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

[Justin39640]

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.

hehe i made a few of those
they tend to crush equally
but like you pointed out scale is important and that you cant deform or break the dominoes randomly changes things too

(i figured out what i did wrong here that made the floor textures do that but i dont want to render for another 10 hours lol)

the NIST model was a 650 foot tall model of building 7
its not scaled down
its a 1/1 model
another reason your g5 aint gonna run it (for that other thread lol)

 

[Z]

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.

Incorrect again. Or are you suggesting the other 1300 units are floating above the rest, adding no weight or force during the impact?

You are completely wrong. It's 1400 units impacting 100 units, followed by 1500 units impacting 100 units, and so on.

Think, man! Think!

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?

When dropped from a height, and when those 9700 units are missing key structural elements (like the cap-piece), and have been weakened by fire and stress - absolutely. Especially when it's simply 97 100 unit segments being crushed by a 1400-unit falling mass.

Elementary.

Suggest you include some elements between your floors and it will become clearer!

...??? Confused. What do you mean, include some elements between floors? You think that will change the fact that 1400 units of mass would crush 100 units, and that each successive 100 unit mass is being affected by the total mass of all the floors above it (such that floor 1 is affected by 11,100 units of falling matter)?

I can only assume you deliberately choose to be this ignorant and obtuse for fun.

 

[Newtons Bit]

Load paths are difficult for truthers to follow. They don't understand that the upper block transfers its weight from the columns (which previously supported it) to one single floor.

It's just plain stupidity, pure and simple.

 

[Heiwa]

Incorrect again. Or are you suggesting the other 1300 units are floating above the rest, adding no weight or force during the impact?

The 1300 other C units - far away and above from the contact area C/A, where 100 C units during impacts 100 A units are being deformed - do not know about what the 100 C units below are doing. When they do, they will start to break apart as you suggest. They become rubble. And rubble cannot impact/destroy A. Please consider that there are vertical elements between the 1300 units transmitting information what the 100 units are up or down to.

 

[tfk]

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?

Answers:

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.

[1A] 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.

[1B] When C2 contact∆s 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!

[2A]. 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)

[2B] 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.

.
You are joking, aren't you??
.

Happy?

.
If my purpose had been to have you embarrass yourself by blowing a trivial mechanics question, then I would be ecstatic.

Let's take that 1 piece at a time.

In your first answer, you say:
.

[1A] 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.

.
Oh really??

A direct, inescapable consequence of this (mistaken) assertion is that "controlled demolitions are impossible". According to you, cutting the supports on the bottom floor of any building is useless. According to you, when the building falls & the bottom floor hits the ground, the whole structure will simply "have its columns deform and then come to rest on the solid surface after a little bounce".

Are you sure you want to back this position??

[Insert 100 videos of controlled demolitions here...]

Look forward to your theory that rubble C2 can destroy an intact structure A.

.
http://www.youtube.com/watch?v=qEbYpts0Onw&NR=1
http://www.youtube.com/watch?v=mknStAMia0Q&NR=1&feature=fvwp

Nah, "rubble" could never do any damage.
___

If you simply answer a couple of questions honestly in my next post, I believe that we can put this whole issue to bed.

With a formal proof.

Tom

 

[Heiwa]

A

.
You are joking, aren't you??
.

.
If my purpose had been to have you embarrass yourself by blowing a trivial mechanics question, then I would be ecstatic.

Let's take that 1 piece at a time.

In your first answer, you say:
.

.
Oh really??

A direct, inescapable consequence of this (mistaken) assertion is that "controlled demolitions are impossible". According to you, cutting the supports on the bottom floor of any building is useless. According to you, when the building falls & the bottom floor hits the ground, the whole structure will simply "have its columns deform and then come to rest on the solid surface after a little bounce".

Are you sure you want to back this position??

[Insert 100 videos of controlled demolitions here...]


.
http://www.youtube.com/watch?v=qEbYpts0Onw&NR=1
http://www.youtube.com/watch?v=mknStAMia0Q&NR=1&feature=fvwp

Nah, "rubble" could never do any damage.
___

If you simply answer a couple of questions honestly in my next post, I believe that we can put this whole issue to bed.

With a formal proof.

Tom

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]

.

A direct, inescapable consequence of this (mistaken) assertion is that "controlled demolitions are impossible". According to you, cutting the supports on the bottom floor of any building is useless. According to you, when the building falls & the bottom floor hits the ground, the whole structure will simply "have its columns deform and then come to rest on the solid surface after a little bounce".

