April 4, 2009 update: With the publication of a paper in a peer-reviewed journal proving that active thermitic materials were discovered in the WTC dust, there is no longer any doubt that a thermite variant was used in the controlled demolitions. The material found was highly energetic and included particles on the nanometer scale. Fleshing out details of where it was placed, and whether conventional thermite or thermate was also used to take out the columns as suggested below, remains to be seen.

See here for links to the paper and associated discussion.

The WTC demolition team used thermate in order to guarantee global collapse
of WTC 1, 2, and 7, whilst simultaneously trying to minimise the evidence of
controlled demolition and promote the fabricated *casus belli* for attacks
on Iraq and Afghanistan: the illusion of "Muslim suicide hijackers / kamikaze
pilots". This analysis aims to ascertain how much thermate would
be required to maintain 3 major pools of molten steel for several weeks, and
whether the amount matches that likely to have been employed. It would certainly take many tons
per building, but at least thermite or thermate is capable of melting steel, in contrast to hydrocarbon combustion yielding CO_{2} and H_{2}O gaseous products in a non-oxygen-enhanced, ventilation-limited compartment fire.

The hollow perimeter columns consisted of three-story (36 feet) long sections that were closed off at each end. And each end featured an access hatch already cut into the inner web (ostensibly to allow access to the four bolts), enabling the thermate to be pumped into the hollow interior. See bottom two images here, Fig. 3.7 at this link, or this image, or several pictures here. With the column splices staggered mid-floor, two adjacent 3-story sections for one-third of the columns could have been accessed from a single floor, and all exterior columns in the targeted impact zone could have been covered by treating every third column over three adjacent floors. Hence, this "fireproofing upgrade" or "measuring and documenting" work could have been limited to the offices of a single company if need be, such as Marsh (WTC 1) and Fuji Bank (WTC 2).

Ideally, the powder would be remotely ignited at the top end of each 3-story section shortly before the intended collapse time. If the treatment had been applied to two 36-feet sections of column, it would have been necessary for the perpetrators to carry out the relatively easy job of planting detonators in another 3 floors above. Marsh occupied 7 or 8 adjacent floors at WTC 1, and Fuji Bank had 4 floors at WTC 2 which would allow a third of columns to be treated over a 72-feet length. If they detonated from the bottom, it would increase the risk of too much molten iron collecting at the bottom and prematurely melting its way through the ¼ inch thick steel in the minutes prior to collapse, as happened with WTC 2 where Fuji Bank had only 4 floors. However, melting 72 feet of every exterior column would have been overkill, especially considering that the perpetrators would have also targeted box columns in the core, down in the basement. They probably omitted some columns and / or mostly limited themselves to 36 feet of column.

The mixture would already contain all the oxygen it needed, and, once ignited, would burn along to the other end. The temperatures reached, and energy available, would be more than ample to melt the perimeter columns. Before this occurred, however, the loss of strength would have had devastating consequences for the building - and its occupants. After global collapse, the thermite reaction would continue to produce molten elemental iron and melt structural steel in the debris pile.

If too much molten iron had collected at the bottom of a section of column and melted its way through too quickly (e.g. the NE corner at around floor 81 on WTC 2 because the demolition team did not want to install detonators on Floors 83 (IQ Financial Systems) and 84 (Euro Brokers), the yellow-orange molten iron would be observed pouring out, much to the criminals' dismay: "You were only supposed to buckle the bloody columns - not shoot the molten iron outside in full view of the cameras! Now we'll have to recruit Frank Greening, James Fetzer and a 'professor' who was in a coma for six years to do a damage limitation exercise." (Irate arch-villain.)

Apart from the hollow exterior columns, the demolition team would have attended to the hollow core box columns in the basement. As the core weakened and the floors sagged in the final few minutes prior to collapse, inward bowing of the thermate-weakened perimeter columns would have ensured failure before any appreciable melting of these columns had occurred, reducing the likelihood of unfortunate mishaps such as the WTC 2 molten iron.

The perpetrators most likely opted for military grade thermate, which is thermite with added barium nitrate and sulfur. Evidence for this includes excessive sulfidation of structural steel, and dust sample analysis which found barium levels peaking at 3,670 ppm (parts per million). The geometric mean from fourteen samples was 533 ppm, which is in line with its natural abundance of 0.05% in the earth's crust. But the highest reading was of 3,670 ppm in sample "WTC 01-16", taken from the junction of Broadway with Dey St and John St, which is a block to the east of WTC 2. On 9/11 the wind was north-westerly, and with the observed molten iron pouring from the northeast corner of WTC 2 after the thermate had melted through the ¼ inch perimeter column steel, barium from the thermate powder could have emerged and been blown eastwards. The most important advantage of thermate is that it is easier to ignite; the thermal effect is also increased.

