Moderated Continuation - Why a one-way Crush down is not possible

I've been reading it for a few years now. I like his writing but I am unmoved by most of his arguments. I regret that you will not tell me exactly what is wrong with ''Smith's Law'. Science should be free flowing like the columns in Part C.


Those columns in Heiwa's mythical part C have been broken at the welds and pushed away. Thirteen falling floors--THE BIG PART--hit ONE floor--THE SMALL PART, crushing it. The process is repeated until the building is gone.
Tfk is an engineer. You know nothing. It is not for you to agree or disagree with him, as you bring no knowledge of the subject to the table. You can either learn from him, or keep doing that voodoo that you do so well.
 
I've been reading it for a few years now. I like his writing but I am unmoved by most of his arguments. I regret that you will not tell me exactly what is wrong with ''Smith's Law'. Science should be free flowing like the columns in Part C.


"Smith's Law" is incoherent gibberish. In non-truther quarters, that would be exactly what is wrong with it. Your bogus "law" starts out by restating what students of basic physics know about equal and opposite reactions (Newton's Third Law), but then goes somewhere strange.
 
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I've been reading it for a few years now. I like his writing but I am unmoved by most of his arguments. I regret that you will not tell me exactly what is wrong with ''Smith's Law'. Science should be free flowing like the columns in Part C.

A few years now? Were you reading tfk before you joined this forum?

The columns weren't free-flowing, that's the problem. They punch through.

Please don't waste my time. I'm not going to get into a big long pointless discussion with either you or Heiwa. Neither one of you has ever shown the ability to admit mistakes or learn new things.

You are a static object.
 
Those columns in Heiwa's mythical part C have been broken at the welds and pushed away. Thirteen falling floors--THE BIG PART--hit ONE floor--THE SMALL PART, crushing it. The process is repeated until the building is gone.
Tfk is an engineer. You know nothing. It is not for you to agree or disagree with him, as you bring no knowledge of the subject to the table. You can either learn from him, or keep doing that voodoo that you do so well.

Put yur hand flat on the middle of f your screen five minutes from now at the 25-min mark exactly and I'll show you voodoo.
 
Put yur hand flat on the middle of f your screen five minutes from now at the 25-min mark exactly and I'll show you voodoo.



I'd prefer it if you'd show me some ability to write coherent English.

Incidentally, thanks for your typically substantive response.
 
Heiwa,
.
Please, don't ask more questions. Concentrate on your paper about a part of rubble elements producing bigger force than a part of same but connected elements.

I understand your unwillingness to answer these questions in public. You clearly have no grasp of the basic concepts. It must be embarrassing for you.

I would not even hazard a guess as to what YOU think about any of these. I will provide the answers that a competent mechanical engineer or physicist would provide.

1. Would a competent engineer agree with the principle of conservation of momentum, as expressed by the Impulse-momentum equation:
∫[F dt] = ∆[m v] ?
Answer: Yes

2A. Would a competent engineer agree that there is no place in that equation where it says " This equation applies only if m is greater than mmin" [some minimum value of mass] ?
Answer: Yes

2B & 2C are therefore irrelevant.

3. Would a competent engineer believe that the TOTAL momentum change of any mass (a monolithic block, a bunch of rubble, a gas, etc) will be precisely equal to the vector sum of the momentum change of all the constituent components of that mass, right down to the atomic, and subatomic, scales?
Answer: Yes

4. In the case of C1 (as built 3 stories) & C2 (same 3 stories, crushed down), the total masses are defined as equal. ASSUMING that both C1 & C2 were moving at the same speed prior to impact, would a competent engineer agree that the TOTAL momentum change that results from the two collisions (C1 to ground vs. C2 to ground) is exactly the same?
Answer: Yes
__

Assume a "reasonable" force vs. time curve. I'd recommend a Weibull distribution, since it can be easily adjusted to represent both hard & soft collisions, while maintaining a constant Cumulative Distribution Function.

5. Would a competent engineer understand that increasing the time duration of any collision, while maintaining the same total impulse, AUTOMATICALLY and INESCAPABLY decreases the average force of that collision?
Answer: Yes

6. Finally, Would a competent engineer understand that ALL of the above is true REGARDLESS of how any of the pieces in any mass in any collision are attached to any other piece(s)??
Answer: Yes

I would have thought that my comment "if you answered 'No' to any of these questions, go back to remedial physics", would have given you a teensy hint...

