Here you go, Opel:

1. The assertion has been repeatedly made that the towers "fell at 'near free-fall' speeds." Video evidence shows that the falls took somewhere between twelve and fifteen seconds. I suggest 9/11 Research has a pretty good collection of such videos and invite anyone with the courage to question their assumptions to examine these videos and see what conclusion they come to, and then examine the following analysis:

Height of the WTC towers: 417m (Source: Wikipedia's article on the World Trade Center, sidebar on the right (for the totally impaired, I have used the facilities of this site to make a link out of the words, "Wikipedia's article" in the previous sentence, hover over those words and right-click on them and you will see the article)).

Acceleration of gravity: 9.80665m/s² (Source: Wikipedia's article on the standard gee, which is taken from the average acceleration of gravity on Earth).

Equation of motion:
t = √(2d/a) Eq. 1
Where,
t is the time taken to move over the distance,
a is the constant acceleration applied over the distance, and
d is the distance.
In natural language, the time taken to fall a distance, for a falling object that starts from rest and is subject to a constant acceleration, is equal to the square root of the quantity, twice the distance divided by the acceleration.
Derivation:
a. Definition of acceleration:
a = (v₂ - v₁)/t Eq. 2
Where,
a is acceleration,
v₁ is the initial velocity of the object,
v₂ is the final velocity of the object, and
t is the time interval between the measurement of the initial and final velocities.
In natural language, the acceleration is equal to the initial velocity subtracted from the final velocity (i.e. the change in velocity) divided by the time.
b. Definition of average velocity:
v = (v₁ + v₂)/2 Eq. 3
Where,
v is the average velocity,
v₁ is the initial velocity, and
v₂ is the final velocity.
In natural language, the average velocity is equal to the sum of the initial and final velocities divided by two.
c. Second definition of average velocity:
v = d/t Eq. 4
Where,
v is the average velocity,
d is the distance, and
t is the time.
In natural language, the average velocity is equal to the distance moved divided by the time to move that distance.
d. Substituting 0 for the initial velocity in Eq. 3:
v = (0 + v₂)/2 Eq. 3a
Therefore,
v = v₂/2 Eq. 3b
In natural language, the average velocity of a falling body beginning from rest (velocity zero) is half the final velocity.
e. Substituting Eq. 3b for the velocity in Eq. 4:
v₂/2 = d/t Eq. 5
Rearranging,
v₂ = 2d/t Eq. 5a
In natural language, the final velocity of a falling body beginning from rest is twice the distance divided by the time.
f. Substituting 0 for the initial velocity in Eq. 2:
a = (v₂ - 0)/t Eq. 2a
Therefore,
a = v₂/t Eq. 2b
In natural language, the acceleration of a falling body beginning from rest is the final velocity divided by the time.
g. Substituting Eq. 5a into Eq. 2b:
a = (2d/t)/t Eq. 6
Therefore,
a = 2d/t² Eq. 6a
Rearranging,
t² = 2d/a Eq. 6b
And finally,
t = √(2d/a) Eq. 6c
Since Eq. 6c is the same as Eq. 1, we see that this is the correct equation to use to calculate the time taken to fall a distance, from a standing start, under a constant acceleration; in other words, if the acceleration is the acceleration of gravity and the object begins from rest, then this equation will give the time it takes to fall, otherwise known as the free-fall time.

Substituting the values above for d and t, we get:
t = √((2 x 417m)/9.80665m/s²)
t = √(834m/9.80665m/s²)
t = √85.04433216235921543034573478201s²
t = 9.2219483929568384583858444860057s
This value is slightly less than the 9.23s I calculated earlier using the figures of 9.80665m/s² and 418m; I am using the figures derived from Wikipedia in order to be precise. If you plug those figures in, you will get 9.23s. With one hundredth of a second's difference, which amounts to an accuracy of just over 1/10 of a percent. This is of little consequence, but feel free to use 9.23s if you wish. The value should be truncated, unless it is to be used in further calculations, because it is far beyond the likely accuracy of the height given; following convention, that value is three significant figures with a maximum error of +/- 1 meter.

We see that this is substantially less than the figure that we can derive for the fall by observing the videos; the most conservative estimate is that it is three seconds less, and the least conservative five. To find how much farther an object in free fall could have fallen in those additional times, we merely rearrange the equation (into it's more familiar form, BTW, solving for distance):
d = at²/2 Eq. 7
Substituting twelve seconds, we get
d = ((9.80665m/s² x 144)/2) = 706.0788m
Substituting fifteen seconds, we get
d = ((9.80665m/s² x 225)/2) = 1103.248125m
So we conclude that the building could have fallen somewhere between one and three-quarters its own height to over two and a half times its own height in the allotted time.

