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9/11 Physics from Non-Experts

Thanks that makes sense. But why doesn't the truth movement mention that information? Why do they leave such details out?
 
Carefulplease said:
Hi Jonnyclueless,

Tests were performed to see if the floor assemblies were compliant with the ASTM E 119 standard for fire resistance. These tests were performed on fire-insulated trusses. They did sag. Some failed to get a two-hour rating.

http://wtc.nist.gov/NISTNCSTAR1-6B.pdf
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The interesting thing is that the 35-foot truss that failed the 2-hour rating was restrained – pinned at each end – as it would be in a real building. Its failure was a surprise to researchers and led to much further research.
 
Welcome to the forums, Johnnyclieless.
Jonnyclueless said:
Thanks that makes sense. But why doesn't the truth movement mention that information? Why do they leave such details out?
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Because the only things holding their fantasies together are lies, omissions, and misrepresentations.
 
Carefulplease said:
I've got pretty much the same figures, unsurprisingly. I don't know how Bazant and Zhou arrived at their estimates. The more recent paper by Bazant, Le, Greening and Benson assumes that the mass is "almost 500,000 tons". They use a distribution (from the top down to the bottom of the bath-tub):

mu(z) = k_0*exp(k_2*z) + k_1

The values of the constants are unspecified. Maybe Apollo20 will inform us.

http://www.democraticunderground.co...mesg&forum=125&topic_id=159282&mesg_id=159611
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For the record, this was inaccurate. This mu function was used from the top to the 81th floor and the rest of the variation was linear all the way down.

Can someone cleverer than I tell me how to solve for k in the expression
c = k*exp(k*z)
where both c and z are numerical constants?
 
Carefulplease; said:
They did perform tests on full-scale short-span (35 feet) trusses. The largest furnaces available to perform fire tests were those of Underwriters Laboratories of Canada in Toronto. They were able to verify the effects of scale through comparing the results for both full-scale and half-scale tests. To build a 60 feet furnace would have been expansive and would have provided diminishing returns.
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What? Another perfectly logical and reasonable explanation that seems almost intuitive? If I was pressed for an answer, and had utterly no clue, this would have been it. This is just another example of how disappointing this whole alleged conspiracy really is. Everything follows from logic and has a reasonable explanation.
 
Carefulplease said:
For the record, this was inaccurate. This mu function was used from the top to the 81th floor and the rest of the variation was linear all the way down.

Can someone cleverer than I tell me how to solve for k in the expression
c = k*exp(k*z)
where both c and z are numerical constants?
Click to expand...

I believe that reduces down by taking the log of both sides. I haven't used logarithms in a long time though.
 
Carefulplease said:
For the record, this was inaccurate. This mu function was used from the top to the 81th floor and the rest of the variation was linear all the way down.

Can someone cleverer than I tell me how to solve for k in the expression
c = k*exp(k*z)
where both c and z are numerical constants?
Click to expand...
and take the ln (lawn) and solve i believe. ln (c) = k^2z i think, been a while
 
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oh yah, the ln(k) is a problem. hmm...that ain't right.

i'm looking at the logarithmic rules but I'm not sure. i forget how to manipulate log func'tn :(

yah, maybe newton's got it, i was going to say maybe its a graphical solution. of course i was also going to say apply Green's theorem as well, so take it with a grain of salt :) i'm looking for my old differentials text book as we speak but don't hold your breath. i been meaning to get it out since the new Bazant/Greening paper...
 
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Can you approximate it using a partial expansion of the infinite series? Then at least you have a polynomial that is easier to work with.

e^x = sigma (n=0 to infinity) of (x^n)/(n!)

A four-term expansion give you something like this:

c = k + (k^2)*z + (k^3)*(z^2)/2 + (k^4)*(z^3)/6
 
I looked at my expansion and my head started to hurt with all those terms, so I simplified even more to a two-term expansion which gives us a quadratic equation.

c = k + z*k^2
or
z*k^2 + k - c = 0

Using the quadratic formula, I can now solve for k:

k = (-1 +/- sqrt(1 - 4*z*(-c)))/(2*z)
or
k = {(-1 + sqrt(1 + 4zc))/2z), (-1 - sqrt(1+4zc))/2z)}


I have no idea what this means by now. I hope you can make something of it, Carefulplease!
 