Tom

PS. I'll answer your "welded cube" post next. True to fashion, you're completely & utterly wrong. But you made up for that by giving an utterly brain-dead stupid example.

PPS. Naaahh, Anders. I'm just kidding you.

Oh, not about your example being "brain-dead stupid". It is. It's an absolutely, utterly stupid example & irrelevant to the collapse, or any of these questions.

I was kidding you about the "... you made up for it ..." part.
:D

Tom
 
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Please, don't ask more questions. Concentrate on your paper about a part of rubble elements producing bigger force than a part of same but connected elements.

Can't answer the questions Heiwa? It appears you and the Bill parrot are into avoiding embarrassment.
 
A few years now? Were you reading tfk before you joined this forum?

The columns weren't free-flowing, that's the problem. They punch through.

Please don't waste my time. I'm not going to get into a big long pointless discussion with either you or Heiwa. Neither one of you has ever shown the ability to admit mistakes or learn new things.

You are a static object.

Sure. We were adversaries on another forum altogether for about two years.
http://www.topix.com/forum/topstories/TSBMT04T49GGG7HFO/p5719

If I make mistakes in this 9/11 thing I always try to correct for them. Otherwise there would be no point . Can't do more than that I'm afraid.
 
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Me joking? Sorry.

Take your IKEA book case, assemble it, use all bolts and nuts, etc, and drop it on the floor. BANG. Quite solid. The force F1 is say 10 units!

Take the same IKEA book case but do not assemble it and drop all the pieces, nuts, and bolts on the floor. Plenty of bang, bang, bangs and plenty of small forces F2 = 1 unit < F1.

Please, do nut suggest that loose pieces or rubble produce a bigger force than an assembly of same pieces.
.
Heiwa is just not getting this. I can't tell of its stubbornness, gamesmanship or just lack of aptitude. But this is for anyone else who is trying to pick apart these concepts. This post & the next one will allow you to really understand well, at a fundamental level, what is going on.

And also where Heiwa is blowing his "analysis".

Heiwa is NOT getting the concept of "compacted". Compacted is approximately equivalent to "broken down, lying flat, still in the box". Dropping one piece, then another, then another is NOT equivalent to compacted.

When you drop a compacted bookcase, you do NOT get "bang, bang, bang". You get one, LOUD "bang". If the components of the bookcase are tightly pressed together, the forces generated can be substantially greater than those generated by the impact of the assembled bookcase.

First, we have to compare the impact of any rigid object versus any flexible object colliding with a rigid surface, like the ground.

We will use a Force versus Time graph to show the evolution of the (equal & opposite) forces that are generated between the colliding bodies.

One thing is absolutely crucial to understand about these curves. In the case of a collision with the ground (as we will discuss below), the total area under each curve is determined BEFORE THE COLLISION ever takes place. Since the final velocity is zero, the final momentum (m*v) is zero, and the change in momentum is completely determined by the initial mass & velocity of the components. For two objects of the same mass, moving at the same speed before the collision and at rest after the collision, the momentum change is exactly the same. It does not matter whether the time interval for this process was long or short.

The analysis is only slightly more complicated for a collision between two moving objects. We won't go into that.

Looking at the curves below, it is easy to see that the main difference between the two collisions is the time duration of the collision. Any flexible object, like a soft rubber ball, draws out the collision in time. Any rigid object, like a baseball, produces a very short duration collision.

Two such collisions are shown in the Force vs. Time Fig. 1 below. Note that the units on both axes are arbitrary and for comparison only.



picture.php

Figure 1. Comparison of the collision of a rigid object versus a flexible object.


Notice that, since the area under the two curves is exactly the same, the simple act of drawing the collision out in time reduces both the peak force and the average force generated by the collision.

Let's apply this insight to the fall of a bookcase. As before, the bookcase has 4 pieces: the shell (the top, bottom, sides & back) and 3 shelves with books. For simplicity, I've made the mass of each piece be the same.

Using the curves in Fig 1 to describe the bookcase collision with the ground, the flexing of the assembled bookcase results in a "flexible collision", drawn out in time.

The other collisions (the books & shelves hitting the ground, and the packaged bookcase hitting the ground) are considered rigid, short duration collisions.

There are 3 cases to consider:

Figure 2 shows the force versus time graph that would be obtained for each of the conditions described below.




picture.php

Figure 2. Bookcase collision


Case A (blue line): The bookcase is broken down, flat and disassembled, the books stacked neatly on top, and the whole bundle is wrapped with packing tape into a nice, tight bundle.