The thermite reaction probably involved ferric oxide, also known as iron(III) oxide as shown in the thermochemical equation:

Fe_{2}O_{3} + 2Al ==> Al_{2}O_{3} + 2Fe; deltaH = -851.5 kJ/mol.

(See the Wiki entry for thermite.) The alternative magnetite iron(II,III) thermite reaction releases a lower amount of energy per unit mass of thermite powder compared to the iron(III) version, but is often employed as it is cheaper. It is improbable that the psychopathic crooks would have jeopardized a trillion-dollar crime of the century simply to save on materials. In any case, they probably got the thermate from the US or a similar elective dictatorship oligarchy masquerading as "Western democracy" satellite state at a knock-down (pun not intended) price.

By 851.5 kJ/mol in the above equation, the mole would refer to one (gram-) mole of ferric oxide. (See the D P Grimmer paper on the possible use of thermite: *Thus, 849 kJ of energy are released for every g-mole-equivalent (mol) of Fe2O3 that reacts with 2 mol of Al.*) Adding the atomic weights of Fe (55.847), Al (26.9815) and O (say, 16), there is 159.694 grams of Fe_{2}O_{3} and 53.963 grams of aluminum powder to release 851.5 kJ, with the products consisting of 101.963 grams of Al_{2}O_{3} and 111.694 grams of elemental iron. The reactants are in the solid state, and the products in the liquid state; the 851.5 kJ is the difference between the products' and reactants' heat of formation after allowing for the phase change to the liquid state. So at 3.985 MJ/kg, the heat of reaction is similar to that of TNT and only some 1/10th that of hydrocarbons such as kerosene. Flame temperatures in hydrocarbon combustion are lower than temperatures attained in thermite reactions, because of all the nitrogen in the mix that also requires heating in the absence of oxygen enhancement, and because of the endothermic dissociation reactions (e.g. CO_{2} goes back to CO and ½O_{2} at high temperatures), and due to the higher specific heats of the products H_{2}O and CO_{2} compared to iron.

The density of iron(III) oxide is 5.24 g/cc; the density of aluminum is 2.7 g/cc. In the above equation, the 213.657 grams of thermite reactants, if the powder were infinitely compacted, would have a volume of 159.694 g divided by 5.24 g/cc, plus 53.963 g divided by 2.7 g/cc, which is 30.476 plus 19.986 = 50.462 cc. Since its mass is 213.657 grams, its density is 4.234 g/cc.

Thermate-TH3 is generally 68.7% thermite, 29% barium nitrate, 2% sulfur, and 0.3% binder by weight. Barium nitrate has a density of only 3.24 g/cc, which would slightly lower the density of thermate compared to thermite, although the increased thermal effect would help to compensate. Sulfur has a nominal density of 2 g/cc, but in crystalline form can be as high as 5.4 g/cc. Disregarding the 0.3% binder and allowing only 2 g/cc for the sulfur, we have 68.9% thermite, 29.1% barium nitrate, and 2% sulfur by weight. For every 213.657 grams of thermite which takes up a volume of 50.462 cc, there is 90.24 grams of barium nitrate with a volume of 27.85 cc, and 6.2 grams of sulfur with a volume of 3.1 cc. The total mass for every 213.657 g of thermite is 310.097 g, and the total volume is 81.412 cc; hence, the mean density of the infinitely compacted thermate mixture is 3.809 g/cc.

Thermite powder packing densities of 82% have evidently been achieved; let us suppose 80% was attained for the thermate. Thus, the mean density of the thermate powder is 0.8 * 3.809 = 3.047 g/cc.

The iron(III) or hematite (rust)-based thermite reaction yields molten elemental iron as 52.28% of the products by weight. Since the products' mass equals that of the reactants, 100 tons of the iron(III) form of thermite would produce 52.28 tons of molten iron. (100 tons of the iron(II,III) thermite would yield 55.2 tons of iron. The energy is released by oxidising the aluminum; reducing the iron is endothermic. So with more iron and less aluminum oxide in the products, the iron(II,III) thermite releases less energy per unit weight of powder.)

With the thermate mixture consisting of only 68.7% thermite after the barium nitrate, sulfur and binder has been accounted for, the amount of molten iron produced from the iron(III) version becomes 0.687 * 0.5228 = 0.3592 of the total by weight. So 100 tons of thermate would produce 35.92 tons of molten iron. In addition to this would be any structural steel that it had been in thermal contact with and melted.

The perimeter columns of WTC 1 and 2 were fabricated from steel plates that were only about ¼ inch thick at the higher floors. The flanges were both 13.5 inches long; the inner web was 15.75 inches in length, and the length of the outer web was 14 inches less the thickness of both flanges, making it 13.5 inches long. (See NIST interim report, Appendix E, Fig. E-3.) So the total cross-sectional area is 0.25 times (13.5 + 13.5 + 15.75 + 13.5) = 14.06 square inches.

