It looks like you're using an Ad Blocker.

Please white-list or disable AboveTopSecret.com in your ad-blocking tool.

Thank you.

 

Some features of ATS will be disabled while you continue to use an ad-blocker.

 

"Inside Job": Hidden energy in reports by Prof. Bazant, Dr. Greening and D. Thomas

page: 10
5
<< 7  8  9   >>

log in

join
share:

posted on Nov, 20 2011 @ 04:15 PM
link   

Originally posted by Akareyon

We can't even agree what an average is


Average what?


and that E=p*V


Ok, I agree with you.


and now you try to tell me that you'd throw the "inevitability theory" overboard if I'd just post some fancy diagram


Nope. It would need to stand up to scrutiny too.


proving the change in tension per floor


Tension? Are you trying to equate this with PE?


that ultimately MUST go along with the crush-down.


Well if it's gravitational PE, then yeah, it's gonna go DOWN. It MUST go down.


No, really, sorry, I don't believe this discussion is going anywhere soon. I suggest we all stay friends and keep up our respect for each other, even if I won't change my mind about the laws of nature and you don't have to change your mind about the inevitability of gravitational progressive collapses. Is that okay for you?


SO then quit discussing it here and wasting your time.

Get cracking on that collapse program !!

Show NIST/Bazant/ASCE/CTBUH/etc.... that they are all fools !!!

Be a friggin' hero of the twoof movement !!!



posted on Nov, 20 2011 @ 04:48 PM
link   

Originally posted by Joey Canoli

and that E=p*V


Ok, I agree with you.



You do? I would like to hear your explanation. It seems to me the only situation where (average) energy density as result of pressure makes any sense is in a fluid. How does it apply to building collapses?



posted on Nov, 20 2011 @ 05:03 PM
link   

Originally posted by -PLB-

Originally posted by Joey Canoli

and that E=p*V


Ok, I agree with you.



You do? I would like to hear your explanation. It seems to me the only situation where (average) energy density as result of pressure makes any sense is in a fluid. How does it apply to building collapses?


I'm just humoring the poor confused soul....



posted on Nov, 20 2011 @ 05:43 PM
link   

Originally posted by Joey Canoli
SO then quit discussing it here and wasting your time.

Get cracking on that collapse program !!

Show NIST/Bazant/ASCE/CTBUH/etc.... that they are all fools !!!

Be a friggin' hero of the twoof movement !!!
See, Joey, that's not what I'm here for. Bazant and some of the other people at NIST and ASCE know exactly what they have done. They will have to answer a greater judge.

And I don't want to be the friggin' hero of anyone. I discovered for myself a truth important enough to share. To share, not to shove it down people's throats. I consider my job done. Anyone with a spark of intellect and the knowledge how to handle an internet browser can now study this discussion, verify the points made, dissect the arguments put forward, enhance his studies, follow the thought experiments, refine the formulae and come to his own conclusions, based on his own experience and expertise. For some, it may be a reason to rethink their bias - if not now, maybe tomorrow. Others may have their gut instinct confirmed.

One way or the other.

To me, discussing this topic with one or two people who know what they are talking about was an important lesson I had to learn and I am very grateful for their input.

But you reveal an interesting aspect of the debate, of which this discussion is but a small filament in the loom of threads. In some people's eyes, this seems to be some sort of game - who is right and who is not; a debate for the sake of a debate, a psychological battle against the patience and goodwill of honest seekers of peace, and love, and truth. I have seen such on both sides; left and right, truthers and debunkers, creationists and darwinists, feminists and patriachalists, capitalists and communists, sceptics and spiritualists. Most of them just repeat what they are being told by some authority figure. Those are the people who would insist the moon is made of cheese if just somebody with a space suit would stick a Camembert into the camera and say he brought it back from his mission. Not many base their opinions and arguments on empirical studies or personal experience. Few care for the truth. Personally, I prefer a heated debate with someone with a different opinion that is based on sound logic over an agreement with someone who doesn't have the smallest clue what he's talking about. But, what shall I think about a discussion about cars with someone who knows everything about motors but, when challenged in his belief that cars don't need breaks suddenly pretends he never heard that cars have wheels and goes like "Wheels? What wheels?"...



posted on Nov, 20 2011 @ 06:00 PM
link   

Originally posted by Akareyon

I have found no explanation whatsoever for the "optimistic" assumption that the very structure that held itself up against several fires, one plane impact, one huge kerosene explosion, one bomb in the basement and several storms should suddenly be "doomed" out of the blue under its own weight...

