Question for Apollo20

[Oxigen]

Yep, Stand by my statements. No planes and no tapes. Love the Patriot Act though!

 

[Oxigen]

And I'm not a guy!

 

[Thabiguy]

This is where you are wrong. The crushed debris and the upper part of the towers are, according to Bazant, traveling downwards and crushing the lower structure as a single body. Bazant theorizes that, every time a floor is crushed, the crushed debris adds its mass to the upper section of floors, which continues to crush the lower structure. I'm afraid you don't understand Bazant's theory. There aren't three bodies in his equation, there are two.

Sir, if there are only two bodies in question, i.e. bottom part, and the crushed debris+upper part, then you cannot talk of force exerted by crushed debris on the upper part, because they are part of a single body. If you want to talk about destruction of the upper part by the crushed debris, then you must consider three bodies.

I do not discuss Bazant's theory here. I address the assertions that you have made, i.e. that the crushed debris will destroy the bottom part of the tower at the same rate as the upper part of the tower. You claimed that this must be so, because of Newton's third law. I have explained to you, in several different ways, why your assertion is wrong, and you yourself have explained why you cannot use Newton's third law to arrive to such incorrect conclusion. Your assertion considers three bodies, not two, and therefore also does my explanation. At no point does Bazant's theory come into this.

Anybody can make an incorrect assertion. Why cannot you simply admit that your assertion was wrong?

Bazant doesn't get it both ways (and neither do you). He can't include the mass of the upper section of floors to in the force equation that destroys the lower structure without considering the upper section of floors a part of the body that is crushing the lower structure, and thus be acted upon by Newton's Third Law.

Sir, in our earlier example of a person standing on the roof of a building, you yourself have (correctly) included the mass of the person in determining the force exerted by the building on the ground. Please make an effort to understand why this is correct, and why the force exerted by the building on the ground equals F=gm, where m is the weight of the building and the person - not just the weight of the building.

Not including the mass of the attached object X2 (the person, or the upper part of the tower) in calculation of the force exerted by the assembly X1 & X2 is equivalent to saying that there is no force between X1 and X2 - i.e. that the objects are not attached, and X2 cannot get crushed by X1. This would be an actual error.

The crushed debris is assumed to be fully compacted, so the equal and opposite upward force has to be absorbed by the upper floors.

Sir, please, make an effort to understand that this statement is equivalent to claiming that when a person stands on top of a fully compacted heap of debris, he must absorb "equal and opposite" upward force as the force exerted by the ground on the fully compacted heap of debris.

You yourself have explained why you cannot use Newton's third law to conclude this. I have provided you with an equation, F2=F1*m2/(m1+m2), that clearly shows that this is not so, and even allows you to calculate the exerted force. (If you think this equation is incorrect, please explain why you think so, or even better, post a correct equation.)

How can you, after all this, still not see that what you claim contradicts both physics and reality?

Note to competent engineers and physicists: If I am wrong in anything I'm saying here, please speak up and point out where I'm wrong. I will gladly admit an error if I made one.

 

[beachnut]

Yep, Stand by my statements. No planes and no tapes. Love the Patriot Act though!

No facts either. It flows very well, you could use it in your thesis, no planes, no tapes, no facts.

 

[Thabiguy]

This is where you are wrong. The crushed debris and the upper part of the towers are, according to Bazant, traveling downwards and crushing the lower structure as a single body. Bazant theorizes that, every time a floor is crushed, the crushed debris adds its mass to the upper section of floors, which continues to crush the lower structure. I'm afraid you don't understand Bazant's theory. There aren't three bodies in his equation, there are two.

In yet another desperate attempt to explain this to you, I will humor you and analyze the situation as if the crushed debris and upper part are a single, compact body. I'll call it the "upper body".

As the upper body crushes the bottom part of the tower, force is exerted on it - opposite but equal to the force with which the upper body crushes the bottom part of the tower.

How does this force affect the upper floors?

When force is applied to an edge of a solid body, it is not felt throughout the whole body. If you make a virtual slice through the body in question, and insert a virtual sensor to measure the force with which the body is pushed together (or pulled apart) at that particular point, you will realize that the local force decreases with increasing distance from the point where the force is applied.