Case B (green line): The bookshelf is assembled with the books on it. When the bookcase hits the ground, the whole frame and shelves all flex like crazy, but all components manage to withstand the impact without breaking.

Case C (red line): The bookcase is assembled with books on it. When the bookcase hits the ground, each of the shelf support brackets break when the load has grown to 0.05 (force units). This means that the total force generated by each shelf will rise to that level during the initial impact, just like in Case B. But suddenly each of the shelves will break free and the force component from the shelves will drop to zero.

The important thing to realize is that, once again, the total area under each of these curves is the same. That area is determined before the collision occurs by the initial mass & velocity of the parts. It has nothing whatsoever to do with the stiffness or flexibility of the components, or for that matter, whether they are joined into one piece or in a million pieces.

The time duration of the impact is determined by the flexibility of the various components, and by the spatial distribution of the parts. As you can see, if parts are separated in space, then their collisions are also separated in time. As shown in Case C, fractured components can result in collisions that are separated in time.

If these principles are starting to make sense, then you're approaching a deep understanding of how collisions really work.

I'll save that one for the next (& last) post in this series. It'll be a fun one.

Tom
 
.
Heiwa is just not getting this. I can't tell of its stubbornness, gamesmanship or just lack of aptitude. But this is for anyone else who is trying to pick apart these concepts. This post & the next one will allow you to really understand well, at a fundamental level, what is going on.

And also where Heiwa is blowing his "analysis".

Heiwa is NOT getting the concept of "compacted". Compacted is approximately equivalent to "broken down, lying flat, still in the box". Dropping one piece, then another, then another is NOT equivalent to compacted.

When you drop a compacted bookcase, you do NOT get "bang, bang, bang". You get one, LOUD "bang". If the components of the bookcase are tightly pressed together, the forces generated can be substantially greater than those generated by the impact of the assembled bookcase.

First, we have to compare the impact of any rigid object versus any flexible object colliding with a rigid surface, like the ground.

We will use a Force versus Time graph to show the evolution of the (equal & opposite) forces that are generated between the colliding bodies.

One thing is absolutely crucial to understand about these curves. In the case of a collision with the ground (as we will discuss below), the total area under each curve is determined BEFORE THE COLLISION ever takes place. Since the final velocity is zero, the final momentum (m*v) is zero, and the change in momentum is completely determined by the initial mass & velocity of the components. For two objects of the same mass, moving at the same speed before the collision and at rest after the collision, the momentum change is exactly the same. It does not matter whether the time interval for this process was long or short.

The analysis is only slightly more complicated for a collision between two moving objects. We won't go into that.

Looking at the curves below, it is easy to see that the main difference between the two collisions is the time duration of the collision. Any flexible object, like a soft rubber ball, draws out the collision in time. Any rigid object, like a baseball, produces a very short duration collision.

Two such collisions are shown in the Force vs. Time Fig. 1 below. Note that the units on both axes are arbitrary and for comparison only.



[qimg]http://www.internationalskeptics.com/forums/picture.php?albumid=176&pictureid=1377[/qimg]
Figure 1. Comparison of the collision of a rigid object versus a flexible object.


Notice that, since the area under the two curves is exactly the same, the simple act of drawing the collision out in time reduces both the peak force and the average force generated by the collision.

Let's apply this insight to the fall of a bookcase. As before, the bookcase has 4 pieces: the shell (the top, bottom, sides & back) and 3 shelves with books. For simplicity, I've made the mass of each piece be the same.

Using the curves in Fig 1 to describe the bookcase collision with the ground, the flexing of the assembled bookcase results in a "flexible collision", drawn out in time.

The other collisions (the books & shelves hitting the ground, and the packaged bookcase hitting the ground) are considered rigid, short duration collisions.

There are 3 cases to consider:

Figure 2 shows the force versus time graph that would be obtained for each of the conditions described below.




[qimg]http://www.internationalskeptics.com/forums/picture.php?albumid=176&pictureid=1378[/qimg]
Figure 2. Bookcase collision


Case A (blue line): The bookcase is broken down, flat and disassembled, the books stacked neatly on top, and the whole bundle is wrapped with packing tape into a nice, tight bundle.

Case B (green line): The bookshelf is assembled with the books on it. When the bookcase hits the ground, the whole frame and shelves all flex like crazy, but all components manage to withstand the impact without breaking.