Loss of capacity?


In the meantime, my experiments were worthless because paper loops are more stable than steel...

Not exactly an accurate characterization. The paper loops in question dissipated more energy as a function of the imposed load than steel columns would. There's a scaling issue in there which can't be overlooked - it seems ridiculous until you consider a structure three feet high versus one a quarter mile high - but even that is not the issue here. The issue is that the load displacement response of paper loops is increasing resistive force with increasing compaction, whereas for steel columns it's just the opposite. Since you understand that the work done in compaction is the integral of the load displacement curve, I'm sure you can see how a flat or slightly rising curve involves much more work than one with a narrow peak in front and greatly diminished There's absolutely no insinuation that paper is stronger than steel.


What kind of circular logic is that? The potential energy didn't go up there for free, but it stayed up there for free, no tensile strength or resistive force needed, thank you, it's just in the weakest links.

Once up there, it stayed up there for free energy-wise, but to think it did so with no resistive force is way off.



And naaw, we cannot derive an average for the tension in the towers...

Tension may come into play when loads are redistributed following component failure but otherwise tension is not a factor in primary vertical support members in an intact tower - compression is. Sure, the difference is "only" numeric sign (direction), but I'll tell you in no uncertain terms: if there are primary vertical members going into tension in a tower, you've got a serious problem.



...but if Bazant and Greening say each floor's resistance was negligible...

They say nothing of the sort.


...it's hewn in stone and not debatable because he's the expert.

In light of the fact he didn't even say it, what does that do to this assertion?

I don't think you understand that the very same engineering mechanics which allows structures like this to be built also dictates the circumstances under which they'll fail.
edit on 20-11-2011 by IrishWristwatch because: (no reason given)



posted on Nov, 20 2011 @ 06:34 PM
link   

Originally posted by Akareyon
But you reveal an interesting aspect of the debate, of which this discussion is but a small filament in the loom of threads. In some people's eyes, this seems to be some sort of game - who is right and who is not; a debate for the sake of a debate, a psychological battle against the patience and goodwill of honest seekers of peace, and love, and truth. I have seen such on both sides; left and right, truthers and debunkers, creationists and darwinists, feminists and patriachalists, capitalists and communists, sceptics and spiritualists. Most of them just repeat what they are being told by some authority figure.

This is true and not just trivially so, it's profoundly so. There are exceptions, and that's refreshing when it happens. I'll tell you what I think and maybe you'll learn something which causes an adjustment, and vice versa. We don't have to believe the same thing or march in lockstep. I just happen to think, in your particular case, that you really do have the aptitude (most don't) to understand the argument being made and incorporate that into your worldview.

In that last statement, did I claim Bazant was right and you were wrong? No, I didn't. I said you can "incorporate" the knowledge gleaned from having confronted the argument. But I have to be frank. At this point, I can see there are parts of his argument you don't understand and, while you're under no obligation to play the argument according to his rules, in order to make headway you have to at least put forth an alternate argument which forces a contradiction with his. Then we can say that both of these arguments can't be true simultaneously and the resolution can only come from more in-depth examination of both. That hasn't happened yet, in my opinion.

It's also clear you're coming from a position of intense disbelief, evident from the flavor of mild ridicule contained in some of your statements. That's fine, but should you ultimately be wrong by way of not fully grasping what you argue against, you are making much ado about nothing. I think even you have to agree that making a transition from your current position is not going to come easy. If you are correct, you've got a fairly difficult road ahead making your case to the world at large. I admit that Bazant practically controls JEM, and his status in the field is such that there are few who'd be inclined to challenge him unless they had a very solid case. But the papers were peer-reviewed and published in a widely read journal. BLGB definitely did get kicked back by the reviewers at least once for correction. You're not the only person who has gone over it with a fine-toothed comb.

