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# Yep, It's Thermite! So Much for the "Oxygen" Excuse

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posted on Jun, 9 2009 @ 08:51 PM

Originally posted by exponent
We're ignoring vertical velocity, taking purely the horizontal component, gravity will not accelerate it, so the only acceleration after ejection is friction from the air.

This is correct.

The ejected matter travels through a medium, air. The air will produce a frictional forces against the matter's motion. Although the effect would be relatively small, over a short distance, it is still enough to decay the motion.

The main initial, impulse force is the one that gives the matter its kick start velocity in the horizontal component. Once the mass is on it's way, it's not subjected to any further acceleration.

I agree with exponent's results.

posted on Jun, 9 2009 @ 08:53 PM

Originally posted by tezzajw
I agree with exponent's results.

Thank you, although I find it funny that bsbray has so many flags for an incorrect post. Either people don't like me, people do like him, or people vote on partisan lines. That or I have committed a False Trichotomy fallacy!

posted on Jun, 9 2009 @ 09:35 PM

You weren't clear enough in your response.

Just so I can see that you know what you're doing, I give these three variables and solve the equation for "a":

d=150m (total lateral distance traveled)
t=9s (total time elapsed)
v0=0 (started from rest)

Just so we can be clear we are in agreement about the math, that the math is correct.

Also the lateral distance we assume, 150m, or a little less than 500 ft., is accurate and in fact conservative of some of the ejected debris.

The 9 seconds is derived from the vertical free-fall time from the top of a tower the debris had before hitting the ground, thus also based on fact. But if you want you can increase it to 10 or 12 seconds, to give yourself more time and to require less energy.

posted on Jun, 9 2009 @ 09:40 PM

Originally posted by exponent
If you assume that the object accelerates from rest over a period of 9 seconds, its average acceleration is as you say. But the object did not do this.

For the third time, yes, I know this, I said it in my first post myself.

It accelerated from rest in an exceedingly short period of time, and then travelled 150m in 9 seconds

Once again, why not, but it doesn't make a difference in the formula I was using.

No, this is absolutely incorrect. You are not working out the 'net force' here, you are making claims about the Kinetic Energy content of the object when it has travelled 150m.

No, I was finding the net force, F, exerted on the object horizontally. I said this in my first post:

Originally posted by bsbray11
Technically, in mechanics, a force is defined as a change in momentum or also a mass multiplied by an acceleration, or F=ma.

A piece of debris weighing a ton (a modest weight compared to a lot of debris ejected from the towers), or 907 kg, traveling over 500 feet or about 150 meters laterally, takes a large force to produce.

That was the first thing I posted about all this. Nowhere do I talk in terms of the kinetic energy of the object itself. ONLY the force and work done to displace it horizontally, and to find that force I had to use a number of other variables. The velocity specifically was one I was NOT considering, but had to find mathematically to find the net force. So I used a formula. You do realize what a formula is used for, right?

By using a formula based on a continuous acceleration, you ignore the fact that the final velocity is marginally smaller than the ejection velocity.

I wasn't talking about the initial or final velocities! Stop saying "continuous acceleration," you're giving me a headache, you MEAN to say a constant acceleration. YES, I am assuming a constant acceleration, but it is THE constant acceleration that is the single number equivalent of the sum of all accelerations during the 9 second period of time! What don't you understand about that? It is CORRECT to use that acceleration for the entire period of the 9 seconds, because NO OTHER NUMBER will accurately describe the sum of all accelerations over the entire 9 seconds. Again, what do you not understand about that? Are you reading my words? And then thinking about them?

They are wrong, the distance travelled is 150m, the time taken is 9s, so therefore the speed is 16.7m/s.

I wasn't trying to find speed.... for Christ's sake, you've accused me of incorrectly trying to find the final velocity, a constant velocity, and now a speed. You aren't even arguing consistently. I said from the very beginning I was finding the FORCE, F. You don't even know what you are saying.

