and the boobs they just keep coming.
listen "removebush" they already debated all your nonsense over and over again. do a search of this site. here is just one post that addresses your free fall fallacy
from PVT1863's post
http://www.internationalskeptics.com/forums/showpost.php?p=2195537&postcount=236
The total collapse of the buildings was inevitable because of the immense weight of the upper floors that were falling. In order to give you an idea of how much force was involved here are a few calculations... I was the original author of at least part of that calculation. I posted it on the IMDB board for
World Trade Center about six months ago. The thread has since been deleted. I didn't realize it had been saved by anyone or reposted elsewhere.
I have good news and bad news. The bad news is that the equation I used was wrong. I realized this after the thread was deleted, so I didn't point it out. I might have if I would have realized it would turn up again. I also had a factor of ten error that appears to have been corrected, though this error did not carry through to the force ratios since it was applied equally to the collapsing force and the normal force.
The good news is that this error caused me to severely
underestimate the force ratio necessary to stop the falling floors.
The correct equation for the force is:
F = (-1/2) * (v^2 * m)/(dstop)
Substituting for v, it becomes:
F = -(h * m * g)/(dstop)
F = force
v = velocity with which the upper floors strike the undamaged one
m = mass of the falling floors
dstop = the distance over which the falling floors are decelerated (the distance of deflection)
h = the distance the upper floors fell
g = the acceleration due to gravity
Here is the derivation (I figured I should show this since I screwed up last time).
The floor must come to rest, so the final velocity, vf, must equal zero. If an acceleration, a, is constantly applied to an object with an initial velocity vi, then
vf = vi + a*t
Since vf = 0 for the object to stop, we can set it to zero and solve for t.
0 = vi + a*t
-vi = a*t
t = (-vi)/(a) Call this equation (1).
So the time it takes to stop a moving object through the constant application of an acceleration is (-vi)/(a). The negative sign is there because the acceleration must be in the direction opposite that of the initial velocity if the object is to stop.
Basic kinetics tells us that the final position, xf, of an object is a function of its initial position xi, its initial velocity vi, its acceleration a, and the time duration t.
xf = xi + vi*t + (1/2)*a*t^2
If we define the original position to be x=0, then xi=0.
xf = 0 + vi*t + (1/2)*a*t^2
Now we substitute equation (1) in for t and rearrange.
xf = 0 + vi*(-vi/a) + (1/2)*a*(-vi/a)^2
xf = -(vi^2)/a + (1/2)*vi^2/a
xf = -(1/2)*(vi^2)/a Call this equation (2)
Next we address the most basic equation of Newtonian physics, F = ma
F = m*a
a = F/m
We plug this in for a in equation (2) and rearrange to get
xf = -(1/2)*(vi^2)/(F/m)
F = -(1/2)*(vi^2*m)/(xf)
In this equation, xf is the distance it takes to stop and vi is the initial velocity. In our particular application xf=dstop and vi=v. So
F = -(1/2)*(v^2*m)/(dstop) Call this equation (3)
This is sufficient to solve, but it would require the calculation of v. For the sake of simplicity, we can insert the calculation of the initial velocity based on free-fall into this equation. It is known that the time it takes to free-fall a distance h from rest is
t = sqrt((2*h)/g)
It is also known that the final velocity, vf, of a free-falling object that has no initial velocity is
vf = g*t
Rearranged, that is
t = vf/g
When we substitute that into the equation for time of free-fall and rearrange, we get
vf/g = sqrt((2*h)/g)
vf = g * sqrt((2*h)/g)
vf = sqrt((2*h*g^2)/g)
vf = sqrt(2*h*g)
This gives us the velocity of a falling object that had no initial velocity after falling a distance of h. Since the v we use in our calculation as the velocity of the falling floors is equal to the velocity of the floors after falling from their original position at rest, vf=v. So
v = sqrt(2*h*g)
We can now plug this into equation (3) and rearrange to get
F = -(1/2)*([sqrt(2*h*g)]^2*m)/(dstop)
F = -(1/2)*(2*h*g*m)/dstop
F = -(h*g*m)/dstop
Now we can start plugging in values.
h = -10 feet; the damaged floors traveled 10 feet downwards
g = -32.2 ft/s^2; the acceleration due to gravity
m = 3,387,916 lbm; the approximate mass of 12 floors
dstop = 1/6 foot; the 2 inches I used for the allowable deceleration region
F = -(h*g*m)/dstop
F = -[(-10 feet)*(-32.2ft/s^2)*(3387916 lbm)]/(1/6 foot)
F = 6,505,713,457 lbs
That is six and a half
billion pounds of force (it appears my original calcuation underestimated the force by a factor of approximately six). This is the force that the undamaged floor would have to impart on the collapsing floors during the entire duration of the deceleration (while the falling floors traveled those two inches)
If we use the same calculation of the normal load on one floor, we get approximately nine million tons per floor. If all of the force I calculated above falls on one floor, the ratio of the force needed to stop them to the normal load on that floor is:
~6,500,000,000/~9,000,000 = about 722.