(On the lower floors, the flanges were probably about twice the web thickness, with the spandrel plates of intermediate grade. If the webs were only an eighth of an inch thick, the cross-section was only 10.4 ins^2. They may not have gone down below ¼ inch thickness, so we'll assume a 14.06 ins^2 cross-section.)

A one-foot length of exterior column within the impact zone would have a volume of 12 ins * 14.06 ins^2 = 168.72 ins^3. Taking the density of steel at 7860 kg/m^3 or 0.284 lb/ins^3, its mass would be 47.92 lbs = 21.73 kg. The cross-section of the internal space within the columns was about 13.5 inches x 11 inches = 148.5 ins^2, so a 1 foot length could contain a 12 ins * 148.5 ins^2 = 1782 cubic inches = 0.0292 cubic meters of the thermate powder mixture. Given the density of the thermate powder at 3,047 kg/m^3, the mass of 0.0292 m^3 of this powder would be about 89 kg. The iron(III) thermite can release about 4 MJ/kg. If the barium nitrate and sulfur contributed nothing, the energy released would be 0.687 * 4 = 2.748 MJ/kg, or 89 * 2.748 = 244.57 MJ per one-foot length of column.

Each kg of steel takes an average of about 550 J/kg.K to raise the temperature over the whole range from 20 to 1,500 C (and that's assuming too high a melting point, given the effects of the sulfur). After adding some 250 kJ/kg for the latent heat of fusion, the total energy required to melt one kilogram is 1,480 * 550 + 250,000 = 1.064 MJ. So to melt a 21.73 kg one-foot section would require 21.73 * 1.064 = 23.12 MJ. The energy available from the thermate powder along a one-foot length of column is at least 244.57 MJ, leaving a surplus of 221.45 MJ. So there is plenty of energy to melt the entire column, and with a relatively small proportion of the total heat release escaping to the environment, temperatures would not fall significantly below the 2,500 C or so that the thermite reaction is capable of attaining.

Another way of looking at that is that the conservatively-estimated energy from the thermate reaction divided by the energy required to melt the steel (2.748 MJ/kg divided by 1.064 MJ/kg) = a ratio of 2.583, and the density of the thermate divided by the density of the steel (3,047 kg/m^3 divided by 7,860 kg/m^3) = a ratio of 0.3877, which annul each other. We are then left with the ratio of the cross-section of the hollow area available to the thermate divided by the cross-section of the steel: 148.5 ins^2 / 14.06 ins^2 = a factor of more than 10.

In order to melt the steel this factor, the hollow cross-section divided by the steel cross-section, would need to be somewhat greater than 1. As the proportion of heat released that is pulled away from the products becomes greater, the temperature attained by the products will decrease. However, in order to collapse the building, it would not be necessary to melt the steel, merely weaken it sufficiently. If we stick with a steel cross-section of 14.06 ins^2 and a powder-packing density of 0.8, but lower the hollow cross-section to that of the steel, the steel volume over a one-foot length remains at 12 ins x 14.06 ins^2 = 168.72 ins^3, and at 0.284 lbs/ins^3 its mass is 47.92 lb = 21.73 kg. The thermate powder also has a volume of 168.72 ins^3 = 2,765 cc. Its density is 3.047 g/cc, so its mass is 2,765 * 3.047 = 8.425 kg. If 68.7% of that is thermite with a calorific value of 4 MJ/kg, the total energy available is 0.687 * 8.425 * 4 = 23.15 MJ. Taking half of the energy out of the products the temperature would remain a little over the mean of ambient and adiabatic values (specific heat rises with temperature, so supplying half of the energy will initially produce more than half of the temperature rise. The adiabatic temperature for the thermate reaction products would be at least 2,500 C, so we assume somewhat over 1,250 C, and possibly enough to keep the iron molten. Half of the 23.15 MJ, say, 11.58 MJ is applied to the 21.73 kg of steel. Assuming an average of 550 J/kg.K over the range, the temperature rise in the steel would be 11,580,000 / (21.73 * 550) = a 969 degrees K or C increase, enough to induce collapse if applied to enough columns.

So in the general case, in order to melt the steel, the hollow cross-section required for the thermate powder need not be much greater than the cross-section of the steel, and in order to sufficiently weaken the steel in order to induce global collapse, the cross-sections merely need be about equal. Any significant surplus, as there probably was in the case of the targeted impact zone perimeter columns, ensured that the thermate would have no problem in melting all the structural steel along the length of column it was enclosed by. Once the reaction had started, there would be no way of stopping it; the perpetrators would have preferred to have stopped it as soon as global collapse ensued, so as to reduce the tell-tale smoking gun signs of a controlled demolition. Signs that included steel members in the debris pile that appeared to have been partly evaporated in extraordinarily high temperatures. And that was WTC 7, which was not hit by any aircraft!

