The height of the top floor of the towers was 1,348 ft (411.0 m).
The height of the root of the towers was 1,368 ft (417.0 m).
The height of the antenna/spire of the towers was 1,731.9 ft (527.9 m)
source:
http://en.wikipedia.org/wiki/Wtc
Given the approximate acceleration at the towers would be 9.8 m/s^2 we can calculate both the time it would take to fall, and the velocity at any given point.
Using the formulas available on this page:
http://www.grc.nasa.gov/WWW/K-12/airplane/mofall.html we generate the following (bolded lines are of greatest interest):
[table=""]Time|Accel - m/sec^2|Dist - meters
0.0|9.8|0.0
1.0|9.8|4.9
2.0|19.6|19.6
3.0|29.4|44.1
4.0|39.2|78.4
5.0|49.0|122.5
6.0|58.8|176.4
7.0|68.6|240.1
8.0|78.4|313.6
9.0|88.2|396.9
9.1|
89.2|
405.8
9.2|
90.2|
414.7
9.3|
91.1|
423.8
9.4|92.1|433.0
9.5|93.1|442.2
9.6|94.1|451.6
9.7|95.1|461.0
9.8|96.0|470.6
9.9|97.0|480.2
10.0|98.0|490.0
10.1|99.0|499.8
10.2|100.0|509.8
10.3|
100.9|
519.8
10.4|
101.9|
530.0
10.5|102.9|540.2
10.6|103.9|550.6
10.7|104.9|561.0
10.8|105.8|571.5
10.9|106.8|582.2
11.0|107.8|592.9| || |[/table]
From this we can see that from the top floor it would take 9.1-9.2 seconds to reach the ground and the final velocity would be 89.2-90.2 m/sec.
From the root it would take 9.2-9.3 seconds to reach the ground and the final velocity would be 90.2-91.1 m/sec.
From the top of the spire/antenna it would take 10.3-10.4 seconds to reach the ground and the final velocity would be 100.9-101.9 m/sec.
Estimates on this vary do to visibility problems and which points are being used. I will use the values you provided in this post
http://www.internationalskeptics.com/forums/showthread.php?postid=2206211#post2206211 and go with 12-16 seconds.
[table=""]Reported Time|Calculated Time|%
12.0|9.1|131.9
12.0|9.2|130.4
12.0|9.3|129.0
12.0|10.3|116.5
12.0|
10.4|
115.4
13.0|9.1|142.9
13.0|9.2|141.3
13.0|9.3|139.8
13.0|10.3|126.2
13.0|10.4|125.0
14.0|9.1|153.8
14.0|9.2|152.2
14.0|9.3|150.5
14.0|10.3|135.9
14.0|10.4|134.6
15.0|9.1|164.8
15.0|9.2|163.0
15.0|9.3|161.3
15.0|10.3|145.6
15.0|10.4|144.2
16.0|
9.1|
175.8
16.0|9.2|173.9
16.0|9.3|172.0
16.0|10.3|155.3
16.0|10.4|153.8| || |[/table]
We can see that the towers took somewhere between 115.4% to 175.8% free-fall time to complete collapsing. Personally, I do not consider 15.4%-75.8% margin of difference to qualify as being labeled "near".
http://www.house.gov/science/hot/wtc/wtc-report/WTC_ch2.pdf Section
2.2.1.1
American Airlines Flight 11 struck the north face of WTC 1 approximately between the 94th and
98th floors
2.2.1.5
Construction of WTC 1 resulted in the storage of more than 4x10^11 joules of potential energy over the
1,368-foot height of the structure. Of this, approximately 8x10^9 joules of potential energy were stored in the
upper part of the structure, above the impact floors, relative to the lowest point of impact.
2.2.2.1
United Airlines Flight 175 struck the south face of WTC 2 approximately between the 78th and 84th
floors.
For WTC 1, the top 12 floors of the tower translates into 8x10^9 joules of the total 4x10^11 joules. So, the top ~10.9% of WTC 1 contained ~2% of the entire PE of WTC 1. Extrapolating this on to WTC 2 (since the above mentioned report does not specify the amount PE contained above the WTC 2 impact point) we get the following:
WTC 2 => top 26 floors => ~23.6% of WTC 2.
If ~10.9% of WTC 1 translates into 8x10^9 joules PE
Then ~23.6% of WTC 2 translates into N joules PE
Therefore 10.9/8*10^9 = 23.6/N
=> 10.9*N/8*10^9 = 23.6
=> 10.9*N = 23.6*(8*10^9)
=> N = 23.6*(8*10^9)/10.9
=> N = 17321100917.431192660550458715596
=> N = 17.3*10^9 joules PE
=> ~34.7% of the entire PE of WTC 2
What does this mean? It means for WTC 1, that ~2% (8x10^9 joules) of PE was converted to KE almost instantaneously upon structural failure at floors 94-98. It means for WTC 2, that ~34.7% (17.3*10^9 joules) of PE was converted to KE almost instantaneously upon structural failure at floors 78-84.
If you feel this is insufficient to explain the damage seen, please provide your maths showing why.
<snip>