AZCat said:
I looked at my expansion and my head started to hurt with all those terms, so I simplified even more to a two-term expansion which gives us a quadratic equation.

c = k + z*k^2
or
z*k^2 + k - c = 0

Using the quadratic formula, I can now solve for k:

k = (-1 +/- sqrt(1 - 4*z*(-c)))/(2*z)
or
k = {(-1 + sqrt(1 + 4zc))/2z), (-1 - sqrt(1+4zc))/2z)}


I have no idea what this means by now. I hope you can make something of it, Carefulplease!
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Thanks for the hard work AZCat. I fear the two-term expansion might deviate too much in the range I am interested with. I might try Newtons Bit poor-man's numerical method and tell you how much yours deviates ;)
 
Carefulplease said:
Thanks for the hard work AZCat. I fear the two-term expansion might deviate too much in the range I am interested with. I might try Newtons Bit poor-man's numerical method and tell you how much yours deviates ;)
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If you give me an idea of the values you will be using, I can give an estimate of the error of the approximation. The range of possible "k" values and the value of "z" will determine the magnitude of the error.
 
GregoryUrich said:
The correct mass and proper sequence of "crush up" first makes a huge difference in the timings. If Bazant is wrong about the amount of ejected debris this also makes a significant difference. I still think 20% is very low. I think we will find that the gravity driven collapse time will end up being around 20 seconds.
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Based on the height of the rubble pile within the tower footprint and assuming that the stories were squashed down to about 10 to 15% of their original height, a total mass shedding of 20% of the tower mass isn't very low. 20% to 30% shedding is reasonable. When the shedding is higher than 30%, it becomes difficult to explain how the top of the pile could be near the bowtie level, unless the debris in the pit is considered to be loosely compacted.

The collapse duration is relatively insensitive to mass shedding. I've written a Greening type of model that accounts for a constant amount of mass shedding per story impact during crush-down. A total shedding of 30% of the tower mass adds about one second to the collapse duration compared with no shedding. That doesn't put the predicted collapse duration beyond the observed duration.
 
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GregoryUrich said:
The correct mass and proper sequence of "crush up" first makes a huge difference in the timings.
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The mass doesn't make a huge difference in collapse duration unless the energy dissipated per story, E1, approaches the GPE released per story. In Greening's model, E1 is about one-third of the GPE in order for the model to predict the same fall rate as was observed during the first four seconds of collapse. In order for the collapse duration to start to becoming sensitive to changes in mass, the mass would need to be about one-third of its value for the GPE released per story to approach Greening's estimate of E1. Or, E1 would have to be about three times higher than Greening's estimate for the same mass.

The tower mass was probably less than the initial estimates of Greening and BZ. It's also possible that E1 was overestimated by Greening and BZ. A better approach is to consider the ratio of E1 to the amount of mass that is falling. There will be some ratio of E1/mass such that the observed collapse rate agrees most closely with the predicted rate of the collapse model.
 
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shagster; said:
Based on the height of the rubble pile within the tower footprint and assuming that the stories were squashed down to about 10 to 15% of their original height, a total mass shedding of 20% of the tower mass isn't very low. 20% to 30% shedding is reasonable. When the shedding is higher than 30%, it becomes difficult to explain how the top of the pile could be near the bowtie level, unless the debris in the pit is considered to be loosely compacted.

The collapse duration is relatively insensitive to mass shedding. I've written a Greening type of model that accounts for a constant amount of mass shedding per story impact during crush-down. A total shedding of 30% of the tower mass adds about one second to the collapse duration compared with no shedding. That doesn't put the predicted collapse duration beyond the observed duration.
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Hey shagster, welcome. I was curious about your thoughts on the communition of Zone C to Zone B are for WTC 1 and 2 respectively. (not just the concrete, the whole upper mass). Second, what effect does the free standing core have on the amount of mass "shed", and hence the collapse duration, in your model? As I think you are aware, it is still my contention that the lateral forces on the exterior by Zone B preclude it (the exterior) from contributing to the available energy of Zones B and C. These two combined would certainly represent much more than 30% of the mass, while still allowing it to remain relatively close to the perimeter.
 
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