None of this may be THAT important to you. If not, you may never get to the point where you can meet this argument at a sufficient level to confirm or refute it. You can then retain your beliefs, but you need to recognize they are just that - beliefs.



posted on Nov, 20 2011 @ 08:29 PM
link   

Originally posted by Akareyon

Bazant and some of the other people at NIST and ASCE know exactly what they have done. They will have to answer a greater judge.


No planers say the same thing - they will pay........


I discovered for myself a truth important enough to share. To share, not to shove it down people's throats.


And here's the money shot : if you TRULY believed that you were right, and that 9/11 WAS an inside job, you'd do the work necessary.

But this is what I see from you : you think you know the material well enough to make a pronouncement about how Bazant, etc have committed a fraud on the public. Believe me, you do not.

You might be earnest in your beliefs. You might be just a typical troll.

No way to tell.



posted on Nov, 21 2011 @ 09:57 AM
link   
Irish, I'm serious and not ridiculing you when I say that I very much enjoy talking to someone with your insight into the work of Bazant, Greening and Benson. I sense no arrogance or dogmatism in what you write, and you have met me with great friendlyness and patience. In the words of Voltaire, I don't agree with you but I'd fight for your right to speak your mind. It is for this reason that I will make another attempt to explain WHY I believe the towers were brought down, and why I think that logic and the laws of nature endorse my belief. For that, please try (please do!) to understand, even though I'm wrong - just as I understand the global collapse theory although I think it's wrong (even though I've been accused for not understanding).

First: our experiments. There is still no model for a stack of stuff that compresses itself under a portion of it's own accumulating weight. All we have is a hole with my silhouette in the floor slabs of a tower made of ice and a row of domino bricks that were built to collapse.

Second: Fig. 4a-c in BV'07. I promise I will remain within the boundaries of Bazants paper from now on. If somebody builds something, he will make sure that the maxwell line is way above any m*g that can be expected. If it is below m*g, it will crush. With a FoS of 2, the maxwell line will even be two times m*g. The area between the maxwell line and m*g is the energy needed to crush one floor. This E_crush=(maxwellline-m*g) * u. So, if maxwelline=2*m*g, then E_crush=(2*m*g-m*g)*u, so E_crush=m*g*u. This is the energy needed to crush one floor. It is no different from Φ(u) in Eq. 3. The only difference between Φ(u) and E_crush is that Φ(u) is the integral of the load-displacement curve from F(u) minus m*g*u, that is, the area between the curve for F(u) and the line of the load force, while E_crush is simply the "average" rectangle that stands for the same phenomenon. The greater the load force m*g, the smaller both E_c and Φ(u) - the energy needed to crush one floor. Why? Because the area between the maxwell line and m*g gets "compressed", it becomes smaller. Therefor, it does not matter if the curve for F(u) has a huge peak at the left hand side and then touches the ordinate (as F(u) for pasta or steel columns would), or if it takes a slope up at the right hand side (as F(u) for a paper loop would). In Bazants words: "what matters is energy, not the strength, nor stiffness." What matters is the area between F(u) and m*g.

If now something impacts one floor with kinetic energy (K) greater than E_crush (or Φ(u)), K will be diminished by E_crush. E_crush will diminish K so long until K=0. The only way for K not to become 0 is to make sure that E_crush (or Φ(u)) is very small. There are two ways to achieve that: either drag the maxwell line down or push m*g up.

In the event of 12 floors dropping from 3.7m height, it is obvious that m*g must be huge and well above the maxwell line. So the area between F(u) and m*g, Φ(u) that is, is very small and K is diminished just a little. So, what is K? It is m*g*h. Usually, h is zero - nothing dropping, so the kinetic energy is zero. If it drops from 3.7m, it is much greater. But - is it actually dropping from 3.7m?

