Please tell me these are not serious questions.
Question for the twoofers about why NIST is wrong
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Please tell me these are not serious questions.
Gravy
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Holy crap, Wizard, this isn't about "balance." Take the last sentence out of that post. Is Steven Jones less wrong because the Bush administration also does not respect real science?
Remember you demanding backup for my claims about Oklahoma City, me providing them, and you instantly dismissing them. That's you, Wizard, in the real world. Maybe it's time for you to take a break from making 50 posts a day on this forum. Maybe it's time to step back and reflect on the heartfelt advice that you've been given here.
Gravy I conceded to the oklahoma evidence you gave.
The Almond
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They are serious questions. Please answer them. Alternatively, you may read my argument and refrain from making further blanket statments such as:
If the argument is bad, crazy or based on false rationalizations, it should be very simple to point them out to the satisfaction of all the skeptics here.
Gravy
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You immediately rejected it, and you had to be told by several people to READ it. You accepted it after that admonition.
I appreciate that you've changed your mind about some beliefs here. That's far more than most conspiracy believers have done (or admitted to) on these forums. It's your adversarial approach that baffles me. It's not the questions or statements, it's the way you phrase them. When you're shown time and time again that you're in a place where people respect learning, and where you're shown time and time again that your assumptions are wrong, maybe it's time to prick up your ears.
Gravy
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See my last post.
Ok so I would like you to shed some light on this for me twoofers. Why is the NIST report wrong? Using actual scientific data and physics, why is it not possible for the towers to collapse the way NIST said that they did? What specifically did they mess up on? Here are some unacceptable answers:
"I just simply don't believe that it could happen that way"
"I don't buy that government influenced report."
"I disagree with what the report said."
"There is no way the towers could collapse from just fire."
Acceptable answers:
"I just simply don't believe that it could happen that way BECAUSE...(insert scientific data/physics/evidence)
"I don't buy that government influenced report BECAUSE...(insert scientific data/physics/evidence)."
"I disagree with what the report said BECAUSE...(insert scientific data/physics/evidence)."
"There is no way the towers could collapse from just fire BECAUSE...(insert scientific data/physics/evidence)."
What I am looking for is some specific explanations using science as to why NIST got it wrong. What are the reasons that you don't believe a panel of 200 experts with years of experience assigned specifically to investigate the collapses. By reasons, I don't mean "because they work for the government." If they really were paid off to fabricate a report, it should be easily proven wrong using science and physics. So have at it twoofers. Lets hear it.
Note: I am looking for actual data and scientific explanations that could be proven in a peer reviewed journal. I know, that is asking a lot but that is how the real world works. Sorry.
The thread may not be over yet:
http://www.journalof911studies.com/v...estigation.pdf
You have to read NIST to understand why this non degree CT in the Scholars, only an AM, has the motive to make up stuff.
28 pages to rip 10,000 pages?
Have you seen the real report? There are 290 pages just on the Analysis of the Aircraft impact!!! http://wtc.nist.gov/NISTNCSTAR1-2B_Chaps1-8.pdf
And it will help you understand how a simple aircraft damaged the WTC.
Do yourself a favor and look at NIST and see what they did before you buy the snakeoil lies of the Scholars. Look, you will not be sorry about learning and seeing how the Scholars are liars - http://wtc.nist.gov/NISTNCSTAR1-2B_Chaps1-8.pdf
Just an idea, a darn good idea, see what you are trying to bring down before you start burning books and witches like other liars in the past.
Spins
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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.
The towers weighed about 500,000 tons each and the first plane hit between the 94th and 98th floors. So a conservative estimate at the weight of floors above the weakened area is...
((500,000 / 110) * (110 - 98)) = 54,545.45 tons
The second plane hit between the 78th and 84th floors. So again a conservative estimate at the weight of floors above the weakened area is...
((500,000 tons / 110) * (110 - 84)) = 118,181.82 tons
Now lets take the 54,545.45 ton estimate and give you an idea of the force involved as the falling section hit the first floor below the collapse zone...
(54,545.45 tons) * (2000 lbs/ton) = 109,090,900 lbs
So the weight in pounds is about 109 million pounds. Next we convert that to mass by dividing out the acceleration due to gravity.
(109,090,900 lbs) / (32.2 ft/s^2) = 3,387,916.15 lbm
If one body in free-fall strikes another body at rest and both bodies come to rest after a certain displacement the force involved can be calculated with the following equation...
F = -(m1 * vi)/(sqrt[d/g])
...were F is the force, m1 is the mass of the falling object at the time of impact, vi is the velocity of the falling object at the time of impact, d is the deflection, and g is the acceleration due to gravity. If the section falls for 10 feet in free fall (and if we define the upward direction to be positive), then m1 is 3,387,916.15 lbm, vi is -25.3 ft/s, d is -0.167 ft (2 inches of deflection, the floor slabs were 4 inches thick), g is -32.2 ft/s^2.
