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NIST NCSTAR 1A at 45 states: [November 2008]
The slope of the velocity curve is approximately constant between about 1.75 s and 4.0 s, and a good straight line fit to the points in this range (open-circles in Figure 3-15) allowed estimation of a constant downward acceleration during this time interval. This acceleration was 32.2 ft/s2 (9 .81 m/s2 ), equivalent to the acceleration of gravity g.
Originally posted by GhostR1der
We simply have no evidence other than time.
To hazard a guess I'd say a tesla based scalar electromagnetic weapon was used in conjunction with thermal and shaped cutting charges. Tesla weapons can be used in many modes, one of which is a cold explosion or 'implosion'
Originally posted by bsbray11
Drag is resistance provided by air, so this implies there was no resistance provided by air during this period. If there was, it was not able to be measured by NIST.
Originally posted by Joey Canoli
You mean, of course, there was no resistance given by the air..... being forced out of the building's volume, right?
Originally posted by bsbray11
Actually the whole "air being forced out" thing requires energy from the falling building.
Originally posted by Joey Canoli
Originally posted by bsbray11
Actually the whole "air being forced out" thing requires energy from the falling building.
How much?
Show us your awesome physics knowledge.
Drag force on an object is given by the equation
Fd = 0.5*Cd*p*A*v2
= bv2
b is a constant where,
Cd is the drag-coefficient of the object, a dimensionless constant dependent on its shape and determined empirically. The air in the individual floors of the building would move with and can be considered one unit with the structure and therefore ignored in the calculation. Only the bottom-most floor at any one point on the collapse would experience air resistance as it must push air out of the way in order to impact the ground. The drag co-efficient for a long flat plate is 1.98[1] and will be used in this case for simplicity sake.
p is the density of air (assumed near sea level) 1.223Kg/m3
A is the area of the object normal to flow. NIST cites the area of each floor as 2,000,000 ft2 = 185 806.08m2
Therefore,
b = 0.5 * 1.98 * 1.223 * 185 806.08 = 224968.4274816
An object reaches terminal velocity when the drag force acting on the object in the negative equals the force applied to it in the positive, in this case gravity.
Hence at terminal velocity Fd = bv2 = mg
=> VT = SQRT(mg/b)
No figures are available for the mass of WTC7 that I can find, so I will make a very rough estimate based on other published estimates of the mass of WTC1 (5 x 108 Kg). WTC7 was 0.45 times the height of WTC1 and covered comparable floor area, however since there was less steel used in the construction of WTC7, I'll use an arbitrary figure of about a third, placing the mass of WTC7 at 1.7 x 108 Kg
VT
= SQRT(1.7 x 108 * 9.8 / 224968.4274816)
= 86.0551195 m/s
Velocity at time (t) is given by v(t) = VT tanh (gt/VT)
Velocity at time 4.757s
= 86.0551195 * tanh(9.8*4.757/86.0551195)
= 42.53670521674826m/s
Average acceleration across this time
= v/t
= 42.53670521674826/4.757
= 8.94m/s2
The final figure for theoretical collapse acceleration rate of WTC7 in complete free fall in atmosphere and at sea level is 8.94m/s2, which is only a little above the actual observed 8.71m/s2 acceleration rate arrived at from analysis of the CBS footage and using the Emporis height measurement. From this we can imply that the structure provided a negative acceleration, i.e resistive force of approximately 0.23m/s2 to the gravitational collapse.
Originally posted by Joey Canoli
Why would he calculate the drag of the building in freefall through the atmosphere, since that didn't happen?
I note that you were unable to provide your own work, just a cut/paste job........
Originally posted by bsbray11
So you admit there was somehow no air resistance to the falling building.
Originally posted by bsbray11
The air in the individual floors of the building would move with and can be considered one unit with the structure and therefore ignored in the calculation.
Only the bottom-most floor at any one point on the collapse would experience air resistance as it must push air out of the way in order to impact the ground. The drag co-efficient for a long flat plate is 1.98[1] and will be used in this case for simplicity sake.
Originally posted by pteridine
reply to post by bsbray11
The amount of drag due to air would likely be within the error of measurement. You would have to show significant loss to postulate an implosion. If the building is collapsing straight down, the only drag will be shear forces on the outer skin. The geometry is not one of a freely falling body in air. You would have to include the compression of air within the collapsing building. The burst point of windows breaking due to compression of air [the so-called "squibs"] is low enough [a few psi] so as to not require much PV work which leads me to the "error-bar" conclusion.