phunk
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Wouldn't it be 'mg' after, not 0?
Moderated Continuation - Why a one-way Crush down is not possible
- Thread starter Cuddles
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- wtc collapse
Wouldn't it be 'mg' after, not 0?
MIKILLINI
Incromulent Logic
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Justin39640
Illuminator
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this is what i see as a layman
i think anders problem is that he cannot recognize that one piece from the upper part can easily damage many parts of the lower part
anders here just assumes once it makes first contact its irrelevant in the future
bills issue is he/she copy and pastes from morons lol
i think anders problem is that he cannot recognize that one piece from the upper part can easily damage many parts of the lower part
anders here just assumes once it makes first contact its irrelevant in the future
bills issue is he/she copy and pastes from morons lol
Newt,
No problem. Talking about it here is why I brought it up.
My point is that I'm not talking (yet) about the PROCESS by which it got crushed, or the energy that this process will take.
For now, I am talking about the fact that, if you take a structure such as floors of an office building, the debris formed when it has been crushed & compacted has a higher density than the material before it was crushed & compacted.
Or imagine dropping a bookcase full of books (3 shelves, 2' apart) from a height of 32'. (Easy math.) Assume that shell (i.e., top, bottom & sides) of the bookcase happens to weigh the same as each shelf with its books. When it hits the ground, it's doing 32 ft/sec. Imagine three cases:
1. When the book case hits the floor, imagine that the support brackets on each shelf instantly give way.
2. The bookcase is packaged as delivered by IKEA with all the books stacked & wrapped with packing tape on top. A nice tight bundle.
3.The strength of all the components of the bookcase are sufficient that it survives the drop perfectly intact. The shelves & frame flex like crazy, but nothing breaks. Amazing bookcase.
Case 1. The initial impact delivers the momentum of the shell of the bookshelf ONLY. The shelves all break free, and continue towards the floor. The second impact (bottom shelf) happens about (3 ft / 32 ft/sec =) 0.1 second later, the next 0.05 seconds after that, & the last 0.05 seconds after that. (I've assumed the books are 1' high.) In this case, then you have 4 separate impacts, each delivering 1/4th the total momentum of the entire bookcase.
Case 2. This is the equivalent of a "100% crushed & compacted" bookcase. There is no flex, no give, & no opportunity for anything to break, except in the collision itself. This collision delivers the entire momentum to the ground in one instant.
Case 3. This is intermediate between the first two cases. All of the flexing stretches out the impact in time.
In each case, the total momentum that the bookshelf delivers to the ground is identical. (The mass of the bookshelf x it's terminal velocity.) There is a theorem that equates the "impulse" (force integrated over time) to the change in momentum of the bookcase. And Newtons "equal & opposite forces" law says that the force the ground exerts on the bookcase is equal to the force the bookcase delivers to the ground.
On a force (y axis) vs. time (x axis) graph, the impulse is the area under the curve. So, keeping the areas under the curves the same for all three cases, stretching out the duration of the impact in time automatically lowers the peak and average forces.
Parts breaking & subcomponents colliding separately (like the shelves when the supports break), stretches out the collision time a lot. Structures flexing stretches out the time duration to a lesser amount. But both reduce both the max force & average force delivered to the ground.
"Tah dah...!"
Oh, excuse me, academics in the house. "QED".
Later, we'll get into the subtler aspects, like strength of the structural connections prior to crushing, and the strength of the interdigitation after crushing.
Tom
.
No problem. Talking about it here is why I brought it up.
My point is that I'm not talking (yet) about the PROCESS by which it got crushed, or the energy that this process will take.
For now, I am talking about the fact that, if you take a structure such as floors of an office building, the debris formed when it has been crushed & compacted has a higher density than the material before it was crushed & compacted.
Or imagine dropping a bookcase full of books (3 shelves, 2' apart) from a height of 32'. (Easy math.) Assume that shell (i.e., top, bottom & sides) of the bookcase happens to weigh the same as each shelf with its books. When it hits the ground, it's doing 32 ft/sec. Imagine three cases:
1. When the book case hits the floor, imagine that the support brackets on each shelf instantly give way.
2. The bookcase is packaged as delivered by IKEA with all the books stacked & wrapped with packing tape on top. A nice tight bundle.