Yes, it happens! See my latest post in The Heiwa Challenge thread for evidence! http://www.youtube.com/watch?v=UsePU...1&feature=fvwp

But you are changing subject; here we discuss why one-way crush downs are impossible, not the controlled demolitions of, e.g. WTC 1, 2 and 7.

Pls, concentrate on your evidence that disconnected elements/rubble pieces produce bigger impact force(s) than an assembly of same elements. I am really looking forward to that.

 

[GlennB]

Load paths are difficult for truthers to follow. They don't understand that the upper block transfers its weight from the columns (which previously supported it) to one single floor.

It's just plain stupidity, pure and simple.

Well I'm fllbbergasted. Next time a rock lands on my head I'm relying on my burly legs to protect my neck bones. You telling me I'm wrong? I'll give up my gym subscription if so. I only cycle and pump iron to save my long scrawny neck.

And what about those 'human towers' ?

I
? ?
I I I
I I I I
I I I I I
I I I I I I​

Are you seriously suggesting that if Messrs. ? ? fold, those Samoan prop-forwards at the bottom cannot take the strain? * Piffle and balderdash man! I'm pretty sure Newton's 3rd law protects such 'burly citizens', as dear Bill McLaren would have described them.



* As long as they were glued together a bit at the start, Heiwa will be satisfied.

 

[bill smith]

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]

Smith's Law
''Whatever downwards force the moving body exerts on the stationary body of identical construction fixed in the ground is reciprocated by the stationary body equally and oppositely. After that it depends which body is rendered non-viable by mutual erosion first.''


Bearing Smith's aw in mind you will see that whatever force is applied by the bottom floor of part C is met by an equal force of rsistance frpm part A.whether that force be 10 tons or 10,000 tons. This means that the force given back by the top floor of part A is exacty equal to the downwards force imparted by the lowest floor of part C.

Pert A is the stronger of the two parts because it is 10 times the mass of part C, is of similar but more robust construction than part C and because it's 47 vertical columns are anchored 70 feet deep in the ground.

Part C by comparison is is merely the top and lightest 10% of the building, it's vertical columns are anchored in the free-flowing air.

So it is obvious that the two floors that meet do so with a tremendous inpact that will pulverise both of them. But the vertical columns of both parts are still intact and in place. They are assumed to be offset fom each other so the upstanding columns will punch through the next floor that is offered up to them of the descending part C across the length and breadth of the core area . So that floor and any subsequent one is shattered on the anvils of the upstanding columns of part A..

Part C meantime is doing the same to Part A. But let us not forget that part C can never win this war of attrition because it is only one-tenth of the mass of part A and will soon run out of steam through local damage bringing early collapse arrest. Read Snith's law one more time to be sure.

 

[tsig]

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.

Please do not say they produce less.

 

[tsig]

Smith's Law
''Whatever downwards force the moving body exerts on the stationary body of identical construction fixed in the ground is reciprocated by the stationary body equally and oppositely. After that it depends which body is rendered non-viable by mutual erosion first.''


Bearing Smith's aw in mind you will see that whatever force is applied by the bottom floor of part C is met by an equal force of rsistance frpm part A.whether that force be 10 tons or 10,000 tons. This means that the force given back by the top floor of part A is exacty equal to the downwards force imparted by the lowest floor of part C.

Pert A is the stronger of the two parts because it is 10 times the mass of part C, is of similar but more robust construction than part C and because it's 47 vertical columns are anchored 70 feet deep in the ground.

Part C by comparison is is merely the top and lightest 10% of the building, it's vertical columns are anchored in the free-flowing air.

So it is obvious that the two floors that meet do so with a tremendous inpact that will pulverise both of them. But the vertical columns of both parts are still intact and in place. They are assumed to be offset fom each other so the upstanding columns will punch through the next floor that is offered up to them of the descending part C across the length and breadth of the core area . So that floor and any subsequent one is shattered on the anvils of the upstanding columns of part A..

Part C meantime is doing the same to Part A. But let us not forget that part C can never win this war of attrition because it is only one-tenth of the mass of part A and will soon run out of steam through local damage bringing early collapse arrest. Read Snith's law one more time to be sure.[/QUOTE]




I foresee trouble here. Heiwa may not appreciate you taking on lawmaking powers. Beware Bill!

 

[Newtons Bit]

FYI, there's no real structural advantage to a column "anchored 70 feet in the ground" vs. one properly fixed to a foundation at ground level.

The one anchored 70 feet down has somewhat higher deflections but lower stresses.

 

[bill smith]

Smith's Law
''Whatever downwards force the moving body exerts on the stationary body of identical construction fixed in the ground is reciprocated by the stationary body equally and oppositely. After that it depends which body is rendered non-viable by mutual erosion first.''