It was necessary to employ an accessory such as Rudy Giuliani, who would restrict access, ban photographs, and remove incriminating evidence and have it melted down for recycling. NIST had to be supplied with samples that had not been given the thermite treatment; i.e., they would get the members that had been heated by nothing more than the office fires - those panels that never even reached 250 C. But occasionally Rudy would get it wrong - failing to move the partially boiled column from the WTC 7 pile before Dr Jonathan Barnett, a professor of fire protection engineering, got his hands on it. After his initial burst of honesty, Barnett would then be afraid to go against the prevailing view.

*The building* [WTC 7] *had suffered mightily from the fire that raged in it, and it had been wounded by the flying beams falling off the towers.* [Candidate for overstatement of the year.] *But experts said no building like it, a modern, steel-reinforced high-rise, had ever collapsed because of an uncontrolled fire, and engineers have been trying to figure out exactly what happened and whether they should be worried about other buildings like it around the country.* [The
key here is who gets to collect on the building insurance, who controls
security, and are they part of a group with a
long-standing history in deception
and staging false-flag terror attacks.] *...A combination of an uncontrolled fire and the structural damage might have been able to bring the building down, some engineers said. But that would not explain steel members in the debris pile that appear to have been partly evaporated in extraordinarily high temperatures, Dr. Barnett said.* [New York Times, November 29, 2001.]

There are 59 perimeter columns along each side, plus one on each corner bevel, totalling 240 perimeter columns per Tower.
The corner columns were not continuous like the others, but could have been
rigged so we'll count them. The column splices were staggered mid-floor, but if one three-story length had been rigged on all columns, it should have been sufficient to ensure collapse, especially as core columns were almost certainly being taken out as well. 240 columns times 36 feet per column times 148.5 ins^2 hollow cross-section = 240 * 12 * 36 * 148.5 = 15,396,480 cubic inches = 252.3 cubic meters = 252.3 * 3.047
tonnes/m^3 = 768.8 tonnes of thermate. From above, each 100 tonnes of thermate produces 35.92 tonnes of molten elemental iron, so
768.8 * 0.3592 = **276.2 tonnes or 304.4 tons of molten iron**. The molten structural steel would total 14.06 ins^2 * 12 * 36 * 240 = 1,457,741 ins^3
= 23.89 m^3. After multiplying this volume by the density of steel at 7,860
kg/m^3, we obtain **187.8 tonnes or 207 tons of molten structural steel**. That is just for one three-story section of exterior columns per building. Although there were only 47 core columns, the
larger cross-sections - that of the steel, and the hollow area for the thermate
- would concentrate the molten metal and allow it to remain molten for longer.
As we shall see, the molten metal from the basement core columns was not only
dispersed amongst fewer columns, but totalled more than the **464 tonnes of
molten metal** resulting from thermate applied to 240 thirty-six feet sections
of perimeter column.

On the one hand, the more thermate that needed to be installed, the less implausible the theory. On the other hand, the correct demolition theory should account for eyewitness reports such as "massive amounts of molten steel", or "rivers of molten steel", or "21 days after the attack ... molten steel was still running", or "they are finding molten steel" [late October], or "in the first few weeks ... the end of the beam would be dripping molten steel", or "molten steel flowing ... like a foundry or like lava", or "a steel beam ... dripping from the molten steel" in February 2002, or steel from 6 stories underground that was still cherry red 6 weeks after the attacks.

The mass and packing density of thermate suggested above for the perimeter columns, a total of some 768.8 tonnes of thermate or about 3.2 tonnes per 36-feet section of column, was well above that required to sufficiently weaken the steel. The mass of thermate could easily have been reduced by a factor of 5, by a combination of reducing the powder packing density and the volume. Let's suppose the demolition agents opted for 0.64 tonnes per column. And the columns on the side targeted by the aircraft could have been omitted, since half of them would be severed by the impact. So if they treated 180 perimeter columns at 640 kg of thermate per column, the perimeter column requirement totals 115.2 tonnes. Note how FEMA's 403 Ch. 2 p. 2-12 says, "The average thickness of spray-applied fireproofing on the trusses was 3/4 inch. In the mid-1990s, a decision was made to upgrade the fire protection system by applying additional material onto the trusses so as to increase fireproofing thickness to 1-1/2 inches. The fireproofing upgrade was applied to individual floors as they became vacant. By September 11, 2001, a total of 31 stories had been upgraded, including the entire impact zone in WTC1 (floors 94-98), but only the 78th floor in the impact zone in WTC 2 (floors 78-84)."

