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Old 2nd June 2016, 02:00 PM   #1
baldartist
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Failed Truther Paper Fall Time Analysis.

Posted below is a "paper" by truther Marty Beck. I'll just leave this here.




https://lookaside.fbsbx.com/file/WTC...8LAiwOQVf_eitg

Marty Beck, 4/20/2011

Analysis of the Fall Times of the Three World Trade Center Buildings


Abstract:
Explaining the collapse of the three World Trade Center buildings has been an enigma that strains the mind to come to grips with. Those who look at it over and over again try to make sense of what they see. The recorded fall times somehow seem reasonable to many of us when we rely on intuition in thinking about the collapse.

Many citizens are reasoning much like Galileo’s contemporaries must have. They easily assumed a heavier ball falls faster than a lighter ball. Until the experiment. We assume a large mass unleashed could fall at the rate we see. Before Galileo’s experiment they must have wondered why anyone would bother to do the experiment; isn’t it obvious?

Edited by Agatha:  Edited to comply with rule 4. Do not copy and paste large amounts of material available elsewhere on the net, but provide a short extract (one or two short paragraphs) and a link, in this instance https://lookaside.fbsbx.com/file/WTC...8LAiwOQVf_eitg

Last edited by Agatha; 3rd June 2016 at 08:02 AM. Reason: Removed extra space
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Old 2nd June 2016, 02:10 PM   #2
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This part is especially hilarious:

"Weight is the result of mass in earth’s gravity field and the weight pushing down is the force defined as F=ma where m is the mass of the 33 stories and a is the acceleration due to gravity. On earth a is 32 ft/sec2.

So before the collapse was initiated, the weight was 80 million pounds. Once the upper 33 floors start to accelerate at 20 ft/sec2, the resulting force is now proportional to the difference between the 32 ft/sec2 and the 20 ft/sec2. This is 12 ft/sec2 with a resulting force of only 12/32 of what it was before the collapse. "

Thus a falling mass, at impact, exerts less force than it would have done at rest. Uhuh, uhuh. I must experiment tomorrow with stationary bricks on my head vs. falling ones. Or maybe drop some Truthers off a balcony.
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Old 2nd June 2016, 02:20 PM   #3
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Originally Posted by baldartist View Post
So without counting Item 1 and Item 2, we can conclude that 3/4th of strength of the columns were absent/removed to account for the observed fall times. That is 75%. We are told that the most strength the steel could lose if the entire building were on fire is half of the ambient temperature strength. Clearly something is wrong. And not just a little bit wrong. When we account for item 1, the number goes above 7/8th of the strength of the support columns was removed. Those effects will be shown in the next section.
I agree with the highlighted. Not for the reasons the author seems to be thinking, though.

Paraphrasing Babbage, I am not able rightly to apprehend the kind of confusion of ideas that could provoke perpetrating such a "paper".
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Old 2nd June 2016, 02:41 PM   #4
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Quote:
Now each floor is approx 200 ft by 200 ft of 4 inch thick concrete. The highest density I can find published for concrete is 150 pounds per cubic foot. That would be 50 pounds per sq
This seems like an important detail to crosscheck. Why "highest" if I may ask? As far as I understand high rise flooring uses light weight concrete in the floor pans.

I don't trust that number. I bet I could find out before RocketBoy2.0 gets his cast replaced.

Go!
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Old 2nd June 2016, 02:44 PM   #5
baldartist
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You can download the "paper" here:


https://lookaside.fbsbx.com/file/WTC...8LAiwOQVf_eitg

Last edited by baldartist; 2nd June 2016 at 02:45 PM.
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Old 2nd June 2016, 03:05 PM   #6
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http://www.structuremag.org/?p=1163

This random source said 110 to 115. I would check every single datapoint in the paper based on this information.
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Old 2nd June 2016, 04:08 PM   #7
DGM
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Originally Posted by baldartist View Post
You can download the "paper" here:

I can't image why I would bother. I lost interest after the first two paragraphs of the abstract.
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Old 2nd June 2016, 06:06 PM   #8
tfk
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Originally Posted by GlennB View Post
This part is especially hilarious:

"Weight is the result of mass in earth’s gravity field and the weight pushing down is the force defined as F=ma where m is the mass of the 33 stories and a is the acceleration due to gravity. On earth a is 32 ft/sec2.

So before the collapse was initiated, the weight was 80 million pounds. Once the upper 33 floors start to accelerate at 20 ft/sec2, the resulting force is now proportional to the difference between the 32 ft/sec2 and the 20 ft/sec2. This is 12 ft/sec2 with a resulting force of only 12/32 of what it was before the collapse. "

Thus a falling mass, at impact, exerts less force than it would have done at rest. Uhuh, uhuh. I must experiment tomorrow with stationary bricks on my head vs. falling ones. Or maybe drop some Truthers off a balcony.

No, this guys calculation about this is correct, with a clarification.

WHILE the upper mass is descending at an approximately constant 20 ft/sec^2, it is pushing down on the lower mass with a force equal to 12/32s of its weight.

A Force diagram shows this.

There are two forces acting on the upper mass (m):

Fgravity (pulling downward at mg = m 32.2 ft/sec2 = the upper block's weight.
A resisting force pushing upward from the lower mass. We don't automatically know the magnitude of this force, but we can calculate it easily.

