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WTC TWIN TOWER'S POTENTIAL ENERGY OVER ESTIMATED

Chuck Boldwyn, a retired physics and chemistry instructor has uncovered information showing the varying sizes of the structural steel used in the Twin Towers construction. The purpose of this analysis is to correlate this information to determine the various building weights to find a more accurate value for the Twin Tower's potential energy (PE) or gravitational force.

The Twin Tower's PE has often been over estimated when using the general method of calculation. One of the values required for these calculations is the weight of the building. It's been difficult to establish the correct weights of the towers. A popular figure of 500,000 tons per tower is often quoted with the steel content being 200,000 tons a 40% value. This is likely to be a short tons figure and includes all the concrete and steel used in the basements. Others quote much lower amounts.

When researching to find a single towers PE, we found amounts between 396 to 420 G joules to be common. The source quoting a PE of 396 G joules also quoted a structural steel content of 80,000 tons. No building weights were given. The last two values will be used for this model. To calculate a buildings PE only the building weight from the ground floor upwards is required.

The general formulae for a building's PE :-

Weight (kg) x Height (meters) x 9.8 x .5 = Joules

The 9.8 value is the falling speed due to gravity of 9.8 meters/sec

The .5 value is used to find the average height of the building as the higher floors would fall further than the lower floors. This formulae however, assumes all the floors to weigh the same. In the Twin Towers they were not. Many researchers, to find the weight of a single storey, simply divide a tower's weight by 110 (their number of storeys) but this gives a totally wrong answer.

To calculate a tower's above ground weight the values of 396 G joules with their height of 410 meters are used. The 80,000 ton structural steel figure will be used later in this analysis.

The above PE formulae is reconfigured as we already know the quoted PE but not the weight.

                        Joules                   396,000,000,000

Weight = __________________ = _________________ = 197,112 metric tons

                   410 x 9.8 x .5                     2,009

The thickness of the structural steel from ground to level 16 is 5'',from 17 to 32 it reduces to 4'', and further reduces every 16 levels higher until the top 14 levels which had the lowest thickness of 1/4''.

Each tower is made up of a total of 6, 16 storey sections or blocks with different sized structural steel except the highest block which had 14 storeys.

To find the weight of these blocks we first deduct the steel weight from the building weight:-

197,112 – 80,000 = 117,112 Then divide by 110 = 1,064.65 metric tons for each floor with no steel.

To determine the weight of each block without steel content :-

1,064.65 x 16 = 17,034.5 metric tons per 16 levels. There are 6 of these blocks.

For the highest block :-

1,064.65 x 14 = 14,905 metric tons for 14 levels. There is only 1 of these blocks.

Floors     Thickness of     Percentage of     Weight of      Weight of Each     Total Weight

in each   Steel Frames       Structural      Steel used in    Block of Floors   of each block

Block         in inches       Steel Weight     each Block       Without Steel

97-110           1/4                  1.5                  1,200      +       14,905       =       16,105

81-96             1/2                  3.1                  2,480      +       17,034.5    =       19,514.5

65-80              1                    6.4                  5,120      +             ''         =       22,154.5

49-64              2                  12.7                 10,160     +              ''         =       27,194.5

33-48              3                  19.2                 15,360     +              ''         =       32,394.5

17-32              4                  25.2                 20,160     +              ''         =       37,194.5

  0-16              5                  31.9                 25,520     +              ''         =       42,554.5

                                       ______              _______                                     ________

                                        100 %                80,000                                      197,112

The percentage of steel used in each block was determined and listed on the third column. From these values we are able to show these actual amounts in tons as a percentage of 80,000 in column four. These column four steel tonnages are then added to the “weight of each block with out steel” values to find the “total weight of each block”. By adding these we find the total building weight.

These figures are slightly inaccurate. First, the top block has 14 levels while the remainder have 16. Secondly, in the core central columns, box beams are used to the 75th level then H beams from 76 to 110. It is not reported exactly how much less cross sectional area the H beam has compared to a box beam only that it is less. The conclusion one can draw is the upper floor sections would be some what lighter. The Government Agency FEMA has stated the thickness of the steel to be ¼'' at the top and 4'' at ground level. Each of these points would tend to cancel out each other. For this analysis we are going to use the existing figures. Looking at the last column note how much lighter the upper sections are compared to the lower sections. This column is also used in the next table.

