Inertia physics question

[Bill Haywood]

For anyone interested, here is an email I wrote to a scientist and his response. He indicates that the collapsing upper towers hit the initial floor below with force "several to 10s of times" greater than their static weight.

> Dear Sir,

> I’m looking for a bit of information that would be helpful in explaining the tower collapse to resistant minds.
>
After an initial floor gives way, the upper tower drops 3 meters and smacks into the next floor down.
>
I readily accept that the 3 meters of momentum provides enough force to pulverize anything below.
>
But it would be helpful to express this as a number, as in, “the force of the tower falling 3 meters is comparable to 10x the weight of the upper floors before they fell.”
>
> In your Counter Punch article you state:
>
> “F/(m*g) = 1 + dv/(g*dt) = 1 + 0.5/(9.81*0.01) = 6.1,
>
> a load of six times the weight of the upper block”
>
>I just want to be sure. Is that the figure I’m looking for? The force of the upper block falling 3 meters was comparable to 6x its weight?
>
> I know from trying to ask other people that this requires gross simplifications that go against the nature of scientists. But I’m just trying to make a polemical point that is a fair approximation of reality.

Reply from Manuel Garcia of Lawrence Livermore Laboratory:

Yes. To be precise, yes given the assumptions of the example. The assumptions of the example are reasonable simplifications of real life (e.g., time duration of "contact" at 0.01 seconds). For different numbers (say 0.005 seconds) the estimated magnitude of the dynamic force will be different. In real life, with 3D complexity and over the course of the collapse, there will be a range of contact times. Calculating to minute detail would require massive computation. However, from the example we can see that for reasonable estimates of contact times (e.g., a tennis ball on a racket is only a few thousands of a second) we arrive at a dynamic force that is several to 10s of times the static weight of the upper "static load" of the building.

I arrived at 0.01 seconds as a contact time (a characteristic number) for the WTC Towers floors, based on considerations of the time it would take to engage all of their inertia into bodily motion; this is given by the travel times across and through the floor slabs/frameworks by elastic waves (the same as "earthquake" waves). The large size and thickness of the WTC floors adds time (bigger distances) but the stiffness of metal and concrete (as opposed to sound waves in air and water, and even waves in the looser packing of the uppermost layers of the Earth) make for higher speed waves. It all boiled down to 0.01 seconds, say within a factor of 3 either way.

Good luck. I find that some people are genuinely interested to get over their gaps in physics knowledge, while others are just cemented into their fantasies. Don't waste your time on the latter.

http://www.websurdity.com/2007/02/28/unc...-an-inside-job/

Manuel Garcia article: http://www.counterpunch.org/physic11282006.html

[Silent A]

By the way, my calculations above assume a contact time of zero and so the maximum force becomes a function of the deformation of the falling weight and the scale. With something like an iron weight on a spring scale this is reasonable.

However, for something as complex as a falling floor it's not reasonable to assume near instantaneous contact time so the equations the scientist quotes may well dominate the pure elastic deformation equations I used to answer your initial question.

I also wonder why his "dv" is only 0.5 m/s. An object falling 3 m should be travelling at about 7.5 m/s when it hits.

ETA: OK, I read the pertinent part of the article. "dv" is only 0.5 m/s because the falling floor never comes to rest as the lower floor collapses first.

[PairTheBoard]

[ QUOTE ]
I also wonder why his "dv" is only 0.5 m/s. An object falling 3 m should be travelling at about 7.5 m/s when it hits.

ETA: OK, I read the pertinent part of the article. "dv" is only 0.5 m/s because the falling floor never comes to rest as the lower floor collapses first.

[/ QUOTE ]

That dv looks like the reduction in velocity caused by contact with the lower floor before it collapses. Does he show an estimate for the velocity of the upper floors after their 3 meter descent? I'm wondering if that might be less than what we've been figuring because it's not really a free fall. The failing support structures may be gradually deforming over the 3 meters thereby somewhat cushioning the descent.

Also, he mentions that the energy of the falling upper building is being absorbed by "waves" propagating throughout the lower part of the building. I wonder if that might change things. I have no idea how that kind of thing works.

PairTheBoard

[Silent A]

I know what he's talking about but it's hard to explain.

He's talking about the very instant the upper floor hits the lower floor. In this first fraction (which he estimates to be about 0.01 s) the floor looses 0.5 m/s of its speed as the lower floor deflects under the weight and absorbs some of the weight. He doesn't explain how he calculated this to be only 0.5 m/s but it's certainly possible to do - and probably too complicated for this article.

The forces are transmitted through the structure as pressure waves (basically sound waves). The reason he calculates their speed (the speed of sound in concrete) is to show that the time it takes the pressure waves to travel to the columns is less than the the time it takes for the impact to occur. If the time was greater, there would be time for the pressure waves to dissipate their energy and thus only transfer a fraction of the dynamic load to the columns. Since the time is so short, effectively all the load gets transferred and it's pretty safe to say that the columns were not designed for anything close to this kind of force.

So, in the initial impact (0.01 s) the falling floor loses 0.5 m/s of speed and an effective load 6 times the floor weight reaches the columns and they collapse. Once the first floor goes, the rest is inevitable.

[CORed]

The impact force of the dropped object is going to depend on the physical characteristics of the object and the surface you drop it onto. A one pound steel ball dropped onto a concrete floor will exert a lot more force over a much shorter period of time than a one pound teddy bear dropped onto a pillow. I think you are assuming that there is a single number that answers your question. There isn't The momentum and kinetic energy involved are easily calculated, but the impact force isn't without more information.