PLB opened my eyes. K is - potentially! - much greater, it is constantly increased as it crushed through the floors - potential energy being transformed into kinetic energy.

Now please follow the infathomable stupidity of a rambling, mumbling idiot, dive into the madness of a conspiracy shill, jump headfirst into the insanity of someone believing the moon is made of cheese, Elvis lives, and aryan, vegetarian aliens dwell in subterrean caves under Antarctica waiting for their time to come to rule the world with their Vril staffs.

(Hold your breath now)



posted on Nov, 21 2011 @ 10:00 AM
link   
Let the "input energy" be that of 12 floors dropping from h = (98+1)*3.7m height right from the start, in other words: let's lift the 12 floors 3.7 metres, but pretend for a moment that the rest of the tower isn't there anymore. From bottom to ground, there are 366.3 metres. K would be huge upon impact on the ground, wouldn't it? There would be dust and debris all over the place, steel coloums flying everywhere and nothing left of the Top Of The World. Let us say that K_start=0 before it drops and K_ground=m*g*h when it crushes on the ground. The potential energy of block C is K_ground, therefor: E_pot=m*g*h. This is the potential energy of the upper 12 floors, Block C.

Now let's put the rest of the tower back under it. Those 98 stories have suffered nothing yet. We could just put a new roof on it (and be glad we have a wise president who doesn't scream "worldwide war" over a destroyed building). On each floor, m*g is still only half the maxwell line, accordingly, each Φ(u) is still big enough for each floor to support all the floors above. Imagine all the 98 curves for F(u) from Fig. 4c, a perfect, stable building the way it should be, lined up next to another. It would look a little like a twisted sine wave with 98 peaks. This new curve we call F(s), with s=98*u. To keep things neat and easy, we don't account for the gradient in strength, so the line for m*g stays the same, it doesn't move, and so doesn't the maxwell line at 2*m*g. We have changed nothing, we have only lined them up all next to another. Now we can go ahead and calculate Φ(s) - again, it is the area between F(s) and m*g, it is also the area between the maxwell line and m*g, it is: (2*m*g-m*g)*s=m*g*s. Φ(s)=98*Φ(u).

There is one catch, though: we have added one floor between the 98th and the 99th floor consisting of air. So, to the leftmost side of F(s), we must add another u with an F(air) that nearly touches the ordinate. Needless to say that Φ(air) is negative, because its maxwell line is waaay below m*g, it's almost not there. We have a negative area, but right now, we're mathematicians, we could even sqrt(-1) and have imaginary numbers if we need them
No, seriously, this is were Block C gains its momentum and K is increased from zero as it drops through the very first u, diminishing E_pot.

Therefor, now, as we release the upper block, it first gains momentum (a lot of it!) while going through the first u, rolling down F(air). K is here "diminished" by Φ(air), but as Φ(air) is negative, it isn't diminished, but greatly increased (and, of course, E_pot is greatly diminished). After that first u, however, it must climb all those little peaks over s. It will gain momentum when it "rolls" down a peak, and climb another peak, until all K is "gone". When will that be? Easy. When the energy "gone" will be

E_gone = E_pot - (Φ(s)+Φ(air))

E_gone is the energy that goes into deformation, and it is equal to the [potential energy of Block C] MINUS [the area between F(s) and the maxwell line of the 98 stories below] PLUS [the NEGATIVE area between m*g and the "maxwell line" that represents air resistance F(air) for the leftmost part of our 98 lined-up displacement curves].

Let's check: if Φ(air) is zero (no drop), the potential energy of Block C is just as big as Φ(s), the "containing" energy that keeps it up, so their difference is zero, so no energy goes into deformation. If there is a little drop, Φ(air) (the area between F(air) and m*g) becomes "negative", so the part between the parantheses gets SMALLER than E_pot, so E_gone increases - energy goes into deformation.