Case C (red line): The bookcase is assembled with books on it. When the bookcase hits the ground, each of the shelf support brackets break when the load has grown to 0.05 (force units). This means that the total force generated by each shelf will rise to that level during the initial impact, just like in Case B. But suddenly each of the shelves will break free and the force component from the shelves will drop to zero.

The important thing to realize is that, once again, the total area under each of these curves is the same. That area is determined before the collision occurs by the initial mass & velocity of the parts. It has nothing whatsoever to do with the stiffness or flexibility of the components, or for that matter, whether they are joined into one piece or in a million pieces.

The time duration of the impact is determined by the flexibility of the various components, and by the spatial distribution of the parts. As you can see, if parts are separated in space, then their collisions are also separated in time. As shown in Case C, fractured components can result in collisions that are separated in time.

If these principles are starting to make sense, then you're approaching a deep understanding of how collisions really work.

I'll save that one for the next (& last) post in this series. It'll be a fun one.

Tom


Many thanks. That post, as the economist Frank Knight used to say, made a contribution to my education.
 
Well done and easy to understand. My background is math and you gave all these well stated hints earlier about what you planned to state. Thanks for laying everything out so clear and easy to understand.
 
fk:

Your graphs of force shouldn't level out to zero, they should level out to m*g.

You're also leaving out the effects of damping and free-vibration. However you may not want to break everyone's head by introducing that.
 
.
<snip> Very good detail

If these principles are starting to make sense, then you're approaching a deep understanding of how collisions really work.

I'll save that one for the next (& last) post in this series. It'll be a fun one.

Tom

Thanks for the explanation Tom. I understood this from your posts much earlier, but this gives Me more clarification on the details.

Heiwa will write another paragraph or two regurgitating/repeating what he's written before. Parrot Bill will just keep on pretending he himself is an expert.
 
Heiwa,
.


I understand your unwillingness to answer these questions in public. You clearly have no grasp of the basic concepts. It must be embarrassing for you.

I would not even hazard a guess as to what YOU think about any of these. I will provide the answers that a competent mechanical engineer or physicist would provide.

1. Would a competent engineer agree with the principle of conservation of momentum, as expressed by the Impulse-momentum equation:
∫[F dt] = ∆[m v] ?
Answer: Yes

2A. Would a competent engineer agree that there is no place in that equation where it says " This equation applies only if m is greater than mmin" [some minimum value of mass] ?
Answer: Yes

2B & 2C are therefore irrelevant.

3. Would a competent engineer believe that the TOTAL momentum change of any mass (a monolithic block, a bunch of rubble, a gas, etc) will be precisely equal to the vector sum of the momentum change of all the constituent components of that mass, right down to the atomic, and subatomic, scales?
Answer: Yes

4. In the case of C1 (as built 3 stories) & C2 (same 3 stories, crushed down), the total masses are defined as equal. ASSUMING that both C1 & C2 were moving at the same speed prior to impact, would a competent engineer agree that the TOTAL momentum change that results from the two collisions (C1 to ground vs. C2 to ground) is exactly the same?
Answer: Yes
__

Assume a "reasonable" force vs. time curve. I'd recommend a Weibull distribution, since it can be easily adjusted to represent both hard & soft collisions, while maintaining a constant Cumulative Distribution Function.

5. Would a competent engineer understand that increasing the time duration of any collision, while maintaining the same total impulse, AUTOMATICALLY and INESCAPABLY decreases the average force of that collision?
Answer: Yes

6. Finally, Would a competent engineer understand that ALL of the above is true REGARDLESS of how any of the pieces in any mass in any collision are attached to any other piece(s)??
Answer: Yes

I would have thought that my comment "if you answered 'No' to any of these questions, go back to remedial physics", would have given you a teensy hint...

Tom

PS. I'll answer your "welded cube" post next. True to fashion, you're completely & utterly wrong. But you made up for that by giving an utterly brain-dead stupid example.

PPS. Naaahh, Anders. I'm just kidding you.

Oh, not about your example being "brain-dead stupid". It is. It's an absolutely, utterly stupid example & irrelevant to the collapse, or any of these questions.

I was kidding you about the "... you made up for it ..." part.
:D

Tom

What has all this to do with Why a one-way Crush down is not possible? The latter is a structural damage analysis matter where various elements and connections break in an order that can be established using available tools.
 

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