For this reason, when a homogenous building stands on the ground, and the ground exerts force F=gm on it (where m is the mass of the building), the local force at the ground floor will be nearly F1=gm, but only F2=gm/2 in the middle of the building, and will approach zero near the top floor. For this reason, your leg bones have to bear greater forces than your neck vertebrae as you run. For this reason, if you attempt to wedge a finger into a high stack of books you will find this difficult near the bottom, but easy near the top.

When force F is exerted on one edge of an object, the local force felt at a given point P will be equal to F(P) = F*m2/(m1+m2), where m1 is mass of the part of the object between point P and the edge where the force is exerted, and m2 is mass of the rest of the object.

See here to understand why this is so.

So when the upper body crushes the bottom floors with force F(bottom), and equal but opposite force F(bottom) is exerted on it by the bottom floors, the local force at the point where the upper floors meet the crushed debris is F(upper) = F(bottom)*m2/(m1+m2), where m1 is mass of the crushed debris and m2 is mass of the (so far) intact upper floors.

Not only is F(upper) less than F(bottom) - at any time but the very start of collapse - but it decreases relative to F(bottom) as the mass of crushed debris increases.

The crushed debris is assumed to be fully compacted, so the equal and opposite upward force has to be absorbed by the upper floors.

I have shown that this is not the case.

 

[Golden Bear]

When force is applied to an edge of a solid body, it is not felt throughout the whole body. If you make a virtual slice through the body in question, and insert a virtual sensor to measure the force with which the body is pushed together (or pulled apart) at that particular point, you will realize that the local force decreases with increasing distance from the point where the force is applied.

You, sir, cannot be helped.

Here's a better experiment, one you can even do at home. Get a bathroom scale and put it on the ground and turn it on. Put your hand on the scale and push on the scale with 20lbs of force, and see that the bathroom scale reads 20lbs.

Next, put a 10 foot pole on the scale (or a hundred foot pole for that matter, but you might not be able get that one done at your home), with one end on the scale and make the pole perpendicular to the earth, and zero out the scale. Push on the end of the pole opposite the scale with 20lbs of force. Are you seriously contending that the bathroom scale, which is measuring force being applied at the opposite end of this rigid body, will read something less than 20 lbs? I mean, the measurement is being taken some distance from the point where the force is being applied, right? Is the 20lb force being felt throughout the entire body or not?

If you cut the pole in half and put the bathroom scale in between the two sections of the pole, press down on the pole with 20 lbs of force, the bathroom scale will still read 20lbs.

Let's circle this experiment back around now. Leave the pole perpendicular with the ground, but pick up the bathroom scale and put it between your hand and the upper end of the pole. When you push down with 20lbs of force, what will the scale read? You guessed it, 20 lbs. It's the same 20lbs of force no matter which end of the pole you measure it from.

The point here is that no matter where you measure the force being applied to the end of the pole, it will register as 20lbs of force. You're wrong.

I think you're confused. You are correct that, for a building sitting on the earth, the further away you get from the earth, the less gravitational force is being placed on any differential slice of the building you examine. This has nothing to do with where the force is being applied, and everything to do with how much weight is above the building.

It also has nothing to do with the case of two bodies impacting, which is what Bazant is trying to model with the Twin Towers. There is a gravity force accelerating on the upper floors towards the earth during the collapse, no question about it. However, the force I'm talking about is the mechanical work force (ie. the force applied through a distance) that causes the upper floors to decelerate and deform for a period of time after impact, which is not a gravitational force, and thus is not dependent on its distance from the earth.

Finally, consider the case of two homogeneous cube shaped bodies of equal size and mass impacting each other at one face of each cube. Assume that at impact the bodies do not deform. If you measure the force experienced by any differential slice of either cube that is parallel to the face being impacted, the result has to be that the forces are equal for every such differential slice. This is because F=m*a. Every differential slice should have the same mass because the cube is symetrical and homogeneous. Every differential slice must also accelerate at the same rate because the bodies don't deform. (Indeed, every discrete point on each body must have accelerated at the same for the body not to deform.) Solving the equation for each differential slice clearly shows that the force F experienced by each differential slice is equal because the masses and the accelerations are equal.

 

[beachnut]

You, sir, cannot be helped.

Here's a better experiment, one you can even do at home. Get a bathroom scale and put it on the ground and turn it on. Put your hand on the scale and push on the scale with 20lbs of force, and see that the bathroom scale reads 20lbs.