Bearing Smith's aw in mind you will see that whatever force is applied by the bottom floor of part C is met by an equal force of rsistance frpm part A.whether that force be 10 tons or 10,000 tons. This means that the force given back by the top floor of part A is exacty equal to the downwards force imparted by the lowest floor of part C.

Pert A is the stronger of the two parts because it is 10 times the mass of part C, is of similar but more robust construction than part C and because it's 47 vertical columns are anchored 70 feet deep in the ground.

Part C by comparison is is merely the top and lightest 10% of the building, it's vertical columns are anchored in the free-flowing air.

So it is obvious that the two floors that meet do so with a tremendous inpact that will pulverise both of them. But the vertical columns of both parts are still intact and in place. They are assumed to be offset fom each other so the upstanding columns will punch through the next floor that is offered up to them of the descending part C across the length and breadth of the core area . So that floor and any subsequent one is shattered on the anvils of the upstanding columns of part A..

Part C meantime is doing the same to Part A. But let us not forget that part C can never win this war of attrition because it is only one-tenth of the mass of part A and will soon run out of steam through local damage bringing early collapse arrest. Read Snith's law one more time to be sure.

I foresee trouble here. Heiwa may not appreciate you taking on lawmaking powers. Beware Bill![/QUOTE]

Well I once used this statement here on the jref thinking I was paraphrasing Isaac Newton's similar Law but I was aassured by posters no doubt better qualified than I am that I was quite wrong. So I took it upon myself to coin 'Smith's Law' which I still believe to be valid.

 

[tfk]

Heiwa,

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?

Answers:

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.

...

[1B] 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!
...

[2B] 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.

I am only going to address your answers to 1B & 2B, here.

1B. "... the bottom rubble parts of C1 contacts the solid surface but there is no real impact."

2B. "... the rubble parts pile up on the solid surface, F2 just gets slowly bigger; no impact ..."

You seem to be saying that "if an object is below a certain size, it cannot exert a force on something else thru change in its own momentum." Am I interpreting you correctly??

PLEASE ANSWER THESE QUESTIONS.
They are at the very heart of this discussion.

1. Do you agree with the principle of conservation of momentum, as expressed by the Impulse-momentum equation:
∫[F dt] = ∆[m v] ?

2A. Do you agree that there is no place in that equation where it says " This equation applies only if m is greater than m_min" [some minimum value of mass] ?

2B. If you answered "no" to 2A, please explain how it is that gas molecules can propel a rocket or a jet, since that equation defines a rocket's performance.
("m_min" must be less than a single water vapor molecule for the space shuttle to get off the ground.)

2C. If you answered "no" to 2A, do you wish to change your mind in light of 2B?

3. Do you 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?

4. In the case of C1 (as built 3 stories) & C2 (same 3 stories, crushed down), the total masses are defied as equal. ASSUMING that both C1 & C2 were moving at the same speed prior to impact, do you agree that the TOTAL momentum change that results from the collision is exactly the same?

If you have answered "No" to any of the questions above, then you need to take a remedial course in physics and/or mechanics.

If you have answered "Yes" to {1, 2A, 3 & 4}, then you must agree that the total momentum change for the two cases is identically the same.

∆[m_Total v]_C1 = ∆[m_Total v]_C2

And that means that the TOTAL impluses applied to the ground in both collisions are the same:
∫[F dt]_C1 = ∫[F dt]_C2

On a plot of Force (y axis) versus time (x axis), the ∫[F dt] is equal to the area under the curve.

Now, two last questions...

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. Do you 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?

6. Finally, do you 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)??

TomK

 

[Heiwa]

Please do not say they produce less.

An example to convince you; part C consists of 1000 parts each with mass m all connected together by, e.g. welding. Thus part C has mass 1000 m, which can be considered as one big 1000 m mass due to the welded connections between all 1000 parts. The welded connections actually add 'strength' to part C as all 1000 parts are welded together. They form a C-cube with side 10 parts!

Now take part D! It also consists of 1000 parts each with mass m (thus D has total mass 1000 m like C) BUT all these parts are not connected by welding at all. You can try to build a D-cube with side 10 parts with these 1000 parts but try to lift it and drop it. Part D falls apart. Its 1000 parts are not connected. Ok, you put all 1000 D m in a bag and drop it.

I can assure you that the D bag produces a smaller force than the C cube when dropped on a surface.

 

[Z]

The 1300 other C units - far away and above from the contact area C/A, where 100 C units during impacts 100 A units are being deformed - do not know about what the 100 C units below are doing. When they do, they will start to break apart as you suggest. They become rubble. And rubble cannot impact/destroy A. Please consider that there are vertical elements between the 1300 units transmitting information what the 100 units are up or down to.