So if the first undamaged floor had to stop the falling floors by absorbing all of the force they would strike it with in this theoretical situation, the floor would have to impart an upward force approximately 722 times larger than that it had to impart under normal conditions.
Now, of course, this calculation is not accurate. It calculates a theoretical maximum since it assumes no losses and assumes that the entire force of the upper floors, acting as a rigid body,
fall on the lower floors. This is untrue. Some of the force will bear down on the external and core supports. Some of the force will be consumed in the destruction of structures. Some of the weight of the upper floors will not apply to the first floor struck because they will not remain rigidly attached to the lowest floor in the falling region. The acceleration will be slowed by air resistance and deformation of crumpling materials, so it won’t be quite as high as g. Some mass might be lost as the collapsing floors fall.
For the calculation to be truly conservative, all of these would have to be accounted for. I tried to do this in my previous calculation by assigning penalties to certain values which would decrease the total force. I have here changed a few:
-50% reduction in mass. This accounts for the fact that the upper floors might be lighter than the lower ones, that some mass might be lost while falling, and that the upper floors might not be rigidly attached to the lower ones in the collapsing region (and thus their mass won’t contribute). This makes m about 2,700,000 lbm.
-50% reduction in acceleration. This accounts for a slowing
-200% increase in deceleration region. This allows for a slower deceleration. The new value for dstop is 6 inches (.5 feet).
-20% reduction in free-fall distance. This accounts for the fact that the mass will not all fall 10 feet before striking the floor of the first undamaged floor. The new value for h is 8 feet.
If we plug these into our equation, we get
F = -(h*g*m)/dstop
F = -(8ft)*(-16.1ft/s^2)*(-2700000)/-.5
F = 695,520,000 lbs
That is still more than 77 times the normal load on one floor, even with conservative assumptions.
We can even go so far as to calculate the force on a floor if ony the floor directly above it falls on it. We will ignore all the floors which are higher up. If the force of just the floor immediately higher is enough to collapse a floor, then it follows that the collapse will progress downwards indefinitely because the additional force of the upper floors is not needed. So our mass for one floor is 282,326lbm. We will lose 50% of this for the above stated conservatism, giving us an m of 141,163lbm.
F = -(h*g*m)/dstop
F = -(8ft)*(-16.1ft/s^2)*(-141163)/-.5
F = 36,363,588 lbs
That is, on its own, four times the normal load on a floor. So a floor would have to be built with a safety factor of five (four for the falling floor and one for its own weight) to survive the collapse of the above floor, even with some conservative assumptions. And that does not account for the fact that the mass of the falling floors increases as more floors collapse.
DISCLAIMER
For the sake of intellectual integrity, I must point out that this is an unverified back-of-the-envelope calculation. As such, it should not be taken as gospel regarding the forces involved. There are assumptions and values here which I believe are approximately correct, but which I cannot verify specifically. The largest problem is clearly the estimation of the deceleration region. This factor is what makes impact analysis difficult because one has to have a good value for the distance or time over which the impact took place to calculate a good value for the force. I simply chose what I believe to be an overly-large value. If the impacted floor deflected downwards six inches, it is probably a safe assumption that they would have failed. Of course, that, in turn, leads to the assumption that the impacted floor acted rigidly, which is another problematic assumption. I'm simply pointing this out to show that this is not by any means a perfect analysis. But it is still a heck of a lot better than anything I have seen from the CT side.
I merely wanted to estimate so that the sheer magnitude of the forces involved could be demonstrated, and I feel that I have been successful in that endeavor. We are talking about hundreds of millions (or even billions!) of pounds of force when we talk about the collapsing WTC towers. That is not something which is easily comprehended when one approaches the collapse with intuition-based science.
Feel free to discuss, denounce (with good reason), or correct my calculations.[/quote]