Next, put a 10 foot pole on the scale (or a hundred foot pole for that matter, but you might not be able get that one done at your home), with one end on the scale and make the pole perpendicular to the earth, and zero out the scale. Push on the end of the pole opposite the scale with 20lbs of force. Are you seriously contending that the bathroom scale, which is measuring force being applied at the opposite end of this rigid body, will read something less than 20 lbs? I mean, the measurement is being taken some distance from the point where the force is being applied, right? Is the 20lb force being felt throughout the entire body or not?

If you cut the pole in half and put the bathroom scale in between the two sections of the pole, press down on the pole with 20 lbs of force, the bathroom scale will still read 20lbs.

Let's circle this experiment back around now. Leave the pole perpendicular with the ground, but pick up the bathroom scale and put it between your hand and the upper end of the pole. When you push down with 20lbs of force, what will the scale read? You guessed it, 20 lbs. It's the same 20lbs of force no matter which end of the pole you measure it from.

The point here is that no matter where you measure the force being applied to the end of the pole, it will register as 20lbs of force. You're wrong.

I think you're confused. You are correct that, for a building sitting on the earth, the further away you get from the earth, the less gravitational force is being placed on any differential slice of the building you examine. This has nothing to do with where the force is being applied, and everything to do with how much weight is above the building.

It also has nothing to do with the case of two bodies impacting, which is what Bazant is trying to model with the Twin Towers. There is a gravity force accelerating on the upper floors towards the earth during the collapse, no question about it. However, the force I'm talking about is the mechanical work force (ie. the force applied through a distance) that causes the upper floors to decelerate and deform for a period of time after impact, which is not a gravitational force, and thus is not dependent on its distance from the earth.

Finally, consider the case of two homogeneous cube shaped bodies of equal size and mass impacting each other at one face of each cube. Assume that at impact the bodies do not deform. If you measure the force experienced by any differential slice of either cube that is parallel to the face being impacted, the result has to be that the forces are equal for every such differential slice. This is because F=m*a. Every differential slice should have the same mass because the cube is symetrical and homogeneous. Every differential slice must also accelerate at the same rate because the bodies don't deform. (Indeed, every discrete point on each body must have accelerated at the same for the body not to deform.) Solving the equation for each differential slice clearly shows that the force F experienced by each differential slice is equal because the masses and the accelerations are equal.

You are one of the most knowledge challenged 9/11 truth supporter yet. You need to go back to school and try again. I would seek a refund if you paid for your education, if it was a public education, you were the kid who looked out the window, I teach school and have seen the future kids who do not pay attention. But I make them answer the hard questions, so there will not be anyone like you graduating from the schools I teach at. I strive to get kids to think for themselves unlike those who spew 9/11 truth misinformation and lies.

 

[Dave Rogers]

Golden Bear, while you're right about Newton's Third Law, you're wrong about its application to the collapses. It's really very simple. The upper block is falling, therefore the upward reaction force is able to decelerate its fall, which is observed from the fact that the collapse took significantly longer than the time for an object in freefall to fall the same distance. The lower block cannot, however, be accelerated downwards without disruption of the structure; in other words, crushing. Therefore, while some of the upward force on the upper block results in crushing and some in deceleration, all of the downward force on the lower block results in crushing. Since the forces are equal, the lower block must necessarily suffer more crushing than the upper. That's really all there is to it.

Dave

 

[Thabiguy]

You, sir, cannot be helped.

That's the impression I get, yes.

Here's a better experiment, one you can even do at home. Get a bathroom scale and put it on the ground and turn it on. Put your hand on the scale and push on the scale with 20lbs of force, and see that the bathroom scale reads 20lbs.

Next, put a 10 foot pole on the scale (or a hundred foot pole for that matter, but you might not be able get that one done at your home), with one end on the scale and make the pole perpendicular to the earth, and zero out the scale. Push on the end of the pole opposite the scale with 20lbs of force. Are you seriously contending that the bathroom scale, which is measuring force being applied at the opposite end of this rigid body, will read something less than 20 lbs?

Sir, if you place the other end of the pole against a scale that is supported by a floor or a wall and cannot be pushed away, the scale will read 20 lbs. This is because there is not one force on one edge applied to the scale, but two forces: you apply 20 lbs, and the floor/wall applies 20 lbs. The sum of the local effects of these two forces, as you can calculate, is 20 lbs.

