Tuesday, January 16, 2007

9/11 collapse issues

Draft 3 includes some editorial changes, but no physical or mathematical changes.

Preface to Draft 2
This revised post includes data on elastic collisions and not only perfectly inelastic collisions.

The reader may wonder whether my simplifications suffice, but I have attempted to ensure that real-world events could only have taken longer.

In some extremely spare models, the collapse times are close enough to recorded times to make the reader wonder whether there is a case against the government story here.

I suggest that these spare scenarios are too favorable to the government. But, then, there's a faint chance that they are not. So what is needed is further investigation of this matter.

The NIST said it could find no evidence of the use of planted explosives and yet ducked a scientific analysis of, or even informed opinion about, fall times. Obviously such an analysis is potential evidence, meaning the government did not want to find evidence of explosives.

Clearly, the public investigations thus far are woefully inadequate.

See 'Scientists clash over 9/11 collapses' at http://911science.blogspot.com

Two crucial issues confront those concerned with the collapses of each of the twin World Trade Center towers: the symmetry of collapse and how long the collapse took.

I. The issue of symmetry
Each World Trade Center tower had a core of 47 load-bearing columns which supported the weight of the floors and walls. Think of a square table with a square central pole for a support. The table is a floor, the pole is the core. Now imagine a double-decker table where the pole passes through a hole in the first deck. Make an n-decker table with n-1 holes. Now attach outer walls to the table tops (floors).
We see that the weight of the structure is borne by the pole.

The 47 core columns, which surround the elevator and stairwell shaft, are lashed together by cross beams.

The floor slabs are attached to the outer walls and core columns with joists that are intended to bear the load of the floor slab and the materials and persons on it (there are 4 slabs per story, as I recall) and a portion of the exterior wall.

Now, back to our n-deck table. Drop a heavy object onto the top deck. What happens?
If the object is dropped dead center onto the center pole, either the pole buckles or it holds. If the object strikes off center it may crash through that section of the table, driving the table fragment ahead of it, until it hits the next deck and repeats the process. Possibly the object and its accreted table fragments may drive a piece of the bottom deck to the ground.

Now if each table deck is composed of four sections, one would expect that a number of table sections would be unaffected by the sequence of collisions.

So it is necessary for the pole to collapse also in order to have the table decks collapse symmetrically.

Now let's modify the pole and say that it is composed of a group of spaced-apart steel rods that are composed of smaller rods strongly bound at endpoints. These composite rods are driven into the ground and braced with criss-cross steel supports.
Let's drop a group of loosely tied heavy objects onto the center of the top deck.
It's possible these objects break their loose bonds and disperse symmetrically, crashing through every deck pretty much symmetrically. But, how does the pole collapse? Let's consider a single rod. That rod can only lose altitude if compressed downward or if wrenched sideways and then knocked loose from braces and the connector to the rod segment below. In the first case, it seems unlikely there would be enough force to compress all rods. For most rods to fall, each rod component would have to be bent sideways and then snapped loose.

So thinking of a single rod again, we envision a table segment driven down by falling debris but hanging onto the rod, which is wrenched sideways and yanked loose from its link so as to join the debris flow. It and all the other rods must behave this way, overcoming the resistance not only of connectors but of the criss-cross brace system.

So it seems likely that sets of rods -- up to 11 at a time -- must be jerked sideways and driven down by falling debris. That is, a deck quadrant's joists must hold for collections of interlocked core columns long enough to pull those columns over.

In our model, we arranged that the objects would fall symmetrically.
But in the case of each trade center tower, the top block fell asymmetrically: tilting off center.

So it becomes rather problematic that no core columns remained standing above a few floors for either tower.

II. The issue of fall times
Though there has been a degree of official cageyness about the collapse times of the two main towers and building 7, no one suggests a collapse time of more than 15 seconds for either tower, and most records give shorter times. The 9/11 commission gave WTC2's fall time as 10 seconds but the National Insitute for Standards and Technology reportedly came up with a fall time of 12 seconds. I reviewed an ABC videotape of WTC2's collapse and found that the onscreen timer yielded between 14 and 15 seconds -- though internet videos cannot be trusted because of issues of data compression and data reconstruction and also because one cannot be sure the video has not been tampered with. For purposes of calculation, then, we have the reported official limit of 12 seconds and a non-authoritative outside limit of 15 seconds.

