9/11 - WTC/Pentagon/Shanksville m.m - Tråd Nr 2

Inte fåt svar på hur människo delar visa en centi meter små kunde hammnat på tak och andra ställen en bit bort.

In order to allow time for lateral motion, the exterior column(s) that hit WFC 3 were most probably from the upper half of WTC 1. A fall from 1,000 feet to 240 feet would take SQR(2*h/g) = around 6.9 seconds where h = 760 feet and g = 32.17 ft/s^2. In the horizontal plane, a uniform acceleration of 20 m/s^2 for the first second followed by negligible deceleration due to drag for the remaining 5.9 seconds would provide 10 + (5.9 * 20) = 128 metres = 420 feet displacement. At 1,000 feet the WTC 1 perimeter columns, per story, were comprised of:

two flanges of 1/2 x 13.5 x 144 inches each, totalling 1,944 ins^3
one outer web of 1/4 x 13 x 144 inches = 468 ins^3
one inner web of 1/4 x 15.75 x 92 inches = 362 ins^3
one spandrel plate of 3/8 x 40 x 52 inches = 780 ins^3

...totalling 3,554 ins^3 per floor or 10,662 ins^3 = 6.17 ft^3 for a three-floor section which at 490 lb/ft^3 is 3,023 lb (84 pounds per lineal foot) or 1,371 kg. (There is some uncertainty as to the flange thickness; it was known to be only 1/4" at the very highest floors.) The force require to produce an acceleration of 20 m/s^2 in an inertia mass of 1,371 kg is 20 * 1371 = 27,420 N = 6,165 lbf.

The cross-section presented to a wind, per floor, would be 40 x 52 = 2,080 ins^2 for the spandrel plate and 15.75 x 92 = 1,449 ins^2 for the inner web, totalling 3,529 ins^2 per floor or 10,587 ins^2 = 6.83 m^2 for a three-story section of exterior column. (So the required pressure is well under 1 psi.) From the drag equation of

d = Cd * A * r * 0.5 * v^2

we obtain

v = SQR(2 * d / (Cd * A * r))

where r = density of air ~ 1.2 kg/m^3 and assuming a relatively high drag coefficient Cd of 4 / pi ~ 1.27 for a flat plate and d = the previously calculated force of 27,420 N and A = 6.83 m^2 as calculated above. This places the required wind at 72.6 m/s = 162 mph for one second duration. Actual windspeed on the day was up to 10 mph on the ground and up to 20 mph at higher altitude.

Suppose we imagine the collapse initiating at 1,200 feet, and proceeding as per the "pancaking" theory to 1,000 feet. After freely falling 200 feet, the terminal velocity would be SQR(2 * 200 * 32.17 ft/s^2) = 113.4 fps = 77.3 mph. In this theory, there is a small delay due to resistance of the intact building below, but the falling upper section smashes its way through each floor in about 0.1 seconds at the 1,000 feet level. The volume of air per floor is approximately 12 * 200 * 200 feet = 480,000 ft^3. Some will go down, but if the total was forced out through a perimeter of 800 feet by an average height of 6 feet which is an exiting area of 4,800 ft^2, it would (continuing outward) extend for some 100 feet at the end of the 0.1 seconds which is a velocity of 1,000 fps or 682 mph.

Let's set the exiting gases velocity at just 700 fps = 213 m/s, in which case the force acting on the exterior column for 0.1 seconds is given by:

d = Cd * A * r * 0.5 * v^2

= 1.27 * 6.83 * 1.2 * 0.5 * 213^2 ~ 236,000 N

to produce an acceleration of F / m = 236,000 N / 1,371 kg = 172 m/s^2. After 0.1 seconds the velocity of the steel is 17.2 m/s = 38.5 mph, and the horizontal displacement is 0.86 metres. Following another 6.8 seconds at 17.2 m/s the total distance travelled horizontally is 0.86 plus 6.8 * 17.2 ~ 118 metres = 387 feet. The columns have to shear off quickly enough, and the pancaking theory has the problem that the gravitational potential appears to be too low for all the energy sinks, but even this scenario does not appear to rule out the idea that debris could end up a few hundred feet away