OK, just doing a quick calculation and adding it up bear with me...(tapping noises from keys on spreadsheet)...Hmm yeah, OK, I make that a total of one source quoted! I can't believe that because it wouldn't be like you guys to take one snippet of information and use that as a basis to debunk the whole set of data would it?
William,
from a previous post on simple equations.
Displacement (d) = 1/2*
at^2. ________ Velocity (v) =
at
Use 417m for height of building, Time to fall calculation in a vacuum, to street level.
Distance fallen = 417m = 1/2
at^2 (Get rid of the 1/2, multiply both sides by two)
834m =
at^2 ( To find
t, divide both sides by
a, which is 9.8m/sec^2))
(834m)/(9.8m/sec^2) = t^2 ( Meters cancel out by simple division, and sec^2 ends up in the numerator on the left. )
Units match: 834sec^2/9.8 =
t^2 (Take the square root)
sqrt(834sec^2)/9.8) =
t = 9.225 seconds freefall in a vacuum.
But we are not working in a vacuum. A body accelerating in a fluid (air) will accelerate until it reaches terminal velocity, after which point it will fall at a constant velocity. Terminal velocity of a body falling in air. (See discussion at the end) offers a usable range of values between 54 m/s (a high drag body, not aerodynamicallly shaped) to 90 m/s (a streamlined body.) I'll use a skydiver for modeling purposes, since the numbers are easily available, and a skydiver in high drag mode represents an irregularly shaped, non streamlined body falling in air.
For a stream lined body, similar to a skydiver presnting his body to the relative airstream in the most aerodynamic shape possible, or wrapped into a ball, or approaching the streamlined Gyrefalcon in a dive, you reach terminal velocity (v=
at) at roughly 90/9.8 = 9.18 seconds, which is almost the end of the vacuum fall timeline. (9.225 seconds) Are all pieces as streamlined as a falcon or a "tucked" skydiver in a dive: no. If all pieces are streamlined, they hit the ground in roughly 10 seconds. Irregularly shaped chunks of building are not streamlined, so let's use a high drag form, with a terminal velocity of 54 m/s for a "slowest case" fall.
The body reaches terminal velocity om about 5.5 seconds? v = at so v/a = t.
54/9.8 = t
t = 5.5 seconds.
That is just over half of the time of a non decellerated body falling. The v^2 component of air resistance increases from time 0 to time of V(terminal onset). (My problem set up for the differential equation is rusty, with v known and the sum of a and a' (a' being air resistnance in vector opposition to a) being a function of v^2.
d = 1/2
at^2 for how far this drag friendly body falls before reaching terminal velocity. After that, it falls at a constant rate, rather than being accellerated all the way to impact like in the vacuum example.
1/2*9.8*(5.5)^2 = d = 148.225 meters.
After reaching this speed, the non streamlined body falls at 54 m/sec.
417 -148.225 = 268.775 meters to fall.
Divide that by 54 m/sec and you get 4.977 seconds, which we can round up to 5.
So, it takes roughly 5.5 + 5 seconds = 10.5 seconds for a non streamlined body falling against air resistance at sea level to hit the ground from the top of the WTC. The engineers have estimate something near 13 seconds. I expect this is dependent on both their use of integration to account for drag = 0 and v= 0 at time 0, and drag = x at v = Terminal at time = Time of achieving terminal velocity, when a =0. The decision on which coefficient of drag to use can be made mass dependent if you set a = 0 when v = 54 m/s (forces summed) and using a kv^2 convention, yield k (coefficient of drag and other factors) = .00336.
The drag friendly terminal velocity provides a close enough approximation of worst/slowest case, particularly when the dust cloud obscures observation and precludes a lab quality observation and calculation.
In layman's terms, between 9 1/4 to 11 seconds to hit the ground, depending on the shape of the chunk falling.
(Engineer or math majors are invited to demonstrate the simple integral to solve for a more precise t(vterminal) and thus refine my gross estimate of 5.5. seconds as the time to achieve terminal velocity from v = 0 at the top, where a1 = 9.8m.sec^2 and a2 = 0) I am too far from school to give you the precise problem set up, as I am looking to make mass and cross section neutral for the sake of the problem, and combine those and air resistance/drag coefficient in to a generic constant k.)
Wiki discussion of terminal velocity said:
For example, the terminal velocity of a skydiver in a normal free-fall position with a closed parachute is about 195 km/h (120 mph or 54 m/s). This velocity is the asymptotic limiting value of the acceleration process, since the effective forces on the body more and more closely balance each other as it is approached. In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.
Higher speeds can be attained if the skydiver pulls in his limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 89 m/s), which is also the maximum speed of the Peregrine Falcon diving down on its prey.
From UVa "How Things Work" "A person has a terminal velocity of about 200 mph when balled up and about 125 mph (56 - 89 m/s) with arms and feet fully extended to catch the wind."
DR