On top of that, 16.7m/s would NOT be an accurate figure using the average acceleration I gave, like I have said again and again and again. You would actually have the measure the lateral velocity yourself to be able to plug it in as a variable that you know for sure. I found it indirectly using other known figures, in the form of an average acceleration.

To accelerate the object to 16.7m/s, it takes energy 1/2mv^2 or 453.5*16.7^2 or 126KJ.

The problem is that you used the final velocity that my average acceleration implied and used it as if it were a demonstrated fact in your KE figures. I didn't HAVE to measure the final lateral velocity, because I supplied other variables that I DID justify (fall time, and distance traveled), and the only one missing was the unknown variable, the average acceleration (which doesn't imply what the specific accelerations were over time). I still haven't seen how you justify your final lateral velocity, because I certainly never claimed it was valid, so don't point at me. I never even looked at velocities.

The assumption that the object continues to accelerate until impact gives it a higher kinetic energy than the true conditions, which were an initial large acceleration followed by 150m travel at a constant speed (ignoring wind resistance).

1) I was not trying to find the final velocity of the object, 2) an object traveling any velocity has the exact same KE as the same object that has reached that same velocity from any other rate of acceleration, 3) the changes in acceleration over time is AVERAGED TOGETHER in the number I use, for the whole 9 second time period. Please try to understand the things I am saying before you knee-jerk another response. Just sit and think for a second.

[edit on 9-6-2009 by bsbray11]

posted on Jun, 9 2009 @ 09:42 PM

Originally posted by bsbray11
Just so I can see that you know what you're doing, I give these three variables and solve the equation for "a":

d=150m (total lateral distance traveled)
t=9s (total time elapsed)
v0=0 (started from rest)

If you need me to prove that I can rearrange a formula, I'm game.

d = v_0t + 1/2at^2
d = 1/2at^2
2d = at^2
2d/t^2 = a
300/t^2 = a
300/81 = a
a = 3.703(recurring)

Just so we can be clear we are in agreement about the math, that the math is correct.

You have derived the correct result for acceleration based upon your premise, but your premise is faulty.

Here is a graph of your equation's results over time:

Here is mine:

Hopefully you can see why our different approaches converge on the same time and distance, but radically different kinetic energy results. I can obviously elaborate on this if you feel I must.

[edit on 9-6-2009 by exponent]

posted on Jun, 9 2009 @ 09:51 PM

Originally posted by tezzajw
The main initial, impulse force is the one that gives the matter its kick start velocity in the horizontal component. Once the mass is on it's way, it's not subjected to any further acceleration.

I agree with exponent's results.

You mean you agree with what I said first?:

Originally posted by bsbray11
The horizontal component in reality did not accelerate over time, but was imparted with an initial velocity from a force and then just sailed out into the air, only having air to resist it, until it hit the ground.

Very first post of the last page of this thread, you will find me repeating the same thing I am telling you now, to exponent:

www.abovetopsecret.com...

Which just shows how closely you are following along in the thread.

The issue is not how the acceleration changed over time, but the net force exerted on the object to send it out of the building and 150m laterally in 9 seconds. I was the one to bring all this up, and I specifically said Force, with an F. All the math I used to find that force was used correctly, even the average acceleration that I had to find. No one else has any other information on the actual velocities of the ejected debris as far as I have seen. Exponents "results" consisted of taking the average acceleration which we both admit can't be used to find instantaneous velocities or accelerations, and then trying to use it to find an instantaneous velocity.

posted on Jun, 9 2009 @ 09:53 PM

Originally posted by bsbray11
No, I was finding the net force, F, exerted on the object horizontally. I said this in my first post:

You talk about measuring Work, which is a measure of energy, which you have incorrectly derived. In this context in fact W = delta KE.

It is CORRECT to use that acceleration for the entire period of the 9 seconds, because NO OTHER NUMBER will accurately describe the sum of all accelerations over the entire 9 seconds. Again, what do you not understand about that? Are you reading my words? And then thinking about them?