If you apply force from one side only - if the scale is not supported by a floor/wall, and it reads zero before you start pushing - then the scale will read F = 20 * m2 / (m1+m2) lbs, where m1 is the mass of the pole and m2 is the mass of the scale.

Ask a trusted physicist of your choice to confirm this to you. Or better yet, do the experiment and verify that this is the case.

I mean, the measurement is being taken some distance from the point where the force is being applied, right? Is the 20lb force being felt throughout the entire body or not?

No, sir. If you apply 20 lb to one side of the body (and not also to the other side of the body), the force is not felt throughout the entire body. Acceleration is. The force is not.

This is why your neck vertebrae are not compressed with 160 lbs of force that the ground applies to your feet. This is why, if you sit in an accelerating car, you do not feel the acceleration force exerted by the car on the ground, but only feel F2=F1*m2/(m1+m2), where F1 is the acceleration force exerted by the car on the ground, m1 is the mass of the car, and m2 is your mass.

But if you are supported by a stone wall, or in another way held still with respect to the ground as the car begins to accelerate, then you will feel the full force exerted by the car. This is because your support will exert a second force on you.

If you cut the pole in half and put the bathroom scale in between the two sections of the pole, press down on the pole with 20 lbs of force, the bathroom scale will still read 20lbs.

No, sir, it will not. If the other end of the pole is not supported by a wall or a floor as you push, the bathroom scale will not read 20 lbs. It will read F=20 * (0.5*m1+m2) / (m1+m2) lbs, where m1 is the mass of the pole and m2 is the mass of the scale.

Please make an effort to understand these simple facts.

Let's circle this experiment back around now. Leave the pole perpendicular with the ground, but pick up the bathroom scale and put it between your hand and the upper end of the pole. When you push down with 20lbs of force, what will the scale read? You guessed it, 20 lbs. It's the same 20lbs of force no matter which end of the pole you measure it from.

Yes. In this case, the scale will read 20 lbs, because you measure it at the same point where you apply the force.

The point here is that no matter where you measure the force being applied to the end of the pole, it will register as 20lbs of force. You're wrong.

This is not the case, and I urge you to do the experiments that you described (without the scale being supported by a floor or a wall) and find out for yourself.

I think you're confused. You are correct that, for a building sitting on the earth, the further away you get from the earth, the less gravitational force is being placed on any differential slice of the building you examine. This has nothing to do with where the force is being applied, and everything to do with how much weight is above the building.

Sir, if the ground (or another support) does not exert force on the bottom edge of the building, the force exerted between any two parts of the building, as it falls down freely, will be zero.

Please make an effort to understand this.

It also has nothing to do with the case of two bodies impacting, which is what Bazant is trying to model with the Twin Towers. There is a gravity force accelerating on the upper floors towards the earth during the collapse, no question about it. However, the force I'm talking about is the mechanical work force (ie. the force applied through a distance) that causes the upper floors to decelerate and deform for a period of time after impact, which is not a gravitational force, and thus is not dependent on its distance from the earth.

Oh, the humanity.

Finally, consider the case of two homogeneous cube shaped bodies of equal size and mass impacting each other at one face of each cube. Assume that at impact the bodies do not deform.

If that's the case, the cubes will exert an infinite amount of force in an infinitely short amount of time as they impact.

I will assume that you mean that the bodies do decelerate over a finite amount of time, and you measure the effect of the force that the cubes apply to each other as they decelerate upon impact.

If you measure the force experienced by any differential slice of either cube that is parallel to the face being impacted, the result has to be that the forces are equal for every such differential slice. This is because F=m*a.

No, they are not.

The acceleration in any part of the cubes is the same. If you cut through the cubes and measure the force with which the two parts of the cube are pushed together, you will find that it falls down to zero towards the outer edges of the cubes.

This is why, if two cars impact each other, experiencing impact acceleration a, the front of the car will experience force F=a*m1, where m1 is the mass of the entire car, but the rear bumper will only experience force F=a*m2, where m2 is the mass of the rear bumper.

This is why, when two cars decelerate by impacting each other, the front bumpers will be hopelessly deformed, but the rear bumpers will not.