One method to determine a Tower's PE is to calculate each Block's PE separately.

Each block will be numbered 1-7 from the bottom to the top.

Height of each floor 410 = 3.72 meters,   410 - height in meters, 110 - No of storeys

                           _____

                             110

To determine the PE of each block separately, first we find half the height of a 16 level block :-

16 Floors divide by 2 = 8 x 3.72 meters. Therefore 8 Floors are 29.7 meters high (half block height)

Height of the vertical centre of each block above ground was then calculated. See Table Below :-

Once these heights were found the general formulae to find the PE of each block was used :-

Weight (kg) x Height (meters) x 9.8 = Joules

Note that .5 is not used. It is not needed as we have already averaged the height of each block's 16

floors by dividing by 2 as shown above. Block 7 was calculated separately using its 14 levels.

Block   Floors    Vertical Centre    1 Floor     Vertical Centre    Block     Meters      PE

  No    in each    of Block Above     Height    of Block Above    Weight     /sec    G joules

           Block    Ground in Floors   Meters  Ground in Meters    Tons

  7       97-110            103      x      3.72      =      383.1    x    16,105   x   9.8  =   60.46

  6       81-96               88       x        ''         =      327.3    x    19,514   x     ''   =   62.59

  5       65-80               72       x        ''         =      267.8    x    22,154   x     ''   =   58.14

  4       49-64               56       x        ''         =      208.3    x    27,194   x     ''   =   55.51

  3       33-48               40       x        ''         =      148.8    x    32,394   x     ''   =   47.23

  2       17-32               24       x        ''         =        89.2    x    37,194   x     ''   =   32.51

  1         0-16                8        x        ''         =        29.7    x     42,554   x    ''   =   12.38

                                                                                                                      _____

                                                                                                                     328.82

Columns 3-5 determined the height of the centre of each block above the ground in meters. That value is then multiplied by the block's weight and 9.8 to find each blocks PE. These amounts are then added to find the building's PE. Note the smaller variations in the amount of G joule values of the 4 highest blocks despite the height increases. 328.8 G joules = 98.3 kWh = 78.5 tons TNT

Note that the original PE value of 396 G joules is 20% greater than the 328.8 G joules shown in the table above. Naturally, the PE would be higher if the percentage of steel used was lower and vice versa. In this analysis the structural steel content is 40%. Regardless, of what the steel percentage may be, the PE value would always be lower when compared to a value calculated using the general method where all the floor weights are considered the same.

There are large variations quoted when trying to establish the weights of the concrete flooring used in each tower. Two sources quoted 420,000 cubic yards while another source quoted concrete weight at 300,000 tons. Each side of a tower had a width of 64 meters (210 feet), if we multiply 64x64x.1x110 we achieve a figure of 45,056 cubic meters. The 110 figure is the number of levels, the .1 figure is one tenth of a meter which is just under 4'' which is the quoted floor thickness.

With further research we found that a standard cubic meter of concrete weighted 2.4 metric tons, if we multiply this figure by 45,056 we find a value of 108,134 tons. To relate this to a Twin Tower's floor plan we must make deductions for the elevator openings and the fact that light weight concrete was used. It is most likely a figure of 80,000- 90,000 tons is close to the correct amount.

The Towers were built under the New York Building Codes. One requirement was that they needed to be strong enough to support 5 times their own weight. Also, their designers stated the perimeter walls could support 2,000% (20 times) their own weight and if one wall was completely severed including the corners and around the sides, the building could still withstand winds over 100mph.

It's now been established that the Twin Towers weight was substantially less than regularly reported and because they were bottom heavy the PE value was still further reduced as well. There was less gravitational force available to do what was done in the amount of time the destruction took place in. To accept that a lighter weight upper structure was able to pound away at a substantially strong, undamaged and increasingly stronger lower structure in a manner where the lower structure offered virtually no resistance at all, one has to ignore the laws of physics.

                                                                                                      Rod McGuire