- #1
Ackbach
Gold Member
MHB
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$\newcommand{\un}[1]{\,\text{#1}}$
https://www.youtube.com/watch?v=xSVHWiZu8NM
One of my relatives sent me a link to a video (above) claiming that the WTC towers must have had internal demolition in order to fall as fast as they did. I think there are good reasons to oppose the theory, but I thought it as well to do some simple calculations. Here's a model, upon which I invite comment, based on the following assumptions:
1. The collapse started with only the top floor at rest.
2. Each fall occurs with no wind resistance.
3. When a collection of floors impacts the next floor, the collision is perfectly inelastic, and the Conservation of Linear Momentum applies (a simplifying assumption, since the system is actually not closed in the $y$ direction). Also, the target floor offers no inherent resistance to joining the floors above it due to structural weakness as recorded in the NIST investigation.
4. The floors have equal mass.
5. A WTC tower is $415\un{m}$ tall, and $110$ stories. We assume the stories are evenly spaced at $3.77\un{m}$.
With these assumptions, you get (I think) a relatively simple model: free-fall, collision, free-fall, collision, etc. The two relevant equations are the kinematics equation with no time: $v_{n+1}^2 = v_{n}'^2 + 2 a \Delta y$, and the Conservation of Linear Momentum for $n$ floors hitting the next floor: $(n+1) v_{n}' = n v_{n}$, where $v_{n}'$ is the velocity right after the collision, and $v_{n}$ right before. Solving the second equation for $v_{n}'$ yields
$$v_{n}'=\frac{n}{n+1} \, v_{n},$$
and plugging this into the kinematic equation yields
\begin{align*}
v_{n+1}^2&=\left(\frac{n}{n+1} \, v_{n}\right)^{\!\!2}+2 a \Delta y = \left(\frac{n}{n+1}\right)^{\!\!2} v_{n}^2 + 2 a \Delta y, \qquad \text{or} \\
v_{n+1}^2&=\left(\frac{n}{n+1}\right)^{\!\!2} v_{n}^2 + 2 a \Delta y.
\end{align*}
This is a nonlinear recurrence relation. We would like to get $v_{110}$ in terms of $v_{0}'=0$ - essentially knock out all the middle terms. Unfortunately, Wolfram Development Platform cannot solve this in closed form. However, we can simply run this loop and accumulate the times. The time is given as
$$\Delta t = \frac{v_{n+1}-v_{n}'}{a}=\frac{v_{n+1}-\frac{n}{n+1} \, v_n}{a}.$$
Running this $110$ times yields an accumulated time of $15.052\un{s}$. For reference, a pure free-fall would take
$$t=\sqrt{\frac{2 \Delta y}{a}}=\sqrt{\frac{2 (415\un{m})}{9.81\un{m/s}^2}}=9.2\un{s}.$$
So this answer seems reasonable.
As I explained to my relative, the theory in the video suffers (I think) from the following flaws:
1. The system is definitely not closed, since the force of gravity is acting in the direction of travel. It is an "outside force".
2. As the stories collect more and more mass as the collapse progresses, the incoming mass is much greater than the target mass. Think: instead of a ping-pong ball hitting another ping-pong ball when the first story hits the second, it's more like a bowling ball hitting a ping-pong ball as you get down even to the ten-from-top floor. It's actually not clear how many stories began each collapse.
Other considerations:
3. The NIST web page explains why the internal structure offered little to no resistance. The long and short of it is this: temperatures around 1000 deg C due to the jet fuel fires was enough to weaken the structural steel to $10\%$ of its original.
4. The building was designed to withstand significant static forces, which is quite different from withstanding dynamic forces such as one floor collapsing onto another.
5. There is no evidence of any explosions occurring during the collapse.
6. It is apparently NOT true that the top floor, in its entirety, hit the ground $11$ or $9$ seconds after the start of the collapse. Instead, as you get down to the 60th floor of WTC 1 and the 40th floor of WTC 2, there are significant pauses in the collapses, on the order of tens of seconds, before the rest of the buildings collapse. It is true that the outside portions of the top floors hit the ground then, but that's genuine free-fall!