There has been debate as to whether the towers fell at near the free-fall rate. Critics say that such speeds are consistent with controlled demolition and not with top-down gravitational collapse. If one dropped a rock from the top of a trade center tower, the rock could not take less than 9.22 seconds to hit the ground. This calculation neglects air resistance, which in fact has a measurable effect, but which can be reasonably ignored in a simplified scenario [see footnote A].

So a reputed fall time of about 10 seconds for WTC2 would be startlingly suspicious. However, because of problems with the official records, we cannot be sure of that 10-second figure. (See Trade center collapse times: omissions and disparities at http://www.angelfire.com/ult/znewz1/fallrates.html)

So what fall time should be expected for a trade center tower that collapsed as the government alleges?

Let us think of our model above. In that case, we might imagine, without doing the necessary computerized calculations, that the core supports could be snapped sideways rather quickly and that the gathering momentum of "snowballing" debris would result in a rather speedy fall time.

However, this leaves unresolved the issue of how all core columns came to be leveled. A way out would be to suggest that the top block tilted enough so that a considerable amount of mass fell directly onto the core, causing a massive buckling and collapse, which reached down x number of meters. The block fell, with little obstruction through the x meters, before encountering a relatively secure set of core columns again. Upon impact, the scenario was repeated.

This scenario would then explain how the core could have been symmetrically reduced.

It turns out that we can use a simple trick to see whether fall times in this scenario work with respect to observed fall times.

Here is the trick: the strength of the core along some arbitrary number of meters must be enough to hold the entire weight of the building above. Now we know that the weight is simply w = -gm. In that case, the core length in question must have a static force of AT LEAST f = mg. In fact, the core was designed to stand far more force, which is why we needn't worry about the added weight of the jetliner.

Also, the core increases in strength -- in upward force -- as altitude decreases. This is useful, since it further justifies our little trick.

By way of example, picture a standing structure composed of three one-kilogram bricks, A, B, C, of 1 cubic meter each. The normal force at ground zero is n = 3g. At one meter high, n = 2g. Here n is the reaction force of brick A (and the earth) pushing back against the weight of bricks B and C. At h = 2m, n = g and at 3m, n = 0.
Now how much weight could brick A actually hold? This depends on the strength of the material and, if the brick was carved into, say, an arch, the design of the load bearing structure. So we would have f = x2mg, where x is some real number, presumably greater than 1.

Now if f = xmg, then either the acceleration a = xg or there is a mass M such that M = xm. For purposes of calculation, it is immaterial which we choose. So, as we shall see, the choice M = xm is convenient for our purposes.

The trick is to employ the formulas for collision and use the opposed STATIC forces as our initial values. That is |cmg| is greater than or equal to |-mg|, with c some low constant. Since mass means resistance to acceleration, the m represents either the mass of the supported block or the resistance to that block, as in an appropriately designed core segment resisting gravitational acceleration not only with mass but with load distribution effects.

Perfectly inelastic collision
Supposing that the static supporting force equals the weight (f = w), the formula for the inelastic collision of two blocks, A and B, is

va+b = (mava + mbvb)/(ma + mb).

Realizing that we have ma = cmb, we can write, for c = 1,

va+b = ma(va + 0)/2m1

= va/2

For f = 2w, we have

va+b = va/3

and we generally, va+b = va/k

At this juncture, I'm going to cheat a bit in favor of the government. For one thing, I am essentially calculating how long it might take the core to collapse. I am including the mass of the remainder of the building in our considerations and am permitting it to fall relatively unimpeded, though it can be included as part of the force at a height where the core resists.

I am also going to assume, for purposes of calculation, that once the core column segments buckle and lose strength on impact, their masses are all collected up to the height of impact. Though this can't happen in the real world, a more realistic scenario of distributing the mass down the x meters affected can only reduce impact force at the next resistance level and lengthen fall times.

I estimate that WTC2's collapse began 320 meters high. I then posit the block of 100 remaining meters falling through x meters with little resistance until hitting a height where the resistance force is greater than or equal to the static force (weight) of the initial block plus its accreted floors and other building parts.
I have also decided upon making x an average of "free-fall" meters. In reality, such a scenario would yield an irregular number of meters per "free-fall" stage.