You DO think the other units are floating away from the impact area.

My god, how stupid can you possibly be?


Even if the entire 1400 units were loose rubble, they would crush each floor in turn. Ever see tons of dirt dumped on a car? It flattens the car. Ever see tons of snow on a house? It flattens the house.

Do you live in the real world?

 

[Z]

Smith's Law
''Whatever downwards force the moving body exerts on the stationary body of identical construction fixed in the ground is reciprocated by the stationary body equally and oppositely. After that it depends which body is rendered non-viable by mutual erosion first.''[/quote]

Smith was apparently a moron. This is by no means what happens.

Bearing Smith's aw in mind you will see that whatever force is applied by the bottom floor of part C is met by an equal force of rsistance frpm part A.whether that force be 10 tons or 10,000 tons. This means that the force given back by the top floor of part A is exacty equal to the downwards force imparted by the lowest floor of part C.

Pert A is the stronger of the two parts because it is 10 times the mass of part C, is of similar but more robust construction than part C and because it's 47 vertical columns are anchored 70 feet deep in the ground.

Part C by comparison is is merely the top and lightest 10% of the building, it's vertical columns are anchored in the free-flowing air.

So it is obvious that the two floors that meet do so with a tremendous inpact that will pulverise both of them. But the vertical columns of both parts are still intact and in place. They are assumed to be offset fom each other so the upstanding columns will punch through the next floor that is offered up to them of the descending part C across the length and breadth of the core area . So that floor and any subsequent one is shattered on the anvils of the upstanding columns of part A..

Part C meantime is doing the same to Part A. But let us not forget that part C can never win this war of attrition because it is only one-tenth of the mass of part A and will soon run out of steam through local damage bringing early collapse arrest. Read Snith's law one more time to be sure.

As I said, Smith apparently was smoking crack.

 

[Heiwa]

Heiwa,



I am only going to address your answers to 1B & 2B, here.

1B. "... the bottom rubble parts of C1 contacts the solid surface but there is no real impact."

2B. "... the rubble parts pile up on the solid surface, F2 just gets slowly bigger; no impact ..."

You seem to be saying that "if an object is below a certain size, it cannot exert a force on something else thru change in its own momentum." Am I interpreting you correctly??

PLEASE ANSWER THESE QUESTIONS.
They are at the very heart of this discussion.

1. Do you agree with the principle of conservation of momentum, as expressed by the Impulse-momentum equation:
∫[F dt] = ∆[m v] ?

2A. Do you agree that there is no place in that equation where it says " This equation applies only if m is greater than m_min" [some minimum value of mass] ?

2B. If you answered "no" to 2A, please explain how it is that gas molecules can propel a rocket or a jet, since that equation defines a rocket's performance.
("m_min" must be less than a single water vapor molecule for the space shuttle to get off the ground.)

2C. If you answered "no" to 2A, do you wish to change your mind in light of 2B?

3. Do you 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?

4. In the case of C1 (as built 3 stories) & C2 (same 3 stories, crushed down), the total masses are defied as equal. ASSUMING that both C1 & C2 were moving at the same speed prior to impact, do you agree that the TOTAL momentum change that results from the collision is exactly the same?

If you have answered "No" to any of the questions above, then you need to take a remedial course in physics and/or mechanics.

If you have answered "Yes" to {1, 2A, 3 & 4}, then you must agree that the total momentum change for the two cases is identically the same.

∆[m_Total v]_C1 = ∆[m_Total v]_C2

And that means that the TOTAL impluses applied to the ground in both collisions are the same:
∫[F dt]_C1 = ∫[F dt]_C2

On a plot of Force (y axis) versus time (x axis), the ∫[F dt] is equal to the area under the curve.

Now, two last questions...

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. Do you 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?

6. Finally, do you 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)??

TomK

See post above about bag D full of 1000 m loose parts versus cube C of 1000 m parts connected together.

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.

 

[Heiwa]

My god, how stupid can you possibly be?

Ever see tons of snow on a house? It flattens the house.

Do you live in the real world?

Yes! I have seen tons of snow and it doesn't flat the house in the real world. Good for skiing though ... in the winter. Only 6 months to go! But you are off topic! A part C of snow will not one way crush down a part A of snow.

 

[Grizzly Bear]

Yes! I have seen tons of snow and it doesn't flat the house in the real world. Good for skiing though ... in the winter. Only 6 months to go! But you are off topic! A part C of snow will not one way crush down a part A of snow.