Please make an effort to understand that your claims of equal impact forces throughout the entire car completely contradict reality.

Every differential slice should have the same mass because the cube is symetrical and homogeneous. Every differential slice must also accelerate at the same rate because the bodies don't deform. (Indeed, every discrete point on each body must have accelerated at the same for the body not to deform.)

I am talking about the force with which the two parts of the sliced cube are pushed together at the point of the slice. This depends on the mass of the two parts of the sliced cub, and therefore depends on the point where you slice.

Solving the equation for each differential slice clearly shows that the force F experienced by each differential slice is equal because the masses and the accelerations are equal.

Sir, you misapply the equation. You would be able to see that if you actually tried to enumerate it.

Acceleration is uniform throughout the body, yes. The force that you incorrectly calculate using F=a*m, incorrectly using the mass of the differential slice for m, is F=a*epsilon (where epsilon is the infinitesimally small mass of the differential slice), which is infinitesimally small. It should be apparent that the force with which the two parts of the cube are pushed together at the point where you measure it, cannot be infinitesimally small.

The correct value of m to use in the equation F=a*m is the mass that is being accelerated by the force, at the point where you measure the force. You measure the force with which the two parts of the cube are pushed together - i.e. the force with which the inner part of the cube (with mass m1), on whose edge the impact deceleration force is exerted, decelerates the outer part of the cube (with mass m2). This is F = a*m2. Furthermore, a = F0/(m1+m2), where F0 is the force applied to the edge of the cube. This gives F=F0*m2/(m1+m2).

 

[Thabiguy]

What also might help you understand is Example 15 (Hauling a Trailer) on page 30 of this text.

I advise that you first try to understand the calculations and the solution, before you go on making more embarrassingly incorrect assertions about how the force with which parts of the body are pushed together/pulled apart (the tension force) is allegedly constant throughout the body. Pay extra attention to what mass is used as m in the F = a*m equation.

Also, consider reading the entire text. It will do you a world of good.

ETA: Or maybe solution #4 will enlighten you. Notice how the equation I provided, in this case F_AB = F_BA = F * m_B / (m_A+m_B), gives the correct result of 8 N.

If this doesn't help either, I must admit I am beginning to run out of ways to explain this simple stuff.

 

[rwguinn]

What also might help you understand is Example 15 (Hauling a Trailer) on page 30 of this text.

I advise that you first try to understand the calculations and the solution, before you go on making more embarrassingly incorrect assertions about how the force with which parts of the body are pushed together/pulled apart (the tension force) is allegedly constant throughout the body. Pay extra attention to what mass is used as m in the F = a*m equation.

Also, consider reading the entire text. It will do you a world of good.

ETA: Or maybe solution #4 will enlighten you. Notice how the equation I provided, in this case F_AB = F_BA = F * m_B / (m_A+m_B), gives the correct result of 8 N.

If this doesn't help either, I must admit I am beginning to run out of ways to explain this simple stuff.

Thabiguy,
you are arguing with a person who does not comprehend, will never comprehend, and has no desire to comprehend, simple vector addition. Beating your head against the wall because it feels so much better when you quit is self-defeating...

 

[ElMondoHummus]

Thabiguy,

I commend your posts and meticulous explanations in the face of willful misunderstanding on the part of your opponent. Please, continue in this vein, even if you tire of contributing in this thread and on this topic specifically. Posts like yours may not affect the opinion of the entrenched mindset, but they will be illuminating to bystanders and lurkers on the forum.

 

[Major_Tom]

There are many ways to see that the Bazant collapse analysis is founded on false premises.

Consider the most basic modelling of core and perimeter column "buckling" in the following diagram.

The diagram on the far right (c) shows how Bazant-Zhao visualize the perimeter columns "buckling". There is this generally accepted idea that they would form horizontal "creases" and buckling occurred along many perimeter columns along these lines.


Pretty simple.


The largest comprehensive photo album available to the public which specializes only in photos of the perimeter columns as they lay in the rubble is at the link below.


http://www.sharpprintinginc.com/911...p=view&PHPWS_Album_id=12&MMN_position=102:102


These photos collectively capture the TRUE CONDITION of perimeter columns as they were seen undisturbed at the scene of the crime.



Notice that there is no evidence of the creasing and buckling that the Bazant-Zhao model predicts. And these columns are seen AFTER falling and crashing into the ground.