Expressing the difference in terms of time rather than distance, we find that it took somewhere between 22% and 53% longer than free-fall.

So, either way, "near free-fall" is incorrect, according to basic physics.

2. The assertion has been repeatedly made that there was not enough energy to pulverize all the concrete in the towers. This assertion is based on FEMA's calculations (for which they have not given derivations) that the potential energy in the buildings was 4 x 10¹¹J. Not being satisfied with this figure, first because (although potential and kinetic energy should be the same), I have no idea how they derived it, and second because my "seat-of-the-pants" estimate gives a higher figure, I did some research and determined the following fact:
The total mass of each tower was 450,000 metric tonnes. (Source: this article which gives four links to secondary sources that all say the same thing, within their limits of accuracy.)

Now, my first suspicion was that FEMA had failed to account for the contents of the towers in their calculations. So I made an estimate of those contents; rather than repeat that here, I will source that estimate. This document is a review of structural requirements for various types of buildings; on the third line, it gives the design requirements for office space as 50 pounds per square foot (psf). This site states that, "Typical floor-load capacity for an office building is about 50 to 75 pounds per square foot..." (follow the link (just a reminder, I have used this site's facilities to include it in the words, "This site," in the previous sentence) and search for the word, "load"). This document, on page 3 in Table 3, gives live load design requirements of 50 pounds per square foot for office space. I think we are safe concluding that this is the average designed load limits for office space.

So what were the design factors of the WTC towers? This page from 9/11 Research gives figures of from 75 to 150 pounds per square foot. (Please note: I am sourcing this paper for ONE REASON ONLY, and that is to show the designed live load carrying capacity of the WTC towers. Do not bother using material from this paper to attempt to refute that; it is stated there in black and white. You are free to use it to try to disprove my calculations here, but the figures are, again, there in black and white, for anyone to read.) Search on the phrase, "floor live load ratings," and you will find a table that goes floor-by-floor. Then there is this which, although highly critical of the FEMA investigation's early report, nevertheless does not dispute the figure of 50-100 pounds per square foot; here is the original report so that you can verify that Serendipity has not altered it.

OK, so how much loading was there, actually? FEMA has estimated this load at 25% of the design limit of 55 pounds per square foot (follow the link and have a look at page 45). This is an EXTREMELY conservative estimate, and furthermore does not take into account the greater loading of the mechanical floors. We will take this VERY conservative estimate of 13 psf, then, and use it.

How much actual office space was there on a single floor of the WTC? MSN Encarta estimates it at 9 million square feet for both towers- about 4.5 million per tower. Divided by 110 floors, this is about 40,000 square feet per floor. Great Buildings gives an estimate of 40,000 square feet per floor, quoting from Architects on Architecture: New Directions in America, p194-195, by Paul Heyer, Van Nostrand Reinhold Revised edition (September, 1993). Heyer also states that 75% of each floor had been freed for occupancy, and we know that the WTC was 209 feet square, so this yields a more conservative figure of about 33,000 square feet per floor. We'll go with this more conservative figure.

So, multiplying, we get 13 psf * 33,000 sq ft = 429,000 pounds per floor. This includes people, office equipment, partitions, desks, file cabinets, and everything, so there is no need to add more to it to get a rough figure. This is also an extremely conservative figure; we have taken the lowest actual loading figure we can find, and multiplied it by the lowest estimate of the space we can derive. Let me emphasize again, this is EXTREMELY conservative.

OK, now let's derive the total kinetic energy of the fall of the building. First of all, we have 110 stories, and total mass of 450,000 metric tonnes; that gives an estimate of 4090 metric tonnes per floor. Added to this is 429,000 pounds (converting, we get 194.5911267 metric ton; here is a convenient conversion site) for a total of 4285.5 metric tonnes per floor.

OK, now we're going to have to derive the kinetic energy. To do that, we need the final velocity; not the average, mind you, but the final. Rather than use a calculated and therefore potentially inaccurate figure, we'll go back to our height and our acceleration of gravity, and derive it from that. To do this, we'll go to the definition of acceleration:
a = v/t Eq. 8
Where,
a is acceleration,
v is velocity, and
t is time.
In natural language, the acceleration applied is equal to the velocity attained divided by the time to reach that velocity.
Rearranging,
v = at Eq. 8a
Now, we need to define time in terms of distance and acceleration; we don't want to use our earlier calculation of the time, to avoid introducing inaccuracies. So we'll use Eq. 1. Substituting:
v = a(√(2d/a)) Eq.9
The calculation is straightforward:
v = 9.80665m/s² x (√((2 x 417m)/9.80665m/s²)) =
90.436420207790179867929541828687m/s
Rounding, we get 90.4m/s.