The demolition agents targeting the office areas could have been working on the vacated floors, and at night when less people were around, eliminating the need for the daytime fireproofing upgraders to be part of the conspiracy. With the night workers posing as "measuring and documenting" the fireproofing, the regular daytime operatives would not become suspicious when the only evidence of "work" by the night shift consisted of holes having been drilled and blanked off in vermiculite plaster. One possibility would have been nocturnal helicopter flights to deliver thermate to the roof. If each flight could deliver a payload of nearly 10 tonnes per trip, that would permit 15 perimeter columns to be each fitted with 640 kg of thermate, only requiring a dozen landings for all 180 columns on 3 sides.

The crooks aimed to target the Towers as high as possible so that the steel would be a lower grade; at the same time they needed a minimum of ten floors or so above the collapse front, so as to promote the "pancaking theory" or similar such nonsense. This explains why WTC2 was targeted lower down; the demolition agents had access up to the 78th floor, and floors 76 to 78 would have been where thermate was pumped up into the perimeter columns. They had access to the higher floors in WTC1 - where fireproofing was being upgraded - so targeted around floor 98. Even at 0.64 tonnes for 180 columns, 115 tonnes of thermate over three floors seems a lot to have to transport and install, but each WTC floor had some 56 tons of fireproofing. During the upgrade, some of the thicknesses were doubled or more, so if the increase was 28 tons per floor, a mere 5 floors would need more than 115 tons or tonnes to be taken up and sprayed in place. The demolition agents didn't have to worry about gluing the thermate. It just had to be pumped through the access hatches under the "measuring and documenting" pretext.

The above upper estimate of 240 three-story sections of perimeter column yielding some 464 tonnes of molten iron and steel works out at around 1.93 tonnes per column. At 7,860 kg/m^3, the volume of molten metal comes to 0.246 cubic meters per column, i.e. equivalent to a cube of sides 0.627 meters long. To some extent, the molten metal would have pooled together as it ran down through gullies in the debris pile. As mentioned, the quantity of thermate employed for the exterior columns would likely have been considerably less than this upper estimate. But the 47 core box columns, at basement level, were of greater cross-section, and would have produced greater concentrations and longer-lasting pools of molten metal.

Various sizes quoted for the box columns in the lower floors range from 12" x 52" with welded plates up to 7" thick, to 18" x 36" with plates 4" thick. And as high as floors 83 to 86, column 504, for example, was a box column about 22" square. If we take 18" x 36" as the dimensions at the outside edges and assume 4" thick steel, the steel cross-section is 2 * 4 * 36 plus 2 * 4 * (18 - 8) = 368 ins^2. The hollow cross-section is 10" x 28" = 280 ins^2. So if a six-story length of column was filled with thermate, the energy would not quite be sufficient to melt the entire length of column. However, it could easily raise the temperature into the 800 to 1,000 C range, inducing column failure by way of a massive reduction in yield strength. Moreover, the demolition team would have employed a different strategy with the core columns. It would have been more sensible to ignite the thermate from the bottom, hence concentrating the heat on the bottom of the column as the molten iron collected at the bottom. With no television cameras trained on the bottom of the basement, the genocidal criminals could devote their efforts to ensuring collapse, with no worries about leaks of molten iron.

A 12" x 52" column, if those dimensions were at the mid-point and the steel was
7" thick, would have a steel cross-section of 2 * 7 * (52 + 3½) plus 2 * 7 * (12 - 3½) =
896 ins^2, and the hollow cross-section would be 8½" x 48½" = 412.25 ins^2.
Again, the demolition operatives would have ignited from the the bottom,
cramming the hollow volume full with the thermate powder, and this would have
been ignited before the final few minutes of heating the perimeter columns. This
is the best candidate for producing a large pool of molten metal without
requiring that individual pools merge together. Six basement floors is about 72
feet of column. Multiplying by a hollow cross-section of 412 ins^2 obtains
a potential thermate volume of 355,968 ins^3 = 5.83 cubic meters. As above, we
assume a thermate powder packing density of 80% which places the mean density of the thermate powder
at 0.8 * 3.809 = 3.047 g/cc. Hence, the mass of thermate is 5.83 m^3 * 3.047
tonnes/m^3 = 17.76 tonnes (or 19.58 short tons). Ignoring the barium nitrate, sulfur and binder, the
mass of molten iron produced from the iron(iii) thermite is 0.687 * 0.5228 = 0.3592 of the
mass of the thermate. So 17.76 tonnes of thermate would produce **6.379 tonnes or 7.03 tons of molten iron for a six-story length of a single basement core
column**.