Even after all that abuse they STILL don't show the creasing and buckling that Bazant predicts.



We know that the perimeter sections were separated from one another along bolt connections and along some spandrel bolt collections.

We see this in thousands of cases. We know that this type of failure is totally inconsistent with buckling because the perimeter sections were installed "staggered" to increase their strength.


Pointing out the true condition of the perimeter column sections as they were seen in the rubble is alone enough to refute fundamental premises of the Bazant-Zhao model.

 

[T.A.M.]

SO do up a paper that counters or refutes their model, and submit it to the same journal...

Oh wait, i forgot, the big bad NWO will obviously not allow such a work to be published, there fore any attempt to LEGITIMATELY counter with a REAL SCIENTIFIC PAPER would be pointless...right? What a wonderful disclaimer the truth movement has.

TAM

 

[Major_Tom]

A paper?

We are talking about a paper (Bazant's) which relies on perimeter buckling, even though a photographic mapping of the crime scene capturing hundreds of these perimeter column-spandrel plate pre-fabricated sections shows no such buckling.


And you want me to "write a paper" about it?

 

[R.Mackey]

A paper?

We are talking about a paper (Bazant's) which relies on perimeter buckling, even though a photographic mapping of the crime scene capturing hundreds of these perimeter column-spandrel plate pre-fabricated sections shows no such buckling.


And you want me to "write a paper" about it?

Please do us the favor, and don't write another nonsense paper.

There is no problem with Bazant & Zhou. "Buckling" is also known as elastic instability. The word I'd like to point out is elastic. This means the columns can flex, giving up their load and precipitating a collapse, before reaching their elastic limits. After the collapse begins, those same columns may unload and may spring back to their original shape.

I might also point out that a very, very large fraction of the recovered WTC steel does show deformation, sometimes not enough to see easily in a webpage photograph, but enough that you couldn't use it in new construction. Both your premise and your reasoning are false.

 

[Major_Tom]

There is no problem with Bazant & Zhou. "Buckling" is also known as elastic instability. The word I'd like to point out is elastic. This means the columns can flex, giving up their load and precipitating a collapse, before reaching their elastic limits. After the collapse begins, those same columns may unload and may spring back to their original shape.

So where do the perimeter column pre-fabricated sections separate from one another?

Answer: At their column-to-column bolts and along the bolted spandrel plates.

The photo album mentioned above is the single largest and most comprehensive collection of photos focussing only on perimeter pre-fab sections as they were seen at the crime scene.

There is not one small area of the WTC 1,ans 2 rubble area which is not included in the album. This album, therefore, is a comprehansive record of the
ACTUAL CONDITION of the perimeter columns after the "collapse".


We do not see buckling. Instead we see a certain kind of failure.


Elastic Buckling to explain the WTC "collapse"?


Along staggered bolt connections?

Seriously?

 

[WildCat]

deleted.

 

[WildCat]

The photo album mentioned above is the single largest and most comprehensive collection of photos focussing only on perimeter pre-fab sections as they were seen at the crime scene.

There is not one small area of the WTC 1,ans 2 rubble area which is not included in the album. This album, therefore, is a comprehansive record of the
ACTUAL CONDITION of the perimeter columns after the "collapse".


We do not see buckling. Instead we see a certain kind of failure.


Elastic Buckling to explain the WTC "collapse"?


Along staggered bolt connections?

Seriously?

So which pics depict cutter charges or thermite?

 

[R.Mackey]

We do not see buckling. Instead we see a certain kind of failure.


Elastic Buckling to explain the WTC "collapse"?


Along staggered bolt connections?

Seriously?

Seriously.

Flexure in the columns -- below their elastic limit -- places extraordinary stress on the connections. Enough to shear bolted connections. The connections in this case are weaker than the columns themselves, otherwise we would expect to see necking or crimping in the columns.

There is also fracture to consider. The progressive collapse mechanism is one of repeated and severe impacts, and the stresses in question tend to be waves, not static loads. This biases the failures even more strongly towards the bolts. Solid columns are much more tolerant of stress waves than bolts, due to the lack of interfaces and thus wave reflections inside the material.

Seriously.

I strongly recommend you study statics, strength of materials, and solid mechanics before claiming you know better than Drs. Bazant and Zhou. You most certainly do not.