity as the fraction of m2 turned to dust. Although the factor σ should depend upon the kinetic energy of m1, it is assumed for simplicity that σ is a constant since so much of the towers turned to dust.

To represent the effect of external forces indirectly, let a virtual mass m0 collide with m1 at time t2 with velocity v0>0 so that the force of collision is equal in intensity to that necessary to overcome the sum of forces holding m2 intact and in place. Denote the velocity of the virtual mass m0=m0 after the collision by v0. Then by the conservation of momentum,
m0v0 + m1v1 + Σl > 1 ml vl = m0v0 + m1v1 + m2v2

Thus, with Δp0=m0[v0-v0], the velocity of the falling mass after the collision is determined by the equations above according to:
v1 = [m1v1 + Δp0] / [m1 + (1-σ)m2]

(As an alternative to these algebraic formulations, let Δp0δ(t-t2) denote the external upward force exerted on m1 as resistances holding mass m2 intact and in place are overcome, and integrate Newton's Law as indicated in the Introduction.)

Summary. The formulas used in the WTC Model are the following. Given the time t1 at which the mass m1 began to fall, given the velocity u1<0 at which it began to fall, and given the height h which it will fall before colliding with the next mass, the next collision time t2 is given by:
t2 = t1 + [u1 + √(u12 + 2gh)]/g
Just before the collision, the mass m1 has velocity
v1 = -√(u12 + 2gh).
Just after the collision, the combined mass m1=m1+(1-σ)m2 has the velocity:
v1 = [m1v1+Δp0] / [m1 + (1-σ)m2]
where σ is the fraction of the mass m2 which is turned to dust and Δp0 is the upward momentum resulting as resistances holding the mass m2 intact and in place are overcome by the falling mass m1. If v1 above is positive the collapse stops. The momentum Δp0 is quantified as Δp0 = Nm2√(2gh) in terms of the minimum number N of floors, each also with mass m2, necessary to fall from a height h and overcome the resistances holding the mass m2 intact and in place.

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