I hope I haven't lost you already because I'm talking of negative energy. I know it's not professional, but I'm trying to explain as good as a layman can, and I enclose a diagram to show what I mean:



Let's try and make K=0. E_gone = E_pot - (Φ(s)+Φ(air)). So, collapse will arrest WHEN [the energy "gone" (into deformation)] IS EQUAL TO [the potential ("potentially kinetic") energy] MINUS [the area between the looong sinus-like load displacement curve and the maxwell line] PLUS [the "negative" area at the leftmost part of our curve where it drops in freefall].

Of course, all this is utterly false. It is even complete rubbish! And in the next post, I'll explain why.



posted on Nov, 21 2011 @ 10:02 AM
link   
It's plain obvious. Φ(s) (the area between the sinus-like F(s) and g*m), is so huge (even with all the negative intervals where F(s) sinks below the m*g line representing the brittleness) that Φ(air), the energy "gained" (substracted from E_pot) during freefall, would fit tenfold into it. Collapse would be arrested after just a few stories, and that's not what we want. We want to bring the towers down. How do we achieve that?


There are numerous ways.

We could drag the line for m*g up for F(s) so that the area of the 98 peaks of F(s) above m*g are almost as huge as the area of the valleys beneath m*g. That would make sure that Φ(s) is so small that when added with the negative area from Φ(air), there's almost nothing left in the parantheses of the E_gone = E_pot - (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone - complete destruction. How do we drag up the line for m*g, closer to the maxwell line?

Another way could be to "cap" the peaks above m*g. That would drag the maxwell line down, closer to m*g, thus diminish Φ(s) until almost nothing is left in the parantheses of the E_gone = E_pot - (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone - complete destruction. How do we cap the peaks of F(s) above m*g?

Another way could be to "dig" the valleys deeper below m*g. That would drag the maxwell line down, closer to m*g, thus diminish Φ(s) until almost nothing is left in the parantheses of the E_gone = E_pot - (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone - complete destruction. How do we dig the valleys of F(s) below m*g deeper?

Another way could be to increase Φ(air) so that the "negative" area between m*g and air's maxwell line is so huge that it is larger than Φ(s) until almost nothing is left in the parantheses of the E_gone = E_pot - (Φ(s)+Φ(air)) equation, hardly anything is substracted from E_pot, so that as much of E_pot as possible goes into E_gone - complete destruction. How do we increase Φ(air)?

Now one may argue that steel is very brittle and more like pasta than like paper, that is, the curve looks different. To explain, Fig. PASTA is enclosed. Here, I left m*g where it is and only changed the shape of the load displacement curve. Accordingly, the maxwell line drops down closer to m*g.



As we can see, Φ(air) remains the same, only Φ(s) is getting real small now because of all those deep valleys under m*g. Suddenly, global collapse becomes much inevitablerer! It will take a lot of s until all Φ(air) is used up in Φ(s).

However, a good engineer would account for the the brittleness of the pasta, wouldn't he? Of course he would, he wants a FoS of 2 at least, and so he'll make sure that m*g, the forces acting, remain within the boundaries of sanity. Otherwise, everything in the building would be screaming for destruction! Just a little breeze and it would all be crumbling down, but we're expecting cyclones and storms and office fires! So, obviously, he would pull m*g down by putting less mass on top, which, again, would look like this:



(Or he would just use more pasta).

This is what I think I understand and therefor believe that global progressive disproportional collapse is not a new law of nature, but has to be planned.
edit on 21-11-2011 by Akareyon because: (no reason given)



posted on Nov, 21 2011 @ 10:34 AM
link   
I see that, at least for now, you dropped E=p*V, which I think is good, as it didn't make much sense to me.


Originally posted by Akareyon
Second: Fig. 4a-c in BV'07. I promise I will remain within the boundaries of Bazants paper from now on. If somebody builds something, he will make sure that the maxwell line is way above any m*g that can be expected. If it is below m*g, it will crush. With a FoS of 2, the maxwell line will even be two times m*g.


This it not correct though. FoS only applies to a static situation. You can have a FoS of 10 but still have a maxwell line below m*g. It depends on the design of your structure and the used materials.