We'll use the standard kinetic energy equation:
e = mv²/2 Eq. 10
Where,
e is kinetic energy,
m is mass, and
v is velocity.
In natural language, kinetic energy is half the quantity the mass times the square of the velocity.

Plugging in our figures from above, we get:
e = ((4,285,500kg x (90.436420207790179867929541828687m/s)²)/2) = 17,525,008,205.775J
Now, this is the kinetic energy for the topmost average floor falling from the topmost height. But of course, not all the floors fell from that height; in fact, they fell from different and lesser heights, and therefore had less kinetic energy. This is not the average kinetic energy, but the maximum. Let's do the same calculation for a floor that was halfway down the tower; we can do this simply by reducing the height to half, to derive the velocity, and then using this figure in the kinetic energy equation:
v = a(√(2d/a)) = 9.80665m/s(√((2 x 208.5m)/9.80665m/s²)) = 63.948205995164555515260416192403m/s
e = mv²/2 = ((4,285,500kg x (63.948205995164555515260416192403m/s)²)/2) =
8,762,504,102.8875J

Looking at this, we find that the second figure is exactly half the first. We've found a serendipitous relation: the kinetic energy of a falling object varies directly as the height from which it is dropped. Thus, we know that if something falls from half the height, it has half the kinetic energy; a quarter the height is a quarter the energy, but most importantly, 1/110 the height is 1/110 the energy. Thus, we can divide the kinetic energy by 110 (which, for the mathematically challenged, is the same as multiplying by 1/110) and use one of those for the first floor, two for the second, three for the third, and so forth, until we get to 110 for the 110th floor.

Now, how can we figure the total of 1 through 110? Simple.

Say we take all the whole numbers from 1 to 100. There is a trick that we can use: add 1 to 100, 2 to 99, 3 to 98, and so forth, until we get to adding 50 to 51. Each of these pairs is equal to 101, and there are 50 pairs. (100 items taken by pairs is 50 pairs, IOW.) Thus, the sum of all the numbers from 1 to 100 is 101 x 50 = 5050. Does this work if the difference is not 1? Yes. Let's take all the numbers divisible by 5 between 1 and 100: 5, 10, 15, and so forth. We get 100 and 5, 95 and 10, 90 and 15, and so forth until we get to 55 and 50. Adding these up, we get ten pairs that are 105 each (figure it out for yourself), for a total of 1,050. Feel free to do it the hard way.

So for 110 stories, we get (110 + 1) + (109 + 2)... = (55 + 56); there are 55 pairs, and each sums to 111. That's 6105. The kinetic energy is proportional to the floor number, so we take our maximum figure and divide by the number of floors to get the increment per floor; that gives 159,318,256.41613636363636363636364J per floor. Then we multiply by 6150 to get the total, and that gives 979807276959.23863636363636363636J, rounding, about 980GJ, for the total energy. Nearly a terajoule (a trillion joules).

Now, the FEMA figure was 4 x 10¹¹J; that's 400,000,000,000J, about 400GJ. So we can see that that figure is off by a factor of more than two; twice that would be 800GJ, which is still less than our 980GJ.

Converting to kilowatt-hours (kWh), which is the figure Hoffman uses, at 1J = 1Ws, which implies that 1Wh = 3,600J and 1kWh = 3,600,000J, we get 272,168.68804423295454545454545455kWh, 272,000kWh in round numbers.

So how much concrete will that pulverize? Well, according to Hoffman, 135,000kWh will crush all the concrete in the WTC. I'll take that figure, thank you very much. That means that we have about twice the energy needed to pulverize all the concrete in the whole building. And I'm going to note that that's from the kinetic energy alone; we haven't included anything that burned, neither the jet fuel nor the office contents. This is ONLY the kinetic energy, and it's a kinetic energy figure that we arrived at using VERY conservative numbers.

3.Hoffman estimates that there was not enough energy to drive the pyroclastic clouds.

First of all, I have to point out that Hoffman's figure for the kinetic energy of the collapse, being derived from FEMA's figures, is off by a factor of 2.

Second, Hoffman clearly fails to note that the kinetic energy turns to heat after it crushes the concrete; thus, the energy budget to expand the cloud must include the energy that crushes the concrete. Instead, Hoffman subtracts this energy, as if it just disappears after the concrete is crushed.