Yes, and I have clearly shown why you are wrong, you claim not to be discussing kinetic energy, nor the final velocity, but by trying to express work, you must inherently be talking about them, as in this situation work = 1/2m * delta v^2

The problem is that you used the final velocity that my average acceleration implied and used it as if it were a demonstrated fact in your KE figures. I didn't HAVE to measure the final lateral velocity, because I supplied other variables that I DID justify

You derive the kinetic energy of the final state, which is immediately dependent on the final velocity. It is inescapable if you desire to talk about Work in any fashion.

I still haven't seen how you justify your final lateral velocity, because I certainly never claimed it was valid, so don't point at me. I never even looked at velocities.

It is the distance travelled divided by the time taken to travel, the average speed.

Please try to understand the things I am saying before you knee-jerk another response. Just sit and think for a second.

Luckily, I have a very good understanding of the simple motion equations, which is why I was able to spot that you had overestimated the energy content of the object by a factor of 3.

Let me state my position clearly:
You agree that the speed the object travels after ejection is approximately constant. If it is a constant speed over a finite distance it can be discovered by d/t or 150/9 or 16.7m/s. Given this speed, kinetic energy is calculated by this formula: E_k = 1/2mv^2. Substituting in the actual values gives is E_k = 453.5*16.7^2 = 126KJ.

If I am wrong, please show where I am wrong, what value have I gotten incorrect, or how does Work not equal Kinetic Energy? Please do not simply insult me and claim I don't understand physics, this was one of the first things we ever learned and it is incredibly easy to apply.

posted on Jun, 9 2009 @ 09:53 PM

Originally posted by exponent
You have derived the correct result for acceleration based upon your premise, but your premise is faulty.

How is the premise faulty in the actual equation?

Is it any one of the variables I use? Surely the equation itself is reliable.

Once again, I am trying to find FORCE, not velocities or instantaneous accelerations.

posted on Jun, 9 2009 @ 09:54 PM

Originally posted by bsbray11
Exponents "results" consisted of taking the average acceleration which we both admit can't be used to find instantaneous velocities or accelerations, and then trying to use it to find an instantaneous velocity.

No, the average acceleration has no bearing on my results, my results are based off the required speed to travel 150 metres in 9 seconds, nothing else. The acceleration I gave was a ballpark to indicate just how small (relatively) the force needed was, 75KN or so.

posted on Jun, 9 2009 @ 09:56 PM

Originally posted by bsbray11
How is the premise faulty in the actual equation?

Is it any one of the variables I use? Surely the equation itself is reliable.

Once again, I am trying to find FORCE, not velocities or instantaneous accelerations.

The premise is faulty in that you claim to be able to calculate work, and work = kinetic energy, kinetic energy is dependent on velocity. Your calculation of force is irrelevant, it has no bearing nor any real meaning that I can see. Surely you don't dispute the upper section of the towers can exert a few thousand Newtons of force?

posted on Jun, 9 2009 @ 10:20 PM

Originally posted by exponent
Thank you, although I find it funny that bsbray has so many flags for an incorrect post. Either people don't like me, people do like him, or people vote on partisan lines. That or I have committed a False Trichotomy fallacy!
It's just like life, we all vote on partisan lines - don't we? - should we?

Everyone stuffs up now and again. I've stuffed up a few posts on ATS before. Meh... As if ATS gets peer reviewed, anyway. It's just like reading the NIST report, in here.

I'm happy to agree with calculations that are correct, whether they are truther calculations or not. I don't see a problem with what you've done, exponent. I would have done it the same way.

The average acceleration across the time interval is zero, with the exception of the minor frictional retardation.

[edit on 9-6-2009 by tezzajw]

posted on Jun, 9 2009 @ 10:47 PM

Originally posted by exponent
The premise is faulty in that you claim to be able to calculate work, and work = kinetic energy, kinetic energy is dependent on velocity.