Thoughts?
https://www.youtube.com/watch?v=xSVHWiZu8NM
One of my relatives sent me a link to a video (above) claiming that the WTC towers must have had internal demolition in order to fall as fast as they did. I think there are good reasons to oppose the theory, but I thought it as well to do some simple calculations. Here's a model, upon which I invite comment, based on the following assumptions:
1. The collapse started with only the top floor at rest.
2. Each fall occurs with no wind resistance.
3. When a collection of floors impacts the next floor, the collision is perfectly inelastic, and the Conservation of Linear Momentum applies (a simplifying assumption, since the system is actually not closed in the $y$ direction). Also, the target floor offers no inherent resistance to joining the floors above it due to structural weakness as recorded in the NIST investigation.
4. The floors have equal mass.
5. A WTC tower is $415\un{m}$ tall, and $110$ stories. We assume the stories are evenly spaced at $3.77\un{m}$.
With these assumptions, you get (I think) a relatively simple model: free-fall, collision, free-fall, collision, etc. The two relevant equations are the kinematics equation with no time: $v_{n+1}^2 = v_{n}'^2 + 2 a \Delta y$, and the Conservation of Linear Momentum for $n$ floors hitting the next floor: $(n+1) v_{n}' = n v_{n}$, where $v_{n}'$ is the velocity right after the collision, and $v_{n}$ right before. Solving the second equation for $v_{n}'$ yields
$$v_{n}'=\frac{n}{n+1} \, v_{n},$$
and plugging this into the kinematic equation yields
\begin{align*}
v_{n+1}^2&=\left(\frac{n}{n+1} \, v_{n}\right)^{\!\!2}+2 a \Delta y = \left(\frac{n}{n+1}\right)^{\!\!2} v_{n}^2 + 2 a \Delta y, \qquad \text{or} \\
v_{n+1}^2&=\left(\frac{n}{n+1}\right)^{\!\!2} v_{n}^2 + 2 a \Delta y.
\end{align*}
This is a nonlinear recurrence relation. We would like to get $v_{110}$ in terms of $v_{0}'=0$ - essentially knock out all the middle terms. Unfortunately, Wolfram Development Platform cannot solve this in closed form. However, we can simply run this loop and accumulate the times. The time is given as
$$\Delta t = \frac{v_{n+1}-v_{n}'}{a}=\frac{v_{n+1}-\frac{n}{n+1} \, v_n}{a}.$$
Running this $110$ times yields an accumulated time of $15.052\un{s}$. For reference, a pure free-fall would take
$$t=\sqrt{\frac{2 \Delta y}{a}}=\sqrt{\frac{2 (415\un{m})}{9.81\un{m/s}^2}}=9.2\un{s}.$$
So this answer seems reasonable.
As I explained to my relative, the theory in the video suffers (I think) from the following flaws:
1. The system is definitely not closed, since the force of gravity is acting in the direction of travel. It is an "outside force".
2. As the stories collect more and more mass as the collapse progresses, the incoming mass is much greater than the target mass. Think: instead of a ping-pong ball hitting another ping-pong ball when the first story hits the second, it's more like a bowling ball hitting a ping-pong ball as you get down even to the ten-from-top floor. It's actually not clear how many stories began each collapse.
Other considerations:
3. The NIST web page explains why the internal structure offered little to no resistance. The long and short of it is this: temperatures around 1000 deg C due to the jet fuel fires was enough to weaken the structural steel to $10\%$ of its original.
4. The building was designed to withstand significant static forces, which is quite different from withstanding dynamic forces such as one floor collapsing onto another.
5. There is no evidence of any explosions occurring during the collapse.
6. It is apparently NOT true that the top floor, in its entirety, hit the ground $11$ or $9$ seconds after the start of the collapse. Instead, as you get down to the 60th floor of WTC 1 and the 40th floor of WTC 2, there are significant pauses in the collapses, on the order of tens of seconds, before the rest of the buildings collapse. It is true that the outside portions of the top floors hit the ground then, but that's genuine free-fall!
Thoughts?