Using the values above we get a force of 1,190,208,026.08 lbs.
Now to find the normal load of a floor we can make a rough estimate by dividing the weight of the towers by 110, obviously I'm assuming the weight is evenly distributed and I'm also not taking into account the weight of the core etc, so again I'm over estimating. CT'ers often make the mistake of assuming each floor held up all the other floors above, this is wrong each floor did not hold up the floors above the core and exterior columns that ran past each floor held up the rest of the building. An analogy would be a ladder, each rung on the ladder does not hold up the rungs above; the rungs are held up by the columns on either side of the rung.
((500,000 tons) * (2000 lbs/ton) / 110 floors) = 9,090,909.09 lbs
So now to find the ratio of the force from the falling section to the force normally held by each floor is...
1,190,208,026.08 / 9,090,909.09 = 130.92
...so in order for the floor beneath the collapse zone to have withstood the impact of the falling floors above it would have to have been able to carry a load 130.92 times larger than what it normally did.
Now lets make more conservative estimates, resistance caused the collapsing section to fall with an acceleration half that of gravity (vi = -17.9 ft/s), 20% of the mass was lost during the collapse (m1 = 2,710,332.92 lbm), deflection was 4 inches the thickness of a floor slab (d = -0.333 ft). Using these values we get a force of 477,069,322.09 lbs, now lets assume half the impact force was carried to the core and exterior columns so the final value is 238,534,661.045 lbs. So once again lets get the ratio of the force normally held by each floor...
238,534,661.045 / 9,090,909.09 = 26.24
...that is still 26.24 times the normal load on a floor. So even when very conservative assumptions are made we end up with a force on the first undamaged floor much higher than the floor would have been designed to withstand, and remember I used the smaller collapsing section.
Maths isn't my strong point so I apologise for any errors.
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Matthew Best
Penultimate Amazing
Well, maths isn't my strong point either, but you seem to make a very strong case.
Lurker
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Gravy,
Thanks for your forebearance on Mechanical Engineers (me being one!). But I will state that ME's don't really get into structural analysis at any level compared to the Civil Engineers. We have a course each in Statics, Mechanics of Materials, Dynamics, Kinematics, and Metallurgy but those are the only classes peripherally related to WTC collapse theory. This all pales in comaprison to what Structural Engineers take in this area.
I guess that is a lot more than what CT'ers know though. I have a good friend in Japan right now that subscribes to the CD theory. His big claim is that it is obvious even to the non-expert that WTCs had to come down via CD. It is quite frustrating trying to show him that his "common sense" may be wrong and is no substitute for expertise and training.
Lurker
Thanks for your forebearance on Mechanical Engineers (me being one!). But I will state that ME's don't really get into structural analysis at any level compared to the Civil Engineers. We have a course each in Statics, Mechanics of Materials, Dynamics, Kinematics, and Metallurgy but those are the only classes peripherally related to WTC collapse theory. This all pales in comaprison to what Structural Engineers take in this area.
I guess that is a lot more than what CT'ers know though. I have a good friend in Japan right now that subscribes to the CD theory. His big claim is that it is obvious even to the non-expert that WTCs had to come down via CD. It is quite frustrating trying to show him that his "common sense" may be wrong and is no substitute for expertise and training.
Lurker
The Almond
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So it looks like I'm not going to get a response from you about my post?
pvt1863
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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.
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The Almond
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The equation and derivation look good from a theoretical standpoint. It makes, quite conclusively, the point Wizard has been harping about. The towers had no ability to stop, dampen or absorb the collapse itself once initiation started. That's why it's valid to assume that once the collapse begins, it will lead to total destruction of the towers.
Spins
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Yes, I actually PM'd you about a month after you posted that message asking you if you minded me using the calculations you posted but you never reply'd. You are correct the original post has been deleted from imdb, along with all those old posts of mine, anyways this was the post of yours I saved for reference...
Thanks for the new calculations.pvt1863 said:
Spins
Muse
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Yes, this is exactly what I was trying to show that once the collapse started the forces involved were so immense that "global collapse" was inevitable. Sadly I don't think it'll change the mind of your average hardcore CT'er who is blind to physics, science, and expert opinion.
A W Smith
Philosopher
OK I read it three times now and my head hurts. One thing I cant see addressed. The collapse is happening in two directions. As the floors below the impact zone collapse so do the floors above especially considering the joist bearing plates were not designed for upward lift. So doesn't the top third or whatever part of the tower accelerate even faster thereby increasing the momentum? Another thing...If you knew the cross section of the perimeter and core columns at the base in weight per foot and that of the top story in weight per foot and then average them . and the weight per linear foot of floor joist. And given that the weight per square foot of concrete plus say half the designed live load of 100 PSF Couldn't somebody come up with a reasonable proportion % of the weight of what collapsed and that of what fell away? Of course this assumes the change in weight per foot of columns is linear from top to bottom which is probably incorrect because of wind load design and other dynamics,
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