Work is also defined by a force times a distance, and using that other equation I didn't need to find instantaneous velocities or accelerations. A constant force accelerating a one-ton mass at 3.7m/s^2 will travel 150m in 9 seconds just the same, and that whole force would be applied the whole distance as far as I can tell.

Your calculation of force is irrelevant, it has no bearing nor any real meaning that I can see. Surely you don't dispute the upper section of the towers can exert a few thousand Newtons of force?

It is an abstract figure but I was mainly just trying to make a point that there was real energy behind these things flying everywhere and it wasn't just from leaning and rotating about a fulcrum. And I have a hard time imagining what might happen inside one of the buildings that might send out a multi-ton chunk of steel several hundred feet laterally, let alone upwards of 80% of the total mass.

posted on Jun, 9 2009 @ 11:09 PM

Originally posted by bsbray11
Work is also defined by a force times a distance, and using that other equation I didn't need to find instantaneous velocities or accelerations. A constant force accelerating a one-ton mass at 3.7m/s^2 will travel 150m in 9 seconds just the same, and that whole force would be applied the whole distance as far as I can tell.

Indeed to get the result you did not require the velocity in the equation, but whether you realised this or not it was being implicitly used in your calculation. The problem is not that the distance or time changes, it is that in your equation the section is travelling at > 30m/s when it impacts, when in reality it would have been travelling at around 17m/s, as kinetic energy scales to the square of velocity, this results in your estimate being much higher than in reality.

It is an abstract figure but I was mainly just trying to make a point that there was real energy behind these things flying everywhere and it wasn't just from leaning and rotating about a fulcrum.

That's a fair comment, there were quite a few large sections which hinged off, you can see some in the Bob and Bri video, but a good accounting of the collapse behaviour is extremely hard to do.

And I have a hard time imagining what might happen inside one of the buildings that might send out a multi-ton chunk of steel several hundred feet laterally, let alone upwards of 80% of the total mass.

Well lets remember that the speed required is only 16.7m/s, which is under 2 seconds of gravitational acceleration. The physical mechanism may be unintuitive, but that is simply because we never experience anything like this in our everyday life, and so we have no real point of comparison.

edit: It's also worthy to point out that you can use W=Fd with my calculation too. I calculate a required acceleration of 83.3m/s/s, and using your above equation we can find that the distance travelled is 1.67m and the force applied is 75.5KN, giving us W=75583.33*1.67 = 126KJ. Within rounding errors as before.

[edit on 9-6-2009 by exponent]

posted on Jun, 9 2009 @ 11:44 PM

Originally posted by exponent
The problem is not that the distance or time changes, it is that in your equation the section is travelling at > 30m/s when it impacts, when in reality it would have been travelling at around 17m/s, as kinetic energy scales to the square of velocity, this results in your estimate being much higher than in reality.

Ok, forget energy, and forget work, I should have never mentioned that either. The acceleration I assumed to find that force doesn't contain any specific information about real instantaneous values, only an equivalent, like I've been saying. And I intentionally used other variables that I knew to avoid having assume any instantaneous velocities. I was originally just making the point that a non-gravitational force would have to be present, some significant amount of force, and furthermore it would have to be applied to the majority of the mass of either tower.

That's a fair comment, there were quite a few large sections which hinged off, you can see some in the Bob and Bri video, but a good accounting of the collapse behaviour is extremely hard to do.

I've seen perimeter columns laid out still connected, but most of them weren't, and they were all over the place:

All of those lateral displacements added up would be a significant amount of energy, that goes out of the collapse system and doesn't come back. And with it, goes lots of mass, no longer able to contribute to a collapse.

Well lets remember that the speed required is only 16.7m/s, which is under 2 seconds of gravitational acceleration.