3.The strength of all the components of the bookcase are sufficient that it survives the drop perfectly intact. The shelves & frame flex like crazy, but nothing breaks. Amazing bookcase.
Case 1. The initial impact delivers the momentum of the shell of the bookshelf ONLY. The shelves all break free, and continue towards the floor. The second impact (bottom shelf) happens about (3 ft / 32 ft/sec =) 0.1 second later, the next 0.05 seconds after that, & the last 0.05 seconds after that. (I've assumed the books are 1' high.) In this case, then you have 4 separate impacts, each delivering 1/4th the total momentum of the entire bookcase.
Case 2. This is the equivalent of a "100% crushed & compacted" bookcase. There is no flex, no give, & no opportunity for anything to break, except in the collision itself. This collision delivers the entire momentum to the ground in one instant.
Case 3. This is intermediate between the first two cases. All of the flexing stretches out the impact in time.
In each case, the total momentum that the bookshelf delivers to the ground is identical. (The mass of the bookshelf x it's terminal velocity.) There is a theorem that equates the "impulse" (force integrated over time) to the change in momentum of the bookcase. And Newtons "equal & opposite forces" law says that the force the ground exerts on the bookcase is equal to the force the bookcase delivers to the ground.
On a force (y axis) vs. time (x axis) graph, the impulse is the area under the curve. So, keeping the areas under the curves the same for all three cases, stretching out the duration of the impact in time automatically lowers the peak and average forces.
Parts breaking & subcomponents colliding separately (like the shelves when the supports break), stretches out the collision time a lot. Structures flexing stretches out the time duration to a lesser amount. But both reduce both the max force & average force delivered to the ground.
"Tah dah...!"
Oh, excuse me, academics in the house. "QED".
Later, we'll get into the subtler aspects, like strength of the structural connections prior to crushing, and the strength of the interdigitation after crushing.
Tom
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phunk,
No, he's got it right.
Other than the earth (which does it with gravity), in order for two parts to exert forces on each other, they have to touch each other. Either directly, or thru an intermediate structure called a "load path".
Before the collision & after the collision, parts aren't touching each other, so they exert no force on each other.
One important point is that parts can exert forces on each other in collisions that are MUCH greater than their weight. The weight of something is called it's "static load". When one thing hits another in a collision, the force that one exerts on the other is called a "dynamic load". This load can be MUCH greater than the static load.
Carefully set a brick on a glass table: No problem. Drop the brick on the glass table: big problem. The dynamic load can be much (10x, 20x, 100x) bigger than the static one, depending on the speed and "rigidity" of the impact.
Tom
newton3376
The Truth Movement.....still not at 1%
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Excellent post Tom. I think I see what you are doing...starting with general concepts and then progressing into the specifics.....
The bookshelf was an excellent example to illustrate some basic concepts about momentum and force etc. Anyway...thanks for the answer....Im curious to see your next post...
MIKILLINI
Incromulent Logic
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It seems?? So you're not certain??
What has more impact? A brick resting on your head? Or a brick dropped on your head? Which is dynamic?
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I have no idea what tfk is trying to prove. Re question a brick resting on a head provides no impact at all. It is resting = no impact. If you drop it and it contacts something, e.g. head, suggest wearing a helmet that easily absorbs the impact and deflects the brick unless brick just bounces on helmet.
bill smith
Philosopher
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Pay no attention to him. He is just fishing for an angle. I like the brick thing though. I am thinking of two composite glass bricks built around steel fremes. One brick is 90 inches tall and the one to be dropped on it is 10 inches tall and is of simillar but somewhat lighter construction.
Im also imagining what would happen if the 10 inch brick was crushed, broken and disintegrating before it contacted the lower brick.
alienentity
Illuminator
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I like the brick thing too. It describes your thought process very well.
bill smith
Philosopher
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Very profound Alienentity. 'thinking like a brick'. That might become a common phrase don't you think ? lol
Questions:
1. Do you, or do you not agree that, if dropped onto a solid surface, the peak force of a crushed and compacted mass of 3 stories of WTC would generate a higher AVERAGE force on the solid surface than the same pre-crushed structure?
2. Do you, or do you not agree that, if dropped onto a solid surface, the peak force of a crushed and compacted mass of 3 stories of WTC would generate a higher PEAK force on the solid surface than the same pre-crushed structure?