If the thermite is 68.7% of the thermate and yields 4 MJ/kg, the total energy
released is 17,760 kg * 0.687 * 4 MJ = 48.80 GJ. Assuming a 1.064 GJ/tonne
requirement to melt steel, the potential mass that could be melted would be 48.8
/ 1.064 = **45.86 tonnes or 50.55 tons of structural steel per column**. A 72 feet length
of such a column, with a steel cross-section of 896 ins^2, would be 448 cubic
feet = 12.69 cubic meters of steel = 7.86 * 12.69 = 99.74 tonnes of steel. So a
reasonable proportion of this column could have been melted. With ignition at
the bottom, the reaction would melt the bottom of the column, severely weaken
the remainder, and proceed to melt tonnes of the supporting grillage beneath.
See this link for a diagram
of a WTC grillage, which distributed the load of each core column through to
a 2.1 meter thick concrete slab and on to the bedrock.

As mentioned above, a proportion of the heat of reaction must remain locked in
the combustion products. But if there was 6 tonnes of molten iron and 30 tonnes
of molten steel for each of 47 columns, **the total for all core columns would
amount to 282 tonnes of molten iron and 1,410 tonnes of molten steel**. The
perpetrators would not have needed to target all core columns, and considering
that some columns had a smaller hollow cross-section, 36 tonnes of molten metal
for a single column is probably nearer to the maximum than the mean. But
the basement core columns would contribute the lion's share of the molten
steel, and being divided over a smaller number of columns, these pools would be
much more enduring.

The molten metal would run onto the 2.1 meter thick concrete slab. With the melting point of concrete being 1,800 C or above, the molten iron could melt some of this for a time, then persisting in the recess at 1,800 C through to 1,530 C. This analysis of the WTC specifications concluded that the grillages consisted of some 62 tons of steel per column. If there was just 20 tonnes of molten structural steel and 6 tonnes of molten iron, and we assume a density of 7,000 kg/m^3 for the molten metal, the volume would be 26 / 7 = 3.714 m^3 which if forming a depth of 1 meter would have an area of 3.714 m^2 and could have equal sides of 1.927 meters = 6 feet 4 inches. The mean spacing between core columns was about 18 feet. If the molten metal from 4 adjacent core columns had run off the concrete slab and pooled together on the bedrock, the volume would be 104 / 7 = 14.857 m^3. At a depth of 1 meter, such a pool could have horizontal dimensions of 3.854 m x 3.854 m = 12.64 feet square.

In the general case, if the density is 7,000 kg/m^3, then the mass of a depth d in meters of molten iron and steel with an area of 1 m^2 is equal to 7,000 d kg. Assuming 725 J/kg.K for the sensible heat capacity at high temperatures, 250 kJ/kg for the latent heat of fusion, an initial temperature of 2,500 C and a melting point of 1,500 C, then the total energy transfer required in order to solidify is given by:

E (joules) = 7,000 * d * [250,000 + (725 * 1,000)]

leading to:

E (joules) = 6.825*10^9 * d *[equation 1]*

The rate of heat conduction Q / t from the molten metal is related to the thermal conductivity k of the surrounding medium, the area of heat transfer A and the temperature gradient (T1 - T2) / x where T1 - T2 is the difference in temperatures at two points and x is the distance between those points, according to:

Q / t (watts) = k * A * (T1 - T2) / x
*[equation 2]*

A is taken as 1 m^2. We want to determine the time to solidify for a range of molten pool depths, and the mean temperature T1 of the molten pool (assuming quicker cooling at higher temperatures) is taken to be 1,900 C. At the other end of the distance x in the surrounding insulating medium, T2 is assumed to be midway between ambient (20 C) and T1 temperature, hence T2 is 960 C and the temperature difference T1 - T2 is 940 degrees Kelvin. The thermal conductivity of the insulating debris pile is taken as 0.9 W/m.K (see below). Dividing equation 1 by equation 2, energy divided by power, obtains time:

t (seconds) = 6.825*10^9 * d * x / (940 * 0.9) which is close enough to:

t (seconds) = 8,067,000 * d * x *[equation 3]*

The thermal diffusivity of the surroundings will help us estimate a suitable value for x. If we suppose that the debris pile adjacent to the molten steel is approximated by the characteristics of lightweight concrete, with a thermal conductivity k of 0.9 W/m.K, a density p of 1,750 kg/m^3 and a specific heat c of 800 J/kg.K, its thermal diffusivity alpha is given by:

alpha = k / (p * c) = 6.429*10^-7 square meters per second.