Gravity acts on every single atom simultaneously. Even 16 meters in a second is pretty damned fast for something weighing a ton. And in reality we are talking about masses weighing over 20 tons moving that fast:

Bump the ton up to a realistic 20 tons and you'll see I was still being very conservative in my estimate.

So I was also being conservative when I assumed it fell from the very top of a tower, instead of further down from where it was actually ejected.

I also said you could increase the fall time. All of this still adds up taking all the severed columns into account, and maybe more importantly it takes mass away.

The physical mechanism may be unintuitive, but that is simply because we never experience anything like this in our everyday life, and so we have no real point of comparison.

Something else that's counter-intuitive to me is the fact that all this ejected debris didn't just come from a side where the collapses tended to lean, and where things tended to be under the more stress. It went out in all directions.

I calculate a required acceleration of 83.3m/s/s, and using your above equation we can find that the distance travelled is 1.67m and the force applied is 75.5KN, giving us W=75583.33*1.67 = 126KJ. Within rounding errors as before.

If you want to be that specific with it then you might as well figure drag into account and see what it would take to compensate for it.

Or a realistic mass for that matter.

[edit on 9-6-2009 by bsbray11]

posted on Jun, 10 2009 @ 12:00 AM

Originally posted by bsbray11
Ok, forget energy, and forget work, I should have never mentioned that either. The acceleration I assumed to find that force doesn't contain any specific information about real instantaneous values, only an equivalent, like I've been saying. And I intentionally used other variables that I knew to avoid having assume any instantaneous velocities. I was originally just making the point that a non-gravitational force would have to be present, some significant amount of force, and furthermore it would have to be applied to the majority of the mass of either tower.

I still don't understand this last statement, why does the force have to be non gravitational? Gravity will accelerate parts of the towers downwards, and when they are resisted by the lower structure, this will push parts out sideways. The energy does get used up and the mass does get depleted as you have said, but the question is does enough energy get used to slow or stop the collapse? The most rigorous calculations done so far are those in BLBG and this has been pretty rigorously reviewed at this point. Even if you dislike Dr Greening, there are still 3 other authors on that paper, and debunkers get criticised all the time for mentioning Jones when there are other authors on his paper.

Something else that's counter-intuitive to me is the fact that all this ejected debris didn't just come from a side where the collapses tended to lean, and where things tended to be under the more stress. It went out in all directions.

Actually as you can see in one of the images you posted, the majority of exterior column debris was spread out in 4 distinct lines perpendicular to the sides of each tower. Still, the reason that the collapse was global is that as soon as it began, elements in all areas of each floor level were experiencing forces they were not designed for, or not sufficiently designed for. Truss seats were experiencing gigantic overloads, perimeter columns were being pushed outwards after being disconnected from their only bracing, and core columns were subject to eccentric impacts. The best word to describe the collapse and the ensuing behaviour is 'chaotic'. There is enough energy to allow many types of motion and failure and so the result was a violent global collapse.

posted on Jun, 10 2009 @ 12:17 AM

Originally posted by exponent
I still don't understand this last statement, why does the force have to be non gravitational?

Even if the energy is ultimately provided by gravity, you have to admit something else is happening along the way. Whether you think it's a rotation that abruptly ends but some chunks break loose and keep flying, or the debris slides down an incline that gives it lateral momentum, it isn't purely gravitational energy at work, there are other things involved. And as those "other things" are needed to explain more and more of the total mass being thrown out, I think the possibilities become more limited, but nonetheless I think it defeats investigation to just call it all gravity from the start.

The energy does get used up and the mass does get depleted as you have said, but the question is does enough energy get used to slow or stop the collapse?

Shouldn't any loss of energy slow it by definition?

The most rigorous calculations done so far are those in BLBG and this has been pretty rigorously reviewed at this point.

For some reason I doubt they took mass being lost over the sides into account, or the energy required to displace all that mass laterally. But I could be wrong. I know Greening never took that into account in his earlier 1-dimensional arithmetic exercise of a model. Either way the whole process boils down to what you put in, what you leave out, and how much energy you think what you put in should really take up, and they play with a lot of room in those numbers.