Anwers:
1. So we have two structures; one, C1, is 3 floors high with columns between the floors and the other, C2, is same as C1 but floors and columns have been compressed into a thin layer of rubble. C1 and C2 is dropped on a solid surface.
When C1 contacts the solid surface, evidently the lowest floor contacts the solid surface, BANG, and applies a force F1 on the solid surface. The solid surface applies a force -F1 on C1. The columns between floors in C1 deform, etc, and C1 comes to rest on the solid surface after a little bounce.
When C2 contacts the solid surface, evidently the bottom rubble parts of C1 contacts the solid surface but there is no real impact. The remaining rubble parts then just pile up on the other rubble parts on ground. The force F2 that the rubble applies on the solid surface is small.
So it would appear F1>F2!
2. When C1 contacts the solid surface, F1, reaches a peak value pretty quickly (depending on the flexibility of the C1 assembly) and after a while F1 is reduced until equilibrium is reinstated (at end impact and deformation of C1)
When C2 contacts the solid surface, i.e. the rubble parts pile up on the solid surface, F2 just gets slowly bigger; no impact, no deformation of C2, just equilibrium during complete process.
So again F1>F2.
Happy?
Look forward to your theory that rubble C2 can destroy an intact structure A.
Z
Variable Constant
And that is why you fail to understand what happened in the collapse. We're not talking a tiny, light brick dropped on a huge heavy brick. We're talking about a composite glass and steel structure that is composed of 14 levels being dropped on another composite glass and steel structure composed of 97 levels (or 96 or whatever).
Think, Bill! Think! The bottom section wasn't a single, monolithic, 97-story brick. It was another composite structure composed of 97 stacked bricks, essentially. Why can you not grasp the idea that you're talking about falling debris with 14 times the mass of a floor impacting first with one floor, then another, then another? It's not a 10--> 90 impact; it's a 14--> 1 impact, followed by a 15--> 1 impact, followed by a 16--> 1 impact, and so forth until the bottom. Hence, a crush-down. It is inevitable.
Say each level has a mass of 100 units.
Now, each level in turn has to be dealt with as a single level on the bottom, because the bottom is not moving and has no momentum. However, the levels on the top have to be treated as a total mass, as the entire top is falling and, most likely, breaking apart into component pieces.
So the initial impact on the bottom section, which occurs to floor 97, is 1400 units impacting a level which has only about 100 units of mass. 1400 units of glass and steel come plowing into a glass and steel structure of only 100 units - destruction ensues.
Now level 96 receives impact from 1500 units of mass. Then level 95 with 1600 units of mass, and so forth.
Bill, do you see what happens here?
BILL! I'm asking you a question, please respond to this post!!!
OK, simple challenge: Make a domino tower of 97 floors height, with room on each floor for ten LEGO men to stand inside. Then, however many dominoes it takes to make one floor, multiply that by 14. Then measure off a few floors of height, and drop those 14 floors worth of dominoes on the 97 floor structure, and see what happens.
OF course, the mass proportions are way off - another fact you twoofers can never seem to grasp is that all factors don't scale the same way, so this model is actually highly inaccurate. But it might illustrate the fact that if you drop 14 times the mass of a single floor onto a structure composed of floors, you're likely to destroy the entire structure.
Im also imagining what would happen if the 10 inch brick was crushed, broken and disintegrating before it contacted the lower brick.[/QUOTE]
.
Cripes, Anders. I wouldn't be particularly proud of this.
Newton's getting it already. And he's a young EE.
And, even if I weren't getting it, I'd keep it to myself, go home & work a little harder until I did. Rather than announce my own shortcomings to the rest of the class...
Tom
Locknar
Sum of all evils tm
Several off-topic posts have been moved to AAH...don't let this happen to you! Make sure your posts are on topic, civil & polite.
Replying to this modbox in thread will be off topic Posted By: Locknar
Sorry, only 100 units impacting 100 units. Replace the latter 100 units with solid ground and wonder what happens. Right, the other 1300 units, that have impacted nothing, break apart ... as you suggest.
Now, replace the solid ground with a tower of 9700 units stacked on top of each other. These 9700 units previously carried 1400 units. You do not really believe that the 1400 units can crush 9700 units, do you?
Suggest you include some elements between your floors and it will become clearer!