If a bar of semi-infinite length with insulated sides is "instantly" heated at one end, the distance x in meters at which half the temperature rise occurs is given by:

x = 2 * z * SQR(alpha * t)

where z is the value required to yield a Gauss error function of 0.5, alpha is the thermal diffusivity in square meters per second, and t is the elapsed time in seconds. Since erf(0.477) is close to 0.5, the distance is related to time according to:

x = 2 * 0.477 * SQR(6.429*10^-7 * t)

At 1 week (604,800 seconds), for example, x = 0.595 meters, and a quadrupling of t to 4 weeks would double x. Since x is related to the square root of the elapsed time, the mean value of x over a time t is two-thirds of its value at the end of the time t. (Integrating the function y = x^(1/2), we obtain the integral (2/3)*x^(3/2)+c, and the average height of the curve from 0 to any given point on the x-axis is found to be 2/3 of its height at the given point x. Here, y corresponds to the distance 'x', the dependent variable, which we want to determine given the independent variable x which corresponds to time t.)

So, finding the square root of the 6.429*10^-7 to move it outside the brackets and inserting the 2/3 correction for a mean value of x, we have:

x = 2 * 0.477 * (2/3) * 0.0008018 * SQR(t)

which is:

x = 0.00051 * SQR(t) *[equation 4]*

We bring back equation 3 and use equation 4 to substitute for x:

t (seconds) = 8,067,000 * d * x *[equation 3]*

t = 8,067,000 * d * 0.00051 * SQR(t) ===> SQR(t) = 4,114 * d ===> t (seconds) = 16,925,000 * d^2

leading to:

t (weeks) = 27.98 * d^2

A correcting factor of 2 is now introduced, to allow for heat escaping up or down. The required pool becomes twice as deep, so the height 'd' in the above equations, that determines the quantity of energy which must flow in a particular direction, is half of the true pool depth:

t = 27.98 * (0.5*d)^2

which is near enough to:

t (weeks) = 7 * d^2 (meters)

or d (meters) = SQR(t/7) (weeks)

This should be about right for a pool with horizontal dimensions large in relation to its vertical dimensions. So if molten metal from four adjacent core columns had merged together to form a pool 1 m x 3.854 m x 3.854 m, as described above, horizontal losses at the middle square meter of the horizontal area would be relatively low, hence a vertical height of 1 meter would be associated with a solidification time of almost 7 weeks. And the mean spacing between core columns was only about 18 feet or 5.5 meters. The formula for predicting the point at which half the temperature increase is attained in a semi-infinite bar assumes that the other end of the bar is cold. Heat transferred in a horizontal direction would have been met by heat from adjacent columns; consequently horizontal temperature gradients would be lower than those in a vertical direction. In general, horizontal heat losses for the WTC molten pools in the basement - whether cube-shaped, cuboid or even spherical - would have been relatively low.

For another approach, consider that the thermate contained in the hollow volume of 72 feet of a single basement core box column could release some 48.8 GJ. From equation [2] above, assuming k = 0.9 and a temperature difference T1 - T2 of 940 Kelvin, the energy outflow per square meter would be at a rate of:

Q/tA = 846 / x

From equation [4], if we input t = 1 week (604,800 seconds), Q/tA works out at 2.133 kW/m^2 (equation [4] has allowed for the 2/3 correction to obtain a mean x of 0.3966 meters). At t = 7 weeks, Q/tA goes down to 806 W/m^2. At 14 weeks it has reduced by a further factor of SQR(2) to 570 W/m^2. At an estimate of a typical heat transfer rate of 1 kW/m^2 (that is with the molten pool surrounded by insulating debris; radiated losses could be at some 1,000 times the intensity), this would correspond to a time of 48,800,000 seconds or 80.7 weeks, divided by the area in square meters, for the energy to totally flow out. If only half the energy has to be transferred to solidify the molten pool, then a time of 5 weeks would correspond to a conducting area of 8.07 m^2. In the example above of a pool comprised of 6 tonnes of molten iron and 20 tonnes of structural steel, the volume would be some 26 / 7 = 3.714 m^3, and a cuboid pool could be 1 meter deep by 1.927 meters by 1.927 meters. The surface area would amount to 2 * 3.714 + 4 * 1.927 = 15.136 cubic meters. This is nearly twice the required 8.07 m^2, but does not allow for horizontal losses being at a lower rate. If two 1.927 m by 1.927 m top and bottom boundaries transmitted energy at a rate of 1 kW/m^2 and heat losses for four 1 m by 1.927 m sides were only 0.5 kW/m^2, the total works out at 11.282 kW; an average of 0.745 kW/m^2. Moreover, a cube-shaped pool of volume 3.714 m^3 would have a surface area of 14.39 m^2; and a spherical pool would have a volume of only 11.598 m^2.