[edit on 10-6-2009 by bsbray11]

posted on Jun, 10 2009 @ 12:45 AM

Originally posted by bsbray11
Even if the energy is ultimately provided by gravity, you have to admit something else is happening along the way. Whether you think it's a rotation that abruptly ends but some chunks break loose and keep flying, or the debris slides down an incline that gives it lateral momentum, it isn't purely gravitational energy at work, there are other things involved. And as those "other things" are needed to explain more and more of the total mass being thrown out, I think the possibilities become more limited, but nonetheless I think it defeats investigation to just call it all gravity from the start.

Yes and no, ultimately it is all gravitational energy, but you are correct in saying that some local interactions occur which result in these effects.

Shouldn't any loss of energy slow it by definition?

Yes, I should have been more accurate in my statement, what I meant was to slow the collapse past the observed collapse time.

For some reason I doubt they took mass being lost over the sides into account, or the energy required to displace all that mass laterally. But I could be wrong.

They do

I know Greening never took that into account in his earlier 1-dimensional arithmetic exercise of a model. Either way the whole process boils down to what you put in, what you leave out, and how much energy you think what you put in should really take up, and they play with a lot of room in those numbers.

That's true, but even so one would expect if the towers were demolished with explosives, for the difference between the observed variables and theoretical variables to be irreconcilable. As it is everything seems to match up well with the numbers we have to hand, from aircraft impact speed to fire temperatures to collapse times, there is good science supporting all of it.

I would also like to ask that you acknowledge that I was not simply picking on truthers in my previous posts, I did identify that you had incorrectly identified a value for work of 500KJ which was not valid, and in fact the true number for the work done in your hypothetical example was approximately 125KJ.

posted on Jun, 10 2009 @ 02:23 AM
exponent, I also agree with some of your calculations after further examinations, independent research and gaining a better understanding of the laws of Physics. I will now continue exploring and educating myself in the world of Physics and controlled demolition.

Then again, I still feel confident in stating that Prof Jones and others who are working on this eagerly awaited new paper will be supported in time by many more world renowned scientists/Physicists and other relevant specialists around the globe.

You obviously know Physics, done your homework and have studied extensively.

However, have you thought about why some members are very concerned when you told everyone that you are a spokesman for NIST?

I think you should keep your distance from NIST, especially regarding WTC 7 and continue with your own calculations and studies on controlled demolitions, with the help of people who you can trust to be neutral and have NO connections with NIST.

If you, me and everyone else really are committed in discovering the truth involving 9/11, we all have to admit the Fact that a lot of mistakes were made before (especially the 1993 bombing), on and after that day and explore all of them extensively.

Cheers.

[edit on 10-6-2009 by Skyline666]

[edit on 10-6-2009 by Skyline666]

posted on Jun, 10 2009 @ 05:43 AM

posted by jprophet420

The NIST report did not in fact determine the cause of the collapses. It determined that the pancake collapse could have been initiated under the circumstances with manipulated data.

posted by mmiichael

I'm a spokesman for NIST, but don't find their revisions suspicious nor do I think there was withholding of evidence to conceal significant data.

Well that certainly explains a lot and is not a bit surprising. Many government loyalists seem to be spokesmen for the 9-11 perps also.

posted on Jun, 10 2009 @ 06:23 AM

Originally posted by SPreston

posted by mmiichael

I'm a spokesman for NIST, but don't find their revisions suspicious nor do I think there was withholding of evidence to conceal significant data.

Well that certainly explains a lot and is not a bit surprising. Many government loyalists seem to be spokesmen for the 9-11 perps also.

Sorry.

Typo.

I'm NOT a spokesman for NIST.

Not American. Not qualified.

Though I would have thought clear from the context.

Mike

[edit on 10-6-2009 by mmiichael]

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