Third, Hoffman either assumes that the concrete was heated to over 740C (1020K), a Fahrenheit temperature of over 1,300 degrees, or that the "moisture content" of the concrete "vaporized;" but neither of these could be the case. In the first case, we would have seen the instant ignition of any burnable material that was touched by the cloud, and anyone caught in any portion of it would have been instantly burned alive, and this is not what happened according to the testimentary and photographic evidence. In the second case, unfortunately, the water in concrete is not free water; it chemically combines with the cement that is used to make the concrete (concrete is a mixture of cement and rocks that are bound by the cement, yielding a higher-strength material than the cement alone can make).

Fourth, and most telling, Hoffman neglects to analyze where the enormous amount of energy he details could have come from. His lowest estimate is 2.6 million kWh. This is a vast amount of energy. It is, in fact, 9,360,000,000,000J, 9.36TJ. At 239.0057361 US short tons of TNT equivalent per TJ, this is 2,237 US short tons of TNT equivalent; over TWO KILOTONS. A two-kiloton explosion in the middle of the WTC complex? Check this site out. You can see the overpressure blast effect of a 2kt weapon on the city of San Franciso. Choose "automobile" as the delivery method for the most conservative estimate. Note the panhandle of Golden Gate Park on the left of the picture. Residents will be aware that the panhandle is eight blocks wide; if you don't believe it, call up Mapquest and check for yourself. The area in which buildings would be destroyed by the blast wave is the blue circle. It's almost TEN BLOCKS WIDE. That means that if the energy that Hoffman calculates had been contained in that cloud, it would have knocked over every building within five blocks in every direction! We wouldn't even be discussing how 7 WTC collapsed; it would have been completely destroyed, along with fifteen or twenty other buildings!

Obviously, the actual energy estimate must be much more conservative; 200,000 to 300,000 kWh is a much more reasonable estimate of the energy involved.

4. Various estimates have repeatedly been made that there was not enough energy to cause pools of molten metal under the foundations, and have questioned where the fires that burned under the rubble came from.

From the above estimates, we can see that we have 429,000 pounds of burnable material on each floor. About 80% of one floor's material was consumed; we'll just elminate one floor and use 109 floors' worth. There were two sky lobbies, and two machinery floors. That will subtract another four floors, leaving 105. The ground floor was a couple of stories high, so we'll subtract another two, leaving 103. The top few floors were machinery floors, so we'll subtract another three leaving us 100.

The material likely didn't all wind up under the rubble; there were numerous reports that made it obvious that a substantial amount was dispersed in the debris cloud; papers floating around and so forth. To account for this, we'll subtract half of the material, so after we've multiplied the weight per floor by 100 floors, we'll use half that figure. We get 42,900,000 pounds, and dividing by two gives 21,450,000 pounds. Dividing by two thousand to get tons, we get 10,725 tons.

Now, how can we derive an energy figure from this? I looked for quite a while, and came up with this site, which gives the energy available from burning garbage (which is a pretty good estimate of the contents of the floors where the plane struck- think about it for a moment). This energy content is given as 25% (1/4) of the energy content of the equivalent amount of coal. The same site has another page that has a convenient "coal energy calculator" on it. Now, how much of that office content is actually burnable? I have previously (in a prior post) used a figure of 33%; I'll be even more conservative and reduce that to 25%. So of all the contents, only 25% by weight is burnable paper, cloth, wood, and plastic; the remainder is unburnable metal.

25% of 10,725 tons is 2,681.25 tons; and 25% of that is 670.3125 tons. Plugging this in to our trusty calculator, we get 14677584.8MJ, or 14.7TJ. This is an ENORMOUS amount of energy; it's equivalent to 3,508.0269598 tons of TNT. 3.5 KILOTONS! But remember: this energy is released over time. There isn't much oxygen down in the rubble, and more doesn't come in very fast, so it takes weeks for all of this material to burn. All ten thousand tons of it. And it doesn't all burn; a fair bit of it never burns, because it either gets put out when it's dug up, or because the heat never reaches it, or because there isn't enough oxygen to burn it when it's hot enough. But this is more than enough heat, even released over weeks, to liquify many thousands of tons of metal.

Finally, we've also got an answer from this material for where any extra energy beyond the 200,000 kWh we got from the kinetic energy of the collapse that's required to drive the cloud came from: if only 1% of this material burned on the way down, it would have given us the energy equivalent of an extra 35 tons of explosives, an additional 41,000kWh of energy. A few percent would add up to another 100,000kWh, easily enough to meet our energy requirements.

This should be sufficient to put to rest many wild speculations regarding what happened with the towers. Please let me know if you spot any errors.