For ease of calculation, I rounded off all numbers at the second decimal place. The round-off error then results in several instances where fall times for particular stages are given as equal when they differ slightly. However, since we are dealing with only 320 meters, the roundoff effect shouldn't affect the results enough to change the implications.

Kinematic free-fall equations yield:

vimpact = (vo2 + 19.6x)1/2

t = (vi - vo)/9.8

For two levels of resistance above ground, I use heights of 320-108, 320-108-106. There are another 106 meters to the ground.

Below is a table for x = 108, 106, 106; k = 2

Stage 1
vo = 0
vi = 28
t = 4.69

Stage 2
vo = 14
vi = 47.68
t = 3.44

Stage 3
vo = 23.84
vi = 51.44
t = 2.82

Stage 4
The final stage models the roof collapsing 100 meters to the ground with negligible resistance
vo = 51.44
t = 1.68

Total fall time: 12.68 seconds

I omit the remaining tables
x = 108, 106, 106, k = 3

Total fall time: 12.7 seconds
x = 108, 106, 106, k = 3.5

Total fall time: 13.73 seconds
x = 80, k = 2

Total fall time: 13.08 seconds
x = 80, k = 3

Total fall time: 14.62 seconds
x = 80, k = 3.5

Total fall time: 14.92 seconds
x = 40, k = 2

Total fall time: 16.85 seconds
x = 40, k = 3

Total fall time: 19.5 seconds

Now these estimates must be regarded as extreme lower bounds because serious sources of resistance have been ignored. Also, assuming symmetrical collapse and leveling of the core, the last calculation is the most reasonable of the set.

Now if we accept NIST's reported time estimate for WTC2, then a new investigation into the cause of collapse is undeniably warranted.

If we accept my non-authoritative 15-second estimate, the NIST theory still has only limited plausibility because their model could only conceivably work only when our parameters are far too generous.

Elastic collision
When one mass is at rest, the elastic collision [see footnote B] formula, with i indicating pre-collision and f post-collision, is

v1f = v1i(m1 - m2)/(m1 + m2)

v2f = v1i(2m1/(m1 + m2)
We now examine x = 80 meters and m1 = m2.

In this case the post-collision velocity of block 2 is simply that of block 1 prior to collision. The post-collision velocity of block 1 is 0 (in our free-fall frame of reference).

So for stage one's drop of 80 meters, we get 4.04 seconds and for stages 2 through 4 we get 3(1.67) seconds, and for the final stage when we time the free-fall of the roof to the ground from 100 meters we have 1.57 seconds for a total of 10.62 seconds. This "billiard-ball" scenario is, of course, far too generous to the government.
With m1 = m2 and x = 40, the total collapse time is 13.14 seconds. Again, this scenario seems much too generous.
With m1 = m2 and x = 20, the total time is 17.11 seconds.

Now suppose we posit m2 = 2m1. In that case,

v1f = v1i(m - 2m)/(m + 2m) = -2/3v1i

v2f = v1i(2m/4m) = 1/2v1i

We see that block 2 has the same velocity as the entangled block in the inelastic case where m2 = 2m1. As the upward momentum of the recoiling upper block cannot increase the lower block's speed, we know that this model cannot produce total collapse times that are less than those for our inelastic model. [See footnote B]

Footnote A:
Since the distance is short, I calculated air resistance as kv for the upper block falling through 320 meters of air only. Obviously, this isn't realistic. For one thing, the "block" falls in various pieces, and smaller masses tend to have more air resistance.

But, anyway, for the record:

WTC7 had 770,000 tons of steel. That weight plus, I guessed, 10 percent for other materials and content is 847,000 tons. These figures lead to an estimated weight for the top block of 3.81*10^8 Newtons.

The linear differential equation I used was

v = (mg/k)(1 - e-kt/m)

For k = 2, 3 and 30, roundoff values are all 79.19 m/s. The free-fall value is 79.18 m/s. So fall time for the bottom of the block rounds off to 8.08s in both cases. The actual difference is about a thousandth of a second.

Footnote B
The coefficient of restitution e quantifies the elasticity of the system thus:

v2f - v1f = -e(v2i - v1i

When e = 1, a collision has no inelasticity; when e = 0, a collision is perfectly inelastic. Values of e between 0 and 1 give real-world degrees of elasticity.
However, in our models, setting 0 < e < 1 cannot yield bounds lower than those listed.