The area between the maxwell line and m*g is the energy needed to crush one floor.


Also not correct. The maxwell line itself determines the energy consumed when crushing one story. In other words, the energy to crush one story, divided by the height of one floor, results in F_c, or:

F_c = E_crush/h

And E_crush is determined by taking the integral of F(u). The area under F(u) is equal to the area under F_c. (I think by the way that F(z) in figure 4 should be F(u)). Note that this is almost literally explained in Bazants paper.

See en.wikipedia.org... for more information.

The difference between the maxwell line and mg gives the excess kinetic energy.
edit on 21-11-2011 by -PLB- because: typo



posted on Nov, 21 2011 @ 11:12 AM
link   
We're putting a stable and static structure under E_pot, remember? : )

Originally posted by -PLB-
Also not correct. The maxwell line itself determines the energy consumed when crushing one story. In other words, the energy to crush one story, divided by the height of one floor, results in F_c, or:

F_c = E_crush/h

And E_crush is determined by taking the integral of F(u). The area under F(u) is equal to the area under F_c. (I think by the way that F(z) in figure 4 should be F(u)). Note that this is almost literally explained in Bazants paper.
It is already loaded, though (except for the topmost floor), hence, we substract m*g from F_c (because it's a little easier now to crush it). I didn't want to cheat but be as "optimistic" as possible towards collapse ;-)

But that's not the point, really. Let the first peaks of F(s) be somewhat twisted and deformed to account for the fact that there are a lot of chaotic processes going on in the impact zone. Makes Φ(s) a little smaller, that is true.



posted on Nov, 21 2011 @ 12:04 PM
link   

Originally posted by Akareyon
We're putting a stable and static structure under E_pot, remember? : )


Static structures do not exists, just static situations. F_c does not exist in a static situation.


It is already loaded, though (except for the topmost floor), hence, we substract m*g from F_c (because it's a little easier now to crush it). I didn't want to cheat but be as "optimistic" as possible towards collapse ;-)

But that's not the point, really. Let the first peaks of F(s) be somewhat twisted and deformed to account for the fact that there are a lot of chaotic processes going on in the impact zone. Makes Φ(s) a little smaller, that is true.


I am not following you. m*g and F_c are two different forces, one in a static situation, the other in a dynamic situation. They do not occur simultaneously. Can you explain in more detail what you mean?


edit on 21-11-2011 by -PLB- because: (no reason given)



posted on Nov, 21 2011 @ 12:13 PM
link   

Originally posted by -PLB-
They do not occur simultaneously.
Then there must be something wrong either with my eyes or with Figs. 3 and 4a-c in BV'07 where F_c is designated as "maxwell line".



posted on Nov, 21 2011 @ 01:13 PM
link   
reply to post by Akareyon
 


A short summery:

F(u) is the actual force that occurs during crushing a floor (it is the only real force).
F_c is the average force over a complete floor. This force does not actually exists.
mg is the force in a static situation. This force does not exist during the crushing of a floor.

Of course there are intersection points, where the actual force is equal to either mg or f_c. But this only happens for infinitesimal points.



posted on Nov, 21 2011 @ 02:28 PM
link   
reply to post by -PLB-
 



This force does not actually exists.
Now we agree. Sorry, I didn't know what it was called. But it should exist, that's what I'm saying, it was there just yesterday.



posted on Nov, 21 2011 @ 02:42 PM
link   
reply to post by Akareyon
 


How about this, explain in your own words what F(u), F_c, and mg represent and how they relate to the forces in an actual collapse.



posted on Nov, 21 2011 @ 02:56 PM
link   

Originally posted by -PLB-
reply to post by Akareyon
 


How about this, explain in your own words what F(u), F_c, and mg represent and how they relate to the forces in an actual collapse.
I tried to, but E=p*V didn't make sense to me either.

//edit: wait, I'll explain in Irish's words:

www.youtube.com...
edit on 21-11-2011 by Akareyon because: (no reason given)



new topics

top topics



 
5
<< 7  8  9   >>

log in

join