It is clear that the observed times of molten streams and pools persisting for a matter of weeks, with plenty of cherry red (about 745 C) steel or iron observed six weeks after the collapses, and some molten metal running and dripping at even later times, is consistent with a total mass of molten metal in the order of several hundred tonnes per building, and a number of individual pools ranging from tens of tonnes up to a hundred tonnes or more for pools that had merged. In contrast, consider the amount of aluminum that might have been melted in the office fires scenario of the official conspiracy theory. For a start, it is probable that the mean temperature at any given point, even within the fires zone, was below the melting point of aluminum. The study commissioned by Silverstein Properties placed the temperatures at 750 F to 1300 F (about 400 C to 700 C). NIST concluded that temperatures of 1,000 C would have been sustained for only 20 minutes or so before local combustibles had been consumed and the fires moved on to another section of the floor or to other floors. The rest of the time, temperatures would have been as low as 500 C, which would make the mean temperature about 600 C for a given point in the fire zone. A method outlined here for estimating the mean temperature from the heat flux predicted an average temperature of 627 C.

If we suppose that an aluminum alloy has a melting point temperature of only 600 C, and it attains this at collapse time minus 20 minutes, then in the final 20 minutes prior to collapse it just happens to be in a 1,000 C zone where local combustibles are burning fiercely, the possible absorption in W/m^2 is at a rate according to the Stefan - Boltzmann law:

P/a = e * alpha * (Te^4 - Ta^4)

... where e is the emissivity of the absorber or radiator/emitter, alpha is the Stefan - Boltzmann constant 5.67*10^-8 W/m^2.K^4, and Te and Ta are the absolute temperatures of the emitter and absorber respectively. Aluminum has a low emissivity of only about 0.1 (hence its use as a reflector in infra red heaters). Assuming Te = 1,273 K and Ta = 873 K, we have:

P/a = 0.1 * 5.67*10^-8 * (1273^4 - 873^4) = 11.597 kW/m^2.

Hence, in 20 minutes, the possible absorption is 11,597 * 20 * 60 = 13.92 MJ. With the latent heat of fusion of aluminum at some 397 kJ/kg and taking the density at 2,700 kg/m^3, the amount of aluminum that could be melted after 20 minutes would be 13,920 / 397 = 35.06 kg per square meter, which would have a volume of 35.06 / 2,700 = 0.013 cubic meters. So this would correspond to a 1.3 cm or ½ inch depth of molten aluminum. From the above equation for steel, if we input d = 0.013 m into:

t (weeks) = 7 * d^2 (meters)

... then the predicted time to solidify is only 11.9 minutes. On the one hand, half the temperature difference would be only some 300 Kelvin for the aluminum compared to 940 degrees for the molten iron and steel; on the other hand, the fires would have barely heated the aluminum beyond the melting point, and it would be necessary to take out a mere 400 kJ/kg or so compared to some 975 kJ/kg for the iron and steel.

Given a total floor area of 63.4 m by 63.4 m = 4,019 m^2 and taking the proportion of a floor subjected to 1,000 C temperatures in the final 20 minutes of a 102-minute fire as 0.2, if the aluminum had conveniently shredded itself into "magic bullets" in order to dislodge as much fireproofing as possible and then managed to reinvent itself as ½ inch thick sheet distributed over the entire 804 m^2 area spaced over the entire zone of final 1,000 C temperatures, the maximum molten mass would be 804 * 0.03506 = 28.2 tonnes per floor. This would not have happened, but there is a remote possibility of several tonnes of aluminum melting per floor. However, concentrations of molten aluminum would be much too low to produce pools with a depth significantly greater than a centimeter or two. Any aluminum pools would have solidified in minutes, rather than weeks.

For those who still haven't deduced who did 9/11, here are a few hints. Let's suppose there is a middle-aged lady, who has been happily married for 29 years. The first husband dies or disappears, and husband number two takes out a massive $3+ billion insurance policy on her life. It is subsequently found that the first husband's removal from the scene is linked to a friend and business partner of the new husband. A mere six weeks later, the lady dies in mysterious circumstances in an accident that would not be fatal unless the laws of physics, chemistry and biology were revised - e.g., a small piece of cardboard falls on her head from a height of one foot. Any detective worth their salt should be suspicious, to say the least.

Bullet wounds to the chest are then discovered, and five assassins - who are linked to husband #2 and his associates - are arrested after being found dancing and celebrating with high fives and with traces of gunpowder on their fingers. "Ah", the skeptics cry, "a human body is comprised of some highly reactive elements. You have potassium, sodium, calcium, magnesium, iron, plenty of water, and oxygen in the air. Moreover, the body is heated to about twenty degrees above ambient. Isn't it quite possible that some sort of exothermic reaction occurred, forming the observed bullet holes?" The skeptics then conclude that the death was due to natural causes, after her head was hit by a piece of cardboard at 5 mph. Everyone else marks them down as kooks.

Of course, the aluminothermic reaction-based demolitions had been in the planning process for years, and the perpetrators had always intended to promote the "natural thermite reactions" theories when folk became suspicious (one reason for the choice of cutting agent).

*(Above article) Revised 31 March, 2007* /
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