Fresultant = fgravity - fresisting

The gravity force (aka, "weight) and the resultant force are positive because they are downward.
The resisting force is negative, because it is upwards.

We know that the RESULTANT downward acceleration is 20 m/sec2. Therefore the resultant force acting on this body must be the mass of the body times its resultant acceleration.

Fresultant = m aresultant.

Equating the two resultant forces, and rearranging the terms gives:

fresisting = fgravity - Fresultant

fresisting = fgravity - m aresultant

fresisting = m g - m aresultant

fresisting = m (g - aresultant)

The resisting force upward is equal to the mass times (g - a).

In this case, fresisting = m (32 - 20) = 12 m/sec2 m Kg
__

By Newton's 2nd law, the force that the lower block exerts on the upper is exactly equal to the force the upper body exerts on the lower one ... again, WHILE it is falling at this acceleration.

As counter-intuitive as it may seem, YES, while it is falling, it is exerting LESS force on the lower body than when it was stationary. The loads on the lower floors' columns dropped, decreased in magnitude, while the building was collapsing.

The lower body can not stop it because of load path & the destruction to the topmost floor before the upper mass even arrives at that floor.
__

Later in the fall, the resistance likely rose to equal the (increasing) weight of the upper block, and then it likely reached an equilibrium: terminal velocity.

Two factors caused it to reach terminal velocity: 1) as it picked up speed, it was hitting more floors per second, and it had to force the air out of the room space faster & faster.
__

In any impact, the full momentum / kinetic energy of the falling block can be turned into an enormous force, capable of fracturing the components trying to stop its fall. In this case, the force that the falling body could exert on some resisting member could have risen to many, many times its own weight, depending on how fast the structure tries to bring the upper block to a halt.

__

Scanning over the intro, there seems no need to read the rest of it.

Anybody have any idea what this guy's credentials are?
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Old 2nd June 2016, 08:54 PM   #9
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The paper, so far as I can see, is based entirely on the simple, but colossal, error of assuming that the structural strength of a collapsing column remains constant throughout its collapse. I realise that there are people insufficiently intelligent to appreciate that this is not the case, but I don't think there's much to be done for them above and beyond meeting their immediate care needs.

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Old 2nd June 2016, 09:11 PM   #10
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Quote:
3) The strength of the columns themselves accounts for the other force decelerating the falling mass of floors. This can be seen as a continuous force resisting the collapse. We will calculate the loss in strength of the support columns to account for the fall times.
The columns can only resist the movement of mass that is coupled to that column. Once trusses have sheared away from connections to perimeter, or core belt truss, forces generated by that falling mass are no longer coupled to columns.

With mass falling and destroying floors and truss connections, most of the dynamic forces generated simply bypass the columns.
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Old 3rd June 2016, 04:41 AM   #11
baldartist
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Originally Posted by tfk View Post
No, this guys calculation about this is correct, with a clarification.

WHILE the upper mass is descending at an approximately constant 20 ft/sec^2, it is pushing down on the lower mass with a force equal to 12/32s of its weight.

A Force diagram shows this.

There are two forces acting on the upper mass (m):

Fgravity (pulling downward at mg = m 32.2 ft/sec2 = the upper block's weight.
A resisting force pushing upward from the lower mass. We don't automatically know the magnitude of this force, but we can calculate it easily.

Fresultant = fgravity - fresisting

The gravity force (aka, "weight) and the resultant force are positive because they are downward.
The resisting force is negative, because it is upwards.

We know that the RESULTANT downward acceleration is 20 m/sec2. Therefore the resultant force acting on this body must be the mass of the body times its resultant acceleration.

Fresultant = m aresultant.

Equating the two resultant forces, and rearranging the terms gives:

fresisting = fgravity - Fresultant

fresisting = fgravity - m aresultant

fresisting = m g - m aresultant

fresisting = m (g - aresultant)

The resisting force upward is equal to the mass times (g - a).

In this case, fresisting = m (32 - 20) = 12 m/sec2 m Kg
__

By Newton's 2nd law, the force that the lower block exerts on the upper is exactly equal to the force the upper body exerts on the lower one ... again, WHILE it is falling at this acceleration.

As counter-intuitive as it may seem, YES, while it is falling, it is exerting LESS force on the lower body than when it was stationary. The loads on the lower floors' columns dropped, decreased in magnitude, while the building was collapsing.

The lower body can not stop it because of load path & the destruction to the topmost floor before the upper mass even arrives at that floor.
__

Later in the fall, the resistance likely rose to equal the (increasing) weight of the upper block, and then it likely reached an equilibrium: terminal velocity.

Two factors caused it to reach terminal velocity: 1) as it picked up speed, it was hitting more floors per second, and it had to force the air out of the room space faster & faster.
__

In any impact, the full momentum / kinetic energy of the falling block can be turned into an enormous force, capable of fracturing the components trying to stop its fall. In this case, the force that the falling body could exert on some resisting member could have risen to many, many times its own weight, depending on how fast the structure tries to bring the upper block to a halt.

__

Scanning over the intro, there seems no need to read the rest of it.

Anybody have any idea what this guy's credentials are?
He claims to be an electrical engineer. He quotes Robert Karol a lot. When chatting with him, he has a hard time with word salad and is often confusing momentum and KE.
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